# 5.13: Discrete -Time Systems in the Frequency Domain

Learning Objectives

• As with analog linear systems, we need to find the frequency response of discrete-time systems.

As with analog linear systems, we need to find the frequency response of discrete-time systems. We used impedances to derive directly from the circuit's structure the frequency response. The only structure we have so far for a discrete-time system is the difference equation. We proceed as when we used impedances: let the input be a complex exponential signal. When we have a linear, shift-invariant system, the output should also be a complex exponential of the same frequency, changed in amplitude and phase. These amplitude and phase changes comprise the frequency response we seek. The complex exponential input signal is

$x(n)=Xe^{i2\pi fn}$

Note that this input occurs for all values of No need to worry about initial conditions here. Assume the output has a similar form:

$y(n)=Ye^{i2\pi fn}$

Plugging these signals into the fundamental difference equation, we have

$Ye^{i2\pi fn}=a_{1}Ye^{i2\pi f(n-1)}+...+a_{p}Ye^{i2\pi f(n-p)}+b_{0}Xe^{i2\pi fn}+b_{1}Xe^{i2\pi f(n-1)}+...+b_{q}Xe^{i2\pi f(n-q)}$

The assumed output does indeed satisfy the difference equation if the output complex amplitude is related to the input amplitude by

$Y=\frac{b_{0}+b_{1}e^{-(i2\pi f)}+...+b_{q}e^{-(i2\pi qf)}}{1-a_{1}e^{-(i2\pi f)-...-a_{p}e^{-(i2\pi pf)}}}X$

This relationship corresponds to the system's frequency response or, by another name, its transfer function. We find that any discrete-time system defined by a difference equation has a transfer function given by

$H(e^{i2\pi f})=\frac{b_{0}+b_{1}e^{-(i2\pi f)}+...+b_{q}e^{-(i2\pi qf)}}{1-a_{1}e^{-(i2\pi f)-...-a_{p}e^{-(i2\pi pf)}}}$

Furthermore, because any discrete-time signal can be expressed as a superposition of complex exponential signals and because linear discrete-time systems obey the Superposition Principle, the transfer function relates the discrete-time Fourier transform of the system's output to the input's Fourier transform.

$Y(e^{i2\pi f})=X(e^{i2\pi f})H(e^{i2\pi f})$

Example $$\PageIndex{1}$$:

The frequency response of the simple IIR system

$H(e^{i2\pi f})=\frac{b}{1-ae^{-(i2\pi f)}}$

This Fourier transform occurred in a previous example; the exponential signal spectrum portrays the magnitude and phase of this transfer function. When the filter coefficient a is positive, we have a lowpass filter; negative a results in a highpass filter. The larger the coefficient in magnitude, the more pronounced the lowpass or highpass filtering.

Example $$\PageIndex{1}$$:

The length-q boxcar filter has the frequency response

$H(e^{i2\pi f})=\frac{1}{q}\sum_{m=0}^{q-1}e^{-(i2\pi fm)}$

This expression amounts to the Fourier transform of the boxcar signal.

There we found that this frequency response has a magnitude equal to the absolute value of dsinc(πf). See the length-10 filter's frequency response. We see that boxcar filters--length-q signal averagers--have a lowpass behavior, having a cutoff frequency of 1/q.

Exercise $$\PageIndex{1}$$

Suppose we multiply the boxcar filter's coefficients by a sinusoid:

$b_{m}=\frac{1}{q}\cos (2\pi f_{0}m)$

Use Fourier transform properties to determine the transfer function. How would you characterize this system: Does it act like a filter? If so, what kind of filter and how do you control its characteristics with the filter's coefficients?

Solution

It now acts like a bandpass filter with a center frequency of and a bandwidth equal to twice of the original lowpass filter.

These examples illustrate the point that systems described (and implemented) by difference equations serve as filters for discrete-time signals. The filter's order is given by the number p of denominator coefficients in the transfer function (if the system is IIR) or by the number q of numerator coefficients if the filter is FIR. When a system's transfer function has both terms, the system is usually IIR, and its order equals p regardless of q. By selecting the coefficients and filter type, filters having virtually any frequency response desired can be designed. This design flexibility can't be found in analog systems. In the next section, we detail how analog signals can be filtered by computers, offering a much greater range of filtering possibilities than is possible with circuits.

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