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Electronic applications
Electronic applications

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2.1 Frequency-dependent gain

In Figure 4, the box in the middle represents a device with frequency-dependent gain cap g of omega – in other words, a filter. Here the symbol omega is being used to represent angular frequency, measured in radians per second.

Described image
Figure 4 Filter with input amplitude cap v sub in and output amplitude cap v sub out

A sinusoidal input signal with amplitude cap v sub in is applied to the filter, and the output is a sinusoidal signal with amplitude cap v sub out. The voltage gain of the filter at the frequency in question is cap v sub out divided by cap v sub in – in other words, it is the ratio of the output voltage amplitude to the input voltage amplitude. So, at any particular frequency,

cap v sub out divided by cap v sub in equals cap g

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 one divided by four (or 0.25) means that the amplitude of the output is one-quarter that of the input.

The above gain is referred to as ‘voltage gain’ because gain is sometimes expressed as a ratio of output and input powers. In this course this way of expressing gain will be referred to as power gain. As power ratios can be expressed in decibels, so power gains are almost invariably given in decibels.

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.

Although two different types of gain have been mentioned (voltage gain and power gain), you will often just see the word gain 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.

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.

Activity 1

Timing: Allow about 5 minutes
  • a.A passive filter has an input signal of  v sub in of t equals 10 postfix times sine of 200 times t volts. The steady-state output is v sub out of t equals two postfix times sine of 200 times t minus 0.6 times pi volts. What is the gain as a voltage ratio?
  • b.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  v sub out of t equals five postfix times sine of 100 times t minus 0.2 times pi volts. What is the new gain as a voltage ratio?


  • a.Here cap v sub in, the amplitude of v sub in of t, is 10 V and cap v sub out, the amplitude of v sub out of t, is 2 V. Therefore the gain is two cap v divided by 10 cap v, or 0.2.
  • b.cap v sub in is still 10 V and cap v sub out is now 5 V. Therefore the gain is now five cap v divided by 10 cap v, or 0.5.

In the next section you will discover the characteristics of an ideal filter.