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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.

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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

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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