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Fourier Transform and the Transfer Function

Module by: Don Johnson

Summary: Correlates the unit response with the frequency and time domains.

Because we are interested in actual computations rather than analytic calculations, we must consider the details of the discrete Fourier transform. To compute the length- NN DFT, we assume that the signal has a duration less than or equal to NN. Because frequency responses have an explicit frequency-domain specification in terms of filter coefficients, we don't have a direct handle on which signal has a Fourier transform equaling a given frequency response. Finding this signal is quite easy. First of all, note that the discrete-time Fourier transform of a unit sample equals one for all frequencies. Because of the input and output of linear, shift-invariant systems are related to each other by Yⅇⅈ2πf=Hⅇⅈ2πfXⅇⅈ2πf Y 2 f H 2 f X 2 f , a unit-sample input results in the output's Fourier transform equaling the system's transfer function.

Exercise 1

This statement is a very important result. Derive it yourself.

Solution 1

The DTFT of the unit sample equals a constant (equaling 1). In this case, the Fourier transform of the output equals the transfer function.

In the time-domain, the output for a unit-sample input is known as the system's unit-sample response, and is denoted by hn h n . Combining the frequency-domain and time-domain interpretations of a linear, shift-invariant system's unit-sample response, we have that hn h n and the transfer function are Fourier transform pairs in terms of the discrete-time Fourier transform.

Discrete-time Fourier Transform

hn↔Hⅇⅈ2πf ↔ h n H 2 f (1)

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