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    <title>RSS feed for Electronic applications</title>
    <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-0</link>
    <description>This RSS feed contains all the sections in Electronic applications</description>
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    <language>en-gb</language><lastBuildDate>Wed, 15 Jul 2020 10:19:12 +0100</lastBuildDate><pubDate>Wed, 15 Jul 2020 10:19:12 +0100</pubDate><dc:date>2020-07-15T10:19:12+01:00</dc:date><dc:publisher>The Open University</dc:publisher><dc:language>en-gb</dc:language><dc:rights>Copyright © 2020 The Open University</dc:rights><cc:license>Copyright © 2020 The Open University</cc:license><item>
      <title>Introduction</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-0</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;The modern world would not be able to function without electronic systems. Using a variety of teaching material, including videos and interactive activities, this free course, &lt;i&gt;Electronic applications&lt;/i&gt;, will show you how electronic systems can be found everywhere in communications, control and signal processing. It focusses on electronic filters, particularly digital filters.&lt;/p&gt;&lt;p&gt;Note that the interactive activities have been designed to work in the Firefox and Chrome browsers, so you will need to use one of these browsers if you want to access the interactive content.&lt;/p&gt;&lt;p&gt;This OpenLearn course is an adapted extract from the Open University course &lt;span class="oucontent-linkwithtip"&gt;&lt;a class="oucontent-hyperlink" href="http://www.open.ac.uk/courses/modules/t312"&gt;T312 &lt;i&gt;Electronics: signal processing, control and communications&lt;/i&gt;&lt;/a&gt;&lt;/span&gt;&lt;/p&gt;</description>
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    <dc:title>Introduction</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;The modern world would not be able to function without electronic systems. Using a variety of teaching material, including videos and interactive activities, this free course, &lt;i&gt;Electronic applications&lt;/i&gt;, will show you how electronic systems can be found everywhere in communications, control and signal processing. It focusses on electronic filters, particularly digital filters.&lt;/p&gt;&lt;p&gt;Note that the interactive activities have been designed to work in the Firefox and Chrome browsers, so you will need to use one of these browsers if you want to access the interactive content.&lt;/p&gt;&lt;p&gt;This OpenLearn course is an adapted extract from the Open University course &lt;span class="oucontent-linkwithtip"&gt;&lt;a class="oucontent-hyperlink" href="http://www.open.ac.uk/courses/modules/t312"&gt;T312 &lt;i&gt;Electronics: signal processing, control and communications&lt;/i&gt;&lt;/a&gt;&lt;/span&gt;&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>Learning outcomes</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section---learningoutcomes</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;After studying this course, you should be able to:&lt;/p&gt;&lt;ul&gt;&lt;li&gt;&lt;p&gt;understand the mathematical representations and techniques for manipulating of signals in the time and frequency domains&lt;/p&gt;&lt;/li&gt;&lt;li&gt;&lt;p&gt;explain the application, benefits and limitations of communications, control and signal processing techniques in real world applications&lt;/p&gt;&lt;/li&gt;&lt;li&gt;&lt;p&gt;select and apply appropriate techniques to the analysis of time-varying signals represented in both the time and frequency domain&lt;/p&gt;&lt;/li&gt;&lt;li&gt;&lt;p&gt;use a digital filter to remove Gaussian noise from a signal.&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section---learningoutcomes</guid>
    <dc:title>Learning outcomes</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;After studying this course, you should be able to:&lt;/p&gt;&lt;ul&gt;&lt;li&gt;&lt;p&gt;understand the mathematical representations and techniques for manipulating of signals in the time and frequency domains&lt;/p&gt;&lt;/li&gt;&lt;li&gt;&lt;p&gt;explain the application, benefits and limitations of communications, control and signal processing techniques in real world applications&lt;/p&gt;&lt;/li&gt;&lt;li&gt;&lt;p&gt;select and apply appropriate techniques to the analysis of time-varying signals represented in both the time and frequency domain&lt;/p&gt;&lt;/li&gt;&lt;li&gt;&lt;p&gt;use a digital filter to remove Gaussian noise from a signal.&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>1 Electronics everywhere</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-1</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Begin by considering the following situation:&lt;/p&gt;&lt;div class="oucontent-quote oucontent-s-box"&gt;&lt;blockquote&gt;&lt;p&gt;You are sitting at home watching the television, and a woman in a diving suit, surrounded by sharks is speaking directly to you, live on television from the Caribbean (Figure 1).&lt;/p&gt;&lt;/blockquote&gt;&lt;/div&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/d86faba4/t312_openlearn_fig01.tif.jpg" alt="Described image" width="512" height="257" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133050052064"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 1&lt;/b&gt; Swimming with sharks&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133050052064&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133050052064"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;It is possible to take this type of situation for granted, but when thinking about it more closely you might wonder how this is possible. How can someone swimming underwater in the Caribbean be seen and heard by you in your house in the UK? &lt;/p&gt;&lt;p&gt;In this course you will start thinking about everything that’s involved in these electronic systems – from signal processing to control and communications – which allow you to watch a live broadcast from the Caribbean. &lt;/p&gt;&lt;p&gt;If you have studied some electronics before, you may be familiar with analogue components such as resistors, capacitors and inductors, as well as more complex components such as the operational amplifier. You may also know something about digital circuits, which include logic gates, microprocessors and even software. And you may be aware of some of the fundamental principles that describe the way that voltages and currents in a circuit are related, including Ohm’s law and Kirchhoff’s voltage and current laws. &lt;/p&gt;&lt;p&gt;Whilst all of this is necessary to understand electronics, it doesn’t really explain how electronics allows you to watch a live broadcast from the Caribbean. In part, this is due to the sheer scale of electronic systems. A typical computer in 2020 has around 10&amp;#xA0;000&amp;#xA0;000&amp;#xA0;000 (or 10&amp;#xA0;billion) transistors. Nobody could sit down and design such a computer armed with Ohm’s and Kirchhoff’s laws alone. Electronic systems have to be broken down into subsystems, and these subsystems in turn have to be broken down into further subsystems. Go far enough and you’ll find that you can explain what is going on using Ohm’s and Kirchhoff’s laws, but that’s a long way down. &lt;/p&gt;&lt;p&gt;At a higher level, when designing subsystems, descriptions like &amp;#x2018;the control subsystem’ or &amp;#x2018;the communications subsystem’ are used. These are designed using principles that describe the function of the subsystem. An electronics engineer would use these system-level principles to design the subsystem, then implement it electronically. &lt;/p&gt;&lt;p&gt;So, what are the systems and subsystems that allow you to watch someone swimming with sharks live on television?&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-1</guid>
    <dc:title>1 Electronics everywhere</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Begin by considering the following situation:&lt;/p&gt;&lt;div class="oucontent-quote oucontent-s-box"&gt;&lt;blockquote&gt;&lt;p&gt;You are sitting at home watching the television, and a woman in a diving suit, surrounded by sharks is speaking directly to you, live on television from the Caribbean (Figure 1).&lt;/p&gt;&lt;/blockquote&gt;&lt;/div&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/d86faba4/t312_openlearn_fig01.tif.jpg" alt="Described image" width="512" height="257" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133050052064"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 1&lt;/b&gt; Swimming with sharks&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133050052064&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133050052064"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;It is possible to take this type of situation for granted, but when thinking about it more closely you might wonder how this is possible. How can someone swimming underwater in the Caribbean be seen and heard by you in your house in the UK? &lt;/p&gt;&lt;p&gt;In this course you will start thinking about everything that’s involved in these electronic systems – from signal processing to control and communications – which allow you to watch a live broadcast from the Caribbean. &lt;/p&gt;&lt;p&gt;If you have studied some electronics before, you may be familiar with analogue components such as resistors, capacitors and inductors, as well as more complex components such as the operational amplifier. You may also know something about digital circuits, which include logic gates, microprocessors and even software. And you may be aware of some of the fundamental principles that describe the way that voltages and currents in a circuit are related, including Ohm’s law and Kirchhoff’s voltage and current laws. &lt;/p&gt;&lt;p&gt;Whilst all of this is necessary to understand electronics, it doesn’t really explain how electronics allows you to watch a live broadcast from the Caribbean. In part, this is due to the sheer scale of electronic systems. A typical computer in 2020 has around 10 000 000 000 (or 10 billion) transistors. Nobody could sit down and design such a computer armed with Ohm’s and Kirchhoff’s laws alone. Electronic systems have to be broken down into subsystems, and these subsystems in turn have to be broken down into further subsystems. Go far enough and you’ll find that you can explain what is going on using Ohm’s and Kirchhoff’s laws, but that’s a long way down. &lt;/p&gt;&lt;p&gt;At a higher level, when designing subsystems, descriptions like ‘the control subsystem’ or ‘the communications subsystem’ are used. These are designed using principles that describe the function of the subsystem. An electronics engineer would use these system-level principles to design the subsystem, then implement it electronically. &lt;/p&gt;&lt;p&gt;So, what are the systems and subsystems that allow you to watch someone swimming with sharks live on television?&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>1.1 Three key subsystems</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-1.1</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;In order to watch someone swimming with sharks live on television somebody or something must be operating the camera and recording the sounds. The images and sounds are converted into electronic signals; these signals are then sent as electromagnetic waves, via satellite, to your satellite dish at home. This then delivers the electronic signals to your television, which converts them back into images and sounds. As you can see, there are several subsystems involved: &lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;sensors – converting images and sounds into electronic signals&lt;/li&gt;&lt;li&gt;processors – converting the electronic signals into a form that can be stored and transmitted as radio waves&lt;/li&gt;&lt;li&gt;communication – actually transmitting the waves, and receiving them at the other end&lt;/li&gt;&lt;li&gt;processors – converting the waves back into images and sounds&lt;/li&gt;&lt;li&gt;display – your television.&lt;/li&gt;&lt;/ul&gt;&lt;p&gt;These components can be broadly categorised into signal processing and communications. &lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;&lt;b&gt;Signal processing&lt;/b&gt; explains how signals are manipulated electronically to filter out noise and to alter the signals so that they can be communicated. &lt;/li&gt;&lt;li&gt;&lt;b&gt;Communications&lt;/b&gt; looks at how the communication subsystems work, showing how the electronic signals are converted to radio waves for transmission and reception. &lt;/li&gt;&lt;/ul&gt;&lt;p&gt;Are there any other possible sub-systems?&lt;/p&gt;&lt;p&gt;Yes, there are many, but the one other major area of electronics is &lt;b&gt;control&lt;/b&gt;. In the scenario above, it’s not obvious how control comes into it, however there are probably control systems involved, carrying out tasks such as the auto-focusing function on the cameras. &lt;/p&gt;&lt;p&gt;Another and perhaps more obvious way you could introduce control would be to make your viewing experience more interactive. What if you could control the camera remotely so that you decide what you want to see? Is that so far-fetched? There are already systems which allow you to control your home from your phone – you can adjust the heating, switch on lights, and even see who is at the door. So why not have the full interactive experience of seeing and hearing what you want to see and hear by controlling the live camera equipment through your television? &lt;/p&gt;&lt;p&gt;What has been described here so far is mainly for entertainment. However, the same principles can be applied to other systems, such as a Mars rover (Figure&amp;#xA0;2). This is a vehicle that travels around the surface of Mars semi-autonomously, collecting samples and analysing them, then sending the data back to Earth. Such a system is clearly making use of signal processing and communication, but it is also using control subsystems to move the vehicle around. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/69ea6893/t312_openlearn_fig02.tif.jpg" alt="Described image" width="512" height="409" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049999712"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 2&lt;/b&gt; Mars Rover&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049999712&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049999712"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Systems that use signal processing, control and communications are not just found on other planets. On Earth there are similar systems, including semi-autonomous delivery robots that can bring goods to homes or places of work (Figure&amp;#xA0;3). &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/6a9a09e4/581915f9/t312_delivery_robot.tif.jpg" alt="Described image" width="512" height="341" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049989680"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 3&lt;/b&gt; Delivery robot&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049989680&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049989680"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;For the rest of this course you will focus on one of these three subsystems: signal processing. You will look at some of the basic principles of signal processing and how it is implemented. &lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-1.1</guid>
    <dc:title>1.1 Three key subsystems</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;In order to watch someone swimming with sharks live on television somebody or something must be operating the camera and recording the sounds. The images and sounds are converted into electronic signals; these signals are then sent as electromagnetic waves, via satellite, to your satellite dish at home. This then delivers the electronic signals to your television, which converts them back into images and sounds. As you can see, there are several subsystems involved: &lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;sensors – converting images and sounds into electronic signals&lt;/li&gt;&lt;li&gt;processors – converting the electronic signals into a form that can be stored and transmitted as radio waves&lt;/li&gt;&lt;li&gt;communication – actually transmitting the waves, and receiving them at the other end&lt;/li&gt;&lt;li&gt;processors – converting the waves back into images and sounds&lt;/li&gt;&lt;li&gt;display – your television.&lt;/li&gt;&lt;/ul&gt;&lt;p&gt;These components can be broadly categorised into signal processing and communications. &lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;&lt;b&gt;Signal processing&lt;/b&gt; explains how signals are manipulated electronically to filter out noise and to alter the signals so that they can be communicated. &lt;/li&gt;&lt;li&gt;&lt;b&gt;Communications&lt;/b&gt; looks at how the communication subsystems work, showing how the electronic signals are converted to radio waves for transmission and reception. &lt;/li&gt;&lt;/ul&gt;&lt;p&gt;Are there any other possible sub-systems?&lt;/p&gt;&lt;p&gt;Yes, there are many, but the one other major area of electronics is &lt;b&gt;control&lt;/b&gt;. In the scenario above, it’s not obvious how control comes into it, however there are probably control systems involved, carrying out tasks such as the auto-focusing function on the cameras. &lt;/p&gt;&lt;p&gt;Another and perhaps more obvious way you could introduce control would be to make your viewing experience more interactive. What if you could control the camera remotely so that you decide what you want to see? Is that so far-fetched? There are already systems which allow you to control your home from your phone – you can adjust the heating, switch on lights, and even see who is at the door. So why not have the full interactive experience of seeing and hearing what you want to see and hear by controlling the live camera equipment through your television? &lt;/p&gt;&lt;p&gt;What has been described here so far is mainly for entertainment. However, the same principles can be applied to other systems, such as a Mars rover (Figure 2). This is a vehicle that travels around the surface of Mars semi-autonomously, collecting samples and analysing them, then sending the data back to Earth. Such a system is clearly making use of signal processing and communication, but it is also using control subsystems to move the vehicle around. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/69ea6893/t312_openlearn_fig02.tif.jpg" alt="Described image" width="512" height="409" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049999712"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 2&lt;/b&gt; Mars Rover&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049999712&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049999712"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Systems that use signal processing, control and communications are not just found on other planets. On Earth there are similar systems, including semi-autonomous delivery robots that can bring goods to homes or places of work (Figure 3). &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/6a9a09e4/581915f9/t312_delivery_robot.tif.jpg" alt="Described image" width="512" height="341" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049989680"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 3&lt;/b&gt; Delivery robot&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049989680&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049989680"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;For the rest of this course you will focus on one of these three subsystems: signal processing. You will look at some of the basic principles of signal processing and how it is implemented. &lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2 Signal processing</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Signal processing is a branch of electronics concerned not just with the properties of signals, but also with the properties of the devices and systems that carry the signals. The objective of signal processing is to optimise the recovery of some particular aspect of the signal that is of most interest, or to optimise the use of a communication medium. &lt;/p&gt;&lt;p&gt;Signal processing usually means filtering a signal. This could be to reduce or remove interference; it might mean changing the signal so that the communication channel can be used more economically or more efficiently; or it might mean processing a signal so that it can be sampled satisfactorily in an analogue-to-digital converter. &lt;/p&gt;&lt;div class="oucontent-internalsection"&gt;
&lt;h2 class="oucontent-h2 oucontent-internalsection-head"&gt;What is filtering?&lt;/h2&gt;
&lt;p&gt;In the context of electronic signals, filtering means altering the signal so that some aspects of the signal are removed while other parts of the signal remain. In this section you will learn about the difference between &amp;#x2018;ideal’ filters and real filters. You will also look at different types of filters and their characteristics and a type of graph used to show the frequency-dependent gain of a filter. &lt;/p&gt;
&lt;p&gt;One of the most common uses of filters is to reduce the intrusion of &lt;i&gt;unwanted&lt;/i&gt; signals, or noise, into &lt;i&gt;wanted&lt;/i&gt; signals. A common example of filtering is radio and television tuning. Here, the antenna picks up multiple transmissions being broadcast on different frequencies; the tuning circuit ideally passes the wanted broadcast to the output and blocks all the others, which are on different frequencies from the wanted broadcast. &lt;/p&gt;
&lt;p&gt;Filters can be analogue or digital. Analogue filters use components such as inductors, capacitors, resistors, and sometimes &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary" class="oucontent-glossaryterm-notfound"&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;operational amplifiers (op-amps)&lt;/span&gt;&lt;/a&gt;. Digital filters are basically computers and achieve their filtering effects through mathematical operations on the sample values of a digitised signal. You will look at these more closely in Section 3. In this section, however, you will focus only on analogue filters. &lt;/p&gt;
&lt;/div&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2</guid>
    <dc:title>2 Signal processing</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Signal processing is a branch of electronics concerned not just with the properties of signals, but also with the properties of the devices and systems that carry the signals. The objective of signal processing is to optimise the recovery of some particular aspect of the signal that is of most interest, or to optimise the use of a communication medium. &lt;/p&gt;&lt;p&gt;Signal processing usually means filtering a signal. This could be to reduce or remove interference; it might mean changing the signal so that the communication channel can be used more economically or more efficiently; or it might mean processing a signal so that it can be sampled satisfactorily in an analogue-to-digital converter. &lt;/p&gt;&lt;div class="oucontent-internalsection"&gt;
&lt;h2 class="oucontent-h2 oucontent-internalsection-head"&gt;What is filtering?&lt;/h2&gt;
&lt;p&gt;In the context of electronic signals, filtering means altering the signal so that some aspects of the signal are removed while other parts of the signal remain. In this section you will learn about the difference between ‘ideal’ filters and real filters. You will also look at different types of filters and their characteristics and a type of graph used to show the frequency-dependent gain of a filter. &lt;/p&gt;
&lt;p&gt;One of the most common uses of filters is to reduce the intrusion of &lt;i&gt;unwanted&lt;/i&gt; signals, or noise, into &lt;i&gt;wanted&lt;/i&gt; signals. A common example of filtering is radio and television tuning. Here, the antenna picks up multiple transmissions being broadcast on different frequencies; the tuning circuit ideally passes the wanted broadcast to the output and blocks all the others, which are on different frequencies from the wanted broadcast. &lt;/p&gt;
&lt;p&gt;Filters can be analogue or digital. Analogue filters use components such as inductors, capacitors, resistors, and sometimes &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary" class="oucontent-glossaryterm-notfound"&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;operational amplifiers (op-amps)&lt;/span&gt;&lt;/a&gt;. Digital filters are basically computers and achieve their filtering effects through mathematical operations on the sample values of a digitised signal. You will look at these more closely in Section 3. In this section, however, you will focus only on analogue filters. &lt;/p&gt;
&lt;/div&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.1 Frequency-dependent gain</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.1</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;In Figure 4, the box in the middle represents a device with frequency-dependent gain &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="5643590c6c73fa99ad1af505b6427addd3d4da2e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_1d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 2206.0 1119.0820" width="37.4539px"&gt;

&lt;desc id="eq_dea10bfd_1d"&gt;cap g of omega&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; – in other words, a filter. Here the symbol &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="f9d4d2552f37ca2f0b559ff2c54a7b5345cf4388"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_2d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 627.0 530.0915" width="10.6453px"&gt;

&lt;desc id="eq_dea10bfd_2d"&gt;omega&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is being used to represent angular frequency, measured in radians per second. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/8a631f95/t312_openlearn_fig04.tif.jpg" alt="Described image" width="512" height="128" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049952576"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 4&lt;/b&gt; Filter with input amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_3d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_3d"&gt;cap v sub in&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and output amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_4d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_4d"&gt;cap v sub out&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049952576&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049952576"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;A sinusoidal input signal with amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_5d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_5d"&gt;cap v sub in&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is applied to the filter, and the output is a sinusoidal signal with amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_6d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_6d"&gt;cap v sub out&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. The &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048963936" class="oucontent-glossaryterm" data-definition="For a sinusoidal input and output, voltage gain is the ratio of the output voltage’s amplitude to that of the input voltage. It has no units." title="For a sinusoidal input and output, voltage gain is the ratio of the output voltage’s amplitude to th..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;voltage gain&lt;/span&gt;&lt;/a&gt; of the filter at the frequency in question is &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="cec405e9bb8187ef230b8e404de308f1af76a777"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_7d" height="39px" role="math" style="vertical-align: -15px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 2080.4 2297.0631" width="35.3215px"&gt;

&lt;desc id="eq_dea10bfd_7d"&gt;cap v sub out divided by cap v sub in&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; – in other words, it is the ratio of the output voltage amplitude to the input voltage amplitude. So, at any particular frequency, &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b51097a1ff3c81180c4eb69079afb05e967150ea"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_8d" height="39px" role="math" style="vertical-align: -15px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 4209.9 2297.0631" width="71.4765px"&gt;

&lt;desc id="eq_dea10bfd_8d"&gt;cap v sub out divided by cap v sub in equals cap g&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;The voltage gain can be expressed simply as a number or fraction (or decimal). For example, a gain of 2 means that the output amplitude is twice the input amplitude. A gain of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="3e925096cff54129be65e4d688e475a76943ba75"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_9d" height="25px" role="math" style="vertical-align: -8px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1001.2839 717.1 1472.4763" width="12.1751px"&gt;

&lt;desc id="eq_dea10bfd_9d"&gt;one divided by four&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; (or 0.25) means that the amplitude of the output is one-quarter that of the input. &lt;/p&gt;&lt;p&gt;The above gain is referred to as &amp;#x2018;voltage gain’ because gain is sometimes expressed as a ratio of output and input &lt;i&gt;powers&lt;/i&gt;. In this course this way of expressing gain will be referred to as &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048972400" class="oucontent-glossaryterm" data-definition="The ratio of output power to input power. It is usually expressed in decibels (dB). A power gain of 0 dB means that the output power is the same as the input power. A power gain of 3 dB (or, more exactly, 3.0103 dB) means that the output power is double the input power." title="The ratio of output power to input power. It is usually expressed in decibels (dB). A power gain of ..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;power gain&lt;/span&gt;&lt;/a&gt;. As power ratios can be expressed in decibels, so power gains are almost invariably given in &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048993968" class="oucontent-glossaryterm" data-definition="A logarithmic way of expressing a power ratio. For powers P1 and P2, their ratio in decibels is defined as 10 log10 (P1/P2). The symbol for decibels is dB. Strictly the decibel is not a unit, as any ratio must be a pure (that is, dimensionless) number. However, it is often regarded as a unit." title="A logarithmic way of expressing a power ratio. For powers P1 and P2, their ratio in decibels is defi..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;decibels&lt;/span&gt;&lt;/a&gt;. &lt;/p&gt;&lt;p&gt;For the time being you will only consider sinusoidal inputs and outputs, as these have a single frequency. This limitation to a single frequency helps to clarify what a filter does. In practice, though, a filter would typically operate on a complex waveform consisting of many frequency components. In such a case, the inputs and outputs would themselves be functions of frequency. &lt;/p&gt;&lt;div class="oucontent-studynote oucontent-s-gradient oucontent-s-box &amp;#10;        oucontent-s-noheading&amp;#10;      "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;div class="oucontent-inner-box"&gt;&lt;p&gt;Although two different types of gain have been mentioned (voltage gain and power gain), you will often just see the word &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048982048" class="oucontent-glossaryterm" data-definition="In amplification, a measure of how many times the input signal amplitude is increased. It is generally measured as the ratio between the input signal amplitude and the output signal level. If a gain value is given as just a number (i.e. with no units), the gain is likely to be a ratio of voltages; if the value is given in decibels, it is a ratio of powers. See also voltage gain and power gain." title="In amplification, a measure of how many times the input signal amplitude is increased. It is general..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;gain&lt;/span&gt;&lt;/a&gt; used by itself. If the gain is given as an ordinary numerical value (such as 2, 10, 3000 or 0.001), voltage gain is almost invariably indicated. If the numerical value is in decibels, power gain is being represented. &lt;/p&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;The output of a filter differs from the input not only in amplitude but (usually) also in phase. You will look more closely at the question of phase later in the course. However, for now you will continue to focus on gain. Complete Activity 1 to test your understanding so far.&lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 1 &lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;A passive filter has an input signal of &amp;#xA0;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c3cfe2cc3ff26fd86f660a7f4b81fa77f5386b5d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_10d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 8313.7 1119.0820" width="141.1517px"&gt;

&lt;desc id="eq_dea10bfd_10d"&gt;v sub in of t equals 10 postfix times sine of 200 times t&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; volts. The steady-state output is &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="6b30bf9a4051fdcb656cf012d7b413ac3c9d3344"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_11d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 11964.0 1119.0820" width="203.1272px"&gt;

&lt;desc id="eq_dea10bfd_11d"&gt;v sub out of t equals two postfix times sine of 200 times t minus 0.6 times pi&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; volts. What is the gain as a voltage ratio? &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The input to the filter in part&amp;#xA0;(a) remains unchanged in amplitude, but its frequency changes. The steady-state output is now found to be &amp;#xA0;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="68d80e4142ae5a14268251cb13c1fff6c58b5b79"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_12d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 11964.0 1119.0820" width="203.1272px"&gt;

&lt;desc id="eq_dea10bfd_12d"&gt;v sub out of t equals five postfix times sine of 100 times t minus 0.2 times pi&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; volts. What is the new gain as a voltage ratio? &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Here &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_13d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_13d"&gt;cap v sub in&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, the amplitude of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="065acf14618ad32bed5671c6fd8cc82d6e5ae9eb"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_14d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 2340.8 1119.0820" width="39.7426px"&gt;

&lt;desc id="eq_dea10bfd_14d"&gt;v sub in of t&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, is 10&amp;#xA0;V and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_15d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_15d"&gt;cap v sub out&lt;/desc&gt;
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&lt;path d="M383 58Q327 -10 256 -10H249Q124 -10 105 89Q104 96 103 226Q102 335 102 348T96 369Q86 385 36 385H25V408Q25 431 27 431L38 432Q48 433 67 434T105 436Q122 437 142 438T172 441T184 442H187V261Q188 77 190 64Q193 49 204 40Q224 26 264 26Q290 26 311 35T343 58T363 90T375 120T379 144Q379 145 379 161T380 201T380 248V315Q380 361 370 372T320 385H302V431Q304 431 378 436T457 442H464V264Q464 84 465 81Q468 61 479 55T524 46H542V0Q540 0 467 -5T390 -11H383V58Z" id="eq_dea10bfd_15MJMAIN-75" stroke-width="10"/&gt;
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&lt;/defs&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_15MJMATHI-56" y="0"/&gt;
&lt;g transform="translate(588,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_15MJMAIN-6F"/&gt;
 &lt;use transform="scale(0.707)" x="505" xlink:href="#eq_dea10bfd_15MJMAIN-75" y="0"/&gt;
 &lt;use transform="scale(0.707)" x="1066" xlink:href="#eq_dea10bfd_15MJMAIN-74" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, the amplitude of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="87d273bc0f88f8ee654d4bf3c1342a3061a55dfb"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_16d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 2776.4 1119.0820" width="47.1383px"&gt;

&lt;desc id="eq_dea10bfd_16d"&gt;v sub out of t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M173 380Q173 405 154 405Q130 405 104 376T61 287Q60 286 59 284T58 281T56 279T53 278T49 278T41 278H27Q21 284 21 287Q21 294 29 316T53 368T97 419T160 441Q202 441 225 417T249 361Q249 344 246 335Q246 329 231 291T200 202T182 113Q182 86 187 69Q200 26 250 26Q287 26 319 60T369 139T398 222T409 277Q409 300 401 317T383 343T365 361T357 383Q357 405 376 424T417 443Q436 443 451 425T467 367Q467 340 455 284T418 159T347 40T241 -11Q177 -11 139 22Q102 54 102 117Q102 148 110 181T151 298Q173 362 173 380Z" id="eq_dea10bfd_16MJMATHI-76" stroke-width="10"/&gt;
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&lt;path d="M383 58Q327 -10 256 -10H249Q124 -10 105 89Q104 96 103 226Q102 335 102 348T96 369Q86 385 36 385H25V408Q25 431 27 431L38 432Q48 433 67 434T105 436Q122 437 142 438T172 441T184 442H187V261Q188 77 190 64Q193 49 204 40Q224 26 264 26Q290 26 311 35T343 58T363 90T375 120T379 144Q379 145 379 161T380 201T380 248V315Q380 361 370 372T320 385H302V431Q304 431 378 436T457 442H464V264Q464 84 465 81Q468 61 479 55T524 46H542V0Q540 0 467 -5T390 -11H383V58Z" id="eq_dea10bfd_16MJMAIN-75" stroke-width="10"/&gt;
&lt;path d="M27 422Q80 426 109 478T141 600V615H181V431H316V385H181V241Q182 116 182 100T189 68Q203 29 238 29Q282 29 292 100Q293 108 293 146V181H333V146V134Q333 57 291 17Q264 -10 221 -10Q187 -10 162 2T124 33T105 68T98 100Q97 107 97 248V385H18V422H27Z" id="eq_dea10bfd_16MJMAIN-74" stroke-width="10"/&gt;
&lt;path d="M94 250Q94 319 104 381T127 488T164 576T202 643T244 695T277 729T302 750H315H319Q333 750 333 741Q333 738 316 720T275 667T226 581T184 443T167 250T184 58T225 -81T274 -167T316 -220T333 -241Q333 -250 318 -250H315H302L274 -226Q180 -141 137 -14T94 250Z" id="eq_dea10bfd_16MJMAIN-28" stroke-width="10"/&gt;
&lt;path d="M26 385Q19 392 19 395Q19 399 22 411T27 425Q29 430 36 430T87 431H140L159 511Q162 522 166 540T173 566T179 586T187 603T197 615T211 624T229 626Q247 625 254 615T261 596Q261 589 252 549T232 470L222 433Q222 431 272 431H323Q330 424 330 420Q330 398 317 385H210L174 240Q135 80 135 68Q135 26 162 26Q197 26 230 60T283 144Q285 150 288 151T303 153H307Q322 153 322 145Q322 142 319 133Q314 117 301 95T267 48T216 6T155 -11Q125 -11 98 4T59 56Q57 64 57 83V101L92 241Q127 382 128 383Q128 385 77 385H26Z" id="eq_dea10bfd_16MJMATHI-74" stroke-width="10"/&gt;
&lt;path d="M60 749L64 750Q69 750 74 750H86L114 726Q208 641 251 514T294 250Q294 182 284 119T261 12T224 -76T186 -143T145 -194T113 -227T90 -246Q87 -249 86 -250H74Q66 -250 63 -250T58 -247T55 -238Q56 -237 66 -225Q221 -64 221 250T66 725Q56 737 55 738Q55 746 60 749Z" id="eq_dea10bfd_16MJMAIN-29" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_16MJMATHI-76" y="0"/&gt;
&lt;g transform="translate(490,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_16MJMAIN-6F"/&gt;
 &lt;use transform="scale(0.707)" x="505" xlink:href="#eq_dea10bfd_16MJMAIN-75" y="0"/&gt;
 &lt;use transform="scale(0.707)" x="1066" xlink:href="#eq_dea10bfd_16MJMAIN-74" y="0"/&gt;
&lt;/g&gt;
&lt;g transform="translate(1622,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_16MJMAIN-28" y="0"/&gt;
 &lt;use x="394" xlink:href="#eq_dea10bfd_16MJMATHI-74" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, is 2&amp;#xA0;V. Therefore the gain is &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="020fa7ff8b8c5384e5bff463b0a620e724641825"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_17d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 2375.0 2179.2650" width="40.3232px"&gt;

&lt;desc id="eq_dea10bfd_17d"&gt;two cap v divided by 10 cap v&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M109 429Q82 429 66 447T50 491Q50 562 103 614T235 666Q326 666 387 610T449 465Q449 422 429 383T381 315T301 241Q265 210 201 149L142 93L218 92Q375 92 385 97Q392 99 409 186V189H449V186Q448 183 436 95T421 3V0H50V19V31Q50 38 56 46T86 81Q115 113 136 137Q145 147 170 174T204 211T233 244T261 278T284 308T305 340T320 369T333 401T340 431T343 464Q343 527 309 573T212 619Q179 619 154 602T119 569T109 550Q109 549 114 549Q132 549 151 535T170 489Q170 464 154 447T109 429Z" id="eq_dea10bfd_17MJMAIN-32" stroke-width="10"/&gt;
&lt;path d="M114 620Q113 621 110 624T107 627T103 630T98 632T91 634T80 635T67 636T48 637H19V683H28Q46 680 152 680Q273 680 294 683H305V637H284Q223 634 223 620Q223 618 313 372T404 126L490 358Q575 588 575 597Q575 616 554 626T508 637H503V683H512Q527 680 627 680Q718 680 724 683H730V637H723Q648 637 627 596Q627 595 515 291T401 -14Q396 -22 382 -22H374H367Q353 -22 348 -14Q346 -12 231 303Q114 617 114 620Z" id="eq_dea10bfd_17MJMAIN-56" stroke-width="10"/&gt;
&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_17MJMAIN-31" stroke-width="10"/&gt;
&lt;path d="M96 585Q152 666 249 666Q297 666 345 640T423 548Q460 465 460 320Q460 165 417 83Q397 41 362 16T301 -15T250 -22Q224 -22 198 -16T137 16T82 83Q39 165 39 320Q39 494 96 585ZM321 597Q291 629 250 629Q208 629 178 597Q153 571 145 525T137 333Q137 175 145 125T181 46Q209 16 250 16Q290 16 318 46Q347 76 354 130T362 333Q362 478 354 524T321 597Z" id="eq_dea10bfd_17MJMAIN-30" stroke-width="10"/&gt;
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&lt;/g&gt;
&lt;g transform="translate(60,-713)"&gt;
 &lt;use xlink:href="#eq_dea10bfd_17MJMAIN-31"/&gt;
 &lt;use x="505" xlink:href="#eq_dea10bfd_17MJMAIN-30" y="0"/&gt;
 &lt;use x="1260" xlink:href="#eq_dea10bfd_17MJMAIN-56" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, or 0.2. &lt;/li&gt;&lt;li class="oucontent-markeroutside"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_18d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_18d"&gt;cap v sub in&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 648Q52 670 65 683H76Q118 680 181 680Q299 680 320 683H330Q336 677 336 674T334 656Q329 641 325 637H304Q282 635 274 635Q245 630 242 620Q242 618 271 369T301 118L374 235Q447 352 520 471T595 594Q599 601 599 609Q599 633 555 637Q537 637 537 648Q537 649 539 661Q542 675 545 679T558 683Q560 683 570 683T604 682T668 681Q737 681 755 683H762Q769 676 769 672Q769 655 760 640Q757 637 743 637Q730 636 719 635T698 630T682 623T670 615T660 608T652 599T645 592L452 282Q272 -9 266 -16Q263 -18 259 -21L241 -22H234Q216 -22 216 -15Q213 -9 177 305Q139 623 138 626Q133 637 76 637H59Q52 642 52 648Z" id="eq_dea10bfd_18MJMATHI-56" stroke-width="10"/&gt;
&lt;path d="M69 609Q69 637 87 653T131 669Q154 667 171 652T188 609Q188 579 171 564T129 549Q104 549 87 564T69 609ZM247 0Q232 3 143 3Q132 3 106 3T56 1L34 0H26V46H42Q70 46 91 49Q100 53 102 60T104 102V205V293Q104 345 102 359T88 378Q74 385 41 385H30V408Q30 431 32 431L42 432Q52 433 70 434T106 436Q123 437 142 438T171 441T182 442H185V62Q190 52 197 50T232 46H255V0H247Z" id="eq_dea10bfd_18MJMAIN-69" stroke-width="10"/&gt;
&lt;path d="M41 46H55Q94 46 102 60V68Q102 77 102 91T102 122T103 161T103 203Q103 234 103 269T102 328V351Q99 370 88 376T43 385H25V408Q25 431 27 431L37 432Q47 433 65 434T102 436Q119 437 138 438T167 441T178 442H181V402Q181 364 182 364T187 369T199 384T218 402T247 421T285 437Q305 442 336 442Q450 438 463 329Q464 322 464 190V104Q464 66 466 59T477 49Q498 46 526 46H542V0H534L510 1Q487 2 460 2T422 3Q319 3 310 0H302V46H318Q379 46 379 62Q380 64 380 200Q379 335 378 343Q372 371 358 385T334 402T308 404Q263 404 229 370Q202 343 195 315T187 232V168V108Q187 78 188 68T191 55T200 49Q221 46 249 46H265V0H257L234 1Q210 2 183 2T145 3Q42 3 33 0H25V46H41Z" id="eq_dea10bfd_18MJMAIN-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_18MJMATHI-56" y="0"/&gt;
&lt;g transform="translate(588,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_18MJMAIN-69"/&gt;
 &lt;use transform="scale(0.707)" x="283" xlink:href="#eq_dea10bfd_18MJMAIN-6E" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is still 10&amp;#xA0;V and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_19d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_19d"&gt;cap v sub out&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 648Q52 670 65 683H76Q118 680 181 680Q299 680 320 683H330Q336 677 336 674T334 656Q329 641 325 637H304Q282 635 274 635Q245 630 242 620Q242 618 271 369T301 118L374 235Q447 352 520 471T595 594Q599 601 599 609Q599 633 555 637Q537 637 537 648Q537 649 539 661Q542 675 545 679T558 683Q560 683 570 683T604 682T668 681Q737 681 755 683H762Q769 676 769 672Q769 655 760 640Q757 637 743 637Q730 636 719 635T698 630T682 623T670 615T660 608T652 599T645 592L452 282Q272 -9 266 -16Q263 -18 259 -21L241 -22H234Q216 -22 216 -15Q213 -9 177 305Q139 623 138 626Q133 637 76 637H59Q52 642 52 648Z" id="eq_dea10bfd_19MJMATHI-56" stroke-width="10"/&gt;
&lt;path d="M28 214Q28 309 93 378T250 448Q340 448 405 380T471 215Q471 120 407 55T250 -10Q153 -10 91 57T28 214ZM250 30Q372 30 372 193V225V250Q372 272 371 288T364 326T348 362T317 390T268 410Q263 411 252 411Q222 411 195 399Q152 377 139 338T126 246V226Q126 130 145 91Q177 30 250 30Z" id="eq_dea10bfd_19MJMAIN-6F" stroke-width="10"/&gt;
&lt;path d="M383 58Q327 -10 256 -10H249Q124 -10 105 89Q104 96 103 226Q102 335 102 348T96 369Q86 385 36 385H25V408Q25 431 27 431L38 432Q48 433 67 434T105 436Q122 437 142 438T172 441T184 442H187V261Q188 77 190 64Q193 49 204 40Q224 26 264 26Q290 26 311 35T343 58T363 90T375 120T379 144Q379 145 379 161T380 201T380 248V315Q380 361 370 372T320 385H302V431Q304 431 378 436T457 442H464V264Q464 84 465 81Q468 61 479 55T524 46H542V0Q540 0 467 -5T390 -11H383V58Z" id="eq_dea10bfd_19MJMAIN-75" stroke-width="10"/&gt;
&lt;path d="M27 422Q80 426 109 478T141 600V615H181V431H316V385H181V241Q182 116 182 100T189 68Q203 29 238 29Q282 29 292 100Q293 108 293 146V181H333V146V134Q333 57 291 17Q264 -10 221 -10Q187 -10 162 2T124 33T105 68T98 100Q97 107 97 248V385H18V422H27Z" id="eq_dea10bfd_19MJMAIN-74" stroke-width="10"/&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_19MJMATHI-56" y="0"/&gt;
&lt;g transform="translate(588,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_19MJMAIN-6F"/&gt;
 &lt;use transform="scale(0.707)" x="505" xlink:href="#eq_dea10bfd_19MJMAIN-75" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is now 5&amp;#xA0;V. Therefore the gain is now &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="37cf167ffedf637208d6fb6500e099408581202c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_20d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 2375.0 2179.2650" width="40.3232px"&gt;

&lt;desc id="eq_dea10bfd_20d"&gt;five cap v divided by 10 cap v&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M114 620Q113 621 110 624T107 627T103 630T98 632T91 634T80 635T67 636T48 637H19V683H28Q46 680 152 680Q273 680 294 683H305V637H284Q223 634 223 620Q223 618 313 372T404 126L490 358Q575 588 575 597Q575 616 554 626T508 637H503V683H512Q527 680 627 680Q718 680 724 683H730V637H723Q648 637 627 596Q627 595 515 291T401 -14Q396 -22 382 -22H374H367Q353 -22 348 -14Q346 -12 231 303Q114 617 114 620Z" id="eq_dea10bfd_20MJMAIN-56" stroke-width="10"/&gt;
&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_20MJMAIN-31" stroke-width="10"/&gt;
&lt;path d="M96 585Q152 666 249 666Q297 666 345 640T423 548Q460 465 460 320Q460 165 417 83Q397 41 362 16T301 -15T250 -22Q224 -22 198 -16T137 16T82 83Q39 165 39 320Q39 494 96 585ZM321 597Q291 629 250 629Q208 629 178 597Q153 571 145 525T137 333Q137 175 145 125T181 46Q209 16 250 16Q290 16 318 46Q347 76 354 130T362 333Q362 478 354 524T321 597Z" id="eq_dea10bfd_20MJMAIN-30" stroke-width="10"/&gt;
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&lt;g transform="translate(60,-713)"&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, or&amp;#xA0;0.5. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;In the next section you will discover the characteristics of an ideal filter.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.1</guid>
    <dc:title>2.1 Frequency-dependent gain</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;In Figure 4, the box in the middle represents a device with frequency-dependent gain &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="5643590c6c73fa99ad1af505b6427addd3d4da2e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_1d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 2206.0 1119.0820" width="37.4539px"&gt;

&lt;desc id="eq_dea10bfd_1d"&gt;cap g of omega&lt;/desc&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; – in other words, a filter. Here the symbol &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="f9d4d2552f37ca2f0b559ff2c54a7b5345cf4388"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_2d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 627.0 530.0915" width="10.6453px"&gt;

&lt;desc id="eq_dea10bfd_2d"&gt;omega&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is being used to represent angular frequency, measured in radians per second. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/8a631f95/t312_openlearn_fig04.tif.jpg" alt="Described image" width="512" height="128" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049952576"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 4&lt;/b&gt; Filter with input amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_3d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_3d"&gt;cap v sub in&lt;/desc&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and output amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_4d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_4d"&gt;cap v sub out&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049952576&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049952576"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;A sinusoidal input signal with amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_5d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_5d"&gt;cap v sub in&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is applied to the filter, and the output is a sinusoidal signal with amplitude &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_6d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_6d"&gt;cap v sub out&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. The &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048963936" class="oucontent-glossaryterm" data-definition="For a sinusoidal input and output, voltage gain is the ratio of the output voltage’s amplitude to that of the input voltage. It has no units." title="For a sinusoidal input and output, voltage gain is the ratio of the output voltage’s amplitude to th..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;voltage gain&lt;/span&gt;&lt;/a&gt; of the filter at the frequency in question is &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="cec405e9bb8187ef230b8e404de308f1af76a777"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_7d" height="39px" role="math" style="vertical-align: -15px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 2080.4 2297.0631" width="35.3215px"&gt;

&lt;desc id="eq_dea10bfd_7d"&gt;cap v sub out divided by cap v sub in&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; – in other words, it is the ratio of the output voltage amplitude to the input voltage amplitude. So, at any particular frequency, &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b51097a1ff3c81180c4eb69079afb05e967150ea"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_8d" height="39px" role="math" style="vertical-align: -15px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 4209.9 2297.0631" width="71.4765px"&gt;

&lt;desc id="eq_dea10bfd_8d"&gt;cap v sub out divided by cap v sub in equals cap g&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;The voltage gain can be expressed simply as a number or fraction (or decimal). For example, a gain of 2 means that the output amplitude is twice the input amplitude. A gain of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="3e925096cff54129be65e4d688e475a76943ba75"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_9d" height="25px" role="math" style="vertical-align: -8px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1001.2839 717.1 1472.4763" width="12.1751px"&gt;

&lt;desc id="eq_dea10bfd_9d"&gt;one divided by four&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; (or 0.25) means that the amplitude of the output is one-quarter that of the input. &lt;/p&gt;&lt;p&gt;The above gain is referred to as ‘voltage gain’ because gain is sometimes expressed as a ratio of output and input &lt;i&gt;powers&lt;/i&gt;. In this course this way of expressing gain will be referred to as &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048972400" class="oucontent-glossaryterm" data-definition="The ratio of output power to input power. It is usually expressed in decibels (dB). A power gain of 0 dB means that the output power is the same as the input power. A power gain of 3 dB (or, more exactly, 3.0103 dB) means that the output power is double the input power." title="The ratio of output power to input power. It is usually expressed in decibels (dB). A power gain of ..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;power gain&lt;/span&gt;&lt;/a&gt;. As power ratios can be expressed in decibels, so power gains are almost invariably given in &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048993968" class="oucontent-glossaryterm" data-definition="A logarithmic way of expressing a power ratio. For powers P1 and P2, their ratio in decibels is defined as 10 log10 (P1/P2). The symbol for decibels is dB. Strictly the decibel is not a unit, as any ratio must be a pure (that is, dimensionless) number. However, it is often regarded as a unit." title="A logarithmic way of expressing a power ratio. For powers P1 and P2, their ratio in decibels is defi..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;decibels&lt;/span&gt;&lt;/a&gt;. &lt;/p&gt;&lt;p&gt;For the time being you will only consider sinusoidal inputs and outputs, as these have a single frequency. This limitation to a single frequency helps to clarify what a filter does. In practice, though, a filter would typically operate on a complex waveform consisting of many frequency components. In such a case, the inputs and outputs would themselves be functions of frequency. &lt;/p&gt;&lt;div class="oucontent-studynote oucontent-s-gradient oucontent-s-box 
        oucontent-s-noheading
      "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;div class="oucontent-inner-box"&gt;&lt;p&gt;Although two different types of gain have been mentioned (voltage gain and power gain), you will often just see the word &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048982048" class="oucontent-glossaryterm" data-definition="In amplification, a measure of how many times the input signal amplitude is increased. It is generally measured as the ratio between the input signal amplitude and the output signal level. If a gain value is given as just a number (i.e. with no units), the gain is likely to be a ratio of voltages; if the value is given in decibels, it is a ratio of powers. See also voltage gain and power gain." title="In amplification, a measure of how many times the input signal amplitude is increased. It is general..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;gain&lt;/span&gt;&lt;/a&gt; used by itself. If the gain is given as an ordinary numerical value (such as 2, 10, 3000 or 0.001), voltage gain is almost invariably indicated. If the numerical value is in decibels, power gain is being represented. &lt;/p&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;The output of a filter differs from the input not only in amplitude but (usually) also in phase. You will look more closely at the question of phase later in the course. However, for now you will continue to focus on gain. Complete Activity 1 to test your understanding so far.&lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 1 &lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;A passive filter has an input signal of  &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c3cfe2cc3ff26fd86f660a7f4b81fa77f5386b5d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_10d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 8313.7 1119.0820" width="141.1517px"&gt;

&lt;desc id="eq_dea10bfd_10d"&gt;v sub in of t equals 10 postfix times sine of 200 times t&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; volts. The steady-state output is &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="6b30bf9a4051fdcb656cf012d7b413ac3c9d3344"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_11d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 11964.0 1119.0820" width="203.1272px"&gt;

&lt;desc id="eq_dea10bfd_11d"&gt;v sub out of t equals two postfix times sine of 200 times t minus 0.6 times pi&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; volts. What is the gain as a voltage ratio? &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The input to the filter in part (a) remains unchanged in amplitude, but its frequency changes. The steady-state output is now found to be  &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="68d80e4142ae5a14268251cb13c1fff6c58b5b79"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_12d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 11964.0 1119.0820" width="203.1272px"&gt;

&lt;desc id="eq_dea10bfd_12d"&gt;v sub out of t equals five postfix times sine of 100 times t minus 0.2 times pi&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; volts. What is the new gain as a voltage ratio? &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Here &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_13d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_13d"&gt;cap v sub in&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_14d"&gt;v sub in of t&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, is 10 V and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_15d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_15d"&gt;cap v sub out&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_16d"&gt;v sub out of t&lt;/desc&gt;
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&lt;path d="M94 250Q94 319 104 381T127 488T164 576T202 643T244 695T277 729T302 750H315H319Q333 750 333 741Q333 738 316 720T275 667T226 581T184 443T167 250T184 58T225 -81T274 -167T316 -220T333 -241Q333 -250 318 -250H315H302L274 -226Q180 -141 137 -14T94 250Z" id="eq_dea10bfd_16MJMAIN-28" stroke-width="10"/&gt;
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&lt;path d="M60 749L64 750Q69 750 74 750H86L114 726Q208 641 251 514T294 250Q294 182 284 119T261 12T224 -76T186 -143T145 -194T113 -227T90 -246Q87 -249 86 -250H74Q66 -250 63 -250T58 -247T55 -238Q56 -237 66 -225Q221 -64 221 250T66 725Q56 737 55 738Q55 746 60 749Z" id="eq_dea10bfd_16MJMAIN-29" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_16MJMATHI-76" y="0"/&gt;
&lt;g transform="translate(490,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_16MJMAIN-6F"/&gt;
 &lt;use transform="scale(0.707)" x="505" xlink:href="#eq_dea10bfd_16MJMAIN-75" y="0"/&gt;
 &lt;use transform="scale(0.707)" x="1066" xlink:href="#eq_dea10bfd_16MJMAIN-74" y="0"/&gt;
&lt;/g&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_16MJMAIN-28" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, is 2 V. Therefore the gain is &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="020fa7ff8b8c5384e5bff463b0a620e724641825"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_17d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 2375.0 2179.2650" width="40.3232px"&gt;

&lt;desc id="eq_dea10bfd_17d"&gt;two cap v divided by 10 cap v&lt;/desc&gt;
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 &lt;use xlink:href="#eq_dea10bfd_17MJMAIN-31"/&gt;
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 &lt;use x="1260" xlink:href="#eq_dea10bfd_17MJMAIN-56" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, or 0.2. &lt;/li&gt;&lt;li class="oucontent-markeroutside"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="b39d806227b5b904b0f8a354d8ac5922d7e8676d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_18d" height="16px" role="math" style="vertical-align: -3px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1284.8 942.3849" width="21.8136px"&gt;

&lt;desc id="eq_dea10bfd_18d"&gt;cap v sub in&lt;/desc&gt;
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&lt;path d="M69 609Q69 637 87 653T131 669Q154 667 171 652T188 609Q188 579 171 564T129 549Q104 549 87 564T69 609ZM247 0Q232 3 143 3Q132 3 106 3T56 1L34 0H26V46H42Q70 46 91 49Q100 53 102 60T104 102V205V293Q104 345 102 359T88 378Q74 385 41 385H30V408Q30 431 32 431L42 432Q52 433 70 434T106 436Q123 437 142 438T171 441T182 442H185V62Q190 52 197 50T232 46H255V0H247Z" id="eq_dea10bfd_18MJMAIN-69" stroke-width="10"/&gt;
&lt;path d="M41 46H55Q94 46 102 60V68Q102 77 102 91T102 122T103 161T103 203Q103 234 103 269T102 328V351Q99 370 88 376T43 385H25V408Q25 431 27 431L37 432Q47 433 65 434T102 436Q119 437 138 438T167 441T178 442H181V402Q181 364 182 364T187 369T199 384T218 402T247 421T285 437Q305 442 336 442Q450 438 463 329Q464 322 464 190V104Q464 66 466 59T477 49Q498 46 526 46H542V0H534L510 1Q487 2 460 2T422 3Q319 3 310 0H302V46H318Q379 46 379 62Q380 64 380 200Q379 335 378 343Q372 371 358 385T334 402T308 404Q263 404 229 370Q202 343 195 315T187 232V168V108Q187 78 188 68T191 55T200 49Q221 46 249 46H265V0H257L234 1Q210 2 183 2T145 3Q42 3 33 0H25V46H41Z" id="eq_dea10bfd_18MJMAIN-6E" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_18MJMATHI-56" y="0"/&gt;
&lt;g transform="translate(588,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_18MJMAIN-69"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is still 10 V and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="7087a1d7164a02d73b1e89a1e14b15c6099ed437"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_19d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 1720.4 1001.2839" width="29.2093px"&gt;

&lt;desc id="eq_dea10bfd_19d"&gt;cap v sub out&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 648Q52 670 65 683H76Q118 680 181 680Q299 680 320 683H330Q336 677 336 674T334 656Q329 641 325 637H304Q282 635 274 635Q245 630 242 620Q242 618 271 369T301 118L374 235Q447 352 520 471T595 594Q599 601 599 609Q599 633 555 637Q537 637 537 648Q537 649 539 661Q542 675 545 679T558 683Q560 683 570 683T604 682T668 681Q737 681 755 683H762Q769 676 769 672Q769 655 760 640Q757 637 743 637Q730 636 719 635T698 630T682 623T670 615T660 608T652 599T645 592L452 282Q272 -9 266 -16Q263 -18 259 -21L241 -22H234Q216 -22 216 -15Q213 -9 177 305Q139 623 138 626Q133 637 76 637H59Q52 642 52 648Z" id="eq_dea10bfd_19MJMATHI-56" stroke-width="10"/&gt;
&lt;path d="M28 214Q28 309 93 378T250 448Q340 448 405 380T471 215Q471 120 407 55T250 -10Q153 -10 91 57T28 214ZM250 30Q372 30 372 193V225V250Q372 272 371 288T364 326T348 362T317 390T268 410Q263 411 252 411Q222 411 195 399Q152 377 139 338T126 246V226Q126 130 145 91Q177 30 250 30Z" id="eq_dea10bfd_19MJMAIN-6F" stroke-width="10"/&gt;
&lt;path d="M383 58Q327 -10 256 -10H249Q124 -10 105 89Q104 96 103 226Q102 335 102 348T96 369Q86 385 36 385H25V408Q25 431 27 431L38 432Q48 433 67 434T105 436Q122 437 142 438T172 441T184 442H187V261Q188 77 190 64Q193 49 204 40Q224 26 264 26Q290 26 311 35T343 58T363 90T375 120T379 144Q379 145 379 161T380 201T380 248V315Q380 361 370 372T320 385H302V431Q304 431 378 436T457 442H464V264Q464 84 465 81Q468 61 479 55T524 46H542V0Q540 0 467 -5T390 -11H383V58Z" id="eq_dea10bfd_19MJMAIN-75" stroke-width="10"/&gt;
&lt;path d="M27 422Q80 426 109 478T141 600V615H181V431H316V385H181V241Q182 116 182 100T189 68Q203 29 238 29Q282 29 292 100Q293 108 293 146V181H333V146V134Q333 57 291 17Q264 -10 221 -10Q187 -10 162 2T124 33T105 68T98 100Q97 107 97 248V385H18V422H27Z" id="eq_dea10bfd_19MJMAIN-74" stroke-width="10"/&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_19MJMATHI-56" y="0"/&gt;
&lt;g transform="translate(588,-150)"&gt;
 &lt;use transform="scale(0.707)" xlink:href="#eq_dea10bfd_19MJMAIN-6F"/&gt;
 &lt;use transform="scale(0.707)" x="505" xlink:href="#eq_dea10bfd_19MJMAIN-75" y="0"/&gt;
 &lt;use transform="scale(0.707)" x="1066" xlink:href="#eq_dea10bfd_19MJMAIN-74" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is now 5 V. Therefore the gain is now &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="37cf167ffedf637208d6fb6500e099408581202c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_20d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 2375.0 2179.2650" width="40.3232px"&gt;

&lt;desc id="eq_dea10bfd_20d"&gt;five cap v divided by 10 cap v&lt;/desc&gt;
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&lt;path d="M96 585Q152 666 249 666Q297 666 345 640T423 548Q460 465 460 320Q460 165 417 83Q397 41 362 16T301 -15T250 -22Q224 -22 198 -16T137 16T82 83Q39 165 39 320Q39 494 96 585ZM321 597Q291 629 250 629Q208 629 178 597Q153 571 145 525T137 333Q137 175 145 125T181 46Q209 16 250 16Q290 16 318 46Q347 76 354 130T362 333Q362 478 354 524T321 597Z" id="eq_dea10bfd_20MJMAIN-30" stroke-width="10"/&gt;
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 &lt;use x="755" xlink:href="#eq_dea10bfd_20MJMAIN-56" y="0"/&gt;
&lt;/g&gt;
&lt;g transform="translate(60,-713)"&gt;
 &lt;use xlink:href="#eq_dea10bfd_20MJMAIN-31"/&gt;
 &lt;use x="505" xlink:href="#eq_dea10bfd_20MJMAIN-30" y="0"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, or 0.5. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;In the next section you will discover the characteristics of an ideal filter.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.2 Gain functions of ideal filters</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.2</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Figure 5 shows some common types of &amp;#x2018;ideal’ filter. Ideal filters are sometimes characterised as &amp;#x2018;brick-wall’ filters because graphs of their gain functions have perfectly horizontal or vertical lines. In practice, such brick-wall gain functions can never be achieved. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/65562369/t312_openlearn_fig05.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049880768"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 5&lt;/b&gt;  Four types of ideal filter: (a) low-pass filter; (b) high-pass filter; (c) band-pass filter; (d) band-stop (notch) filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049880768&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049880768"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Notice that each of the four types of filter has a name summarising what it does. For example, the low-pass filter (Figure&amp;#xA0;5(a)) passes all frequencies below the cut-off frequency &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="cb07dbc0fc3146a086dfd8bdc6b1806b67736221"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_21d" height="18px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 912.5 1060.1830" width="15.4926px"&gt;

&lt;desc id="eq_dea10bfd_21d"&gt;f sub c&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and blocks all frequencies above it. &lt;/p&gt;&lt;p&gt;In all the filters, a frequency band where the signal is passed is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048974096" class="oucontent-glossaryterm" data-definition="The band or bands of frequencies passed by a filter with least attenuation, or no attenuation. Frequencies outside the passband are cut off, or stopped, by the filter. Passband is the counterpart of stop band." title="The band or bands of frequencies passed by a filter with least attenuation, or no attenuation. Frequ..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;passband&lt;/span&gt;&lt;/a&gt;, and a frequency band where the signal is blocked is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048967328" class="oucontent-glossaryterm" data-definition="The band or bands of frequencies stopped, or cut off, by a filter. The counterpart of the passband." title="The band or bands of frequencies stopped, or cut off, by a filter. The counterpart of the passband."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;stop band&lt;/span&gt;&lt;/a&gt;. All the filters in Figure&amp;#xA0;5 have one or more passbands, one or more stop bands, and one or more cut-off frequencies. &lt;/p&gt;&lt;p&gt;In all the passbands shown, the voltage gain is 1, but this is a convention for this type of diagram. The actual passband gain depends on various factors such as whether the filter is passive (that is, consists only of passive components, such as resistors, capacitors and inductors) or active (that is, includes amplification as well as passive components). &lt;/p&gt;&lt;p&gt;Similarly, all the stop bands are shown with a voltage gain of 0. In practice, the gain is likely to be above 0. However, provided the stop-band gain is several orders of magnitude below the passband gain, the term &amp;#x2018;stop band’ is reasonable. &lt;/p&gt;&lt;p&gt;In the next section you will see what is meant by interference and noise, which is what is trying to be removed from a signal using a filter. &lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.2</guid>
    <dc:title>2.2 Gain functions of ideal filters</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Figure 5 shows some common types of ‘ideal’ filter. Ideal filters are sometimes characterised as ‘brick-wall’ filters because graphs of their gain functions have perfectly horizontal or vertical lines. In practice, such brick-wall gain functions can never be achieved. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/65562369/t312_openlearn_fig05.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049880768"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 5&lt;/b&gt;  Four types of ideal filter: (a) low-pass filter; (b) high-pass filter; (c) band-pass filter; (d) band-stop (notch) filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049880768&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049880768"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Notice that each of the four types of filter has a name summarising what it does. For example, the low-pass filter (Figure 5(a)) passes all frequencies below the cut-off frequency &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="cb07dbc0fc3146a086dfd8bdc6b1806b67736221"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_21d" height="18px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 912.5 1060.1830" width="15.4926px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and blocks all frequencies above it. &lt;/p&gt;&lt;p&gt;In all the filters, a frequency band where the signal is passed is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048974096" class="oucontent-glossaryterm" data-definition="The band or bands of frequencies passed by a filter with least attenuation, or no attenuation. Frequencies outside the passband are cut off, or stopped, by the filter. Passband is the counterpart of stop band." title="The band or bands of frequencies passed by a filter with least attenuation, or no attenuation. Frequ..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;passband&lt;/span&gt;&lt;/a&gt;, and a frequency band where the signal is blocked is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048967328" class="oucontent-glossaryterm" data-definition="The band or bands of frequencies stopped, or cut off, by a filter. The counterpart of the passband." title="The band or bands of frequencies stopped, or cut off, by a filter. The counterpart of the passband."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;stop band&lt;/span&gt;&lt;/a&gt;. All the filters in Figure 5 have one or more passbands, one or more stop bands, and one or more cut-off frequencies. &lt;/p&gt;&lt;p&gt;In all the passbands shown, the voltage gain is 1, but this is a convention for this type of diagram. The actual passband gain depends on various factors such as whether the filter is passive (that is, consists only of passive components, such as resistors, capacitors and inductors) or active (that is, includes amplification as well as passive components). &lt;/p&gt;&lt;p&gt;Similarly, all the stop bands are shown with a voltage gain of 0. In practice, the gain is likely to be above 0. However, provided the stop-band gain is several orders of magnitude below the passband gain, the term ‘stop band’ is reasonable. &lt;/p&gt;&lt;p&gt;In the next section you will see what is meant by interference and noise, which is what is trying to be removed from a signal using a filter. &lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.3 Types of interference</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.3</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;In the same way that Figure&amp;#xA0;5 shows simplified models of filters, there exist simplified models of the type of signals and noise we might want to apply filters to. &lt;/p&gt;&lt;p&gt;A common type of interference is adjacent channel interference, in which the interfering signal is in a frequency band above or below that of the wanted signal, as shown in Figure&amp;#xA0;6. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/7e13395b/t312_openlearn_fig06.tif.jpg" alt="Described image" width="512" height="115" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049851632"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 6&lt;/b&gt; Adjacent channel interference: (a) below wanted signal; (b) above wanted signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049851632&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049851632"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In each case, an appropriate filter can be used to reduce the interference. For example, you can see that in the case of Figure&amp;#xA0;6(b), where the adjacent channel interference is above the wanted signal, a low-pass filter can be used, with the passband coinciding with the wanted signal and the stop band coinciding with the interference. With this arrangement an ideal filter could, in principle, remove the interference altogether. (However, as you will see later, the reality is somewhat different.) &lt;/p&gt;&lt;p&gt;Life is trickier when the signal and interference overlap in frequency, as in the narrowband interference and wideband interference shown in Figure&amp;#xA0;7. Here, no form of filtering can give us what we would like, which is a noise-free signal with no adverse effect on the signal. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/e82d0905/t312_openlearn_fig07.tif.jpg" alt="Described image" width="512" height="115" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049839552"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 7&lt;/b&gt;  (a) Narrowband interference; (b) wideband interference &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049839552&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049839552"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Note that, just as brick-wall filters are unachievable in practice, the brick-wall frequency bands and signal strengths of these various types of interference are not achievable in practice. In real-world situations, the boundaries are less clearly defined. &lt;/p&gt;&lt;p&gt;Now complete Activity 2 and apply the correct filter to the interference type.&lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 2&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;For each of the following types of interference, suggest a suitable filter to improve the signal-to-noise ratio, and say how the passbands and stop bands should be arranged. Explain any drawbacks. &lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Narrowband interference&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;Wideband interference&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;For narrowband interference you can use a band-stop filter, with the stop band centred on the interference and of an equal width. This could in principle remove the interference. The drawback is that the stop band also removes some signal power. &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;When you have wideband interference, a suitable remedy is to use a band-pass filter with the passband centred on the wanted signal and equal in width to the bandwidth of the signal. However, although this gives the best signal-to-noise ratio, it cannot remove all the interference. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Having seen the characteristics of ideal filters and the sorts of interference that you want to remove, the next section will look you what real filters are like.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.3</guid>
    <dc:title>2.3 Types of interference</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;In the same way that Figure 5 shows simplified models of filters, there exist simplified models of the type of signals and noise we might want to apply filters to. &lt;/p&gt;&lt;p&gt;A common type of interference is adjacent channel interference, in which the interfering signal is in a frequency band above or below that of the wanted signal, as shown in Figure 6. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/7e13395b/t312_openlearn_fig06.tif.jpg" alt="Described image" width="512" height="115" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049851632"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 6&lt;/b&gt; Adjacent channel interference: (a) below wanted signal; (b) above wanted signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049851632&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049851632"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In each case, an appropriate filter can be used to reduce the interference. For example, you can see that in the case of Figure 6(b), where the adjacent channel interference is above the wanted signal, a low-pass filter can be used, with the passband coinciding with the wanted signal and the stop band coinciding with the interference. With this arrangement an ideal filter could, in principle, remove the interference altogether. (However, as you will see later, the reality is somewhat different.) &lt;/p&gt;&lt;p&gt;Life is trickier when the signal and interference overlap in frequency, as in the narrowband interference and wideband interference shown in Figure 7. Here, no form of filtering can give us what we would like, which is a noise-free signal with no adverse effect on the signal. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/e82d0905/t312_openlearn_fig07.tif.jpg" alt="Described image" width="512" height="115" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049839552"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 7&lt;/b&gt;  (a) Narrowband interference; (b) wideband interference &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049839552&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049839552"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Note that, just as brick-wall filters are unachievable in practice, the brick-wall frequency bands and signal strengths of these various types of interference are not achievable in practice. In real-world situations, the boundaries are less clearly defined. &lt;/p&gt;&lt;p&gt;Now complete Activity 2 and apply the correct filter to the interference type.&lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 2&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;For each of the following types of interference, suggest a suitable filter to improve the signal-to-noise ratio, and say how the passbands and stop bands should be arranged. Explain any drawbacks. &lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Narrowband interference&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;Wideband interference&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;For narrowband interference you can use a band-stop filter, with the stop band centred on the interference and of an equal width. This could in principle remove the interference. The drawback is that the stop band also removes some signal power. &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;When you have wideband interference, a suitable remedy is to use a band-pass filter with the passband centred on the wanted signal and equal in width to the bandwidth of the signal. However, although this gives the best signal-to-noise ratio, it cannot remove all the interference. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Having seen the characteristics of ideal filters and the sorts of interference that you want to remove, the next section will look you what real filters are like.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.4 First-order filters</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.4</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;In addition to the filter categories already introduced (low-pass, band-pass, etc.), filters are categorised by their &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048976144" class="oucontent-glossaryterm" data-definition="A numerical classification for filters (e.g. &amp;#x2018;first order’, &amp;#x2018;second order’, &amp;#x2018;third order’, etc.). The order is determined by the differential equation of the filter. For a first-order filter, the highest differential coefficient in the equation is first-order (e.g. dv/dt); for a second-order filter, the highest differential coefficient is second-order (e.g. d2v/dt2). The higher the order, the steeper the roll-off and the sharper the cut-off between passband and stop band. Increasing the order by one adds 20 dB/decade to the filter’s roll-off." title="A numerical classification for filters (e.g. &amp;#x2018;first order’, &amp;#x2018;second order’, &amp;#x2018;third order’, etc.). Th..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;order&lt;/span&gt;&lt;/a&gt;. The order of a filter is determined by the form of the &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048990576" class="oucontent-glossaryterm" data-definition="A mathematical equation in which one or more terms contains a mathematically differentiated variable." title="A mathematical equation in which one or more terms contains a mathematically differentiated variable..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;differential equation&lt;/span&gt;&lt;/a&gt; governing the filter’s behaviour. The simplest type of filter, with the simplest equation, is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048987296" class="oucontent-glossaryterm" data-definition="As applied to a filter, the simplest type of filter, having in its passive form a single reactive element (a capacitor or an inductor) and a roll-off of 20 dB/decade, or 6 dB/octave. As applied to a differential equation, such an equation in which the main variable is differentiated once. Any system that can be modelled with such a differential equation would be referred to as a first-order system." title="As applied to a filter, the simplest type of filter, having in its passive form a single reactive el..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;first-order&lt;/span&gt;&lt;/a&gt; filter. &lt;b&gt;Higher-order&lt;/b&gt; filters are more complex than first-order filters, both in their circuitry and in the differential equation that governs them. The higher the order, the more effective the filter. &lt;/p&gt;&lt;p&gt;An example of a first-order filter is the simple circuit in Figure&amp;#xA0;8.&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/23ef598c/t312_openlearn_fig08.tif.jpg" alt="Described image" width="512" height="187" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049813360"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 8&lt;/b&gt;  First-order filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049813360&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049813360"&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 3&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Which of the four categories of filter shown in Figure&amp;#xA0;5 does the filter in Figure&amp;#xA0;8 belong to? You should be able to work it out from the behaviour of the capacitor at low frequencies and high frequencies. Explain your answer.&lt;/p&gt;
&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/65562369/t312_openlearn_fig05.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049798896"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 5 (repeated)&lt;/b&gt;  Four types of ideal filter: (a) low-pass filter; (b) high-pass filter; (c) band-pass filter; (d) band-stop (notch) filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049798896&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049798896"&gt;&lt;/a&gt;&lt;/div&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;It is a low-pass filter. &lt;/p&gt;
&lt;p&gt;At 0 Hz, or DC, the capacitor is open circuit. Under those circumstances, all the input voltage would appear on the output. At high frequencies, the capacitor becomes increasingly like a short circuit, and the output voltage decreases as frequency increases. Therefore low frequencies are passed and high frequencies are (to some extent) blocked. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Low-pass and high-pass filters can be first-order, second-order, third-order, and so on. However, band-pass filters and band-stop filters must be second-order or higher, although it is possible to achieve their effect by combining two first-order filters. &lt;/p&gt;&lt;p&gt;Despite being the simplest type, a first-order filter is not simple to analyse mathematically, as its behaviour is governed by a first-order differential equation. This is what the voltage gain function of the filter in Figure&amp;#xA0;8 turns out to be: &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a8c76e09278d2d40cea9969d5a7bd4035605d33b"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_22d" height="57px" role="math" style="vertical-align: -33px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 12047.0 3357.2461" width="204.5364px"&gt;

&lt;desc id="eq_dea10bfd_22d"&gt;equation sequence cap g equals cap v sub out divided by cap v sub in equals one divided by Square root of one plus open omega times cap r times cap c close squared&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;You could plot a graph of the gain function against frequency from this equation, but the details would depend on the choice of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="97a59cbf4a47ad1d1ecc1bb0a57622c087b17a7f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_23d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 764.0 824.5868" width="12.9713px"&gt;

&lt;desc id="eq_dea10bfd_23d"&gt;cap r&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c7ee6a8a0b0501ca0380eabacbd2361c02628cd3"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_24d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 765.0 824.5868" width="12.9883px"&gt;

&lt;desc id="eq_dea10bfd_24d"&gt;cap c&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. However, irrespective of the values chosen for &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="97a59cbf4a47ad1d1ecc1bb0a57622c087b17a7f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_25d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 764.0 824.5868" width="12.9713px"&gt;

&lt;desc id="eq_dea10bfd_25d"&gt;cap r&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c7ee6a8a0b0501ca0380eabacbd2361c02628cd3"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_26d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 765.0 824.5868" width="12.9883px"&gt;

&lt;desc id="eq_dea10bfd_26d"&gt;cap c&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, the general shape and certain other properties of the gain function would be the same for any first-order low-pass filter. Therefore, rather than show how the gain function changes with frequency for particular values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="97a59cbf4a47ad1d1ecc1bb0a57622c087b17a7f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_27d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 764.0 824.5868" width="12.9713px"&gt;

&lt;desc id="eq_dea10bfd_27d"&gt;cap r&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c7ee6a8a0b0501ca0380eabacbd2361c02628cd3"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_28d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 765.0 824.5868" width="12.9883px"&gt;

&lt;desc id="eq_dea10bfd_28d"&gt;cap c&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, in the next section you will look at a &amp;#x2018;normalised’ version of the gain function. (Normalisation is the presentation of information in a generalised way that can easily be adapted to specific cases.) &lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.4</guid>
    <dc:title>2.4 First-order filters</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;In addition to the filter categories already introduced (low-pass, band-pass, etc.), filters are categorised by their &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048976144" class="oucontent-glossaryterm" data-definition="A numerical classification for filters (e.g. ‘first order’, ‘second order’, ‘third order’, etc.). The order is determined by the differential equation of the filter. For a first-order filter, the highest differential coefficient in the equation is first-order (e.g. dv/dt); for a second-order filter, the highest differential coefficient is second-order (e.g. d2v/dt2). The higher the order, the steeper the roll-off and the sharper the cut-off between passband and stop band. Increasing the order by one adds 20 dB/decade to the filter’s roll-off." title="A numerical classification for filters (e.g. ‘first order’, ‘second order’, ‘third order’, etc.). Th..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;order&lt;/span&gt;&lt;/a&gt;. The order of a filter is determined by the form of the &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048990576" class="oucontent-glossaryterm" data-definition="A mathematical equation in which one or more terms contains a mathematically differentiated variable." title="A mathematical equation in which one or more terms contains a mathematically differentiated variable..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;differential equation&lt;/span&gt;&lt;/a&gt; governing the filter’s behaviour. The simplest type of filter, with the simplest equation, is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048987296" class="oucontent-glossaryterm" data-definition="As applied to a filter, the simplest type of filter, having in its passive form a single reactive element (a capacitor or an inductor) and a roll-off of 20 dB/decade, or 6 dB/octave. As applied to a differential equation, such an equation in which the main variable is differentiated once. Any system that can be modelled with such a differential equation would be referred to as a first-order system." title="As applied to a filter, the simplest type of filter, having in its passive form a single reactive el..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;first-order&lt;/span&gt;&lt;/a&gt; filter. &lt;b&gt;Higher-order&lt;/b&gt; filters are more complex than first-order filters, both in their circuitry and in the differential equation that governs them. The higher the order, the more effective the filter. &lt;/p&gt;&lt;p&gt;An example of a first-order filter is the simple circuit in Figure 8.&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/23ef598c/t312_openlearn_fig08.tif.jpg" alt="Described image" width="512" height="187" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049813360"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 8&lt;/b&gt;  First-order filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049813360&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049813360"&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 3&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Which of the four categories of filter shown in Figure 5 does the filter in Figure 8 belong to? You should be able to work it out from the behaviour of the capacitor at low frequencies and high frequencies. Explain your answer.&lt;/p&gt;
&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/65562369/t312_openlearn_fig05.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049798896"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 5 (repeated)&lt;/b&gt;  Four types of ideal filter: (a) low-pass filter; (b) high-pass filter; (c) band-pass filter; (d) band-stop (notch) filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049798896&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049798896"&gt;&lt;/a&gt;&lt;/div&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;It is a low-pass filter. &lt;/p&gt;
&lt;p&gt;At 0 Hz, or DC, the capacitor is open circuit. Under those circumstances, all the input voltage would appear on the output. At high frequencies, the capacitor becomes increasingly like a short circuit, and the output voltage decreases as frequency increases. Therefore low frequencies are passed and high frequencies are (to some extent) blocked. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Low-pass and high-pass filters can be first-order, second-order, third-order, and so on. However, band-pass filters and band-stop filters must be second-order or higher, although it is possible to achieve their effect by combining two first-order filters. &lt;/p&gt;&lt;p&gt;Despite being the simplest type, a first-order filter is not simple to analyse mathematically, as its behaviour is governed by a first-order differential equation. This is what the voltage gain function of the filter in Figure 8 turns out to be: &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a8c76e09278d2d40cea9969d5a7bd4035605d33b"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_22d" height="57px" role="math" style="vertical-align: -33px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 12047.0 3357.2461" width="204.5364px"&gt;

&lt;desc id="eq_dea10bfd_22d"&gt;equation sequence cap g equals cap v sub out divided by cap v sub in equals one divided by Square root of one plus open omega times cap r times cap c close squared&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;You could plot a graph of the gain function against frequency from this equation, but the details would depend on the choice of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="97a59cbf4a47ad1d1ecc1bb0a57622c087b17a7f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_23d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 764.0 824.5868" width="12.9713px"&gt;

&lt;desc id="eq_dea10bfd_23d"&gt;cap r&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_24d"&gt;cap c&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. However, irrespective of the values chosen for &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="97a59cbf4a47ad1d1ecc1bb0a57622c087b17a7f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_25d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 764.0 824.5868" width="12.9713px"&gt;

&lt;desc id="eq_dea10bfd_25d"&gt;cap r&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c7ee6a8a0b0501ca0380eabacbd2361c02628cd3"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_26d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 765.0 824.5868" width="12.9883px"&gt;

&lt;desc id="eq_dea10bfd_26d"&gt;cap c&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, the general shape and certain other properties of the gain function would be the same for any first-order low-pass filter. Therefore, rather than show how the gain function changes with frequency for particular values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="97a59cbf4a47ad1d1ecc1bb0a57622c087b17a7f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_27d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 764.0 824.5868" width="12.9713px"&gt;

&lt;desc id="eq_dea10bfd_27d"&gt;cap r&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c7ee6a8a0b0501ca0380eabacbd2361c02628cd3"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_28d" height="14px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 765.0 824.5868" width="12.9883px"&gt;

&lt;desc id="eq_dea10bfd_28d"&gt;cap c&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, in the next section you will look at a ‘normalised’ version of the gain function. (Normalisation is the presentation of information in a generalised way that can easily be adapted to specific cases.) &lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.5 Normalised first-order low-pass filters</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.5</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;A graph of gain function against frequency is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048983744" class="oucontent-glossaryterm" data-definition="The response of a system (e.g. a filter) when we input sine waves at different frequencies (but equal amplitude). It tells us how the system will modify the spectrum of any input signal we feed to the system." title="The response of a system (e.g. a filter) when we input sine waves at different frequencies (but equa..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;frequency response&lt;/span&gt;&lt;/a&gt; or a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048997328" class="oucontent-glossaryterm" data-definition="Loosely, a graph of the frequency response of a device or system. Strictly, a pair of graphs showing frequency response and phase response over the same span of frequencies." title="Loosely, a graph of the frequency response of a device or system. Strictly, a pair of graphs showing..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;Bode plot&lt;/span&gt;&lt;/a&gt;. A normalised first-order low-pass frequency response (or Bode plot) is shown in Figure&amp;#xA0;9. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/ad736d54/t312_openlearn_fig09.tif.jpg" alt="Described image" width="512" height="324" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049762400"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 9&lt;/b&gt;  Normalised first-order low-pass frequency response &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049762400&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049762400"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;You can see in the graph that on the horizontal axis, the frequency scale is in a form that is known as logarithmic. What this means is that at equally spaced points instead of having 1, 2, 3 Hz etc., you have 1 , 10, 100 Hz etc. The frequency increases by a factor of 10 at each interval. &lt;/p&gt;&lt;p&gt;Secondly, the vertical axis shows the power gain, but is measured in decibels. This again is a logarithmic measure which is designed to measure the ratio of powers, but is probably more commonly known in the measurement of sound.&lt;/p&gt;&lt;div class="oucontent-studynote oucontent-s-gradient oucontent-s-box &amp;#10;        oucontent-s-noheading&amp;#10;      "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;div class="oucontent-inner-box"&gt;&lt;p&gt;The reason logarithmic axes are used is so that the shape of the graph is very nearly made up of a horizontal straight line in the pass band, and a sloping straight line in the stop band.&lt;/p&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Just from the shape of the graph, it is clear that this gain function of a real filter is very different from the idealised ones that you saw earlier. For example, there is no sharp distinction between the passband and the stop band, and consequently no distinct cut-off frequency. By convention, the frequency at which the power gain drops 3&amp;#xA0;dB below the passband gain (or, in terms of voltage ratio, falls to 0.707&amp;#xA0;times the passband gain) is called the cut-off frequency, and this is true also for an active filter where the passband gain is likely to be other than 1. &lt;/p&gt;&lt;p&gt;In this normalised graph, the frequency axis is the normalised part. Notice that the frequency axis has no units, and the numbers on it are relatively low. (In electronics, such apparently low frequencies are seldom of interest; instead typically frequencies ranging from hundreds of hertz to gigahertz are more often used.) The normalised frequencies have been calculated by dividing actual frequencies by the cut-off frequency. This is why the cut-off point sits at 1 on the normalised frequency axis. &lt;/p&gt;&lt;p&gt;Dividing one frequency by another results in a pure number (that is, one without units), which is why the normalised frequency axis has no units. Similarly, dividing an output voltage by an input voltage produces a gain figure that also has no units. If the gain is converted to a decibel value, then strictly speaking that too has no units, although in Figure 9 &amp;#x2018;dB’ has been added after the number as though it were a unit as a reminder of the logarithmic nature of the function used. &lt;/p&gt;&lt;p&gt;As an example of how to translate normalised frequencies to actual frequencies, suppose a practical low-pass filter had a cut-off frequency of 10&lt;sup&gt;4&lt;/sup&gt;&amp;#xA0;radians per second. Simply multiplying all the numbers on the horizontal axis of the normalised graph by 10&lt;sup&gt;4&lt;/sup&gt; and giving the unit as &amp;#x2018;radians per second’ would transform the graph into one showing gain against actual frequency for that particular filter. If the cut-off frequency were 10&lt;sup&gt;4&lt;/sup&gt;&amp;#xA0;Hz (rather than radians per second), the procedure would be just the same: multiply all the numbers on the axis by 10&lt;sup&gt;4&lt;/sup&gt; and give the unit as Hz. &lt;/p&gt;&lt;p&gt;All first-order low-pass filters have a gain function of this shape and with these slopes. The passband gain might differ if the filter is active (that is, if it incorporates amplification). For first-order passive filters (that is, those without amplification) the passband gain cannot exceed 1. &lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 4&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 10 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;An alternating voltage source is connected to the input of a low-pass filter. What power is drawn from the output of the filter if the voltage source operates: &lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;well below the cut-off frequency of the filter and supplies 0.2&amp;#xA0;W to the input of the filter?&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;at the cut-off frequency of the filter and supplies 0.1&amp;#xA0;W to the input of the filter?&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Well below the cut-off frequency, the gain of the filter is 0&amp;#xA0;dB, which is a power ratio of 1:1. Hence the power drawn from the output of the filter is the same as that supplied at the input, 0.2&amp;#xA0;W. &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;At the cut-off frequency, the power gain is &amp;#x2212;3&amp;#xA0;dB or a half. Hence the output power is half the input power, or 0.05&amp;#xA0;W. The difference between the input power and the output power is also 0.05&amp;#xA0;W, which is dissipated as heat. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Notice that as you move right along the frequency axis, the gain never reaches zero (that is, zero as a voltage ratio rather than as a decibel number). Zero gain would correspond to a negatively infinite number of decibels. &lt;/p&gt;&lt;p&gt;All real first-order filters (as opposed to ideal ones) lack a sharp cut-off frequency; in addition, low-pass filters never fully cut off, although if the gain is low enough you can regard it as having fallen to 0. Higher-order filters can give a sharper cut-off than a first-order filter, but the brick-wall cut-off of an ideal filter can never be achieved in practice. &lt;/p&gt;&lt;p&gt;As you move left along the frequency axis, each division is one-tenth of the one before, and this can continue indefinitely. A logarithmic frequency scale therefore never reaches a frequency of 0. &lt;/p&gt;&lt;p&gt;The steepness of the gain in the stop band is referred to as the filter’s &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048969040" class="oucontent-glossaryterm" data-definition="The steepness of a filter’s attenuation in a stop band. Also, the steepness of the attenuation of any device that produces attenuation (for example, a linear amplifier at the extremes of its operating-frequency range)." title="The steepness of a filter’s attenuation in a stop band. Also, the steepness of the attenuation of an..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;roll-off&lt;/span&gt;&lt;/a&gt;. All first-order filters have a 20&amp;#xA0;dB/decade roll-off. The same roll-off can also be specified as 6&amp;#xA0;dB/octave. An &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048980160" class="oucontent-glossaryterm" data-definition="The span of frequencies covered by a doubling of frequency, or by a halving of frequency. For example, the frequency span from 500 Hz to 1000 Hz is an octave, as is the span from 500 Hz to 250 Hz. In music, the eight notes of a diatonic scale (that is, doh, re, me &amp;#x2026; ti, doh) cover an octave; hence the name &amp;#x2018;octave’ for this span of frequencies." title="The span of frequencies covered by a doubling of frequency, or by a halving of frequency. For exampl..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;octave&lt;/span&gt;&lt;/a&gt; is a term borrowed from music and represents a doubling of frequency. (It is so called because the frequency span in a doubling of frequency is divided into the eight notes of a musical scale.) Higher-order filters have a steeper roll-off. For second-order filters it is 40&amp;#xA0;dB/decade (or 12&amp;#xA0;dB/octave) and for third-order filters it is 60&amp;#xA0;dB/decade (or 18&amp;#xA0;dB/octave). Each successive order adds a further 20&amp;#xA0;dB/decade (or 6&amp;#xA0;dB/octave) to the roll-off. &lt;/p&gt;&lt;p&gt;Note that at the first decade above the cut-off frequency, the gain of a first-order filter is 20&amp;#xA0;dB below the passband gain, not 20&amp;#xA0;dB below the gain at the cut-off frequency. &lt;/p&gt;&lt;p&gt;Now have a go at Activity 5.&lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 5&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 10 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;A first-order low-pass filter has a cut-off frequency of 6.28&amp;#xA0;&amp;#xD7;&amp;#xA0;10&lt;sup&gt;3&lt;/sup&gt;&amp;#xA0;radians per second. &lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;What is the cut-off frequency in hertz? (Round your answer to 2&amp;#xA0;significant figures.)&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;What is the gain at 1&amp;#xA0;kHz? Express your answer in decibels.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;What is the gain at 100&amp;#xA0;kHz? Express your answer in decibels.&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Angular frequency &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="f23fee99bfa7ac9b439a1cd46368dfbc09828e55"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_29d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 3603.6 1001.2839" width="61.1826px"&gt;

&lt;desc id="eq_dea10bfd_29d"&gt;omega equals two times pi times f&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, where &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="d29f199b16a5837455355d0a97ebdef80a4c068b"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_30d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 555.0 1001.2839" width="9.4229px"&gt;

&lt;desc id="eq_dea10bfd_30d"&gt;f&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is the frequency in hertz. So &lt;/p&gt;&lt;p&gt;&amp;#x2003;&amp;#x2002;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="78340ad5b58969e3d810ef93ad3b5d272819aa68"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_31d" height="103px" role="math" style="vertical-align: -47px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0ex; margin-top: 0px;" viewBox="0.0 -3298.3470 8523.6 6066.6025" width="144.7154px"&gt;

&lt;desc id="eq_dea10bfd_31d"&gt;multiline equation line 1 f equals omega divided by two times pi line 2 equation left hand side equals right hand side 6.28 multiplication 10 cubed divided by two times pi Hz line 3 equation left hand side equals right hand side 999.493 times horizontal ellipsis Hz&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;To 2 significant figures, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="d29f199b16a5837455355d0a97ebdef80a4c068b"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_32d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 555.0 1001.2839" width="9.4229px"&gt;

&lt;desc id="eq_dea10bfd_32d"&gt;f&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&amp;#xA0;=&amp;#xA0;10&lt;sup&gt;3&lt;/sup&gt;&amp;#xA0;Hz. So the cut-off frequency is 10&lt;sup&gt;3&lt;/sup&gt;&amp;#xA0;Hz or 1&amp;#xA0;kHz. &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;1 kHz is the cut-off frequency, so the gain here is &amp;#x2212;3&amp;#xA0;dB.&lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;100 kHz is two decades above the cut-off frequency, so the power gain is &amp;#x2212;40&amp;#xA0;dB.&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;As you saw earlier, filters affect the gain and phase of a signal. In the next section you will see how you can show this in the Bode plot.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.5</guid>
    <dc:title>2.5 Normalised first-order low-pass filters</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;A graph of gain function against frequency is called a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048983744" class="oucontent-glossaryterm" data-definition="The response of a system (e.g. a filter) when we input sine waves at different frequencies (but equal amplitude). It tells us how the system will modify the spectrum of any input signal we feed to the system." title="The response of a system (e.g. a filter) when we input sine waves at different frequencies (but equa..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;frequency response&lt;/span&gt;&lt;/a&gt; or a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048997328" class="oucontent-glossaryterm" data-definition="Loosely, a graph of the frequency response of a device or system. Strictly, a pair of graphs showing frequency response and phase response over the same span of frequencies." title="Loosely, a graph of the frequency response of a device or system. Strictly, a pair of graphs showing..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;Bode plot&lt;/span&gt;&lt;/a&gt;. A normalised first-order low-pass frequency response (or Bode plot) is shown in Figure 9. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/ad736d54/t312_openlearn_fig09.tif.jpg" alt="Described image" width="512" height="324" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049762400"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 9&lt;/b&gt;  Normalised first-order low-pass frequency response &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049762400&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049762400"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;You can see in the graph that on the horizontal axis, the frequency scale is in a form that is known as logarithmic. What this means is that at equally spaced points instead of having 1, 2, 3 Hz etc., you have 1 , 10, 100 Hz etc. The frequency increases by a factor of 10 at each interval. &lt;/p&gt;&lt;p&gt;Secondly, the vertical axis shows the power gain, but is measured in decibels. This again is a logarithmic measure which is designed to measure the ratio of powers, but is probably more commonly known in the measurement of sound.&lt;/p&gt;&lt;div class="oucontent-studynote oucontent-s-gradient oucontent-s-box 
        oucontent-s-noheading
      "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;div class="oucontent-inner-box"&gt;&lt;p&gt;The reason logarithmic axes are used is so that the shape of the graph is very nearly made up of a horizontal straight line in the pass band, and a sloping straight line in the stop band.&lt;/p&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Just from the shape of the graph, it is clear that this gain function of a real filter is very different from the idealised ones that you saw earlier. For example, there is no sharp distinction between the passband and the stop band, and consequently no distinct cut-off frequency. By convention, the frequency at which the power gain drops 3 dB below the passband gain (or, in terms of voltage ratio, falls to 0.707 times the passband gain) is called the cut-off frequency, and this is true also for an active filter where the passband gain is likely to be other than 1. &lt;/p&gt;&lt;p&gt;In this normalised graph, the frequency axis is the normalised part. Notice that the frequency axis has no units, and the numbers on it are relatively low. (In electronics, such apparently low frequencies are seldom of interest; instead typically frequencies ranging from hundreds of hertz to gigahertz are more often used.) The normalised frequencies have been calculated by dividing actual frequencies by the cut-off frequency. This is why the cut-off point sits at 1 on the normalised frequency axis. &lt;/p&gt;&lt;p&gt;Dividing one frequency by another results in a pure number (that is, one without units), which is why the normalised frequency axis has no units. Similarly, dividing an output voltage by an input voltage produces a gain figure that also has no units. If the gain is converted to a decibel value, then strictly speaking that too has no units, although in Figure 9 ‘dB’ has been added after the number as though it were a unit as a reminder of the logarithmic nature of the function used. &lt;/p&gt;&lt;p&gt;As an example of how to translate normalised frequencies to actual frequencies, suppose a practical low-pass filter had a cut-off frequency of 10&lt;sup&gt;4&lt;/sup&gt; radians per second. Simply multiplying all the numbers on the horizontal axis of the normalised graph by 10&lt;sup&gt;4&lt;/sup&gt; and giving the unit as ‘radians per second’ would transform the graph into one showing gain against actual frequency for that particular filter. If the cut-off frequency were 10&lt;sup&gt;4&lt;/sup&gt; Hz (rather than radians per second), the procedure would be just the same: multiply all the numbers on the axis by 10&lt;sup&gt;4&lt;/sup&gt; and give the unit as Hz. &lt;/p&gt;&lt;p&gt;All first-order low-pass filters have a gain function of this shape and with these slopes. The passband gain might differ if the filter is active (that is, if it incorporates amplification). For first-order passive filters (that is, those without amplification) the passband gain cannot exceed 1. &lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 4&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 10 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;An alternating voltage source is connected to the input of a low-pass filter. What power is drawn from the output of the filter if the voltage source operates: &lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;well below the cut-off frequency of the filter and supplies 0.2 W to the input of the filter?&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;at the cut-off frequency of the filter and supplies 0.1 W to the input of the filter?&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Well below the cut-off frequency, the gain of the filter is 0 dB, which is a power ratio of 1:1. Hence the power drawn from the output of the filter is the same as that supplied at the input, 0.2 W. &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;At the cut-off frequency, the power gain is −3 dB or a half. Hence the output power is half the input power, or 0.05 W. The difference between the input power and the output power is also 0.05 W, which is dissipated as heat. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Notice that as you move right along the frequency axis, the gain never reaches zero (that is, zero as a voltage ratio rather than as a decibel number). Zero gain would correspond to a negatively infinite number of decibels. &lt;/p&gt;&lt;p&gt;All real first-order filters (as opposed to ideal ones) lack a sharp cut-off frequency; in addition, low-pass filters never fully cut off, although if the gain is low enough you can regard it as having fallen to 0. Higher-order filters can give a sharper cut-off than a first-order filter, but the brick-wall cut-off of an ideal filter can never be achieved in practice. &lt;/p&gt;&lt;p&gt;As you move left along the frequency axis, each division is one-tenth of the one before, and this can continue indefinitely. A logarithmic frequency scale therefore never reaches a frequency of 0. &lt;/p&gt;&lt;p&gt;The steepness of the gain in the stop band is referred to as the filter’s &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048969040" class="oucontent-glossaryterm" data-definition="The steepness of a filter’s attenuation in a stop band. Also, the steepness of the attenuation of any device that produces attenuation (for example, a linear amplifier at the extremes of its operating-frequency range)." title="The steepness of a filter’s attenuation in a stop band. Also, the steepness of the attenuation of an..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;roll-off&lt;/span&gt;&lt;/a&gt;. All first-order filters have a 20 dB/decade roll-off. The same roll-off can also be specified as 6 dB/octave. An &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048980160" class="oucontent-glossaryterm" data-definition="The span of frequencies covered by a doubling of frequency, or by a halving of frequency. For example, the frequency span from 500 Hz to 1000 Hz is an octave, as is the span from 500 Hz to 250 Hz. In music, the eight notes of a diatonic scale (that is, doh, re, me … ti, doh) cover an octave; hence the name ‘octave’ for this span of frequencies." title="The span of frequencies covered by a doubling of frequency, or by a halving of frequency. For exampl..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;octave&lt;/span&gt;&lt;/a&gt; is a term borrowed from music and represents a doubling of frequency. (It is so called because the frequency span in a doubling of frequency is divided into the eight notes of a musical scale.) Higher-order filters have a steeper roll-off. For second-order filters it is 40 dB/decade (or 12 dB/octave) and for third-order filters it is 60 dB/decade (or 18 dB/octave). Each successive order adds a further 20 dB/decade (or 6 dB/octave) to the roll-off. &lt;/p&gt;&lt;p&gt;Note that at the first decade above the cut-off frequency, the gain of a first-order filter is 20 dB below the passband gain, not 20 dB below the gain at the cut-off frequency. &lt;/p&gt;&lt;p&gt;Now have a go at Activity 5.&lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 5&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 10 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;A first-order low-pass filter has a cut-off frequency of 6.28 × 10&lt;sup&gt;3&lt;/sup&gt; radians per second. &lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;What is the cut-off frequency in hertz? (Round your answer to 2 significant figures.)&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;What is the gain at 1 kHz? Express your answer in decibels.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;What is the gain at 100 kHz? Express your answer in decibels.&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Angular frequency &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="f23fee99bfa7ac9b439a1cd46368dfbc09828e55"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_29d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 3603.6 1001.2839" width="61.1826px"&gt;

&lt;desc id="eq_dea10bfd_29d"&gt;omega equals two times pi times f&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, where &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="d29f199b16a5837455355d0a97ebdef80a4c068b"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_30d" height="17px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -765.6877 555.0 1001.2839" width="9.4229px"&gt;

&lt;desc id="eq_dea10bfd_30d"&gt;f&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is the frequency in hertz. So &lt;/p&gt;&lt;p&gt;  &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="78340ad5b58969e3d810ef93ad3b5d272819aa68"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_31d" height="103px" role="math" style="vertical-align: -47px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0ex; margin-top: 0px;" viewBox="0.0 -3298.3470 8523.6 6066.6025" width="144.7154px"&gt;

&lt;desc id="eq_dea10bfd_31d"&gt;multiline equation line 1 f equals omega divided by two times pi line 2 equation left hand side equals right hand side 6.28 multiplication 10 cubed divided by two times pi Hz line 3 equation left hand side equals right hand side 999.493 times horizontal ellipsis Hz&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_32d"&gt;f&lt;/desc&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; = 10&lt;sup&gt;3&lt;/sup&gt; Hz. So the cut-off frequency is 10&lt;sup&gt;3&lt;/sup&gt; Hz or 1 kHz. &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;1 kHz is the cut-off frequency, so the gain here is −3 dB.&lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;100 kHz is two decades above the cut-off frequency, so the power gain is −40 dB.&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;As you saw earlier, filters affect the gain and phase of a signal. In the next section you will see how you can show this in the Bode plot.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.6 The full Bode plot: gain and phase</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.6</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;In Section 2.5, you heard how a full Bode plot would show not only how the gain changes with frequency, but also how the phase difference between output and input changes with frequency. Conventionally the phase plot is put under the gain plot, with their respective frequency axes aligned, as in Figure&amp;#xA0;10. This example is for a first-order low-pass filter. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:410px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/d656b480/t312_openlearn_fig10.tif.jpg" alt="Described image" width="410" height="512" style="max-width:410px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049687808"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 10&lt;/b&gt;  Normalised first-order low-pass frequency response showing gain and phase &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049687808&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049687808"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Figure 10 shows that in the passband, the output is virtually in phase with the input. As frequency increases towards the cut-off frequency (1 on the normalised frequency axis), a phase difference opens between the output and the input, with the output lagging the input (the negative values of phase angle indicate lagging phase). At the cut-off frequency, the phase difference is &amp;#x2212;45&amp;#xB0;. This means that for a sinusoidal input, the output lags the input by 45&amp;#xB0;. By the time you are well into the stop band, the phase difference levels off at &amp;#x2212;90&amp;#xB0;. &lt;/p&gt;&lt;p&gt;That concludes the discussion of first-order filters. To end Section 2 you will consider some commonly found higher-order filters and have a look at their characteristics.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.6</guid>
    <dc:title>2.6 The full Bode plot: gain and phase</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;In Section 2.5, you heard how a full Bode plot would show not only how the gain changes with frequency, but also how the phase difference between output and input changes with frequency. Conventionally the phase plot is put under the gain plot, with their respective frequency axes aligned, as in Figure 10. This example is for a first-order low-pass filter. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:410px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/d656b480/t312_openlearn_fig10.tif.jpg" alt="Described image" width="410" height="512" style="max-width:410px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049687808"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 10&lt;/b&gt;  Normalised first-order low-pass frequency response showing gain and phase &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049687808&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049687808"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Figure 10 shows that in the passband, the output is virtually in phase with the input. As frequency increases towards the cut-off frequency (1 on the normalised frequency axis), a phase difference opens between the output and the input, with the output lagging the input (the negative values of phase angle indicate lagging phase). At the cut-off frequency, the phase difference is −45°. This means that for a sinusoidal input, the output lags the input by 45°. By the time you are well into the stop band, the phase difference levels off at −90°. &lt;/p&gt;&lt;p&gt;That concludes the discussion of first-order filters. To end Section 2 you will consider some commonly found higher-order filters and have a look at their characteristics.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>2.7 Chebyshev and Butterworth filters</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.7</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;The design of higher-order filters is a specialist area, and mathematically complex, so in this section you will look at gain functions of just two celebrated types. The first is the Butterworth filter. The gain function of a Butterworth filter has the familiar flat passband and roll-off you would expect. However, this filter comes in various orders, such as 2, 4, 6, 8 and 10, depending on how steep the roll-off needs to be. &lt;/p&gt;&lt;p&gt;Figure 11 shows the normalised responses of first-order, second-order and eighth-order low-pass Butterworth filters. As the order of the filter increases, it approximates more closely the ideal brick-wall response. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/beecff50/t312_openlearn_fig11.tif.jpg" alt="Described image" width="512" height="266" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049666416"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 11&lt;/b&gt;  Normalised Butterworth magnitude response curves &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049666416&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049666416"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Steeper roll-offs than those of a Butterworth filter can be had if other desirable features of a filter are compromised. The Chebyshev filter has a steeper roll-off, but at the price of a passband response that is not flat. Figure&amp;#xA0;12 compares seventh-order normalised Chebyshev and Butterworth filters. Notice the ripple in the Chebyshev filter’s response in the passband. (There is another type of Chebyshev filter that has a flat response in the passband and ripples in the stop band.) &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/c591e0b1/t312_openlearn_fig12.tif.jpg" alt="Described image" width="512" height="266" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049655328"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 12&lt;/b&gt;  Comparison of seventh-order Butterworth and Chebyshev gain&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049655328&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049655328"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;That concludes the introduction to filters and, in particular, analogue filters. In Section 3 you will learn about digital filtering, which is currently a more popular approach to filtering.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-2.7</guid>
    <dc:title>2.7 Chebyshev and Butterworth filters</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;The design of higher-order filters is a specialist area, and mathematically complex, so in this section you will look at gain functions of just two celebrated types. The first is the Butterworth filter. The gain function of a Butterworth filter has the familiar flat passband and roll-off you would expect. However, this filter comes in various orders, such as 2, 4, 6, 8 and 10, depending on how steep the roll-off needs to be. &lt;/p&gt;&lt;p&gt;Figure 11 shows the normalised responses of first-order, second-order and eighth-order low-pass Butterworth filters. As the order of the filter increases, it approximates more closely the ideal brick-wall response. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/beecff50/t312_openlearn_fig11.tif.jpg" alt="Described image" width="512" height="266" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049666416"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 11&lt;/b&gt;  Normalised Butterworth magnitude response curves &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049666416&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049666416"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Steeper roll-offs than those of a Butterworth filter can be had if other desirable features of a filter are compromised. The Chebyshev filter has a steeper roll-off, but at the price of a passband response that is not flat. Figure 12 compares seventh-order normalised Chebyshev and Butterworth filters. Notice the ripple in the Chebyshev filter’s response in the passband. (There is another type of Chebyshev filter that has a flat response in the passband and ripples in the stop band.) &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/c591e0b1/t312_openlearn_fig12.tif.jpg" alt="Described image" width="512" height="266" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049655328"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 12&lt;/b&gt;  Comparison of seventh-order Butterworth and Chebyshev gain&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049655328&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049655328"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;That concludes the introduction to filters and, in particular, analogue filters. In Section 3 you will learn about digital filtering, which is currently a more popular approach to filtering.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3 Digital signal processing</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Digital signal processing has developed rapidly over the last 50&amp;#xA0;years: digital signal-processing circuits have become faster and cheaper, and memory storage capabilities have increased dramatically. One result of these developments is a migration from analogue to digital circuits for some types of signal processing, and digital filters are an example of this trend. &lt;/p&gt;&lt;p&gt;Digital filters have some advantages over analogue filters. They are programmable, so their operation is determined by a program stored in a processor’s memory. This means a digital filter can easily be changed without affecting the hardware. Also, digital filters are extremely stable with respect to both time and temperature. For complex filters, the hardware requirements are relatively simple and compact in comparison to the equivalent analogue circuitry. &lt;/p&gt;&lt;p&gt;The design of a digital filter is complicated and involves quite advanced mathematics, so software tools that produce a filter design from your specification of filter characteristics are commonly used. However, as you may know from the use of software tools such as circuit simulators, you need to be very wary of using a software design tool to create circuits without having a good understanding of the electrical characteristics of the circuit that you want to create and the parameters used in the design process. A good understanding of how digital filters work will help at every stage of filter design. &lt;/p&gt;&lt;p&gt;For the remainder of the course you will find out about various aspects of filtering a signal digitally. First you will see how a continuous-time signal is converted to produce the digital discrete-time signal used as input to the filter. Then you will find out how mathematical operations applied to the discrete-time signal can remove or diminish unwanted aspects of the signal. &lt;/p&gt;&lt;p&gt;To get started, Figure&amp;#xA0;13 shows the basic set-up of a digital filter. Don’t worry too much about the detail at this point – it will be covered later. As part of the filtering process, the continuous input signal must be sampled and digitised using an analogue-to-digital converter (ADC) to produce a sampled discrete signal. The resulting binary numbers, representing successive sampled values of the input signal, are transferred to the processor, which carries out numerical calculations on them. These calculations typically involve multiplying the input values by constants and adding the products together. If necessary, the results of these calculations, which now represent sampled values of the filtered signal, are output through a digital-to-analogue converter (DAC) to convert the signal back to continuous form. Given that the continuous input signal and the filtered continuous output signal are continuous in time, they are often referred to as continuous-time signals. Similarly, the sampled discrete signal and the digitally filtered signal are often referred to as discrete-time signals. (Here the word &amp;#x2018;discrete’ means &amp;#x2018;consisting of separate parts’ as opposed to the word &amp;#x2018;discreet’ which means &amp;#x2018;unobtrusive’).&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/944e4552/t312_openlearn_fig13.tif.jpg" alt="Described image" width="512" height="151" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049628128"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 13&lt;/b&gt;  Digital filtering of a sampled signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049628128&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049628128"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In the next section you will see an example of a digital filter being used as part of a medical system to illustrate the component parts.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3</guid>
    <dc:title>3 Digital signal processing</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Digital signal processing has developed rapidly over the last 50 years: digital signal-processing circuits have become faster and cheaper, and memory storage capabilities have increased dramatically. One result of these developments is a migration from analogue to digital circuits for some types of signal processing, and digital filters are an example of this trend. &lt;/p&gt;&lt;p&gt;Digital filters have some advantages over analogue filters. They are programmable, so their operation is determined by a program stored in a processor’s memory. This means a digital filter can easily be changed without affecting the hardware. Also, digital filters are extremely stable with respect to both time and temperature. For complex filters, the hardware requirements are relatively simple and compact in comparison to the equivalent analogue circuitry. &lt;/p&gt;&lt;p&gt;The design of a digital filter is complicated and involves quite advanced mathematics, so software tools that produce a filter design from your specification of filter characteristics are commonly used. However, as you may know from the use of software tools such as circuit simulators, you need to be very wary of using a software design tool to create circuits without having a good understanding of the electrical characteristics of the circuit that you want to create and the parameters used in the design process. A good understanding of how digital filters work will help at every stage of filter design. &lt;/p&gt;&lt;p&gt;For the remainder of the course you will find out about various aspects of filtering a signal digitally. First you will see how a continuous-time signal is converted to produce the digital discrete-time signal used as input to the filter. Then you will find out how mathematical operations applied to the discrete-time signal can remove or diminish unwanted aspects of the signal. &lt;/p&gt;&lt;p&gt;To get started, Figure 13 shows the basic set-up of a digital filter. Don’t worry too much about the detail at this point – it will be covered later. As part of the filtering process, the continuous input signal must be sampled and digitised using an analogue-to-digital converter (ADC) to produce a sampled discrete signal. The resulting binary numbers, representing successive sampled values of the input signal, are transferred to the processor, which carries out numerical calculations on them. These calculations typically involve multiplying the input values by constants and adding the products together. If necessary, the results of these calculations, which now represent sampled values of the filtered signal, are output through a digital-to-analogue converter (DAC) to convert the signal back to continuous form. Given that the continuous input signal and the filtered continuous output signal are continuous in time, they are often referred to as continuous-time signals. Similarly, the sampled discrete signal and the digitally filtered signal are often referred to as discrete-time signals. (Here the word ‘discrete’ means ‘consisting of separate parts’ as opposed to the word ‘discreet’ which means ‘unobtrusive’).&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/944e4552/t312_openlearn_fig13.tif.jpg" alt="Described image" width="512" height="151" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049628128"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 13&lt;/b&gt;  Digital filtering of a sampled signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049628128&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049628128"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In the next section you will see an example of a digital filter being used as part of a medical system to illustrate the component parts.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.1 Digital filters</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.1</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;You read about analogue filters in Section 2, so you know that they are electronic circuits made up of components such as resistors, capacitors and inductors connected together to produce the required filtering effect. In comparison, a digital filter uses a digital processor to perform numerical calculations on sampled values of the signal to be filtered. The processor could be a general-purpose computer such as a PC, but it is much more likely to be a specialised digital signal processor (DSP) chip, which is designed to carry out the intensive mathematical operations used in digital signal processing quickly and with low power consumption. The low power consumption is important because it means that purpose-designed DSP chips can be used in mobile devices such as phones and tablets. &lt;/p&gt;&lt;p&gt;A potential use for a digital filter with low power consumption is described by Asgar and Mehrnia (2017), who propose using a digital filter in an electrocardiogram (ECG) heart-monitoring system. Figure&amp;#xA0;14 is a block diagram of the system, taken from their paper. It shows a sensor connected to the surface of the user’s skin to monitor their heart function. This signal is then conditioned. Typical actions here might be to amplify and remove aliases from the signal; then an analogue-to-digital (ADC) converter is used to change the signal from analogue to digital form prior to it being filtered. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/535900df/t312_openlearn_fig14.tif.jpg" alt="Described image" width="512" height="168" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049612160"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 14&lt;/b&gt;  Block diagram of the ECG heart-monitoring system &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049612160&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049612160"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In the paper, the authors describe six different sources that can cause noise to contaminate the measured signal. Three of these sources are as follows: &lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;electrode contact noise, which is due to electrode &amp;#x2018;popping’ or a loose contact with the skin&lt;/li&gt;&lt;li&gt;instrumentation noise, which is due to radio-frequency interference from other equipment (e.g. implanted devices such as pacemakers)&lt;/li&gt;&lt;li&gt;electromyographic (EMG) noise, which is induced by electrical activities of skeletal muscles during periods of contraction.&lt;/li&gt;&lt;/ul&gt;&lt;p&gt;This shows the variety of noise sources that are associated with specific applications and the insights that are needed to understand the noise sources in any filter application. The filtering solution proposed by the authors uses a low-complexity, linear-phase digital filter design. &lt;/p&gt;&lt;p&gt;The next section will describe in more detail the signals that are found in digital systems.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.1</guid>
    <dc:title>3.1 Digital filters</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;You read about analogue filters in Section 2, so you know that they are electronic circuits made up of components such as resistors, capacitors and inductors connected together to produce the required filtering effect. In comparison, a digital filter uses a digital processor to perform numerical calculations on sampled values of the signal to be filtered. The processor could be a general-purpose computer such as a PC, but it is much more likely to be a specialised digital signal processor (DSP) chip, which is designed to carry out the intensive mathematical operations used in digital signal processing quickly and with low power consumption. The low power consumption is important because it means that purpose-designed DSP chips can be used in mobile devices such as phones and tablets. &lt;/p&gt;&lt;p&gt;A potential use for a digital filter with low power consumption is described by Asgar and Mehrnia (2017), who propose using a digital filter in an electrocardiogram (ECG) heart-monitoring system. Figure 14 is a block diagram of the system, taken from their paper. It shows a sensor connected to the surface of the user’s skin to monitor their heart function. This signal is then conditioned. Typical actions here might be to amplify and remove aliases from the signal; then an analogue-to-digital (ADC) converter is used to change the signal from analogue to digital form prior to it being filtered. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/535900df/t312_openlearn_fig14.tif.jpg" alt="Described image" width="512" height="168" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049612160"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 14&lt;/b&gt;  Block diagram of the ECG heart-monitoring system &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049612160&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049612160"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In the paper, the authors describe six different sources that can cause noise to contaminate the measured signal. Three of these sources are as follows: &lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;electrode contact noise, which is due to electrode ‘popping’ or a loose contact with the skin&lt;/li&gt;&lt;li&gt;instrumentation noise, which is due to radio-frequency interference from other equipment (e.g. implanted devices such as pacemakers)&lt;/li&gt;&lt;li&gt;electromyographic (EMG) noise, which is induced by electrical activities of skeletal muscles during periods of contraction.&lt;/li&gt;&lt;/ul&gt;&lt;p&gt;This shows the variety of noise sources that are associated with specific applications and the insights that are needed to understand the noise sources in any filter application. The filtering solution proposed by the authors uses a low-complexity, linear-phase digital filter design. &lt;/p&gt;&lt;p&gt;The next section will describe in more detail the signals that are found in digital systems.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.2 Characteristics of discrete-time and continuous-time signals</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.2</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;A continuous-time signal is shown in Figure&amp;#xA0;15(a). The signal is continuous because it has a value at any instance of time – that is, for any value of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="fbb361b9916a5e93ed248347184df511754bfbfc"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_33d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 366.0 765.6877" width="6.2140px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; from the graph. Most signals in the real world are continuous in time. For example, if you were monitoring the temperature of a room, you would be able to take a measured value of temperature at any time. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/17495c48/t312_openlearn_fig15.tif.jpg" alt="Described image" width="512" height="219" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049593872"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 15 &lt;/b&gt; (a) Continuous-time signal; (b) discrete-time signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049593872&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049593872"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;A discrete-time signal (sometimes referred to as a time-discrete signal or simply a discrete signal) is shown in Figure&amp;#xA0;15(b). In the rest of this course the standard convention of drawing the vertical lines in a discrete-time signal with a round dot on the end will be used; these lines-with-dots are often called &amp;#x2018;lollipops’. The signal in Figure&amp;#xA0;15(b) is discrete because it only has a value at fixed points placed at discrete time intervals &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_35d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

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&lt;desc id="eq_dea10bfd_39d"&gt;n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, such as &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1ae6535d1ba4e21f621719f7183529aa1a68e06c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_40d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_40d"&gt;n equals one&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="534dcec22b20786b3e1936c1aae0028ede8df65d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_41d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_41d"&gt;n equals two&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, etc., but there is no value for the signal at, say, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="739b629e33180e3cc6b1e42bc610d627a2b54a03"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_42d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3236.6 765.6877" width="54.9516px"&gt;

&lt;desc id="eq_dea10bfd_42d"&gt;n equals 1.5&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. Thus &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_43d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_43d"&gt;n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; represents the number of the sample. &lt;/p&gt;&lt;p&gt;It is hard to think of examples of real-world discrete-time signals, since most real-world signals are continuous; however, if you took the temperature reading of a room every day at the same time, the result would be a discrete-time signal. Most discrete-time signals come from sampling continuous-time signals to get them into a digitised form that can be processed by digital computers. &lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 6&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;State whether the following are discrete-time signals or continuous-time signals, giving a reason for each answer:&lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;the wind speed across the blades of a wind turbine&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;the position of a robotic arm as it picks items from a conveyor belt&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;the total distance travelled by the robotic arm each hour over a 24-hour period.&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The wind speed is a continuous-time signal, because you can take a reading at any time.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The robotic arm always has a position – even if it is in a resting position, you know where it is – so this is a continuous-time signal. &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;The total distance travelled by the robotic arm is recorded just once in each hour, so this is a discrete-time signal. Over a 24-hour period there will be 24&amp;#xA0;discrete values recorded. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;In the next section you will learn how a continuous signal is converted to a discrete signal. &lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.2</guid>
    <dc:title>3.2 Characteristics of discrete-time and continuous-time signals</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;A continuous-time signal is shown in Figure 15(a). The signal is continuous because it has a value at any instance of time – that is, for any value of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="fbb361b9916a5e93ed248347184df511754bfbfc"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_33d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 366.0 765.6877" width="6.2140px"&gt;

&lt;desc id="eq_dea10bfd_33d"&gt;t&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, it is possible read a value of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1dd9768d62d8343468cf393bb198947004ee04c6"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_34d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1731.0 1119.0820" width="29.3893px"&gt;

&lt;desc id="eq_dea10bfd_34d"&gt;x of t&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; from the graph. Most signals in the real world are continuous in time. For example, if you were monitoring the temperature of a room, you would be able to take a measured value of temperature at any time. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/17495c48/t312_openlearn_fig15.tif.jpg" alt="Described image" width="512" height="219" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049593872"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 15 &lt;/b&gt; (a) Continuous-time signal; (b) discrete-time signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049593872&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049593872"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;A discrete-time signal (sometimes referred to as a time-discrete signal or simply a discrete signal) is shown in Figure 15(b). In the rest of this course the standard convention of drawing the vertical lines in a discrete-time signal with a round dot on the end will be used; these lines-with-dots are often called ‘lollipops’. The signal in Figure 15(b) is discrete because it only has a value at fixed points placed at discrete time intervals &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_35d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_35d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_35MJMATHI-54" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; seconds apart along the &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="04acab056982ad114ae991b3ac6adc09066b8eae"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_36d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 577.0 530.0915" width="9.7964px"&gt;

&lt;desc id="eq_dea10bfd_36d"&gt;x&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_36MJMATHI-78" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_36MJMATHI-78" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;-axis. &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_37d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_37d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M40 437Q21 437 21 445Q21 450 37 501T71 602L88 651Q93 669 101 677H569H659Q691 677 697 676T704 667Q704 661 687 553T668 444Q668 437 649 437Q640 437 637 437T631 442L629 445Q629 451 635 490T641 551Q641 586 628 604T573 629Q568 630 515 631Q469 631 457 630T439 622Q438 621 368 343T298 60Q298 48 386 46Q418 46 427 45T436 36Q436 31 433 22Q429 4 424 1L422 0Q419 0 415 0Q410 0 363 1T228 2Q99 2 64 0H49Q43 6 43 9T45 27Q49 40 55 46H83H94Q174 46 189 55Q190 56 191 56Q196 59 201 76T241 233Q258 301 269 344Q339 619 339 625Q339 630 310 630H279Q212 630 191 624Q146 614 121 583T67 467Q60 445 57 441T43 437H40Z" id="eq_dea10bfd_37MJMATHI-54" stroke-width="10"/&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_37MJMATHI-54" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is called the sampling interval. Values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_38d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_38d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_38MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_38MJMAIN-5B" stroke-width="10"/&gt;
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&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_38MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_38MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_38MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_38MJMATHI-6E" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; can be found for the integer values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_39d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_39d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_39MJMATHI-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_39MJMATHI-6E" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, such as &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1ae6535d1ba4e21f621719f7183529aa1a68e06c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_40d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_40d"&gt;n equals one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_40MJMAIN-31" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_40MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_40MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_40MJMAIN-31" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="534dcec22b20786b3e1936c1aae0028ede8df65d"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_41d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_41d"&gt;n equals two&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_41MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M56 347Q56 360 70 367H707Q722 359 722 347Q722 336 708 328L390 327H72Q56 332 56 347ZM56 153Q56 168 72 173H708Q722 163 722 153Q722 140 707 133H70Q56 140 56 153Z" id="eq_dea10bfd_41MJMAIN-3D" stroke-width="10"/&gt;
&lt;path d="M109 429Q82 429 66 447T50 491Q50 562 103 614T235 666Q326 666 387 610T449 465Q449 422 429 383T381 315T301 241Q265 210 201 149L142 93L218 92Q375 92 385 97Q392 99 409 186V189H449V186Q448 183 436 95T421 3V0H50V19V31Q50 38 56 46T86 81Q115 113 136 137Q145 147 170 174T204 211T233 244T261 278T284 308T305 340T320 369T333 401T340 431T343 464Q343 527 309 573T212 619Q179 619 154 602T119 569T109 550Q109 549 114 549Q132 549 151 535T170 489Q170 464 154 447T109 429Z" id="eq_dea10bfd_41MJMAIN-32" stroke-width="10"/&gt;
&lt;/defs&gt;
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 &lt;use x="882" xlink:href="#eq_dea10bfd_41MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_41MJMAIN-32" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, etc., but there is no value for the signal at, say, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="739b629e33180e3cc6b1e42bc610d627a2b54a03"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_42d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3236.6 765.6877" width="54.9516px"&gt;

&lt;desc id="eq_dea10bfd_42d"&gt;n equals 1.5&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_42MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M56 347Q56 360 70 367H707Q722 359 722 347Q722 336 708 328L390 327H72Q56 332 56 347ZM56 153Q56 168 72 173H708Q722 163 722 153Q722 140 707 133H70Q56 140 56 153Z" id="eq_dea10bfd_42MJMAIN-3D" stroke-width="10"/&gt;
&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_42MJMAIN-31" stroke-width="10"/&gt;
&lt;path d="M78 60Q78 84 95 102T138 120Q162 120 180 104T199 61Q199 36 182 18T139 0T96 17T78 60Z" id="eq_dea10bfd_42MJMAIN-2E" stroke-width="10"/&gt;
&lt;path d="M164 157Q164 133 148 117T109 101H102Q148 22 224 22Q294 22 326 82Q345 115 345 210Q345 313 318 349Q292 382 260 382H254Q176 382 136 314Q132 307 129 306T114 304Q97 304 95 310Q93 314 93 485V614Q93 664 98 664Q100 666 102 666Q103 666 123 658T178 642T253 634Q324 634 389 662Q397 666 402 666Q410 666 410 648V635Q328 538 205 538Q174 538 149 544L139 546V374Q158 388 169 396T205 412T256 420Q337 420 393 355T449 201Q449 109 385 44T229 -22Q148 -22 99 32T50 154Q50 178 61 192T84 210T107 214Q132 214 148 197T164 157Z" id="eq_dea10bfd_42MJMAIN-35" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_42MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_42MJMAIN-3D" y="0"/&gt;
&lt;g transform="translate(1943,0)"&gt;
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 &lt;use x="505" xlink:href="#eq_dea10bfd_42MJMAIN-2E" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. Thus &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_43d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_43d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_43MJMATHI-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; represents the number of the sample. &lt;/p&gt;&lt;p&gt;It is hard to think of examples of real-world discrete-time signals, since most real-world signals are continuous; however, if you took the temperature reading of a room every day at the same time, the result would be a discrete-time signal. Most discrete-time signals come from sampling continuous-time signals to get them into a digitised form that can be processed by digital computers. &lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 6&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;State whether the following are discrete-time signals or continuous-time signals, giving a reason for each answer:&lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;the wind speed across the blades of a wind turbine&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;the position of a robotic arm as it picks items from a conveyor belt&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;the total distance travelled by the robotic arm each hour over a 24-hour period.&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The wind speed is a continuous-time signal, because you can take a reading at any time.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The robotic arm always has a position – even if it is in a resting position, you know where it is – so this is a continuous-time signal. &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;The total distance travelled by the robotic arm is recorded just once in each hour, so this is a discrete-time signal. Over a 24-hour period there will be 24 discrete values recorded. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;In the next section you will learn how a continuous signal is converted to a discrete signal. &lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.3 Sampling a continuous-time signal</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.3</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;To convert a continuous-time signal into its discrete-time signal equivalent, you need to sample the waveform. To ensure the discrete-time signal contains the full range of frequencies in the continuous-time signal, the continuous-time signal normally needs to be sampled at a rate that is greater than twice the highest frequency component contained in the signal. The lowest sampling frequency that can fully reconstruct the continuous-time signal is called the Nyquist frequency after Harry Nyquist (1889–1976), a Swedish-born American engineer who made important contributions to communications theory. &lt;/p&gt;&lt;p&gt;What happens if the sample rate is equal to or less than twice the highest frequency in the signal – or, to express this another way, if the continuous-time signal is sampled twice or less than twice within each cycle of the highest frequency component contained in the signal? Figure&amp;#xA0;16(a) shows a sine wave (solid line) that is being sampled less than twice each cycle. The samples are represented by blobs. However, another sine wave with a lower frequency can be drawn through these samples. It is shown as a dashed sine wave. This lower-frequency wave is called an &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133049000736" class="oucontent-glossaryterm" data-definition="An error appearing in a sampled signal when the bandwidth of the signal is greater than half the sampling frequency (that is, when the sampling frequency is lower than the Nyquist frequency). Such effects are also referred to as artefacts or ghosts." title="An error appearing in a sampled signal when the bandwidth of the signal is greater than half the sam..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;alias&lt;/span&gt;&lt;/a&gt; of the original sine wave. The way to avoid aliases is to sample more frequently, as in Figure&amp;#xA0;16(b). Here there are twice as many samples as in Figure&amp;#xA0;16(a). Now the alias from (a), also shown dashed in (b), misses some of the samples in (b). There is no waveform of a lower frequency than the sampled waveform that can be fitted to all the samples. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/b116cb05/t312_openlearn_fig16.tif.jpg" alt="Described image" width="512" height="113" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049545168"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 16&lt;/b&gt;  (a) More than one sine wave can be drawn through these samples, so there is a low-frequency alias of the original wave; (b) with more samples, there is no alias that can be fitted to all the samples &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049545168&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049545168"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The way to eliminate the possibility of an alias is either to have greater than two samples per cycle or, conversely, to &amp;#x2018;band-limit’ the signal to make sure that there are no frequencies equal to or higher than one half of the sampling frequency. This is done by including an analogue filter before the sampler in the analogue-to-digital converter that removes all frequencies higher than a certain frequency. As such, a filter is designed to stop any chance of aliasing occurring, it is sometimes called an &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048998992" class="oucontent-glossaryterm" data-definition="A low-pass filter that is able to remove aliasing in sampled signals by cutting all the spectral components that are greater than or equal to half the sampling frequency." title="A low-pass filter that is able to remove aliasing in sampled signals by cutting all the spectral com..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;anti-aliasing filter&lt;/span&gt;&lt;/a&gt;. &lt;/p&gt;&lt;p&gt;After sampling the signal, the samples need to be converted to a digital signal. The next section describes how this is done.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.3</guid>
    <dc:title>3.3 Sampling a continuous-time signal</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;To convert a continuous-time signal into its discrete-time signal equivalent, you need to sample the waveform. To ensure the discrete-time signal contains the full range of frequencies in the continuous-time signal, the continuous-time signal normally needs to be sampled at a rate that is greater than twice the highest frequency component contained in the signal. The lowest sampling frequency that can fully reconstruct the continuous-time signal is called the Nyquist frequency after Harry Nyquist (1889–1976), a Swedish-born American engineer who made important contributions to communications theory. &lt;/p&gt;&lt;p&gt;What happens if the sample rate is equal to or less than twice the highest frequency in the signal – or, to express this another way, if the continuous-time signal is sampled twice or less than twice within each cycle of the highest frequency component contained in the signal? Figure 16(a) shows a sine wave (solid line) that is being sampled less than twice each cycle. The samples are represented by blobs. However, another sine wave with a lower frequency can be drawn through these samples. It is shown as a dashed sine wave. This lower-frequency wave is called an &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133049000736" class="oucontent-glossaryterm" data-definition="An error appearing in a sampled signal when the bandwidth of the signal is greater than half the sampling frequency (that is, when the sampling frequency is lower than the Nyquist frequency). Such effects are also referred to as artefacts or ghosts." title="An error appearing in a sampled signal when the bandwidth of the signal is greater than half the sam..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;alias&lt;/span&gt;&lt;/a&gt; of the original sine wave. The way to avoid aliases is to sample more frequently, as in Figure 16(b). Here there are twice as many samples as in Figure 16(a). Now the alias from (a), also shown dashed in (b), misses some of the samples in (b). There is no waveform of a lower frequency than the sampled waveform that can be fitted to all the samples. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/b116cb05/t312_openlearn_fig16.tif.jpg" alt="Described image" width="512" height="113" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049545168"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 16&lt;/b&gt;  (a) More than one sine wave can be drawn through these samples, so there is a low-frequency alias of the original wave; (b) with more samples, there is no alias that can be fitted to all the samples &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049545168&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049545168"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The way to eliminate the possibility of an alias is either to have greater than two samples per cycle or, conversely, to ‘band-limit’ the signal to make sure that there are no frequencies equal to or higher than one half of the sampling frequency. This is done by including an analogue filter before the sampler in the analogue-to-digital converter that removes all frequencies higher than a certain frequency. As such, a filter is designed to stop any chance of aliasing occurring, it is sometimes called an &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048998992" class="oucontent-glossaryterm" data-definition="A low-pass filter that is able to remove aliasing in sampled signals by cutting all the spectral components that are greater than or equal to half the sampling frequency." title="A low-pass filter that is able to remove aliasing in sampled signals by cutting all the spectral com..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;anti-aliasing filter&lt;/span&gt;&lt;/a&gt;. &lt;/p&gt;&lt;p&gt;After sampling the signal, the samples need to be converted to a digital signal. The next section describes how this is done.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.4 Quantisation of a signal</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.4</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;When a continuous-time signal is sampled, the amplitude of each sampled point undergoes quantisation which means that it is forced to have only certain discrete values. The amplitude of each sample is represented by a digital binary code, and the word length of the code will be a fixed number of digital bits. Representing the amplitude of the samples in this way means the value is represented by a binary number, so it is truncated or quantised to its closest binary equivalent. &lt;/p&gt;&lt;p&gt;A 1-bit binary number can represent two levels because it can only take a value of 0 or a value of 1. A 2-bit binary number represents four levels, where each level takes one of the values 00, 01, 10 or 11. Figure&amp;#xA0;17 shows a discrete-time signal whose values are limited to a 3-bit binary number, which represents eight possible combinations of 0s and 1s – each possible combination is shown on the &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1da1b4654a7a5879426e3ed1980e978bb3478ccf"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_44d" height="12px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 502.0 706.7886" width="8.5231px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;-axis of the figure. Note that each sampled, digitised signal is a binary representation of the analogue value, so in conversion to these binary representations some rounding of the signal values has occurred: some sampled values are just above and some just below the continuous-time signal. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/d0a97c2d/t312_openlearn_fig17.tif.jpg" alt="Described image" width="512" height="436" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049527344"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 17&lt;/b&gt; Sampled and quantised signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049527344&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049527344"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Any binary word will always give an even number of &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048970640" class="oucontent-glossaryterm" data-definition="Conversion of an analogue quantity, which could take any value within a range, to one of a set of discrete values." title="Conversion of an analogue quantity, which could take any value within a range, to one of a set of di..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;quantisation&lt;/span&gt;&lt;/a&gt; levels. In general, a binary word with &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_45d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; quantisation levels – hence a 3-bit word gives 8&amp;#xA0;levels, a 4-bit word gives 16&amp;#xA0;levels, a 5-bit word gives 32&amp;#xA0;levels, etc. &lt;/p&gt;&lt;p&gt;Figure&amp;#xA0;18 shows a sine wave along with a set of quantisation levels. Here the 0 or midpoint of the sine wave occurs at a midpoint between the 011 and 100 levels. The gaps between the quantisation levels are called the quantisation intervals. Sometimes a quantisation level is assigned to the 0 or midpoint, which means that on one side of the 0 there is one more level than on the other side. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/0a34a616/t312_openlearn_fig18.tif.jpg" alt="Described image" width="512" height="170" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049513472"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 18&lt;/b&gt; A sine wave and a set of quantisation levels &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049513472&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049513472"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The larger the quantisation intervals, the more error will be introduced into the sampled signal. The quantisation error for a sample is the difference between the value of the input signal and the resultant quantised signal, with the maximum quantisation error being half the quantisation interval. Quantisation errors can be reduced by increasing the number of levels; however, as the number of levels increases, so does the number of bits needed to represent each sample. &lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 7&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 15 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;An electrocardiogram (ECG) signal contains useful information in frequencies up to 400&amp;#xA0;Hz.&lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;At what rate must the signal be sampled to ensure that no information is lost in sampling?&lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;If a 3-bit quantiser is used to represent the signal, how many bits of data are generated per second if the sampling rate is set to 1000&amp;#xA0;Hz? &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;If the signal range from the ECG extends from +7&amp;#xA0;V to &amp;#x2212;7&amp;#xA0;V, where 7&amp;#xA0;V equates to the highest quantisation level and &amp;#x2212;7&amp;#xA0;V to the lowest quantisation level, what is the quantisation interval and what is the maximum quantisation error in the system? &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;d.&lt;/span&gt;If the quantiser is changed to a 4-bit system, what happens to the maximum quantisation error?&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The signal must be sampled at greater than twice the maximum frequency of the signal, so the minimum sampling rate is 800&amp;#xA0;Hz.&lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;With a 3-bit quantiser each sample point uses three bits of data, so the number of bits of data generated is 3&amp;#xA0;&amp;#xD7;&amp;#xA0;1000&amp;#xA0;=&amp;#xA0;3000&amp;#xA0;bits per second. &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;With a 3-bit quantiser there are a total of eight levels and seven quantisation intervals in the system. If the voltage range covers 14&amp;#xA0;V, then each quantisation interval is 2&amp;#xA0;V. The maximum quantisation error is half this quantisation interval, or 1&amp;#xA0;V. &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;d.&lt;/span&gt;A 4-bit system has 16&amp;#xA0;quantisation levels and 15&amp;#xA0;quantisation intervals, so each quantisation interval is&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;
&lt;p&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="9905201678afdaa68df0c279eaa9d02002b62058"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_47d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 8531.9 2179.2650" width="144.8563px"&gt;

&lt;desc id="eq_dea10bfd_47d"&gt;equation left hand side 14 cap v divided by 15 equals right hand side 0.933 times horizontal ellipsis cap v&lt;/desc&gt;
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&lt;p&gt;This reduces the maximum quantisation error to&lt;/p&gt;
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&lt;desc id="eq_dea10bfd_48d"&gt;0.933 times horizontal ellipsis cap v divided by two equals 0.47 cap v left parenthesis to two s full stop f full stop right parenthesis&lt;/desc&gt;
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&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;In the next section you will return to digital filtering and how they work in the time domain.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.4</guid>
    <dc:title>3.4 Quantisation of a signal</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;When a continuous-time signal is sampled, the amplitude of each sampled point undergoes quantisation which means that it is forced to have only certain discrete values. The amplitude of each sample is represented by a digital binary code, and the word length of the code will be a fixed number of digital bits. Representing the amplitude of the samples in this way means the value is represented by a binary number, so it is truncated or quantised to its closest binary equivalent. &lt;/p&gt;&lt;p&gt;A 1-bit binary number can represent two levels because it can only take a value of 0 or a value of 1. A 2-bit binary number represents four levels, where each level takes one of the values 00, 01, 10 or 11. Figure 17 shows a discrete-time signal whose values are limited to a 3-bit binary number, which represents eight possible combinations of 0s and 1s – each possible combination is shown on the &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1da1b4654a7a5879426e3ed1980e978bb3478ccf"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_44d" height="12px" role="math" style="vertical-align: -4px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 502.0 706.7886" width="8.5231px"&gt;

&lt;desc id="eq_dea10bfd_44d"&gt;y&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;-axis of the figure. Note that each sampled, digitised signal is a binary representation of the analogue value, so in conversion to these binary representations some rounding of the signal values has occurred: some sampled values are just above and some just below the continuous-time signal. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/d0a97c2d/t312_openlearn_fig17.tif.jpg" alt="Described image" width="512" height="436" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049527344"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 17&lt;/b&gt; Sampled and quantised signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049527344&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049527344"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Any binary word will always give an even number of &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048970640" class="oucontent-glossaryterm" data-definition="Conversion of an analogue quantity, which could take any value within a range, to one of a set of discrete values." title="Conversion of an analogue quantity, which could take any value within a range, to one of a set of di..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;quantisation&lt;/span&gt;&lt;/a&gt; levels. In general, a binary word with &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_45d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_45d"&gt;n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; quantisation levels – hence a 3-bit word gives 8 levels, a 4-bit word gives 16 levels, a 5-bit word gives 32 levels, etc. &lt;/p&gt;&lt;p&gt;Figure 18 shows a sine wave along with a set of quantisation levels. Here the 0 or midpoint of the sine wave occurs at a midpoint between the 011 and 100 levels. The gaps between the quantisation levels are called the quantisation intervals. Sometimes a quantisation level is assigned to the 0 or midpoint, which means that on one side of the 0 there is one more level than on the other side. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/0a34a616/t312_openlearn_fig18.tif.jpg" alt="Described image" width="512" height="170" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049513472"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 18&lt;/b&gt; A sine wave and a set of quantisation levels &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049513472&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049513472"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The larger the quantisation intervals, the more error will be introduced into the sampled signal. The quantisation error for a sample is the difference between the value of the input signal and the resultant quantised signal, with the maximum quantisation error being half the quantisation interval. Quantisation errors can be reduced by increasing the number of levels; however, as the number of levels increases, so does the number of bits needed to represent each sample. &lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 7&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 15 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;An electrocardiogram (ECG) signal contains useful information in frequencies up to 400 Hz.&lt;/p&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;At what rate must the signal be sampled to ensure that no information is lost in sampling?&lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;If a 3-bit quantiser is used to represent the signal, how many bits of data are generated per second if the sampling rate is set to 1000 Hz? &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;If the signal range from the ECG extends from +7 V to −7 V, where 7 V equates to the highest quantisation level and −7 V to the lowest quantisation level, what is the quantisation interval and what is the maximum quantisation error in the system? &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;d.&lt;/span&gt;If the quantiser is changed to a 4-bit system, what happens to the maximum quantisation error?&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The signal must be sampled at greater than twice the maximum frequency of the signal, so the minimum sampling rate is 800 Hz.&lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;With a 3-bit quantiser each sample point uses three bits of data, so the number of bits of data generated is 3 × 1000 = 3000 bits per second. &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;c.&lt;/span&gt;With a 3-bit quantiser there are a total of eight levels and seven quantisation intervals in the system. If the voltage range covers 14 V, then each quantisation interval is 2 V. The maximum quantisation error is half this quantisation interval, or 1 V. &lt;/p&gt;&lt;/li&gt;&lt;li class="oucontent-markerinside"&gt;&lt;p class="oucontent-markerpara"&gt;&lt;span class="oucontent-listmarker"&gt;d.&lt;/span&gt;A 4-bit system has 16 quantisation levels and 15 quantisation intervals, so each quantisation interval is&lt;/p&gt;&lt;/li&gt;&lt;/ul&gt;
&lt;p&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="9905201678afdaa68df0c279eaa9d02002b62058"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_47d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 8531.9 2179.2650" width="144.8563px"&gt;

&lt;desc id="eq_dea10bfd_47d"&gt;equation left hand side 14 cap v divided by 15 equals right hand side 0.933 times horizontal ellipsis cap v&lt;/desc&gt;
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&lt;p&gt;This reduces the maximum quantisation error to&lt;/p&gt;
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&lt;desc id="eq_dea10bfd_48d"&gt;0.933 times horizontal ellipsis cap v divided by two equals 0.47 cap v left parenthesis to two s full stop f full stop right parenthesis&lt;/desc&gt;
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&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;In the next section you will return to digital filtering and how they work in the time domain.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.5 Digital filtering in the time domain</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.5</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;A simple form of digital filter is the three-term averaging filter, in which the output value is equal to the average of three successive signal sample values. In Figure&amp;#xA0;19(a), which shows a discrete-time signal applied to a digital three-term averaging filter, the output &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_49d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_49d"&gt;y of n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is given by &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="3e450dc42df13dc84444851cef3cde913ae10f7c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_50d" height="37px" role="math" style="vertical-align: -13px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1413.5773 16937.3 2179.2650" width="287.5649px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;The signals &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_51d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_51d"&gt;x of n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="23cac1b26733e44ab527a86047e61dd4e954d9ca"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_52d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_52d"&gt;x of n minus one&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; are spaced &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_53d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_53d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&amp;#xA0;seconds apart in time, where &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_54d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_54d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is the sampling interval. Similarly, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="23cac1b26733e44ab527a86047e61dd4e954d9ca"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_55d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_55d"&gt;x of n minus one&lt;/desc&gt;
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&lt;g transform="translate(577,0)"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a2dffc0ae301ac7e0547af24c50f8adbf6fd67f7"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_56d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_56d"&gt;x of n minus two&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;g transform="translate(577,0)"&gt;
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 &lt;use x="1110" xlink:href="#eq_dea10bfd_56MJMAIN-2212" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; are spaced &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_57d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_57d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&amp;#xA0;seconds apart. &lt;/p&gt;&lt;p&gt;Figure 19(b) shows the values of the input signal &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_58d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_58d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; that will be applied to the filter input. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/199ec11b/t312_openlearn_fig19.tif.jpg" alt="Described image" width="512" height="167" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049466480"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 19&lt;/b&gt;  (a) Three-term averaging filter; (b) input signal to the three-term averaging filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049466480&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049466480"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_59d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_59d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_59MJMATHI-6E" stroke-width="10"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_60d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_60d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_60MJMATHI-78" stroke-width="10"/&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_60MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_60MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_60MJMATHI-6E" y="0"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; are listed in Table 1. Assume that &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1487dc2dfc958f5069b59b8af08b9a7ae846f0b2"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_61d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3591.6 1119.0820" width="60.9789px"&gt;

&lt;desc id="eq_dea10bfd_61d"&gt;x of n equals zero&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_61MJMATHI-6E" stroke-width="10"/&gt;
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&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_61MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_61MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_61MJMAIN-5D" y="0"/&gt;
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 &lt;use x="2025" xlink:href="#eq_dea10bfd_61MJMAIN-3D" y="0"/&gt;
 &lt;use x="3086" xlink:href="#eq_dea10bfd_61MJMAIN-30" y="0"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; for any &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="2d6197a96f4b44762e233ef8d73769d60ec6eb42"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_62d" height="14px" role="math" style="vertical-align: -2px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 824.5868" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_62d"&gt;n less than negative three&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M694 -11T694 -19T688 -33T678 -40Q671 -40 524 29T234 166L90 235Q83 240 83 250Q83 261 91 266Q664 540 678 540Q681 540 687 534T694 519T687 505Q686 504 417 376L151 250L417 124Q686 -4 687 -5Q694 -11 694 -19Z" id="eq_dea10bfd_62MJMAIN-3C" stroke-width="10"/&gt;
&lt;path d="M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z" id="eq_dea10bfd_62MJMAIN-2212" stroke-width="10"/&gt;
&lt;path d="M127 463Q100 463 85 480T69 524Q69 579 117 622T233 665Q268 665 277 664Q351 652 390 611T430 522Q430 470 396 421T302 350L299 348Q299 347 308 345T337 336T375 315Q457 262 457 175Q457 96 395 37T238 -22Q158 -22 100 21T42 130Q42 158 60 175T105 193Q133 193 151 175T169 130Q169 119 166 110T159 94T148 82T136 74T126 70T118 67L114 66Q165 21 238 21Q293 21 321 74Q338 107 338 175V195Q338 290 274 322Q259 328 213 329L171 330L168 332Q166 335 166 348Q166 366 174 366Q202 366 232 371Q266 376 294 413T322 525V533Q322 590 287 612Q265 626 240 626Q208 626 181 615T143 592T132 580H135Q138 579 143 578T153 573T165 566T175 555T183 540T186 520Q186 498 172 481T127 463Z" id="eq_dea10bfd_62MJMAIN-33" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_62MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_62MJMAIN-3C" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_62MJMAIN-2212" y="0"/&gt;
 &lt;use x="2726" xlink:href="#eq_dea10bfd_62MJMAIN-33" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="e92f1e23b8a707575fb01aea4c807415ba06ef2f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_63d" height="14px" role="math" style="vertical-align: -2px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 824.5868" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_63d"&gt;n greater than five&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_63MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M84 520Q84 528 88 533T96 539L99 540Q106 540 253 471T544 334L687 265Q694 260 694 250T687 235Q685 233 395 96L107 -40H101Q83 -38 83 -20Q83 -19 83 -17Q82 -10 98 -1Q117 9 248 71Q326 108 378 132L626 250L378 368Q90 504 86 509Q84 513 84 520Z" id="eq_dea10bfd_63MJMAIN-3E" stroke-width="10"/&gt;
&lt;path d="M164 157Q164 133 148 117T109 101H102Q148 22 224 22Q294 22 326 82Q345 115 345 210Q345 313 318 349Q292 382 260 382H254Q176 382 136 314Q132 307 129 306T114 304Q97 304 95 310Q93 314 93 485V614Q93 664 98 664Q100 666 102 666Q103 666 123 658T178 642T253 634Q324 634 389 662Q397 666 402 666Q410 666 410 648V635Q328 538 205 538Q174 538 149 544L139 546V374Q158 388 169 396T205 412T256 420Q337 420 393 355T449 201Q449 109 385 44T229 -22Q148 -22 99 32T50 154Q50 178 61 192T84 210T107 214Q132 214 148 197T164 157Z" id="eq_dea10bfd_63MJMAIN-35" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_63MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_63MJMAIN-3E" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_63MJMAIN-35" y="0"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;div class="oucontent-table oucontent-s-normal noborder oucontent-s-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;&lt;b&gt;Table 1&lt;/b&gt;&amp;#xA0;&amp;#xA0;Values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_64d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_64d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_64MJMATHI-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_65d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_65d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_65MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_65MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_65MJMATHI-6E" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_65MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_65MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_65MJMATHI-6E" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; from Figure&amp;#xA0;19(b) &lt;/h2&gt;&lt;div class="oucontent-table-wrapper"&gt;&lt;table&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_66d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_66d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_66MJMATHI-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_66MJMATHI-6E" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;&amp;#x2212;3&lt;/td&gt;
&lt;td&gt;&amp;#x2212;2&lt;/td&gt;
&lt;td&gt;&amp;#x2212;1&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;3&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;5&lt;/td&gt;
&lt;/tr&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_67d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_67d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_67MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_67MJMATHI-6E" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_67MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_67MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_67MJMATHI-6E" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;&amp;#xA0;&amp;#xA0;0&lt;/td&gt;
&lt;td&gt;&amp;#xA0;&amp;#xA0;0&lt;/td&gt;
&lt;td&gt;&amp;#x2212;1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;6&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;/tr&gt;&lt;/table&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;For the output &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_68d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_68d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q21 301 36 335T84 406T158 442Q199 442 224 419T250 355Q248 336 247 334Q247 331 231 288T198 191T182 105Q182 62 196 45T238 27Q261 27 281 38T312 61T339 94Q339 95 344 114T358 173T377 247Q415 397 419 404Q432 431 462 431Q475 431 483 424T494 412T496 403Q496 390 447 193T391 -23Q363 -106 294 -155T156 -205Q111 -205 77 -183T43 -117Q43 -95 50 -80T69 -58T89 -48T106 -45Q150 -45 150 -87Q150 -107 138 -122T115 -142T102 -147L99 -148Q101 -153 118 -160T152 -167H160Q177 -167 186 -165Q219 -156 247 -127T290 -65T313 -9T321 21L315 17Q309 13 296 6T270 -6Q250 -11 231 -11Q185 -11 150 11T104 82Q103 89 103 113Q103 170 138 262T173 379Q173 380 173 381Q173 390 173 393T169 400T158 404H154Q131 404 112 385T82 344T65 302T57 280Q55 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_68MJMATHI-79" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_68MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_68MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_68MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_68MJMATHI-79" y="0"/&gt;
&lt;g transform="translate(502,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_68MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_68MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_68MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; to be calculated, the values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_69d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_69d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_69MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_69MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_69MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_69MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_69MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_69MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_69MJMATHI-6E" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="23cac1b26733e44ab527a86047e61dd4e954d9ca"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_70d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_70d"&gt;x of n minus one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_70MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_70MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z" id="eq_dea10bfd_70MJMAIN-2212" stroke-width="10"/&gt;
&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_70MJMAIN-31" stroke-width="10"/&gt;
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&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_70MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_70MJMATHI-6E" y="0"/&gt;
 &lt;use x="1110" xlink:href="#eq_dea10bfd_70MJMAIN-2212" y="0"/&gt;
 &lt;use x="2115" xlink:href="#eq_dea10bfd_70MJMAIN-31" y="0"/&gt;
 &lt;use x="2620" xlink:href="#eq_dea10bfd_70MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a2dffc0ae301ac7e0547af24c50f8adbf6fd67f7"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_71d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_71d"&gt;x of n minus two&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_71MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_71MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z" id="eq_dea10bfd_71MJMAIN-2212" stroke-width="10"/&gt;
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&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_71MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_71MJMATHI-6E" y="0"/&gt;
 &lt;use x="1110" xlink:href="#eq_dea10bfd_71MJMAIN-2212" y="0"/&gt;
 &lt;use x="2115" xlink:href="#eq_dea10bfd_71MJMAIN-32" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; must be stored in memory and accessible to the processor performing the calculation. The order of a digital filter is the number of previous inputs (stored in the processor’s memory) used to calculate the current output, so this three-term averaging filter is second-order. &lt;/p&gt;&lt;p&gt;For values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="61409ee985763d08063127832b94df50df1c5a00"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_72d" height="14px" role="math" style="vertical-align: -2px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 824.5868" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_72d"&gt;n less than negative one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; the output will be 0, so the calculations below start at &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c4e6c8050e72c7ad2ffdfdbe200285527833f91c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_73d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 765.6877" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_73d"&gt;n equals negative one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;For &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c4e6c8050e72c7ad2ffdfdbe200285527833f91c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_74d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 765.6877" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_74d"&gt;n equals negative one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_74MJMAIN-31" stroke-width="10"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; you can substitute in values to give &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="155de5590673f6915fc4efb9dcfc5c661e4bd0f2"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_75d" height="25px" role="math" style="vertical-align: -8px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1001.2839 17907.7 1472.4763" width="304.0405px"&gt;

&lt;desc id="eq_dea10bfd_75d"&gt;equation sequence y of negative one equals one divided by three multiplication open negative one close plus one divided by three multiplication zero plus one divided by three multiplication zero equals negative one divided by three&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M56 237T56 250T70 270H369V420L370 570Q380 583 389 583Q402 583 409 568V270H707Q722 262 722 250T707 230H409V-68Q401 -82 391 -82H389H387Q375 -82 369 -68V230H70Q56 237 56 250Z" id="eq_dea10bfd_75MJMAIN-2B" stroke-width="10"/&gt;
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&lt;/g&gt;
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 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-33" y="-597"/&gt;
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&lt;/g&gt;
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 &lt;use x="10887" xlink:href="#eq_dea10bfd_75MJMAIN-30" y="0"/&gt;
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&lt;g transform="translate(12397,0)"&gt;
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 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-33" y="-597"/&gt;
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&lt;/g&gt;
 &lt;use x="13558" xlink:href="#eq_dea10bfd_75MJMAIN-D7" y="0"/&gt;
 &lt;use x="14564" xlink:href="#eq_dea10bfd_75MJMAIN-30" y="0"/&gt;
 &lt;use x="15346" xlink:href="#eq_dea10bfd_75MJMAIN-3D" y="0"/&gt;
 &lt;use x="16407" xlink:href="#eq_dea10bfd_75MJMAIN-2212" y="0"/&gt;
&lt;g transform="translate(17190,0)"&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;For &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="e455e2972caf9adeb0db27c0ef70c5147a9247d9"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_76d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_76d"&gt;n equals zero&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;desc id="eq_dea10bfd_77d"&gt;y of zero equals one divided by three multiplication two plus one divided by three multiplication open negative one close plus one divided by three multiplication zero postfix times equation left hand side equals right hand side one divided by three&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;For &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1ae6535d1ba4e21f621719f7183529aa1a68e06c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_78d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_78d"&gt;n equals one&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_79d"&gt;y of one equals one divided by three multiplication four plus one divided by three multiplication two plus one divided by three multiplication open negative one close postfix times equation left hand side equals right hand side five divided by three&lt;/desc&gt;
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 &lt;use x="11204" xlink:href="#eq_dea10bfd_79MJMAIN-D7" y="0"/&gt;
&lt;g transform="translate(12210,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_79MJMAIN-28" y="0"/&gt;
 &lt;use x="394" xlink:href="#eq_dea10bfd_79MJMAIN-2212" y="0"/&gt;
 &lt;use x="1177" xlink:href="#eq_dea10bfd_79MJMAIN-31" y="0"/&gt;
 &lt;use x="1682" xlink:href="#eq_dea10bfd_79MJMAIN-29" y="0"/&gt;
&lt;/g&gt;
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&lt;g transform="translate(15680,0)"&gt;
&lt;g transform="translate(397,0)"&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;Table 2 shows the results of all calculations for &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_80d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_80d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_80MJMAIN-5B" stroke-width="10"/&gt;
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&lt;g transform="translate(502,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_80MJMAIN-5B" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, shown as decimal values to 2&amp;#xA0;significant figures. Note that you can stop at &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="f2278ca81fb5c8db6d6bfd9b3ebe23f08d72e88e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_81d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_81d"&gt;n equals five&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M164 157Q164 133 148 117T109 101H102Q148 22 224 22Q294 22 326 82Q345 115 345 210Q345 313 318 349Q292 382 260 382H254Q176 382 136 314Q132 307 129 306T114 304Q97 304 95 310Q93 314 93 485V614Q93 664 98 664Q100 666 102 666Q103 666 123 658T178 642T253 634Q324 634 389 662Q397 666 402 666Q410 666 410 648V635Q328 538 205 538Q174 538 149 544L139 546V374Q158 388 169 396T205 412T256 420Q337 420 393 355T449 201Q449 109 385 44T229 -22Q148 -22 99 32T50 154Q50 178 61 192T84 210T107 214Q132 214 148 197T164 157Z" id="eq_dea10bfd_81MJMAIN-35" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_81MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_81MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_81MJMAIN-35" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, since above this the output will be 0 again. &lt;/p&gt;&lt;div class="oucontent-table oucontent-s-normal noborder oucontent-s-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;&lt;b&gt;Table 2&lt;/b&gt;&amp;#xA0;&amp;#xA0;Results of calculations for &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_82d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_82d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_82MJMAIN-5B" stroke-width="10"/&gt;
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&lt;g transform="translate(502,0)"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/h2&gt;&lt;div class="oucontent-table-wrapper"&gt;&lt;table&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_83d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_83d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;&amp;#x2212;3&lt;/td&gt;
&lt;td&gt;&amp;#x2212;2&lt;/td&gt;
&lt;td&gt;&amp;#x2212;1&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;3&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;5&lt;/td&gt;
&lt;/tr&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_84d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_84d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;td&gt;&amp;#xA0;&amp;#xA0;0&lt;/td&gt;
&lt;td&gt;&amp;#xA0;&amp;#xA0;0&lt;/td&gt;
&lt;td&gt;&amp;#x2212;1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;6&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;/tr&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_85d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_85d"&gt;y of n&lt;/desc&gt;
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&lt;td&gt;&amp;#xA0;&amp;#xA0;0&lt;/td&gt;
&lt;td&gt;&amp;#xA0;&amp;#xA0;0&lt;/td&gt;
&lt;td&gt;&amp;#x2212;0.33&lt;/td&gt;
&lt;td&gt;0.33&lt;/td&gt;
&lt;td&gt;1.67&lt;/td&gt;
&lt;td&gt;4.00&lt;/td&gt;
&lt;td&gt;4.67&lt;/td&gt;
&lt;td&gt;3.33&lt;/td&gt;
&lt;td&gt;1.33&lt;/td&gt;
&lt;/tr&gt;&lt;/table&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;The resulting output discrete-time waveform is given in Figure&amp;#xA0;20.&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/fe518d14/t312_openlearn_fig20.tif.jpg" alt="Described image" width="512" height="207" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049349776"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 20&lt;/b&gt;  Filter output in response to the input in Figure 19(b) &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049349776&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049349776"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Figure 21 shows the same filter applied to an input &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_86d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; that has more noise in the signal and a longer sequence. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/2c1afd5d/t312_openlearn_fig21.tif.jpg" alt="Described image" width="512" height="301" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049319424"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 21&lt;/b&gt;  Longer sequence filtered by the three-term averaging filter (Wickert, 2011, p. 7) &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049319424&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049319424"&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 8&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Looking at Figure 21, describe the effect that the three-term averaging filter has on the output. What kind of filter is this? &lt;/p&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;The three-term averaging filter has removed the short-term fluctuations in the signal to show the longer-term trend. This is akin to removing higher-frequency noise from a signal, so it is operating like a low-pass filter. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;This digital filter is an example of a system that is both linear and time-invariant, sometimes referred to as an &lt;b&gt;LTI &lt;/b&gt;system. &lt;/p&gt;&lt;p&gt;You can see that the three-term averaging filter is a low-pass filter, but it is difficult to characterise its response. For example, what frequencies is the filter eliminating from the signal? Earlier in this course, you saw how analogue filters can be designed to give a desired frequency response; now you will look at how digital filters can also be designed in the frequency domain. &lt;/p&gt;&lt;p&gt;Filters are usually described in terms that make sense in the frequency domain, e.g. a low pass filter allows the parts of the signal with low frequencies to pass. In the following section you will see how a digital filter is designed in the frequency domain.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.5</guid>
    <dc:title>3.5 Digital filtering in the time domain</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;A simple form of digital filter is the three-term averaging filter, in which the output value is equal to the average of three successive signal sample values. In Figure 19(a), which shows a discrete-time signal applied to a digital three-term averaging filter, the output &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_49d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

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&lt;desc id="eq_dea10bfd_51d"&gt;x of n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="23cac1b26733e44ab527a86047e61dd4e954d9ca"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_52d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_52d"&gt;x of n minus one&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; are spaced &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_53d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_53d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; seconds apart in time, where &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_54d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_54d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; is the sampling interval. Similarly, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="23cac1b26733e44ab527a86047e61dd4e954d9ca"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_55d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_55d"&gt;x of n minus one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_55MJMAIN-5B" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a2dffc0ae301ac7e0547af24c50f8adbf6fd67f7"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_56d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_56d"&gt;x of n minus two&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;g transform="translate(577,0)"&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; are spaced &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="8e8991aaad93401cdfa477204704bb5bad35889c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_57d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 709.0 765.6877" width="12.0375px"&gt;

&lt;desc id="eq_dea10bfd_57d"&gt;cap t&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; seconds apart. &lt;/p&gt;&lt;p&gt;Figure 19(b) shows the values of the input signal &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_58d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_58d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_58MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_58MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_58MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_58MJMAIN-5D" y="0"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; that will be applied to the filter input. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/199ec11b/t312_openlearn_fig19.tif.jpg" alt="Described image" width="512" height="167" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049466480"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 19&lt;/b&gt;  (a) Three-term averaging filter; (b) input signal to the three-term averaging filter &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049466480&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049466480"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_59d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_59d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_59MJMATHI-6E" stroke-width="10"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_60d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_60d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_60MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_60MJMAIN-5B" stroke-width="10"/&gt;
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&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_60MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_60MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_60MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_60MJMATHI-6E" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; are listed in Table 1. Assume that &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="1487dc2dfc958f5069b59b8af08b9a7ae846f0b2"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_61d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3591.6 1119.0820" width="60.9789px"&gt;

&lt;desc id="eq_dea10bfd_61d"&gt;x of n equals zero&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_61MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_61MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_61MJMAIN-5D" stroke-width="10"/&gt;
&lt;path d="M56 347Q56 360 70 367H707Q722 359 722 347Q722 336 708 328L390 327H72Q56 332 56 347ZM56 153Q56 168 72 173H708Q722 163 722 153Q722 140 707 133H70Q56 140 56 153Z" id="eq_dea10bfd_61MJMAIN-3D" stroke-width="10"/&gt;
&lt;path d="M96 585Q152 666 249 666Q297 666 345 640T423 548Q460 465 460 320Q460 165 417 83Q397 41 362 16T301 -15T250 -22Q224 -22 198 -16T137 16T82 83Q39 165 39 320Q39 494 96 585ZM321 597Q291 629 250 629Q208 629 178 597Q153 571 145 525T137 333Q137 175 145 125T181 46Q209 16 250 16Q290 16 318 46Q347 76 354 130T362 333Q362 478 354 524T321 597Z" id="eq_dea10bfd_61MJMAIN-30" stroke-width="10"/&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_61MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_61MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_61MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_61MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
 &lt;use x="2025" xlink:href="#eq_dea10bfd_61MJMAIN-3D" y="0"/&gt;
 &lt;use x="3086" xlink:href="#eq_dea10bfd_61MJMAIN-30" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; for any &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="2d6197a96f4b44762e233ef8d73769d60ec6eb42"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_62d" height="14px" role="math" style="vertical-align: -2px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 824.5868" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_62d"&gt;n less than negative three&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M694 -11T694 -19T688 -33T678 -40Q671 -40 524 29T234 166L90 235Q83 240 83 250Q83 261 91 266Q664 540 678 540Q681 540 687 534T694 519T687 505Q686 504 417 376L151 250L417 124Q686 -4 687 -5Q694 -11 694 -19Z" id="eq_dea10bfd_62MJMAIN-3C" stroke-width="10"/&gt;
&lt;path d="M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z" id="eq_dea10bfd_62MJMAIN-2212" stroke-width="10"/&gt;
&lt;path d="M127 463Q100 463 85 480T69 524Q69 579 117 622T233 665Q268 665 277 664Q351 652 390 611T430 522Q430 470 396 421T302 350L299 348Q299 347 308 345T337 336T375 315Q457 262 457 175Q457 96 395 37T238 -22Q158 -22 100 21T42 130Q42 158 60 175T105 193Q133 193 151 175T169 130Q169 119 166 110T159 94T148 82T136 74T126 70T118 67L114 66Q165 21 238 21Q293 21 321 74Q338 107 338 175V195Q338 290 274 322Q259 328 213 329L171 330L168 332Q166 335 166 348Q166 366 174 366Q202 366 232 371Q266 376 294 413T322 525V533Q322 590 287 612Q265 626 240 626Q208 626 181 615T143 592T132 580H135Q138 579 143 578T153 573T165 566T175 555T183 540T186 520Q186 498 172 481T127 463Z" id="eq_dea10bfd_62MJMAIN-33" stroke-width="10"/&gt;
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&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_62MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_62MJMAIN-3C" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_62MJMAIN-2212" y="0"/&gt;
 &lt;use x="2726" xlink:href="#eq_dea10bfd_62MJMAIN-33" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="e92f1e23b8a707575fb01aea4c807415ba06ef2f"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_63d" height="14px" role="math" style="vertical-align: -2px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 824.5868" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_63d"&gt;n greater than five&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_63MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M84 520Q84 528 88 533T96 539L99 540Q106 540 253 471T544 334L687 265Q694 260 694 250T687 235Q685 233 395 96L107 -40H101Q83 -38 83 -20Q83 -19 83 -17Q82 -10 98 -1Q117 9 248 71Q326 108 378 132L626 250L378 368Q90 504 86 509Q84 513 84 520Z" id="eq_dea10bfd_63MJMAIN-3E" stroke-width="10"/&gt;
&lt;path d="M164 157Q164 133 148 117T109 101H102Q148 22 224 22Q294 22 326 82Q345 115 345 210Q345 313 318 349Q292 382 260 382H254Q176 382 136 314Q132 307 129 306T114 304Q97 304 95 310Q93 314 93 485V614Q93 664 98 664Q100 666 102 666Q103 666 123 658T178 642T253 634Q324 634 389 662Q397 666 402 666Q410 666 410 648V635Q328 538 205 538Q174 538 149 544L139 546V374Q158 388 169 396T205 412T256 420Q337 420 393 355T449 201Q449 109 385 44T229 -22Q148 -22 99 32T50 154Q50 178 61 192T84 210T107 214Q132 214 148 197T164 157Z" id="eq_dea10bfd_63MJMAIN-35" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_63MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_63MJMAIN-3E" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_63MJMAIN-35" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;div class="oucontent-table oucontent-s-normal noborder oucontent-s-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;&lt;b&gt;Table 1&lt;/b&gt;  Values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_64d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_64d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_64MJMATHI-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_64MJMATHI-6E" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_65d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_65d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_65MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_65MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_65MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_65MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_65MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_65MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_65MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_65MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; from Figure 19(b) &lt;/h2&gt;&lt;div class="oucontent-table-wrapper"&gt;&lt;table&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_66d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_66d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_66MJMATHI-6E" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_66MJMATHI-6E" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;−3&lt;/td&gt;
&lt;td&gt;−2&lt;/td&gt;
&lt;td&gt;−1&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;3&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;5&lt;/td&gt;
&lt;/tr&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_67d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_67d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_67MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_67MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_67MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_67MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_67MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_67MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_67MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_67MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;  0&lt;/td&gt;
&lt;td&gt;  0&lt;/td&gt;
&lt;td&gt;−1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;6&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;/tr&gt;&lt;/table&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;For the output &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_68d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_68d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q21 301 36 335T84 406T158 442Q199 442 224 419T250 355Q248 336 247 334Q247 331 231 288T198 191T182 105Q182 62 196 45T238 27Q261 27 281 38T312 61T339 94Q339 95 344 114T358 173T377 247Q415 397 419 404Q432 431 462 431Q475 431 483 424T494 412T496 403Q496 390 447 193T391 -23Q363 -106 294 -155T156 -205Q111 -205 77 -183T43 -117Q43 -95 50 -80T69 -58T89 -48T106 -45Q150 -45 150 -87Q150 -107 138 -122T115 -142T102 -147L99 -148Q101 -153 118 -160T152 -167H160Q177 -167 186 -165Q219 -156 247 -127T290 -65T313 -9T321 21L315 17Q309 13 296 6T270 -6Q250 -11 231 -11Q185 -11 150 11T104 82Q103 89 103 113Q103 170 138 262T173 379Q173 380 173 381Q173 390 173 393T169 400T158 404H154Q131 404 112 385T82 344T65 302T57 280Q55 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_68MJMATHI-79" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_68MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_68MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_68MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_68MJMATHI-79" y="0"/&gt;
&lt;g transform="translate(502,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_68MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_68MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_68MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; to be calculated, the values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_69d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_69d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_69MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_69MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_69MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_69MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_69MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_69MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_69MJMATHI-6E" y="0"/&gt;
 &lt;use x="888" xlink:href="#eq_dea10bfd_69MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="23cac1b26733e44ab527a86047e61dd4e954d9ca"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_70d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_70d"&gt;x of n minus one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_70MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_70MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_70MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z" id="eq_dea10bfd_70MJMAIN-2212" stroke-width="10"/&gt;
&lt;path d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 52T207 57T213 61V578Z" id="eq_dea10bfd_70MJMAIN-31" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_70MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_70MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_70MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_70MJMATHI-6E" y="0"/&gt;
 &lt;use x="1110" xlink:href="#eq_dea10bfd_70MJMAIN-2212" y="0"/&gt;
 &lt;use x="2115" xlink:href="#eq_dea10bfd_70MJMAIN-31" y="0"/&gt;
 &lt;use x="2620" xlink:href="#eq_dea10bfd_70MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; and &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a2dffc0ae301ac7e0547af24c50f8adbf6fd67f7"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_71d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 3480.4 1119.0820" width="59.0909px"&gt;

&lt;desc id="eq_dea10bfd_71d"&gt;x of n minus two&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_71MJMATHI-78" stroke-width="10"/&gt;
&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_71MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_71MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M84 237T84 250T98 270H679Q694 262 694 250T679 230H98Q84 237 84 250Z" id="eq_dea10bfd_71MJMAIN-2212" stroke-width="10"/&gt;
&lt;path d="M109 429Q82 429 66 447T50 491Q50 562 103 614T235 666Q326 666 387 610T449 465Q449 422 429 383T381 315T301 241Q265 210 201 149L142 93L218 92Q375 92 385 97Q392 99 409 186V189H449V186Q448 183 436 95T421 3V0H50V19V31Q50 38 56 46T86 81Q115 113 136 137Q145 147 170 174T204 211T233 244T261 278T284 308T305 340T320 369T333 401T340 431T343 464Q343 527 309 573T212 619Q179 619 154 602T119 569T109 550Q109 549 114 549Q132 549 151 535T170 489Q170 464 154 447T109 429Z" id="eq_dea10bfd_71MJMAIN-32" stroke-width="10"/&gt;
&lt;path d="M22 710V750H159V-250H22V-210H119V710H22Z" id="eq_dea10bfd_71MJMAIN-5D" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_71MJMATHI-78" y="0"/&gt;
&lt;g transform="translate(577,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_71MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_71MJMATHI-6E" y="0"/&gt;
 &lt;use x="1110" xlink:href="#eq_dea10bfd_71MJMAIN-2212" y="0"/&gt;
 &lt;use x="2115" xlink:href="#eq_dea10bfd_71MJMAIN-32" y="0"/&gt;
 &lt;use x="2620" xlink:href="#eq_dea10bfd_71MJMAIN-5D" y="0"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; must be stored in memory and accessible to the processor performing the calculation. The order of a digital filter is the number of previous inputs (stored in the processor’s memory) used to calculate the current output, so this three-term averaging filter is second-order. &lt;/p&gt;&lt;p&gt;For values of &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="61409ee985763d08063127832b94df50df1c5a00"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_72d" height="14px" role="math" style="vertical-align: -2px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 824.5868" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_72d"&gt;n less than negative one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M694 -11T694 -19T688 -33T678 -40Q671 -40 524 29T234 166L90 235Q83 240 83 250Q83 261 91 266Q664 540 678 540Q681 540 687 534T694 519T687 505Q686 504 417 376L151 250L417 124Q686 -4 687 -5Q694 -11 694 -19Z" id="eq_dea10bfd_72MJMAIN-3C" stroke-width="10"/&gt;
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 &lt;use x="882" xlink:href="#eq_dea10bfd_72MJMAIN-3C" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_72MJMAIN-2212" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; the output will be 0, so the calculations below start at &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c4e6c8050e72c7ad2ffdfdbe200285527833f91c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_73d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 765.6877" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_73d"&gt;n equals negative one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;/defs&gt;
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 &lt;use x="882" xlink:href="#eq_dea10bfd_73MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_73MJMAIN-2212" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;For &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c4e6c8050e72c7ad2ffdfdbe200285527833f91c"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_74d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 3231.6 765.6877" width="54.8668px"&gt;

&lt;desc id="eq_dea10bfd_74d"&gt;n equals negative one&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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 &lt;use x="882" xlink:href="#eq_dea10bfd_74MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_74MJMAIN-2212" y="0"/&gt;
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&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; you can substitute in values to give &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="155de5590673f6915fc4efb9dcfc5c661e4bd0f2"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_75d" height="25px" role="math" style="vertical-align: -8px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1001.2839 17907.7 1472.4763" width="304.0405px"&gt;

&lt;desc id="eq_dea10bfd_75d"&gt;equation sequence y of negative one equals one divided by three multiplication open negative one close plus one divided by three multiplication zero plus one divided by three multiplication zero equals negative one divided by three&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M127 463Q100 463 85 480T69 524Q69 579 117 622T233 665Q268 665 277 664Q351 652 390 611T430 522Q430 470 396 421T302 350L299 348Q299 347 308 345T337 336T375 315Q457 262 457 175Q457 96 395 37T238 -22Q158 -22 100 21T42 130Q42 158 60 175T105 193Q133 193 151 175T169 130Q169 119 166 110T159 94T148 82T136 74T126 70T118 67L114 66Q165 21 238 21Q293 21 321 74Q338 107 338 175V195Q338 290 274 322Q259 328 213 329L171 330L168 332Q166 335 166 348Q166 366 174 366Q202 366 232 371Q266 376 294 413T322 525V533Q322 590 287 612Q265 626 240 626Q208 626 181 615T143 592T132 580H135Q138 579 143 578T153 573T165 566T175 555T183 540T186 520Q186 498 172 481T127 463Z" id="eq_dea10bfd_75MJMAIN-33" stroke-width="10"/&gt;
&lt;path d="M630 29Q630 9 609 9Q604 9 587 25T493 118L389 222L284 117Q178 13 175 11Q171 9 168 9Q160 9 154 15T147 29Q147 36 161 51T255 146L359 250L255 354Q174 435 161 449T147 471Q147 480 153 485T168 490Q173 490 175 489Q178 487 284 383L389 278L493 382Q570 459 587 475T609 491Q630 491 630 471Q630 464 620 453T522 355L418 250L522 145Q606 61 618 48T630 29Z" id="eq_dea10bfd_75MJMAIN-D7" stroke-width="10"/&gt;
&lt;path d="M94 250Q94 319 104 381T127 488T164 576T202 643T244 695T277 729T302 750H315H319Q333 750 333 741Q333 738 316 720T275 667T226 581T184 443T167 250T184 58T225 -81T274 -167T316 -220T333 -241Q333 -250 318 -250H315H302L274 -226Q180 -141 137 -14T94 250Z" id="eq_dea10bfd_75MJMAIN-28" stroke-width="10"/&gt;
&lt;path d="M60 749L64 750Q69 750 74 750H86L114 726Q208 641 251 514T294 250Q294 182 284 119T261 12T224 -76T186 -143T145 -194T113 -227T90 -246Q87 -249 86 -250H74Q66 -250 63 -250T58 -247T55 -238Q56 -237 66 -225Q221 -64 221 250T66 725Q56 737 55 738Q55 746 60 749Z" id="eq_dea10bfd_75MJMAIN-29" stroke-width="10"/&gt;
&lt;path d="M56 237T56 250T70 270H369V420L370 570Q380 583 389 583Q402 583 409 568V270H707Q722 262 722 250T707 230H409V-68Q401 -82 391 -82H389H387Q375 -82 369 -68V230H70Q56 237 56 250Z" id="eq_dea10bfd_75MJMAIN-2B" stroke-width="10"/&gt;
&lt;path d="M96 585Q152 666 249 666Q297 666 345 640T423 548Q460 465 460 320Q460 165 417 83Q397 41 362 16T301 -15T250 -22Q224 -22 198 -16T137 16T82 83Q39 165 39 320Q39 494 96 585ZM321 597Q291 629 250 629Q208 629 178 597Q153 571 145 525T137 333Q137 175 145 125T181 46Q209 16 250 16Q290 16 318 46Q347 76 354 130T362 333Q362 478 354 524T321 597Z" id="eq_dea10bfd_75MJMAIN-30" stroke-width="10"/&gt;
&lt;/defs&gt;
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 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-33" y="-597"/&gt;
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&lt;/g&gt;
 &lt;use x="4633" xlink:href="#eq_dea10bfd_75MJMAIN-D7" y="0"/&gt;
&lt;g transform="translate(5639,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_75MJMAIN-28" y="0"/&gt;
 &lt;use x="394" xlink:href="#eq_dea10bfd_75MJMAIN-2212" y="0"/&gt;
 &lt;use x="1177" xlink:href="#eq_dea10bfd_75MJMAIN-31" y="0"/&gt;
 &lt;use x="1682" xlink:href="#eq_dea10bfd_75MJMAIN-29" y="0"/&gt;
&lt;/g&gt;
 &lt;use x="7937" xlink:href="#eq_dea10bfd_75MJMAIN-2B" y="0"/&gt;
&lt;g transform="translate(8720,0)"&gt;
&lt;g transform="translate(342,0)"&gt;
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 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-33" y="-597"/&gt;
&lt;/g&gt;
&lt;/g&gt;
 &lt;use x="9881" xlink:href="#eq_dea10bfd_75MJMAIN-D7" y="0"/&gt;
 &lt;use x="10887" xlink:href="#eq_dea10bfd_75MJMAIN-30" y="0"/&gt;
 &lt;use x="11614" xlink:href="#eq_dea10bfd_75MJMAIN-2B" y="0"/&gt;
&lt;g transform="translate(12397,0)"&gt;
&lt;g transform="translate(342,0)"&gt;
&lt;rect height="60" stroke="none" width="477" x="0" y="220"/&gt;
 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-31" y="638"/&gt;
 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-33" y="-597"/&gt;
&lt;/g&gt;
&lt;/g&gt;
 &lt;use x="13558" xlink:href="#eq_dea10bfd_75MJMAIN-D7" y="0"/&gt;
 &lt;use x="14564" xlink:href="#eq_dea10bfd_75MJMAIN-30" y="0"/&gt;
 &lt;use x="15346" xlink:href="#eq_dea10bfd_75MJMAIN-3D" y="0"/&gt;
 &lt;use x="16407" xlink:href="#eq_dea10bfd_75MJMAIN-2212" y="0"/&gt;
&lt;g transform="translate(17190,0)"&gt;
&lt;g transform="translate(120,0)"&gt;
&lt;rect height="60" stroke="none" width="477" x="0" y="220"/&gt;
 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-31" y="638"/&gt;
 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_75MJMAIN-33" y="-597"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;For &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="e455e2972caf9adeb0db27c0ef70c5147a9247d9"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_76d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_76d"&gt;n equals zero&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_76MJMATHI-6E" stroke-width="10"/&gt;
&lt;path d="M56 347Q56 360 70 367H707Q722 359 722 347Q722 336 708 328L390 327H72Q56 332 56 347ZM56 153Q56 168 72 173H708Q722 163 722 153Q722 140 707 133H70Q56 140 56 153Z" id="eq_dea10bfd_76MJMAIN-3D" stroke-width="10"/&gt;
&lt;path d="M96 585Q152 666 249 666Q297 666 345 640T423 548Q460 465 460 320Q460 165 417 83Q397 41 362 16T301 -15T250 -22Q224 -22 198 -16T137 16T82 83Q39 165 39 320Q39 494 96 585ZM321 597Q291 629 250 629Q208 629 178 597Q153 571 145 525T137 333Q137 175 145 125T181 46Q209 16 250 16Q290 16 318 46Q347 76 354 130T362 333Q362 478 354 524T321 597Z" id="eq_dea10bfd_76MJMAIN-30" stroke-width="10"/&gt;
&lt;/defs&gt;
&lt;g aria-hidden="true" fill="currentColor" stroke="currentColor" stroke-width="0" transform="matrix(1 0 0 -1 0 0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_76MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_76MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_76MJMAIN-30" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; you get &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="fc0307b158b378e5537d5456d4056b94f08e52c6"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_77d" height="25px" role="math" style="vertical-align: -8px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -1001.2839 16508.7 1472.4763" width="280.2880px"&gt;

&lt;desc id="eq_dea10bfd_77d"&gt;y of zero equals one divided by three multiplication two plus one divided by three multiplication open negative one close plus one divided by three multiplication zero postfix times equation left hand side equals right hand side one divided by three&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;desc id="eq_dea10bfd_78d"&gt;n equals one&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_79d"&gt;y of one equals one divided by three multiplication four plus one divided by three multiplication two plus one divided by three multiplication open negative one close postfix times equation left hand side equals right hand side five divided by three&lt;/desc&gt;
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 &lt;use transform="scale(0.707)" x="84" xlink:href="#eq_dea10bfd_79MJMAIN-33" y="-597"/&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;. &lt;/p&gt;&lt;p&gt;Table 2 shows the results of all calculations for &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_80d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_80d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_80MJMAIN-5B" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_80MJMATHI-79" y="0"/&gt;
&lt;g transform="translate(502,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_80MJMAIN-5B" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, shown as decimal values to 2 significant figures. Note that you can stop at &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="f2278ca81fb5c8db6d6bfd9b3ebe23f08d72e88e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_81d" height="13px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -706.7886 2448.6 765.6877" width="41.5728px"&gt;

&lt;desc id="eq_dea10bfd_81d"&gt;n equals five&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M164 157Q164 133 148 117T109 101H102Q148 22 224 22Q294 22 326 82Q345 115 345 210Q345 313 318 349Q292 382 260 382H254Q176 382 136 314Q132 307 129 306T114 304Q97 304 95 310Q93 314 93 485V614Q93 664 98 664Q100 666 102 666Q103 666 123 658T178 642T253 634Q324 634 389 662Q397 666 402 666Q410 666 410 648V635Q328 538 205 538Q174 538 149 544L139 546V374Q158 388 169 396T205 412T256 420Q337 420 393 355T449 201Q449 109 385 44T229 -22Q148 -22 99 32T50 154Q50 178 61 192T84 210T107 214Q132 214 148 197T164 157Z" id="eq_dea10bfd_81MJMAIN-35" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_81MJMATHI-6E" y="0"/&gt;
 &lt;use x="882" xlink:href="#eq_dea10bfd_81MJMAIN-3D" y="0"/&gt;
 &lt;use x="1943" xlink:href="#eq_dea10bfd_81MJMAIN-35" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;, since above this the output will be 0 again. &lt;/p&gt;&lt;div class="oucontent-table oucontent-s-normal noborder oucontent-s-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;&lt;b&gt;Table 2&lt;/b&gt;  Results of calculations for &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_82d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_82d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;path d="M118 -250V750H255V710H158V-210H255V-250H118Z" id="eq_dea10bfd_82MJMAIN-5B" stroke-width="10"/&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_82MJMATHI-6E" stroke-width="10"/&gt;
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&lt;/defs&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_82MJMATHI-79" y="0"/&gt;
&lt;g transform="translate(502,0)"&gt;
 &lt;use x="0" xlink:href="#eq_dea10bfd_82MJMAIN-5B" y="0"/&gt;
 &lt;use x="283" xlink:href="#eq_dea10bfd_82MJMATHI-6E" y="0"/&gt;
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&lt;/g&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/h2&gt;&lt;div class="oucontent-table-wrapper"&gt;&lt;table&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="c577391b03d3366106f2a952b8c35d6b30c6c333"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_83d" height="9px" role="math" style="vertical-align: -1px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -471.1924 605.0 530.0915" width="10.2718px"&gt;

&lt;desc id="eq_dea10bfd_83d"&gt;n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M21 287Q22 293 24 303T36 341T56 388T89 425T135 442Q171 442 195 424T225 390T231 369Q231 367 232 367L243 378Q304 442 382 442Q436 442 469 415T503 336T465 179T427 52Q427 26 444 26Q450 26 453 27Q482 32 505 65T540 145Q542 153 560 153Q580 153 580 145Q580 144 576 130Q568 101 554 73T508 17T439 -10Q392 -10 371 17T350 73Q350 92 386 193T423 345Q423 404 379 404H374Q288 404 229 303L222 291L189 157Q156 26 151 16Q138 -11 108 -11Q95 -11 87 -5T76 7T74 17Q74 30 112 180T152 343Q153 348 153 366Q153 405 129 405Q91 405 66 305Q60 285 60 284Q58 278 41 278H27Q21 284 21 287Z" id="eq_dea10bfd_83MJMATHI-6E" stroke-width="10"/&gt;
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 &lt;use x="0" xlink:href="#eq_dea10bfd_83MJMATHI-6E" y="0"/&gt;
&lt;/g&gt;
&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;−3&lt;/td&gt;
&lt;td&gt;−2&lt;/td&gt;
&lt;td&gt;−1&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;3&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;5&lt;/td&gt;
&lt;/tr&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_84d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_84d"&gt;x of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
&lt;path d="M52 289Q59 331 106 386T222 442Q257 442 286 424T329 379Q371 442 430 442Q467 442 494 420T522 361Q522 332 508 314T481 292T458 288Q439 288 427 299T415 328Q415 374 465 391Q454 404 425 404Q412 404 406 402Q368 386 350 336Q290 115 290 78Q290 50 306 38T341 26Q378 26 414 59T463 140Q466 150 469 151T485 153H489Q504 153 504 145Q504 144 502 134Q486 77 440 33T333 -11Q263 -11 227 52Q186 -10 133 -10H127Q78 -10 57 16T35 71Q35 103 54 123T99 143Q142 143 142 101Q142 81 130 66T107 46T94 41L91 40Q91 39 97 36T113 29T132 26Q168 26 194 71Q203 87 217 139T245 247T261 313Q266 340 266 352Q266 380 251 392T217 404Q177 404 142 372T93 290Q91 281 88 280T72 278H58Q52 284 52 289Z" id="eq_dea10bfd_84MJMATHI-78" stroke-width="10"/&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/td&gt;
&lt;td&gt;  0&lt;/td&gt;
&lt;td&gt;  0&lt;/td&gt;
&lt;td&gt;−1&lt;/td&gt;
&lt;td&gt;2&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;6&lt;/td&gt;
&lt;td&gt;4&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;td&gt;0&lt;/td&gt;
&lt;/tr&gt;&lt;tr&gt;
&lt;td&gt;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="073a9e40caf434f93e69a3e807901249ed035790"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_85d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_85d"&gt;y of n&lt;/desc&gt;
&lt;defs aria-hidden="true"&gt;
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&lt;td&gt;  0&lt;/td&gt;
&lt;td&gt;  0&lt;/td&gt;
&lt;td&gt;−0.33&lt;/td&gt;
&lt;td&gt;0.33&lt;/td&gt;
&lt;td&gt;1.67&lt;/td&gt;
&lt;td&gt;4.00&lt;/td&gt;
&lt;td&gt;4.67&lt;/td&gt;
&lt;td&gt;3.33&lt;/td&gt;
&lt;td&gt;1.33&lt;/td&gt;
&lt;/tr&gt;&lt;/table&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;The resulting output discrete-time waveform is given in Figure 20.&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/fe518d14/t312_openlearn_fig20.tif.jpg" alt="Described image" width="512" height="207" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049349776"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 20&lt;/b&gt;  Filter output in response to the input in Figure 19(b) &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049349776&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049349776"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Figure 21 shows the same filter applied to an input &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="847c2ba9aa403fe146af2062871948290978253e"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_86d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1748.0 1119.0820" width="29.6779px"&gt;

&lt;desc id="eq_dea10bfd_86d"&gt;x of n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; that has more noise in the signal and a longer sequence. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/2c1afd5d/t312_openlearn_fig21.tif.jpg" alt="Described image" width="512" height="301" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049319424"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 21&lt;/b&gt;  Longer sequence filtered by the three-term averaging filter (Wickert, 2011, p. 7) &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049319424&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049319424"&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 8&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Looking at Figure 21, describe the effect that the three-term averaging filter has on the output. What kind of filter is this? &lt;/p&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;The three-term averaging filter has removed the short-term fluctuations in the signal to show the longer-term trend. This is akin to removing higher-frequency noise from a signal, so it is operating like a low-pass filter. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;This digital filter is an example of a system that is both linear and time-invariant, sometimes referred to as an &lt;b&gt;LTI &lt;/b&gt;system. &lt;/p&gt;&lt;p&gt;You can see that the three-term averaging filter is a low-pass filter, but it is difficult to characterise its response. For example, what frequencies is the filter eliminating from the signal? Earlier in this course, you saw how analogue filters can be designed to give a desired frequency response; now you will look at how digital filters can also be designed in the frequency domain. &lt;/p&gt;&lt;p&gt;Filters are usually described in terms that make sense in the frequency domain, e.g. a low pass filter allows the parts of the signal with low frequencies to pass. In the following section you will see how a digital filter is designed in the frequency domain.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.6 Designing a digital filter in the frequency domain</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.6</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Most digital filters are designed in the frequency domain. Input signals are characterised by their frequency spectrum and design filters to modify that spectrum by, for example, removing high-frequency noise with a low-pass filter. &lt;/p&gt;&lt;p&gt;Figure 22 shows the four basic filter structures in the frequency domain. These ideal filters are identical for both analogue and digital filters (you have already seen them earlier in the course in Section 2.2). &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/7ce1bfc5/t312_openlearn_fig22.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049287968"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 22&lt;/b&gt;&amp;#xA0;&amp;#xA0;Ideal filter responses: (a)&amp;#xA0;low-pass; (b)&amp;#xA0;high-pass; (c)&amp;#xA0;band-pass; (d)&amp;#xA0;band-stop (repeat of Figure&amp;#xA0;5) &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049287968&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049287968"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In the design process, the aim is to produce a filter frequency response that best matches the profile of the filter; however, compromises have to be made. You have already seen how the order of the design of the filter affects the roll-off, so decisions are made in the design of these analogue filters to best match the filter to the ideal response. &lt;/p&gt;&lt;p&gt;Figure&amp;#xA0;23 shows some of the characteristics of a typical low-pass digital filter. In comparison to the ideal shape (shown in red), there is a transition region between the passband and stop-band sections, and also a ripple in both the passband and the stop band. These effects can be altered by changing various parameters in the design. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/b15c1352/t312_openlearn_fig23.tif.jpg" alt="Described image" width="512" height="321" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049269280"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 23&lt;/b&gt;  Typical low-pass digital filter response &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049269280&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049269280"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;When designing a digital filter in the frequency domain, software tools are most often used to generate the mathematical expression for the filter. This expression is in the form of a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048992192" class="oucontent-glossaryterm" data-definition="An equation in which all variables have been sampled at fixed intervals, and these variables are multiplied by some coefficient." title="An equation in which all variables have been sampled at fixed intervals, and these variables are mul..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;difference equation&lt;/span&gt;&lt;/a&gt; – an equation involving combinations of samples at specific times. You have already seen a difference equation in this section: the equation &amp;#xA0;&lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="698bbef00acd8ca26929f6e2a8266ae13479b72a"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_87d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_87d"&gt;y of n&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt; for the three-term averaging filter. As another example, the following difference equation contains five terms: &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="a0a03282c629485b7ac145bf2c82cffa4d1b808b"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_89d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 27515.8 1119.0820" width="467.1688px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;In a digital filter design, the number of terms in the difference equation is often referred to as the number of &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048965744" class="oucontent-glossaryterm" data-definition="In a digital filter, the number of taps is the number of terms in the mathematical expression that describes the filter. This expression is given in the form of a difference equation. In digital filter design, the maximum number of taps to be used in the implementation is required as part of the design specification." title="In a digital filter, the number of taps is the number of terms in the mathematical expression that d..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;taps&lt;/span&gt;&lt;/a&gt;, and is specified as a design parameter. A larger number of taps may give a filter design that more closely matches the desired specification; however, more taps will mean that it takes longer to compute the filter outputs. &lt;/p&gt;&lt;p&gt;There are two main classes of LTI (linear time-invariant) digital filter: the finite impulse response (FIR) filter and the infinite impulse response (IIR) filter. An IIR filter requires fewer computations to achieve the same performance as an FIR filter, so has a speed advantage. However, an IIR can have stability issues and also non-linear phase issues (where signals of different frequencies are delayed by different amounts, resulting in a distortion of the output signal). The difference between an FIR filter and an IIR filter is that the FIR filter uses only the filter inputs when generating its output, whereas an IIR filter uses both the filter inputs and past filter outputs – in other words, it uses feedback. The difference equation above is for an FIR filter. The difference equation below has four terms and the final term is an output value, so it is an IIR filter: &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="44e5a458c8094d3ca7abcae2222ce5ff61d95970"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_90d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 20990.2 1119.0820" width="356.3758px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;To calculate the output of an IIR filter, both previous input samples and previous output samples are stored in the processor’s memory. The order of the IIR filter is the larger of the number of input values stored and the number of output values stored. In the above example, the filter is second-order, since two previous input values are stored but only one previous output value. &lt;/p&gt;&lt;p&gt;Before you look at a filter being designed, you need to know a little more about the relationship between the time domain and the frequency domain. You will cover this next.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.6</guid>
    <dc:title>3.6 Designing a digital filter in the frequency domain</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Most digital filters are designed in the frequency domain. Input signals are characterised by their frequency spectrum and design filters to modify that spectrum by, for example, removing high-frequency noise with a low-pass filter. &lt;/p&gt;&lt;p&gt;Figure 22 shows the four basic filter structures in the frequency domain. These ideal filters are identical for both analogue and digital filters (you have already seen them earlier in the course in Section 2.2). &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/7ce1bfc5/t312_openlearn_fig22.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049287968"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 22&lt;/b&gt;  Ideal filter responses: (a) low-pass; (b) high-pass; (c) band-pass; (d) band-stop (repeat of Figure 5) &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049287968&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049287968"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;In the design process, the aim is to produce a filter frequency response that best matches the profile of the filter; however, compromises have to be made. You have already seen how the order of the design of the filter affects the roll-off, so decisions are made in the design of these analogue filters to best match the filter to the ideal response. &lt;/p&gt;&lt;p&gt;Figure 23 shows some of the characteristics of a typical low-pass digital filter. In comparison to the ideal shape (shown in red), there is a transition region between the passband and stop-band sections, and also a ripple in both the passband and the stop band. These effects can be altered by changing various parameters in the design. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/b15c1352/t312_openlearn_fig23.tif.jpg" alt="Described image" width="512" height="321" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049269280"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 23&lt;/b&gt;  Typical low-pass digital filter response &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049269280&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049269280"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;When designing a digital filter in the frequency domain, software tools are most often used to generate the mathematical expression for the filter. This expression is in the form of a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048992192" class="oucontent-glossaryterm" data-definition="An equation in which all variables have been sampled at fixed intervals, and these variables are multiplied by some coefficient." title="An equation in which all variables have been sampled at fixed intervals, and these variables are mul..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;difference equation&lt;/span&gt;&lt;/a&gt; – an equation involving combinations of samples at specific times. You have already seen a difference equation in this section: the equation  &lt;span class="oucontent-inlinemathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="698bbef00acd8ca26929f6e2a8266ae13479b72a"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_87d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 1673.0 1119.0820" width="28.4045px"&gt;

&lt;desc id="eq_dea10bfd_87d"&gt;y of n&lt;/desc&gt;
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&lt;desc id="eq_dea10bfd_88d"&gt;sum with, 3 , summands one divided by three times x of n plus one divided by three times x of n minus one plus one divided by three times x of n minus two&lt;/desc&gt;
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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;In a digital filter design, the number of terms in the difference equation is often referred to as the number of &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048965744" class="oucontent-glossaryterm" data-definition="In a digital filter, the number of taps is the number of terms in the mathematical expression that describes the filter. This expression is given in the form of a difference equation. In digital filter design, the maximum number of taps to be used in the implementation is required as part of the design specification." title="In a digital filter, the number of taps is the number of terms in the mathematical expression that d..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;taps&lt;/span&gt;&lt;/a&gt;, and is specified as a design parameter. A larger number of taps may give a filter design that more closely matches the desired specification; however, more taps will mean that it takes longer to compute the filter outputs. &lt;/p&gt;&lt;p&gt;There are two main classes of LTI (linear time-invariant) digital filter: the finite impulse response (FIR) filter and the infinite impulse response (IIR) filter. An IIR filter requires fewer computations to achieve the same performance as an FIR filter, so has a speed advantage. However, an IIR can have stability issues and also non-linear phase issues (where signals of different frequencies are delayed by different amounts, resulting in a distortion of the output signal). The difference between an FIR filter and an IIR filter is that the FIR filter uses only the filter inputs when generating its output, whereas an IIR filter uses both the filter inputs and past filter outputs – in other words, it uses feedback. The difference equation above is for an FIR filter. The difference equation below has four terms and the final term is an output value, so it is an IIR filter: &lt;/p&gt;&lt;div class="oucontent-equation oucontent-equation-equation oucontent-nocaption"&gt;&lt;span class="oucontent-display-mathml"&gt;&lt;span class="filter_oumaths_equation filter_oumaths_svg" data-ehash="44e5a458c8094d3ca7abcae2222ce5ff61d95970"&gt;&lt;svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" aria-labelledby="eq_dea10bfd_90d" height="19px" role="math" style="vertical-align: -5px; margin-left: 0ex; margin-right: 0ex; margin-bottom: 0px; margin-top: 0px;" viewBox="0.0 -824.5868 20990.2 1119.0820" width="356.3758px"&gt;

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&lt;/svg&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;p&gt;To calculate the output of an IIR filter, both previous input samples and previous output samples are stored in the processor’s memory. The order of the IIR filter is the larger of the number of input values stored and the number of output values stored. In the above example, the filter is second-order, since two previous input values are stored but only one previous output value. &lt;/p&gt;&lt;p&gt;Before you look at a filter being designed, you need to know a little more about the relationship between the time domain and the frequency domain. You will cover this next.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.7 Fourier transforms and the sinc pulse</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.7</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;You saw earlier (Figure 5) that the ideal frequency responses shown in Figure&amp;#xA0;22 are sometimes referred to as brick-wall filters because of the sharp transitions between passbands and stop bands. In other words, they are rectangular functions. However, whilst it is possible to use a rectangular function in the frequency domain to specify the filter, you must perform the calculations to implement the filter in the time domain. To do this, you need to translate between the time and frequency domains; in particular, for a brick-wall filter, you need to know what a rectangular function in the frequency domain looks like in the time domain. For a continuous-time system, you would use a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048985408" class="oucontent-glossaryterm" data-definition="A transformation that extends the concept of the Fourier series to non-periodic signals. It allows us to estimate the spectrum of a signal and perform a frequency analysis." title="A transformation that extends the concept of the Fourier series to non-periodic signals. It allows u..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;Fourier transform&lt;/span&gt;&lt;/a&gt; to do this; for a discrete-time system, you use a corresponding discrete-time Fourier transform. &lt;/p&gt;&lt;p&gt;You can perform mathematical calculations on paper to work out the Fourier transform of a signal in either the time or the frequency domain. Under those circumstances, you would use the formula for either the discrete or the continuous transform, depending on the type of system you were dealing with. However, the majority of Fourier transforms will be carried out by a computer system – even if the system is dealing with continuous-time signals as input and output, the signal processing will be happening in the discrete-time domain of the computer. There is a particular algorithm called the fast Fourier transform (FFT) that is used to carry out Fourier transforms efficiently. Such was the need to perform these calculations at great speed that the FFT was included in a list of the top 10 algorithms of the twentieth century by the IEEE&amp;#xA0;journal &lt;i&gt;Computing in Science &amp;amp; Engineering&lt;/i&gt; in the year 2000 (Dongarra and Sullivan, 2000). &lt;/p&gt;&lt;p&gt;Using the discrete-time Fourier transform, you can see that the time-domain representation of a rectangular function in the frequency domain is the sinc pulse, as shown in Figure&amp;#xA0;24. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/7e238ae3/t312_openlearn_fig24.tif.jpg" alt="Described image" width="512" height="103" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049223520"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 24&lt;/b&gt; Fourier transform pair: a rectangular function in the frequency domain is represented as a sinc pulse in the time domain &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049223520&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049223520"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Mathematically, a sinc pulse or sinc function is defined as sin(x)/x.&lt;/p&gt;&lt;p&gt;Figure 25(a) and Figure 25(b) show a sinc envelope producing an ideal low-pass frequency response. However, there is an issue because the sinc pulse continues to both positive and negative infinity along the time axis. Whilst mathematically you can readily take the Fourier transform of a sinc pulse, it can’t be computed because of the extension to infinity. The obvious solution is to truncate the sinc response as in Figure&amp;#xA0;25(c), so that the ripples no longer extend to infinity. Now that the pulse is finite, it can be shifted so that it only has positive sample numbers. The effects of this in the frequency domain are shown in Figure&amp;#xA0;25(d) – ripples in the passband and the stop band. In essence this shows why you can never have the perfect ideal or &amp;#x2018;brick-wall’ filter.&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/2eabb2af/t312_openlearn_fig25.tif.jpg" alt="Described image" width="512" height="352" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049206176"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 25&lt;/b&gt;  (a) Ideal sinc function in the time domain; (b) frequency response of the ideal sinc function; (c) abruptly truncated sinc function in the time domain; (d) frequency response of the truncated sinc function &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049206176&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049206176"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;A technique for dealing with the truncated sinc is to apply a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048962304" class="oucontent-glossaryterm" data-definition="In signal processing, a mathematical function that is zero-valued outside some chosen interval. When a signal is convolved with a window function, the resultant signal is also zero-valued outside the chosen interval, so it is the original signal viewed through the window function." title="In signal processing, a mathematical function that is zero-valued outside some chosen interval. When..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;window function&lt;/span&gt;&lt;/a&gt; that brings the endpoints of the truncated sinc to zero. Figure&amp;#xA0;26(a) shows a suitable shape of window, Figure&amp;#xA0;26(b) shows the effects of applying the window to the truncated sinc and Figure&amp;#xA0;26(c) shows the resultant frequency response. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/f85b1d2e/t312_openlearn_fig26.tif.jpg" alt="Described image" width="512" height="352" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049189616"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 26&lt;/b&gt;  Using windowing to compensate for a truncated sinc pulse: (a) suitable window shape; (b) window applied to the truncated sinc; (c) resultant frequency response &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049189616&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049189616"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;When a window function is applied, it is effectively &amp;#x2018;multiplied’ with the sinc function. The process used to do this is called &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048995664" class="oucontent-glossaryterm" data-definition="A mathematical operation that combines two signals to produce a third signal. When two signals are convolved, the resultant third signal expresses how the shape of one signal is modified by the other." title="A mathematical operation that combines two signals to produce a third signal. When two signals are c..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;convolution&lt;/span&gt;&lt;/a&gt;. Convolution is outside the scope of this course, but when applied, what is left is where the signals overlap. You can think of this as a &amp;#x2018;view’ through the window function. &lt;/p&gt;&lt;p&gt;There are various shapes of window that can be applied. Two common shapes are the Blackman window and the Hamming window, although there are others (such as Kaiser, Bartlett and Hann). The Hamming window gives a better transition response, but the Blackman window has better stop-band gain and lower passband ripple. &lt;/p&gt;&lt;p&gt;You are now ready to design a digital filter and try it out.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.7</guid>
    <dc:title>3.7 Fourier transforms and the sinc pulse</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;You saw earlier (Figure 5) that the ideal frequency responses shown in Figure 22 are sometimes referred to as brick-wall filters because of the sharp transitions between passbands and stop bands. In other words, they are rectangular functions. However, whilst it is possible to use a rectangular function in the frequency domain to specify the filter, you must perform the calculations to implement the filter in the time domain. To do this, you need to translate between the time and frequency domains; in particular, for a brick-wall filter, you need to know what a rectangular function in the frequency domain looks like in the time domain. For a continuous-time system, you would use a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048985408" class="oucontent-glossaryterm" data-definition="A transformation that extends the concept of the Fourier series to non-periodic signals. It allows us to estimate the spectrum of a signal and perform a frequency analysis." title="A transformation that extends the concept of the Fourier series to non-periodic signals. It allows u..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;Fourier transform&lt;/span&gt;&lt;/a&gt; to do this; for a discrete-time system, you use a corresponding discrete-time Fourier transform. &lt;/p&gt;&lt;p&gt;You can perform mathematical calculations on paper to work out the Fourier transform of a signal in either the time or the frequency domain. Under those circumstances, you would use the formula for either the discrete or the continuous transform, depending on the type of system you were dealing with. However, the majority of Fourier transforms will be carried out by a computer system – even if the system is dealing with continuous-time signals as input and output, the signal processing will be happening in the discrete-time domain of the computer. There is a particular algorithm called the fast Fourier transform (FFT) that is used to carry out Fourier transforms efficiently. Such was the need to perform these calculations at great speed that the FFT was included in a list of the top 10 algorithms of the twentieth century by the IEEE journal &lt;i&gt;Computing in Science &amp; Engineering&lt;/i&gt; in the year 2000 (Dongarra and Sullivan, 2000). &lt;/p&gt;&lt;p&gt;Using the discrete-time Fourier transform, you can see that the time-domain representation of a rectangular function in the frequency domain is the sinc pulse, as shown in Figure 24. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/7e238ae3/t312_openlearn_fig24.tif.jpg" alt="Described image" width="512" height="103" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049223520"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 24&lt;/b&gt; Fourier transform pair: a rectangular function in the frequency domain is represented as a sinc pulse in the time domain &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049223520&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049223520"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Mathematically, a sinc pulse or sinc function is defined as sin(x)/x.&lt;/p&gt;&lt;p&gt;Figure 25(a) and Figure 25(b) show a sinc envelope producing an ideal low-pass frequency response. However, there is an issue because the sinc pulse continues to both positive and negative infinity along the time axis. Whilst mathematically you can readily take the Fourier transform of a sinc pulse, it can’t be computed because of the extension to infinity. The obvious solution is to truncate the sinc response as in Figure 25(c), so that the ripples no longer extend to infinity. Now that the pulse is finite, it can be shifted so that it only has positive sample numbers. The effects of this in the frequency domain are shown in Figure 25(d) – ripples in the passband and the stop band. In essence this shows why you can never have the perfect ideal or ‘brick-wall’ filter.&lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/2eabb2af/t312_openlearn_fig25.tif.jpg" alt="Described image" width="512" height="352" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049206176"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 25&lt;/b&gt;  (a) Ideal sinc function in the time domain; (b) frequency response of the ideal sinc function; (c) abruptly truncated sinc function in the time domain; (d) frequency response of the truncated sinc function &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049206176&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049206176"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;A technique for dealing with the truncated sinc is to apply a &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048962304" class="oucontent-glossaryterm" data-definition="In signal processing, a mathematical function that is zero-valued outside some chosen interval. When a signal is convolved with a window function, the resultant signal is also zero-valued outside the chosen interval, so it is the original signal viewed through the window function." title="In signal processing, a mathematical function that is zero-valued outside some chosen interval. When..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;window function&lt;/span&gt;&lt;/a&gt; that brings the endpoints of the truncated sinc to zero. Figure 26(a) shows a suitable shape of window, Figure 26(b) shows the effects of applying the window to the truncated sinc and Figure 26(c) shows the resultant frequency response. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/f85b1d2e/t312_openlearn_fig26.tif.jpg" alt="Described image" width="512" height="352" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049189616"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 26&lt;/b&gt;  Using windowing to compensate for a truncated sinc pulse: (a) suitable window shape; (b) window applied to the truncated sinc; (c) resultant frequency response &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049189616&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049189616"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;When a window function is applied, it is effectively ‘multiplied’ with the sinc function. The process used to do this is called &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048995664" class="oucontent-glossaryterm" data-definition="A mathematical operation that combines two signals to produce a third signal. When two signals are convolved, the resultant third signal expresses how the shape of one signal is modified by the other." title="A mathematical operation that combines two signals to produce a third signal. When two signals are c..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;convolution&lt;/span&gt;&lt;/a&gt;. Convolution is outside the scope of this course, but when applied, what is left is where the signals overlap. You can think of this as a ‘view’ through the window function. &lt;/p&gt;&lt;p&gt;There are various shapes of window that can be applied. Two common shapes are the Blackman window and the Hamming window, although there are others (such as Kaiser, Bartlett and Hann). The Hamming window gives a better transition response, but the Blackman window has better stop-band gain and lower passband ripple. &lt;/p&gt;&lt;p&gt;You are now ready to design a digital filter and try it out.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.8 A low-pass filter design</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.8</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;You will now look at a real example of low-pass filter design. For this, a software package called Signal Wizard has been used. You are not required to download Signal Wizard, but for information, it is a free package that can be installed on your computer. Details of the Signal Wizard and where to find it are given can be found here: &lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/olink.php?id=103828&amp;amp;targetdoc=Installing+Signal+Wizard+for+use+in+offline+mode" class="oucontent-olink"&gt;Installing Signal Wizard for use in offline mode&lt;/a&gt;.&lt;/p&gt;&lt;p&gt;Figure 27 shows a screenshot from Signal Wizard which shows the input signal waveform both in the time domain (the &amp;#x2018;Time Waveform’) and in the frequency domain (the &amp;#x2018;Frequency Spectrum’). The aim is to remove any frequencies in this waveform above 5000&amp;#xA0;Hz, so a low-pass filter with a 5000&amp;#xA0;Hz cut-off frequency is required. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/3fb8ab87/t312_openlearn_fig27.tif.jpg" alt="Described image" width="512" height="334" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049166256"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 27&lt;/b&gt;  Characteristics of the input signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049166256&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049166256"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The specification of an FIR low-pass filter with a gain of 1 and a cut-off frequency of 5000&amp;#xA0;Hz is entered into the filter design interface. The graphical interface windows in Figure&amp;#xA0;28 all show the &amp;#x2018;brick-wall’ specification in red with the implementation in black. These designs all use a rectangular window, which gives an abruptly truncated sinc function. The first design uses 15&amp;#xA0;taps (Figure&amp;#xA0;28(a)), the second uses 63&amp;#xA0;taps (Figure&amp;#xA0;28(b)) and the third uses 127&amp;#xA0;taps (Figure&amp;#xA0;28(c)). All other parameters are unchanged. As the number of taps in the design increases from the top image to the bottom, the transition zone narrows and the designed filter more closely matches the filter specification. However, the amplitude of the ripples in the passband and the stop band remains unchanged, although the frequency increases as the number of taps increases. &lt;/p&gt;&lt;div class="oucontent-figure oucontent-media-mini"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/558ab49b/t312_openlearn_fig28.tif.jpg" alt="Described image" width="309" height="512" style="max-width:309px;" class="oucontent-figure-image" longdesc="view.php&amp;amp;extra=longdesc_idm46133049154032"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 28&lt;/b&gt;  Digital filter design with a rectangular window: (a) 15 taps; (b) 63 taps; (c) 127 taps &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049154032&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049154032"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The next set of designs, in Figure&amp;#xA0;29, keeps the number of taps at 127 and varies the window function used. The design implemented in Figure&amp;#xA0;29(a) uses a Blackman window, while the design implemented in Figure&amp;#xA0;29(b) uses a Hamming window. You can see that both have reduced the ripples in the passband. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:468px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/017e909e/t312_openlearn_fig29.tif.jpg" alt="Described image" width="468" height="512" style="max-width:468px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049141904"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 29&lt;/b&gt;  Filter designs with different window functions: (a) 127 taps with Blackman window; (b) 127 taps with Hamming window &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049141904&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049141904"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Figure 30 zooms in on sections of the transition band and passband to see the effects in more detail. Comparing Figure&amp;#xA0;30(a) and Figure&amp;#xA0;30(c) shows that the transition zone of the Hamming window is narrower than that of the Blackman window. Comparing Figure&amp;#xA0;30(b) and Figure&amp;#xA0;30(d) shows that the Blackman window has reduced the ripples in the passband more than the Hamming window. These results confirm that the Hamming window gives a better transition response, while the Blackman window has lower passband ripple. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/cf71b31c/t312_openlearn_fig30.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049128016"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 30&lt;/b&gt; Details of the Blackman and Hamming windows: (a) 127 taps with Blackman window, transition zone; (b) 127 taps with Blackman window, passband; (c) 127 taps with Hamming window, transition zone; (d) 127 taps with Hamming window, passband &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049128016&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049128016"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The input signal was filtered using the 127-tap Hamming window design, and the output response is shown in the &amp;#x2018;Time Waveform’ and &amp;#x2018;Frequency Spectrum’ charts in Figure&amp;#xA0;31. The frequency spectrum shows that the filter has indeed taken out the redundant higher-order frequencies above 5000&amp;#xA0;Hz. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/5222c973/t312_openlearn_fig31.tif.jpg" alt="Described image" width="512" height="334" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;amp;extra=longdesc_idm46133049117072"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 31&lt;/b&gt;  Filter response output &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049117072&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049117072"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;When a digital filter such as the one above is being implemented, the mathematical calculations will be defined in a software program, which will then run on some digital hardware. The basic processes taking place in the digital hardware are adding and subtracting; these processes will occur thousands, probably millions of times in the filtering of a sampled signal. &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048988992" class="oucontent-glossaryterm" data-definition="A semiconductor device similar to a microprocessor. Whilst a microprocessor is a general-purpose device, a DSP has been optimised to carry out the computations used for processing discrete signals." title="A semiconductor device similar to a microprocessor. Whilst a microprocessor is a general-purpose dev..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;Digital signal processor (DSP)&lt;/span&gt;&lt;/a&gt; chips are designed specifically to carry out these calculations, and their internal structure (referred to as their architecture) has been optimised to do so – thus it is different from that of a general-purpose processor used in a computer. There are several major DSP chip manufacturers, including Texas Instruments and Analog Devices. &lt;/p&gt;&lt;p&gt;You’ve therefore seen that usually a digital filter is designed using a software package. To finish this course, you will have a chance to explore digital filtering using an interactive resource. The next section introduces this resource.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.8</guid>
    <dc:title>3.8 A low-pass filter design</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;You will now look at a real example of low-pass filter design. For this, a software package called Signal Wizard has been used. You are not required to download Signal Wizard, but for information, it is a free package that can be installed on your computer. Details of the Signal Wizard and where to find it are given can be found here: &lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/olink.php?id=103828&amp;targetdoc=Installing+Signal+Wizard+for+use+in+offline+mode" class="oucontent-olink"&gt;Installing Signal Wizard for use in offline mode&lt;/a&gt;.&lt;/p&gt;&lt;p&gt;Figure 27 shows a screenshot from Signal Wizard which shows the input signal waveform both in the time domain (the ‘Time Waveform’) and in the frequency domain (the ‘Frequency Spectrum’). The aim is to remove any frequencies in this waveform above 5000 Hz, so a low-pass filter with a 5000 Hz cut-off frequency is required. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/3fb8ab87/t312_openlearn_fig27.tif.jpg" alt="Described image" width="512" height="334" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049166256"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 27&lt;/b&gt;  Characteristics of the input signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049166256&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049166256"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The specification of an FIR low-pass filter with a gain of 1 and a cut-off frequency of 5000 Hz is entered into the filter design interface. The graphical interface windows in Figure 28 all show the ‘brick-wall’ specification in red with the implementation in black. These designs all use a rectangular window, which gives an abruptly truncated sinc function. The first design uses 15 taps (Figure 28(a)), the second uses 63 taps (Figure 28(b)) and the third uses 127 taps (Figure 28(c)). All other parameters are unchanged. As the number of taps in the design increases from the top image to the bottom, the transition zone narrows and the designed filter more closely matches the filter specification. However, the amplitude of the ripples in the passband and the stop band remains unchanged, although the frequency increases as the number of taps increases. &lt;/p&gt;&lt;div class="oucontent-figure oucontent-media-mini"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/558ab49b/t312_openlearn_fig28.tif.jpg" alt="Described image" width="309" height="512" style="max-width:309px;" class="oucontent-figure-image" longdesc="view.php&amp;extra=longdesc_idm46133049154032"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 28&lt;/b&gt;  Digital filter design with a rectangular window: (a) 15 taps; (b) 63 taps; (c) 127 taps &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049154032&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049154032"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The next set of designs, in Figure 29, keeps the number of taps at 127 and varies the window function used. The design implemented in Figure 29(a) uses a Blackman window, while the design implemented in Figure 29(b) uses a Hamming window. You can see that both have reduced the ripples in the passband. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:468px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/017e909e/t312_openlearn_fig29.tif.jpg" alt="Described image" width="468" height="512" style="max-width:468px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049141904"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 29&lt;/b&gt;  Filter designs with different window functions: (a) 127 taps with Blackman window; (b) 127 taps with Hamming window &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049141904&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049141904"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Figure 30 zooms in on sections of the transition band and passband to see the effects in more detail. Comparing Figure 30(a) and Figure 30(c) shows that the transition zone of the Hamming window is narrower than that of the Blackman window. Comparing Figure 30(b) and Figure 30(d) shows that the Blackman window has reduced the ripples in the passband more than the Hamming window. These results confirm that the Hamming window gives a better transition response, while the Blackman window has lower passband ripple. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/cf71b31c/t312_openlearn_fig30.tif.jpg" alt="Described image" width="512" height="282" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049128016"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 30&lt;/b&gt; Details of the Blackman and Hamming windows: (a) 127 taps with Blackman window, transition zone; (b) 127 taps with Blackman window, passband; (c) 127 taps with Hamming window, transition zone; (d) 127 taps with Hamming window, passband &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049128016&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049128016"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The input signal was filtered using the 127-tap Hamming window design, and the output response is shown in the ‘Time Waveform’ and ‘Frequency Spectrum’ charts in Figure 31. The frequency spectrum shows that the filter has indeed taken out the redundant higher-order frequencies above 5000 Hz. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/5222c973/t312_openlearn_fig31.tif.jpg" alt="Described image" width="512" height="334" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php&amp;extra=longdesc_idm46133049117072"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 31&lt;/b&gt;  Filter response output &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049117072&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049117072"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;When a digital filter such as the one above is being implemented, the mathematical calculations will be defined in a software program, which will then run on some digital hardware. The basic processes taking place in the digital hardware are adding and subtracting; these processes will occur thousands, probably millions of times in the filtering of a sampled signal. &lt;a href="https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary#idm46133048988992" class="oucontent-glossaryterm" data-definition="A semiconductor device similar to a microprocessor. Whilst a microprocessor is a general-purpose device, a DSP has been optimised to carry out the computations used for processing discrete signals." title="A semiconductor device similar to a microprocessor. Whilst a microprocessor is a general-purpose dev..."&gt;&lt;span class="oucontent-glossaryterm-styling"&gt;Digital signal processor (DSP)&lt;/span&gt;&lt;/a&gt; chips are designed specifically to carry out these calculations, and their internal structure (referred to as their architecture) has been optimised to do so – thus it is different from that of a general-purpose processor used in a computer. There are several major DSP chip manufacturers, including Texas Instruments and Analog Devices. &lt;/p&gt;&lt;p&gt;You’ve therefore seen that usually a digital filter is designed using a software package. To finish this course, you will have a chance to explore digital filtering using an interactive resource. The next section introduces this resource.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.9  Digital filtering in practice</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.9</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;You are now going to use an interactive resource to add Gaussian noise to a noise-free (or &amp;#x2018;clean’) data signal. The noisy data signal is passed to a detector, which determines whether received samples are binary zeros or ones. The added noise causes the detector to make mistakes; hence there are bit errors. Using a finite impulse response (FIR) digital filter, you will &amp;#x2018;clean up’ the noisy data signal and thereby reduce the bit-error rate. &lt;/p&gt;&lt;p&gt;Right-click on the image or link below to open Interactive&amp;#xA0;1 in a new tab or window so you can continue to work through the course and activities with the interactive alongside. The interface consists of four equally sized zones: top left, top right, bottom left and bottom right. Note that depending on the browser you are using, the sliders and other components in the interface may have a different visual appearance from those shown in the screenshots in this document, but the functionality will be the same. &lt;/p&gt;&lt;div id="t312_bl01_fir" class="oucontent-media oucontent-responsive" style="width:512px;"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=thumbnail_idm46133049107728" title="View online activity"&gt;&lt;img alt="" src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/fff2790b/t312_bl01_fir.zip.jpg"/&gt;&lt;/a&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-thumbnaillink oucontent-viewonlineactivity"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=thumbnail_idm46133049107728"&gt;View interactive version&lt;/a&gt;&lt;/div&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Interactive 1&lt;/b&gt; FIR filter&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;a id="back_thumbnail_idm46133049107728"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The top left zone shows the signal, whether it is clean or noisy. When the interactive resource is launched, the noise level is at zero and the signal is clean. The data signal is shown as a single step from 0 to 1, but there are 10&amp;#xA0;000&amp;#xA0;samples for this data signal, half of which are from the 0 part of the signal and half of which are from the 1 part. Noise is added by sliding the &lt;b&gt;Noise level (V)&lt;/b&gt; slider to the right and clicking on &lt;b&gt;Apply&lt;/b&gt;. &lt;/p&gt;&lt;p&gt;The top right zone displays a histogram of the signal. As there is an equal number of zeros and ones in the signal, the probability of each is 0.5. The detected probabilities of zeros and ones in the received signal are shown by the vertical red lines situated at 0 and 1. Without added noise, perfect detection can be achieved. &lt;/p&gt;&lt;p&gt;The top right zone allows the decision level at the detector to be set using the &lt;b&gt;Decision level (V)&lt;/b&gt; slider, and also the degree of low-pass filtering to be set using the &lt;b&gt;FIR filter width (number of samples)&lt;/b&gt; slider. Finally, the top right zone also gives statistics about the number of true and false detections. Figure 32 explains how to interpret this data. Before noise is added, all samples are correctly identified, so &lt;b&gt;False detections&lt;/b&gt; should be at 0 to start with. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/3bc80ef8/t312_openlearn_fig32.tif.jpg" alt="Described image" width="512" height="266" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049091392"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 32&lt;/b&gt;  Interpreting the statistics &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049091392&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049091392"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Note in Figure 32 that if the correct detection of data 1s were to drop below 50%, the deficit would appear as an increase in the false detection of data 0s. Similarly, if the correct detection of data 0s drops below 50%, the deficit should appear as an increase in the false detection of data 1s. Thus the diagonal pairs of statistics in Figure 32 should always add up to 50% – or approximately 50%, given the possibility of rounding errors in the calculation. &lt;/p&gt;&lt;p&gt;The bottom left zone shows the effect of filtering applied to a noisy signal. The bottom right zone is similar to the top right zone, showing a signal histogram and false detection statistics; however, these are after filtering, whereas the top right zone is before filtering. &lt;/p&gt;&lt;p&gt;In the next section you will have a go at adding noise to the data signal in the interactive.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.9</guid>
    <dc:title>3.9  Digital filtering in practice</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;You are now going to use an interactive resource to add Gaussian noise to a noise-free (or ‘clean’) data signal. The noisy data signal is passed to a detector, which determines whether received samples are binary zeros or ones. The added noise causes the detector to make mistakes; hence there are bit errors. Using a finite impulse response (FIR) digital filter, you will ‘clean up’ the noisy data signal and thereby reduce the bit-error rate. &lt;/p&gt;&lt;p&gt;Right-click on the image or link below to open Interactive 1 in a new tab or window so you can continue to work through the course and activities with the interactive alongside. The interface consists of four equally sized zones: top left, top right, bottom left and bottom right. Note that depending on the browser you are using, the sliders and other components in the interface may have a different visual appearance from those shown in the screenshots in this document, but the functionality will be the same. &lt;/p&gt;&lt;div id="t312_bl01_fir" class="oucontent-media oucontent-responsive" style="width:512px;"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=thumbnail_idm46133049107728" title="View online activity"&gt;&lt;img alt="" src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/fff2790b/t312_bl01_fir.zip.jpg"/&gt;&lt;/a&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-thumbnaillink oucontent-viewonlineactivity"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=thumbnail_idm46133049107728"&gt;View interactive version&lt;/a&gt;&lt;/div&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Interactive 1&lt;/b&gt; FIR filter&lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;a id="back_thumbnail_idm46133049107728"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;The top left zone shows the signal, whether it is clean or noisy. When the interactive resource is launched, the noise level is at zero and the signal is clean. The data signal is shown as a single step from 0 to 1, but there are 10 000 samples for this data signal, half of which are from the 0 part of the signal and half of which are from the 1 part. Noise is added by sliding the &lt;b&gt;Noise level (V)&lt;/b&gt; slider to the right and clicking on &lt;b&gt;Apply&lt;/b&gt;. &lt;/p&gt;&lt;p&gt;The top right zone displays a histogram of the signal. As there is an equal number of zeros and ones in the signal, the probability of each is 0.5. The detected probabilities of zeros and ones in the received signal are shown by the vertical red lines situated at 0 and 1. Without added noise, perfect detection can be achieved. &lt;/p&gt;&lt;p&gt;The top right zone allows the decision level at the detector to be set using the &lt;b&gt;Decision level (V)&lt;/b&gt; slider, and also the degree of low-pass filtering to be set using the &lt;b&gt;FIR filter width (number of samples)&lt;/b&gt; slider. Finally, the top right zone also gives statistics about the number of true and false detections. Figure 32 explains how to interpret this data. Before noise is added, all samples are correctly identified, so &lt;b&gt;False detections&lt;/b&gt; should be at 0 to start with. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/3bc80ef8/t312_openlearn_fig32.tif.jpg" alt="Described image" width="512" height="266" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049091392"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 32&lt;/b&gt;  Interpreting the statistics &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049091392&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049091392"&gt;&lt;/a&gt;&lt;/div&gt;&lt;p&gt;Note in Figure 32 that if the correct detection of data 1s were to drop below 50%, the deficit would appear as an increase in the false detection of data 0s. Similarly, if the correct detection of data 0s drops below 50%, the deficit should appear as an increase in the false detection of data 1s. Thus the diagonal pairs of statistics in Figure 32 should always add up to 50% – or approximately 50%, given the possibility of rounding errors in the calculation. &lt;/p&gt;&lt;p&gt;The bottom left zone shows the effect of filtering applied to a noisy signal. The bottom right zone is similar to the top right zone, showing a signal histogram and false detection statistics; however, these are after filtering, whereas the top right zone is before filtering. &lt;/p&gt;&lt;p&gt;In the next section you will have a go at adding noise to the data signal in the interactive.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.10 Adding noise</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.10</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;After familiarising yourself with the interface in Interactive 1, try adding noise in the top left zone. The &lt;b&gt;Noise level (V)&lt;/b&gt; slider is not calibrated, except for having a minimum value of 0 and a maximum of 1. Judging by eye, set the slider to around 0.1&amp;#xA0;V and click on &lt;b&gt;Apply&lt;/b&gt;. You can hear what the noisy signal sounds like by going to &lt;b&gt;Audible signal with noise&lt;/b&gt; and clicking on the &lt;b&gt;Play&lt;/b&gt; (triangle) icon. The top part of the interface should resemble Figure&amp;#xA0;33. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/014a8c70/t312_openlearn_fig33.tif.jpg" alt="Described image" width="512" height="272" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049072400"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 33&lt;/b&gt;  Noise applied to the data signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049072400&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049072400"&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 9&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Why has the display in the top right zone changed? What is the display now showing?&lt;/p&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;The effect of the added noise is to change the voltages representing 0 and 1 randomly around the mean values of 0 and 1. The display shows a histogram of the probabilities of particular signal voltages. Voltages close to the mean are most likely, so the histogram has peaks at 0 and 1; voltages further away are possible, but less likely the further you go from the mean values. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Next you will look at changing the decision level.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.10</guid>
    <dc:title>3.10 Adding noise</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;After familiarising yourself with the interface in Interactive 1, try adding noise in the top left zone. The &lt;b&gt;Noise level (V)&lt;/b&gt; slider is not calibrated, except for having a minimum value of 0 and a maximum of 1. Judging by eye, set the slider to around 0.1 V and click on &lt;b&gt;Apply&lt;/b&gt;. You can hear what the noisy signal sounds like by going to &lt;b&gt;Audible signal with noise&lt;/b&gt; and clicking on the &lt;b&gt;Play&lt;/b&gt; (triangle) icon. The top part of the interface should resemble Figure 33. &lt;/p&gt;&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/014a8c70/t312_openlearn_fig33.tif.jpg" alt="Described image" width="512" height="272" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049072400"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 33&lt;/b&gt;  Noise applied to the data signal &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049072400&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049072400"&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 9&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Why has the display in the top right zone changed? What is the display now showing?&lt;/p&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;The effect of the added noise is to change the voltages representing 0 and 1 randomly around the mean values of 0 and 1. The display shows a histogram of the probabilities of particular signal voltages. Voltages close to the mean are most likely, so the histogram has peaks at 0 and 1; voltages further away are possible, but less likely the further you go from the mean values. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Next you will look at changing the decision level.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.11 Changing the decision level</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.11</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Again, using the interactive in Section 3.9 you will now look at changing the decision level. &lt;/p&gt;&lt;p&gt;The decision level is represented by the right-hand vertical edge of the grey area on the histogram display. Use the &lt;b&gt;Decision level (V)&lt;/b&gt; slider to set the decision level midway between 0 and 1. You will have to judge its position by eye. Click on &lt;b&gt;Apply&lt;/b&gt; to implement the decision level and look at the statistics. &lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 10&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Why are there are false detections?&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;Why are the detection rates (correct and false) practically the same for 0 and 1?&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The effect of noise is to take the signal occasionally over the decision level, so that a 0 is detected as a 1 and vice versa.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The percentages of correct and false detections are virtually the same for 0s and 1s because of the symmetry of the situation. The noise affects the 0s and 1s equally, and the decision level is symmetrically placed between 0 and 1. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;The symmetrical arrangement that you have used so far, with the decision level halfway between the two binary symbols, is typical of much practical implementation of binary signal detection, but looking at an asymmetrical arrangement is instructive. &lt;/p&gt;&lt;p&gt;Use the &lt;b&gt;Decision level (V)&lt;/b&gt; slider to place the decision level asymmetrically between 0 and 1 (that is, much closer to 1 than to 0, or vice versa), then click on &lt;b&gt;Apply&lt;/b&gt;. &lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 11&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 10 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Explain, in general terms, the correct and false detection statistics that have resulted from your asymmetrical placement. You will not be able to give a precise account, but you might be able to explain the relative sizes of the statistics. &lt;/p&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;You may have got something similar to Figure 34, with the decision level fairly close to 1.&lt;/p&gt;
&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/c204f981/t312_openlearn_fig34.tif.jpg" alt="Described image" width="512" height="457" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;amp;extra=longdesc_idm46133049035584"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 34&lt;/b&gt;  Asymmetrical decision level &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;amp;extra=longdesc_idm46133049035584&amp;amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049035584"&gt;&lt;/a&gt;&lt;/div&gt;
&lt;p&gt;Because the decision level is so close to 1, noise-affected 0 signals very rarely go beyond the decision level. This is why false detections of 1 are at 0%, as this statistic reflects 0s that are wrongly detected as 1s. As there are no false detections of 0s, all detections of 0 must be correct, which is why correct detections of 0 are shown at 50%. &lt;/p&gt;
&lt;p&gt;With the decision level close to 1, noise-affected 1s quite often drop below the decision level and are detected as 0s. This is why the false detection rate for 0 is relatively high, at almost 14%. Correspondingly, the correct detection rate for 1 is relatively low, at just above 36%. These two statistics add to approximately 50%. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Now you have explored how adding noise and changing the decision level affects the data signal, to end this course you will apply the FIR filter.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.11</guid>
    <dc:title>3.11 Changing the decision level</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Again, using the interactive in Section 3.9 you will now look at changing the decision level. &lt;/p&gt;&lt;p&gt;The decision level is represented by the right-hand vertical edge of the grey area on the histogram display. Use the &lt;b&gt;Decision level (V)&lt;/b&gt; slider to set the decision level midway between 0 and 1. You will have to judge its position by eye. Click on &lt;b&gt;Apply&lt;/b&gt; to implement the decision level and look at the statistics. &lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 10&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Why are there are false detections?&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;Why are the detection rates (correct and false) practically the same for 0 and 1?&lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The effect of noise is to take the signal occasionally over the decision level, so that a 0 is detected as a 1 and vice versa.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The percentages of correct and false detections are virtually the same for 0s and 1s because of the symmetry of the situation. The noise affects the 0s and 1s equally, and the decision level is symmetrically placed between 0 and 1. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;The symmetrical arrangement that you have used so far, with the decision level halfway between the two binary symbols, is typical of much practical implementation of binary signal detection, but looking at an asymmetrical arrangement is instructive. &lt;/p&gt;&lt;p&gt;Use the &lt;b&gt;Decision level (V)&lt;/b&gt; slider to place the decision level asymmetrically between 0 and 1 (that is, much closer to 1 than to 0, or vice versa), then click on &lt;b&gt;Apply&lt;/b&gt;. &lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 11&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 10 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;p&gt;Explain, in general terms, the correct and false detection statistics that have resulted from your asymmetrical placement. You will not be able to give a precise account, but you might be able to explain the relative sizes of the statistics. &lt;/p&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;p&gt;You may have got something similar to Figure 34, with the decision level fairly close to 1.&lt;/p&gt;
&lt;div class="oucontent-figure" style="width:512px;"&gt;&lt;img src="https://www.open.edu/openlearn/ocw/pluginfile.php/1881285/mod_oucontent/oucontent/95937/1b694830/c204f981/t312_openlearn_fig34.tif.jpg" alt="Described image" width="512" height="457" style="max-width:512px;" class="oucontent-figure-image oucontent-media-wide" longdesc="view.php?id=103828&amp;extra=longdesc_idm46133049035584"/&gt;&lt;div class="oucontent-figure-text"&gt;&lt;div class="oucontent-caption oucontent-nonumber"&gt;&lt;span class="oucontent-figure-caption"&gt;&lt;b&gt;Figure 34&lt;/b&gt;  Asymmetrical decision level &lt;/span&gt;&lt;/div&gt;&lt;/div&gt;&lt;div class="oucontent-longdesclink oucontent-longdesconly"&gt;&lt;a href="https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=103828&amp;extra=longdesc_idm46133049035584&amp;clicked=1"&gt;Long description&lt;/a&gt;&lt;/div&gt;&lt;a id="back_longdesc_idm46133049035584"&gt;&lt;/a&gt;&lt;/div&gt;
&lt;p&gt;Because the decision level is so close to 1, noise-affected 0 signals very rarely go beyond the decision level. This is why false detections of 1 are at 0%, as this statistic reflects 0s that are wrongly detected as 1s. As there are no false detections of 0s, all detections of 0 must be correct, which is why correct detections of 0 are shown at 50%. &lt;/p&gt;
&lt;p&gt;With the decision level close to 1, noise-affected 1s quite often drop below the decision level and are detected as 0s. This is why the false detection rate for 0 is relatively high, at almost 14%. Correspondingly, the correct detection rate for 1 is relatively low, at just above 36%. These two statistics add to approximately 50%. &lt;/p&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Now you have explored how adding noise and changing the decision level affects the data signal, to end this course you will apply the FIR filter.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>3.12 Applying the FIR filter</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.12</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;Use the &lt;b&gt;Decision level (V)&lt;/b&gt; slider to return the decision level to midway between 0 and 1, then click on &lt;b&gt;Apply&lt;/b&gt;. Confirm that the statistics are as would be expected; that is, correct detections of 1s and 0s are at approximately the same percentage. The correct detection percentages do not need to be identical, but try to get them to within about 1% of each other by adjusting the detection level. &lt;/p&gt;&lt;p&gt;You are now going to apply the FIR filter. In the top right zone, set the &lt;b&gt;FIR filter width (number of samples)&lt;/b&gt; slider by eye to around 30–40&amp;#xA0;samples, then click on &lt;b&gt;Apply&lt;/b&gt;. &lt;/p&gt;&lt;div class="&amp;#10;            oucontent-activity&amp;#10;           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 12&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Compare the top right zone (histogram and statistics before filtering) to the bottom right zone (histogram and statistics after filtering). What effect has the filter had on the statistics for correct detection? &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;Explain any difference in detection statistics between those beneath the unfiltered signal and those beneath the filtered signal. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The filter increases the correct detection rate.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The filter reduces the spread of values around the mean, which is shown by the fact that the histogram peaks are narrower than in the unfiltered histogram. The reduced spread of values reduces the probability of sample values crossing the decision level. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Some points to consider as a result of your work with the FIR filter interactive resource are as follows:&lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;Noise causes the actual voltages for 1s and 0s to be distributed around the intended voltage.&lt;/li&gt;&lt;li&gt;The distribution of voltages can cause errors when, for example, the voltage representing a 1 is closer to 0 than to 1.&lt;/li&gt;&lt;li&gt;The wider the distribution (that is, the noisier the signal), the more likely errors become.&lt;/li&gt;&lt;li&gt;Filtering can reduce the width of the distribution of voltages, thereby reducing the error rate.&lt;/li&gt;&lt;/ul&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-3.12</guid>
    <dc:title>3.12 Applying the FIR filter</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;Use the &lt;b&gt;Decision level (V)&lt;/b&gt; slider to return the decision level to midway between 0 and 1, then click on &lt;b&gt;Apply&lt;/b&gt;. Confirm that the statistics are as would be expected; that is, correct detections of 1s and 0s are at approximately the same percentage. The correct detection percentages do not need to be identical, but try to get them to within about 1% of each other by adjusting the detection level. &lt;/p&gt;&lt;p&gt;You are now going to apply the FIR filter. In the top right zone, set the &lt;b&gt;FIR filter width (number of samples)&lt;/b&gt; slider by eye to around 30–40 samples, then click on &lt;b&gt;Apply&lt;/b&gt;. &lt;/p&gt;&lt;div class="
            oucontent-activity
           oucontent-s-heavybox1 oucontent-s-box "&gt;&lt;div class="oucontent-outer-box"&gt;&lt;h2 class="oucontent-h3 oucontent-heading oucontent-nonumber"&gt;Activity 12&lt;/h2&gt;&lt;div class="oucontent-inner-box"&gt;&lt;div class="oucontent-saq-timing"&gt;&lt;span class="accesshide"&gt;Timing: &lt;/span&gt;Allow about 5 minutes&lt;/div&gt;&lt;div class="oucontent-saq-question"&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;Compare the top right zone (histogram and statistics before filtering) to the bottom right zone (histogram and statistics after filtering). What effect has the filter had on the statistics for correct detection? &lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;Explain any difference in detection statistics between those beneath the unfiltered signal and those beneath the filtered signal. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;

&lt;div class="oucontent-saq-answer" data-showtext="Reveal answer" data-hidetext="Hide answer"&gt;&lt;h3 class="oucontent-h4"&gt;Answer&lt;/h3&gt;
&lt;ul class="oucontent-numbered"&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;a.&lt;/span&gt;The filter increases the correct detection rate.&lt;/li&gt;&lt;li class="oucontent-markerdirect"&gt;&lt;span class="oucontent-listmarker"&gt;b.&lt;/span&gt;The filter reduces the spread of values around the mean, which is shown by the fact that the histogram peaks are narrower than in the unfiltered histogram. The reduced spread of values reduces the probability of sample values crossing the decision level. &lt;/li&gt;&lt;/ul&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;p&gt;Some points to consider as a result of your work with the FIR filter interactive resource are as follows:&lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;Noise causes the actual voltages for 1s and 0s to be distributed around the intended voltage.&lt;/li&gt;&lt;li&gt;The distribution of voltages can cause errors when, for example, the voltage representing a 1 is closer to 0 than to 1.&lt;/li&gt;&lt;li&gt;The wider the distribution (that is, the noisier the signal), the more likely errors become.&lt;/li&gt;&lt;li&gt;Filtering can reduce the width of the distribution of voltages, thereby reducing the error rate.&lt;/li&gt;&lt;/ul&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>Conclusion</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-4</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;In this course you have learned about discrete-time signals and the discrete-time systems that use them. In doing so, you have focused on digital filtering and found out why advances in digital computer processing have allowed digital filtering to be used in scenarios where analogue filters would once have been the only viable solution. You have seen how relatively simple averaging filters can remove high-frequency noise, and also that more complex filters are designed. &lt;/p&gt;&lt;p&gt;This completes your study. You should now be able to:&lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;understand how filtering of discrete-time signals can be achieved by mathematical processes such as averaging&lt;/li&gt;&lt;li&gt;understand how mathematical operations applied to a discrete-time signal in the time domain can result in the removal or reduction of unwanted aspects of the signal &lt;/li&gt;&lt;li&gt;explain why filters are designed in the frequency domain, and specify a digital filter to achieve a desired filtering effect.&lt;/li&gt;&lt;/ul&gt;&lt;p&gt;This OpenLearn course is an adapted extract from the Open University course &lt;span class="oucontent-linkwithtip"&gt;&lt;a class="oucontent-hyperlink" href="http://www.open.ac.uk/courses/modules/t312"&gt;T312 &lt;i&gt;Electronics: signal processing, control and communications&lt;/i&gt;&lt;/a&gt;&lt;/span&gt;&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section-4</guid>
    <dc:title>Conclusion</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;In this course you have learned about discrete-time signals and the discrete-time systems that use them. In doing so, you have focused on digital filtering and found out why advances in digital computer processing have allowed digital filtering to be used in scenarios where analogue filters would once have been the only viable solution. You have seen how relatively simple averaging filters can remove high-frequency noise, and also that more complex filters are designed. &lt;/p&gt;&lt;p&gt;This completes your study. You should now be able to:&lt;/p&gt;&lt;ul class="oucontent-bulleted"&gt;&lt;li&gt;understand how filtering of discrete-time signals can be achieved by mathematical processes such as averaging&lt;/li&gt;&lt;li&gt;understand how mathematical operations applied to a discrete-time signal in the time domain can result in the removal or reduction of unwanted aspects of the signal &lt;/li&gt;&lt;li&gt;explain why filters are designed in the frequency domain, and specify a digital filter to achieve a desired filtering effect.&lt;/li&gt;&lt;/ul&gt;&lt;p&gt;This OpenLearn course is an adapted extract from the Open University course &lt;span class="oucontent-linkwithtip"&gt;&lt;a class="oucontent-hyperlink" href="http://www.open.ac.uk/courses/modules/t312"&gt;T312 &lt;i&gt;Electronics: signal processing, control and communications&lt;/i&gt;&lt;/a&gt;&lt;/span&gt;&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>Glossary</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;dl class="oucontent-glossary"&gt;
&lt;dt id="idm46133049000736"&gt;alias&lt;/dt&gt;
&lt;dd&gt;An error appearing in a sampled signal when the bandwidth of the signal is greater than half the sampling frequency (that is, when the sampling frequency is lower than the Nyquist frequency). Such effects are also referred to as artefacts or ghosts.&lt;/dd&gt;
&lt;dt id="idm46133048998992"&gt;anti-aliasing filter&lt;/dt&gt;
&lt;dd&gt;A low-pass filter that is able to remove aliasing in sampled signals by cutting all the spectral components that are greater than or equal to half the sampling frequency.&lt;/dd&gt;
&lt;dt id="idm46133048997328"&gt;Bode plot&lt;/dt&gt;
&lt;dd&gt;Loosely, a graph of the frequency response of a device or system. Strictly, a pair of graphs showing frequency response and phase response over the same span of frequencies.&lt;/dd&gt;
&lt;dt id="idm46133048995664"&gt;convolution&lt;/dt&gt;
&lt;dd&gt;A mathematical operation that combines two signals to produce a third signal. When two signals are convolved, the resultant third signal expresses how the shape of one signal is modified by the other.&lt;/dd&gt;
&lt;dt id="idm46133048993968"&gt;decibel&lt;/dt&gt;
&lt;dd&gt;A logarithmic way of expressing a power ratio. For powers P1 and P2, their ratio in decibels is defined as 10 log10 (P1/P2). The symbol for decibels is dB. Strictly the decibel is not a unit, as any ratio must be a pure (that is, dimensionless) number. However, it is often regarded as a unit.&lt;/dd&gt;
&lt;dt id="idm46133048992192"&gt;difference equation&lt;/dt&gt;
&lt;dd&gt;An equation in which all variables have been sampled at fixed intervals, and these variables are multiplied by some coefficient.&lt;/dd&gt;
&lt;dt id="idm46133048990576"&gt;differential equation&lt;/dt&gt;
&lt;dd&gt;A mathematical equation in which one or more terms contains a mathematically differentiated variable.&lt;/dd&gt;
&lt;dt id="idm46133048988992"&gt;digital signal processor (DSP)&lt;/dt&gt;
&lt;dd&gt;A semiconductor device similar to a microprocessor. Whilst a microprocessor is a general-purpose device, a DSP has been optimised to carry out the computations used for processing discrete signals.&lt;/dd&gt;
&lt;dt id="idm46133048987296"&gt;first-order&lt;/dt&gt;
&lt;dd&gt;As applied to a filter, the simplest type of filter, having in its passive form a single reactive element (a capacitor or an inductor) and a roll-off of 20 dB/decade, or 6 dB/octave.
As applied to a differential equation, such an equation in which the main variable is differentiated once. Any system that can be modelled with such a differential equation would be referred to as a first-order system.&lt;/dd&gt;
&lt;dt id="idm46133048985408"&gt;Fourier transform&lt;/dt&gt;
&lt;dd&gt;A transformation that extends the concept of the Fourier series to non-periodic signals. It allows us to estimate the spectrum of a signal and perform a frequency analysis.&lt;/dd&gt;
&lt;dt id="idm46133048983744"&gt;frequency response&lt;/dt&gt;
&lt;dd&gt;The response of a system (e.g. a filter) when we input sine waves at different frequencies (but equal amplitude). It tells us how the system will modify the spectrum of any input signal we feed to the system.&lt;/dd&gt;
&lt;dt id="idm46133048982048"&gt;gain&lt;/dt&gt;
&lt;dd&gt;In amplification, a measure of how many times the input signal amplitude is increased. It is generally measured as the ratio between the input signal amplitude and the output signal level. If a gain value is given as just a number (i.e. with no units), the gain is likely to be a ratio of voltages; if the value is given in decibels, it is a ratio of powers. See also voltage gain and power gain.&lt;/dd&gt;
&lt;dt id="idm46133048980160"&gt;octave&lt;/dt&gt;
&lt;dd&gt;The span of frequencies covered by a doubling of frequency, or by a halving of frequency. For example, the frequency span from 500 Hz to 1000 Hz is an octave, as is the span from 500 Hz to 250 Hz. In music, the eight notes of a diatonic scale (that is, doh, re, me &amp;#x2026; ti, doh) cover an octave; hence the name &amp;#x2018;octave’ for this span of frequencies.&lt;/dd&gt;
&lt;dt id="idm46133048978320"&gt;operational amplifier&lt;/dt&gt;
&lt;dd&gt;A general-purpose analogue amplifier intended to be used as a component in other electronic circuits, and usually supplied as a multi-pin integrated-circuit device with two inputs and a single output. Typically an op-amp is a differential amplifier (that is, it amplifies the difference between its two inputs) and has an unusably high gain and extremely high input impedance. To give useful and predictable behaviour, external feedback circuitry must be applied. This circuitry determines essential parameters such as input impedance, output impedance, overall gain and frequency response, and also whether the circuit operates as a single-input amplifier or a differential amplifier.&lt;/dd&gt;
&lt;dt id="idm46133048976144"&gt;order&lt;/dt&gt;
&lt;dd&gt;A numerical classification for filters (e.g. &amp;#x2018;first order’, &amp;#x2018;second order’, &amp;#x2018;third order’, etc.). The order is determined by the differential equation of the filter. For a first-order filter, the highest differential coefficient in the equation is first-order (e.g. dv/dt); for a second-order filter, the highest differential coefficient is second-order (e.g. d2v/dt2). The higher the order, the steeper the roll-off and the sharper the cut-off between passband and stop band. Increasing the order by one adds 20 dB/decade to the filter’s roll-off.&lt;/dd&gt;
&lt;dt id="idm46133048974096"&gt;passband&lt;/dt&gt;
&lt;dd&gt;The band or bands of frequencies passed by a filter with least attenuation, or no attenuation. Frequencies outside the passband are cut off, or stopped, by the filter. Passband is the counterpart of stop band.&lt;/dd&gt;
&lt;dt id="idm46133048972400"&gt;power gain&lt;/dt&gt;
&lt;dd&gt;The ratio of output power to input power. It is usually expressed in decibels (dB). A power gain of 0 dB means that the output power is the same as the input power. A power gain of 3 dB (or, more exactly, 3.0103 dB) means that the output power is double the input power.&lt;/dd&gt;
&lt;dt id="idm46133048970640"&gt;quantisation&lt;/dt&gt;
&lt;dd&gt;Conversion of an analogue quantity, which could take any value within a range, to one of a set of discrete values.&lt;/dd&gt;
&lt;dt id="idm46133048969040"&gt;roll-off&lt;/dt&gt;
&lt;dd&gt;The steepness of a filter’s attenuation in a stop band. Also, the steepness of the attenuation of any device that produces attenuation (for example, a linear amplifier at the extremes of its operating-frequency range).&lt;/dd&gt;
&lt;dt id="idm46133048967328"&gt;stop band&lt;/dt&gt;
&lt;dd&gt;The band or bands of frequencies stopped, or cut off, by a filter. The counterpart of the passband.&lt;/dd&gt;
&lt;dt id="idm46133048965744"&gt;taps&lt;/dt&gt;
&lt;dd&gt;In a digital filter, the number of taps is the number of terms in the mathematical expression that describes the filter. This expression is given in the form of a difference equation. In digital filter design, the maximum number of taps to be used in the implementation is required as part of the design specification.&lt;/dd&gt;
&lt;dt id="idm46133048963936"&gt;voltage gain&lt;/dt&gt;
&lt;dd&gt;For a sinusoidal input and output, voltage gain is the ratio of the output voltage’s amplitude to that of the input voltage. It has no units.&lt;/dd&gt;
&lt;dt id="idm46133048962304"&gt;window function&lt;/dt&gt;
&lt;dd&gt;In signal processing, a mathematical function that is zero-valued outside some chosen interval. When a signal is convolved with a window function, the resultant signal is also zero-valued outside the chosen interval, so it is the original signal viewed through the window function.&lt;/dd&gt;
&lt;/dl&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section--glossary</guid>
    <dc:title>Glossary</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;dl class="oucontent-glossary"&gt;
&lt;dt id="idm46133049000736"&gt;alias&lt;/dt&gt;
&lt;dd&gt;An error appearing in a sampled signal when the bandwidth of the signal is greater than half the sampling frequency (that is, when the sampling frequency is lower than the Nyquist frequency). Such effects are also referred to as artefacts or ghosts.&lt;/dd&gt;
&lt;dt id="idm46133048998992"&gt;anti-aliasing filter&lt;/dt&gt;
&lt;dd&gt;A low-pass filter that is able to remove aliasing in sampled signals by cutting all the spectral components that are greater than or equal to half the sampling frequency.&lt;/dd&gt;
&lt;dt id="idm46133048997328"&gt;Bode plot&lt;/dt&gt;
&lt;dd&gt;Loosely, a graph of the frequency response of a device or system. Strictly, a pair of graphs showing frequency response and phase response over the same span of frequencies.&lt;/dd&gt;
&lt;dt id="idm46133048995664"&gt;convolution&lt;/dt&gt;
&lt;dd&gt;A mathematical operation that combines two signals to produce a third signal. When two signals are convolved, the resultant third signal expresses how the shape of one signal is modified by the other.&lt;/dd&gt;
&lt;dt id="idm46133048993968"&gt;decibel&lt;/dt&gt;
&lt;dd&gt;A logarithmic way of expressing a power ratio. For powers P1 and P2, their ratio in decibels is defined as 10 log10 (P1/P2). The symbol for decibels is dB. Strictly the decibel is not a unit, as any ratio must be a pure (that is, dimensionless) number. However, it is often regarded as a unit.&lt;/dd&gt;
&lt;dt id="idm46133048992192"&gt;difference equation&lt;/dt&gt;
&lt;dd&gt;An equation in which all variables have been sampled at fixed intervals, and these variables are multiplied by some coefficient.&lt;/dd&gt;
&lt;dt id="idm46133048990576"&gt;differential equation&lt;/dt&gt;
&lt;dd&gt;A mathematical equation in which one or more terms contains a mathematically differentiated variable.&lt;/dd&gt;
&lt;dt id="idm46133048988992"&gt;digital signal processor (DSP)&lt;/dt&gt;
&lt;dd&gt;A semiconductor device similar to a microprocessor. Whilst a microprocessor is a general-purpose device, a DSP has been optimised to carry out the computations used for processing discrete signals.&lt;/dd&gt;
&lt;dt id="idm46133048987296"&gt;first-order&lt;/dt&gt;
&lt;dd&gt;As applied to a filter, the simplest type of filter, having in its passive form a single reactive element (a capacitor or an inductor) and a roll-off of 20 dB/decade, or 6 dB/octave.
As applied to a differential equation, such an equation in which the main variable is differentiated once. Any system that can be modelled with such a differential equation would be referred to as a first-order system.&lt;/dd&gt;
&lt;dt id="idm46133048985408"&gt;Fourier transform&lt;/dt&gt;
&lt;dd&gt;A transformation that extends the concept of the Fourier series to non-periodic signals. It allows us to estimate the spectrum of a signal and perform a frequency analysis.&lt;/dd&gt;
&lt;dt id="idm46133048983744"&gt;frequency response&lt;/dt&gt;
&lt;dd&gt;The response of a system (e.g. a filter) when we input sine waves at different frequencies (but equal amplitude). It tells us how the system will modify the spectrum of any input signal we feed to the system.&lt;/dd&gt;
&lt;dt id="idm46133048982048"&gt;gain&lt;/dt&gt;
&lt;dd&gt;In amplification, a measure of how many times the input signal amplitude is increased. It is generally measured as the ratio between the input signal amplitude and the output signal level. If a gain value is given as just a number (i.e. with no units), the gain is likely to be a ratio of voltages; if the value is given in decibels, it is a ratio of powers. See also voltage gain and power gain.&lt;/dd&gt;
&lt;dt id="idm46133048980160"&gt;octave&lt;/dt&gt;
&lt;dd&gt;The span of frequencies covered by a doubling of frequency, or by a halving of frequency. For example, the frequency span from 500 Hz to 1000 Hz is an octave, as is the span from 500 Hz to 250 Hz. In music, the eight notes of a diatonic scale (that is, doh, re, me … ti, doh) cover an octave; hence the name ‘octave’ for this span of frequencies.&lt;/dd&gt;
&lt;dt id="idm46133048978320"&gt;operational amplifier&lt;/dt&gt;
&lt;dd&gt;A general-purpose analogue amplifier intended to be used as a component in other electronic circuits, and usually supplied as a multi-pin integrated-circuit device with two inputs and a single output. Typically an op-amp is a differential amplifier (that is, it amplifies the difference between its two inputs) and has an unusably high gain and extremely high input impedance. To give useful and predictable behaviour, external feedback circuitry must be applied. This circuitry determines essential parameters such as input impedance, output impedance, overall gain and frequency response, and also whether the circuit operates as a single-input amplifier or a differential amplifier.&lt;/dd&gt;
&lt;dt id="idm46133048976144"&gt;order&lt;/dt&gt;
&lt;dd&gt;A numerical classification for filters (e.g. ‘first order’, ‘second order’, ‘third order’, etc.). The order is determined by the differential equation of the filter. For a first-order filter, the highest differential coefficient in the equation is first-order (e.g. dv/dt); for a second-order filter, the highest differential coefficient is second-order (e.g. d2v/dt2). The higher the order, the steeper the roll-off and the sharper the cut-off between passband and stop band. Increasing the order by one adds 20 dB/decade to the filter’s roll-off.&lt;/dd&gt;
&lt;dt id="idm46133048974096"&gt;passband&lt;/dt&gt;
&lt;dd&gt;The band or bands of frequencies passed by a filter with least attenuation, or no attenuation. Frequencies outside the passband are cut off, or stopped, by the filter. Passband is the counterpart of stop band.&lt;/dd&gt;
&lt;dt id="idm46133048972400"&gt;power gain&lt;/dt&gt;
&lt;dd&gt;The ratio of output power to input power. It is usually expressed in decibels (dB). A power gain of 0 dB means that the output power is the same as the input power. A power gain of 3 dB (or, more exactly, 3.0103 dB) means that the output power is double the input power.&lt;/dd&gt;
&lt;dt id="idm46133048970640"&gt;quantisation&lt;/dt&gt;
&lt;dd&gt;Conversion of an analogue quantity, which could take any value within a range, to one of a set of discrete values.&lt;/dd&gt;
&lt;dt id="idm46133048969040"&gt;roll-off&lt;/dt&gt;
&lt;dd&gt;The steepness of a filter’s attenuation in a stop band. Also, the steepness of the attenuation of any device that produces attenuation (for example, a linear amplifier at the extremes of its operating-frequency range).&lt;/dd&gt;
&lt;dt id="idm46133048967328"&gt;stop band&lt;/dt&gt;
&lt;dd&gt;The band or bands of frequencies stopped, or cut off, by a filter. The counterpart of the passband.&lt;/dd&gt;
&lt;dt id="idm46133048965744"&gt;taps&lt;/dt&gt;
&lt;dd&gt;In a digital filter, the number of taps is the number of terms in the mathematical expression that describes the filter. This expression is given in the form of a difference equation. In digital filter design, the maximum number of taps to be used in the implementation is required as part of the design specification.&lt;/dd&gt;
&lt;dt id="idm46133048963936"&gt;voltage gain&lt;/dt&gt;
&lt;dd&gt;For a sinusoidal input and output, voltage gain is the ratio of the output voltage’s amplitude to that of the input voltage. It has no units.&lt;/dd&gt;
&lt;dt id="idm46133048962304"&gt;window function&lt;/dt&gt;
&lt;dd&gt;In signal processing, a mathematical function that is zero-valued outside some chosen interval. When a signal is convolved with a window function, the resultant signal is also zero-valued outside the chosen interval, so it is the original signal viewed through the window function.&lt;/dd&gt;
&lt;/dl&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>References</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section---references</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;div class="oucontent-referenceitem"&gt;Asgari, S. and Mehrnia, A. (2017) &amp;#x2018;A novel low-complexity digital filter design for wearable ECG devices’, &lt;i&gt;PLOS ONE&lt;/i&gt;, vol. 12, no. 4 [Online]. Available at https://doi.org/10.1371/journal.pone.0175139 (Accessed 25 March 2019). &lt;/div&gt;
&lt;div class="oucontent-referenceitem"&gt;Dongarra, J. and Sullivan, F. (2000) &amp;#x2018;Guest editors’ introduction: the top 10 algorithms’, &lt;i&gt;Computing in Science &amp;amp; Engineering&lt;/i&gt;, vol.&amp;#xA0;2, no.&amp;#xA0;1, pp.&amp;#xA0;22–3.&lt;/div&gt;
&lt;div class="oucontent-referenceitem"&gt;Wickert, M. (2011) &amp;#x2018;Chapter&amp;#xA0;5 FIR Filters’, ECE&amp;#xA0;2610 &lt;i&gt;Introduction to Signals and Systems&lt;/i&gt; [Online], University of Colorado Colorado Springs. Available at www.eas.uccs.edu/~mwickert/ece2610/lecture_notes/ece2610_chap5.pdf (Accessed 3&amp;#xA0;June 2019).&lt;/div&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section---references</guid>
    <dc:title>References</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;div class="oucontent-referenceitem"&gt;Asgari, S. and Mehrnia, A. (2017) ‘A novel low-complexity digital filter design for wearable ECG devices’, &lt;i&gt;PLOS ONE&lt;/i&gt;, vol. 12, no. 4 [Online]. Available at https://doi.org/10.1371/journal.pone.0175139 (Accessed 25 March 2019). &lt;/div&gt;
&lt;div class="oucontent-referenceitem"&gt;Dongarra, J. and Sullivan, F. (2000) ‘Guest editors’ introduction: the top 10 algorithms’, &lt;i&gt;Computing in Science &amp; Engineering&lt;/i&gt;, vol. 2, no. 1, pp. 22–3.&lt;/div&gt;
&lt;div class="oucontent-referenceitem"&gt;Wickert, M. (2011) ‘Chapter 5 FIR Filters’, ECE 2610 &lt;i&gt;Introduction to Signals and Systems&lt;/i&gt; [Online], University of Colorado Colorado Springs. Available at www.eas.uccs.edu/~mwickert/ece2610/lecture_notes/ece2610_chap5.pdf (Accessed 3 June 2019).&lt;/div&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
    <item>
      <title>Acknowledgements</title>
      <link>https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section---acknowledgements</link>
      <pubDate>Tue, 31 Mar 2020 23:00:00 GMT</pubDate>
      <description>&lt;p&gt;This free course was written by Allan Jones, Bernie Clarke and Phil Picton. It was first published in July 2020.&lt;/p&gt;
&lt;p&gt;Except for third party materials and otherwise stated (see &lt;span class="oucontent-linkwithtip"&gt;&lt;a class="oucontent-hyperlink" href="http://www.open.ac.uk/conditions"&gt;terms and conditions&lt;/a&gt;&lt;/span&gt;), this content is made available under a &lt;a class="oucontent-hyperlink" href="http://creativecommons.org/licenses/by-nc-sa/4.0/deed.en_GB"&gt;Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Licence&lt;/a&gt;.&lt;/p&gt;
&lt;p&gt;The material acknowledged below is Proprietary and used under licence (not subject to Creative Commons Licence). Grateful acknowledgement is made to the following sources for permission to reproduce material in this free course: &lt;/p&gt;
&lt;h2 class="oucontent-h3 oucontent-heading oucontent-basic"&gt;Figures&lt;/h2&gt;
&lt;p&gt;Course image: paulclee / www.pixabay.com &lt;/p&gt;
&lt;p&gt;Figure 1: &amp;#xA9; BBC&lt;/p&gt;
&lt;p&gt;Figure 2: &amp;#xA9; NASA&lt;/p&gt;
&lt;p&gt;Figure 3: Courtesy Starship Technologies &lt;/p&gt;
&lt;p&gt;Figure 14: Adapted from: &lt;a class="oucontent-hyperlink" href="https://doi.org/10.1371/journal.pone.0175139"&gt;https://doi.org/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;10.1371/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;journal.pone.0175139&lt;/a&gt; &lt;/p&gt;
&lt;p&gt;Installing Signal Wizard for use in offline mode PDF, Figure 1: Taken from: &lt;a class="oucontent-hyperlink" href="http://www.signalwizardsystems.com/"&gt;http://www.signalwizardsystems.com/&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;Every effort has been made to contact copyright owners. If any have been inadvertently overlooked, the publishers will be pleased to make the necessary arrangements at the first opportunity.&lt;/p&gt;
&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Don't miss out&lt;/b&gt;&lt;/p&gt;
&lt;p&gt;If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University – &lt;a class="oucontent-hyperlink" href="http://www.open.edu/openlearn/free-courses?utm_source=openlearn&amp;amp;utm_campaign=ol&amp;amp;utm_medium=ebook"&gt;www.open.edu/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;openlearn/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;free-courses&lt;/a&gt;.&lt;/p&gt;</description>
      <guid isPermaLink="true">https://www.open.edu/openlearn/science-maths-technology/electronic-applications/content-section---acknowledgements</guid>
    <dc:title>Acknowledgements</dc:title><dc:identifier>T312_1</dc:identifier><dc:description>&lt;p&gt;This free course was written by Allan Jones, Bernie Clarke and Phil Picton. It was first published in July 2020.&lt;/p&gt;
&lt;p&gt;Except for third party materials and otherwise stated (see &lt;span class="oucontent-linkwithtip"&gt;&lt;a class="oucontent-hyperlink" href="http://www.open.ac.uk/conditions"&gt;terms and conditions&lt;/a&gt;&lt;/span&gt;), this content is made available under a &lt;a class="oucontent-hyperlink" href="http://creativecommons.org/licenses/by-nc-sa/4.0/deed.en_GB"&gt;Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Licence&lt;/a&gt;.&lt;/p&gt;
&lt;p&gt;The material acknowledged below is Proprietary and used under licence (not subject to Creative Commons Licence). Grateful acknowledgement is made to the following sources for permission to reproduce material in this free course: &lt;/p&gt;
&lt;h2 class="oucontent-h3 oucontent-heading oucontent-basic"&gt;Figures&lt;/h2&gt;
&lt;p&gt;Course image: paulclee / www.pixabay.com &lt;/p&gt;
&lt;p&gt;Figure 1: © BBC&lt;/p&gt;
&lt;p&gt;Figure 2: © NASA&lt;/p&gt;
&lt;p&gt;Figure 3: Courtesy Starship Technologies &lt;/p&gt;
&lt;p&gt;Figure 14: Adapted from: &lt;a class="oucontent-hyperlink" href="https://doi.org/10.1371/journal.pone.0175139"&gt;https://doi.org/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;10.1371/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;journal.pone.0175139&lt;/a&gt; &lt;/p&gt;
&lt;p&gt;Installing Signal Wizard for use in offline mode PDF, Figure 1: Taken from: &lt;a class="oucontent-hyperlink" href="http://www.signalwizardsystems.com/"&gt;http://www.signalwizardsystems.com/&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;Every effort has been made to contact copyright owners. If any have been inadvertently overlooked, the publishers will be pleased to make the necessary arrangements at the first opportunity.&lt;/p&gt;
&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Don't miss out&lt;/b&gt;&lt;/p&gt;
&lt;p&gt;If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University – &lt;a class="oucontent-hyperlink" href="http://www.open.edu/openlearn/free-courses?utm_source=openlearn&amp;utm_campaign=ol&amp;utm_medium=ebook"&gt;www.open.edu/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;openlearn/&lt;span class="oucontent-hidespace"&gt; &lt;/span&gt;free-courses&lt;/a&gt;.&lt;/p&gt;</dc:description><dc:publisher>The Open University</dc:publisher><dc:creator>The Open University</dc:creator><dc:type>Course</dc:type><dc:format>text/html</dc:format><dc:language>en-GB</dc:language><dc:source>Electronic applications - T312_1</dc:source><cc:license>Copyright © 2020 The Open University</cc:license></item>
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