The bilateral Z transform is

$$Z(x(k))=\sum_{k=-\infty}^{\infty} x(k) \, z^{-k}$$

where z = A e^{-j ω} for some real number A.

## Motivation for using the Z transform in digital signal processing

The Z transform is a generalization of the discrete-time Fourier transform, which is defined as

$$H(\omega)=\sum_{k=-\infty}^{\infty} x(k) \, e^{-j \, \omega \, k}$$

As with other Fourier transforms, the discrete-time Fourier transform translates a complex valued signal x(k) as a function of time into a set of complex numbers that carry the frequency and phase of each simple wave in the signal. The discrete-time Fourier transform is used when x(k) is discrete, but not of finite duration and not periodic (alternatively, one can use the continuous or the discrete Fourier transform). The discrete-time Fourier transform is often thought of as the logical result of sampling the input to the continuous Fourier transform x(t) into discrete time x(n).

The bilateral Z transform generalizes the discrete-time Fourier transform by introducing the real number A, which represents the magnitude of the transform basis A e^{-j ω k}.

$$H(\omega)=\sum_{k=-\infty}^{\infty} x(k) \, z^{-k}=\sum_{k=-\infty}^{\infty} x(k) \, A \, e^{-j \, \omega \, k}$$

Alternatively, the discrete-time Fourier transform is simply the Z transform with A = 1. When A = 1, the transform basis becomes the numbers e^{-j ω k}, which in the complex plane form a circle of radius 1 around the origin. Thus, we can say that the discrete-time Fourier transform is the Z transform evaluated on the unit circle.

The motivation behind using the discrete-time Fourier transform in digital signal processing is that it allows us to use a discrete signal, but a continuous set of frequencies and phases. The motivation behind using the Z transform in digital signal processing rather than the discrete-time Fourier transform is only that it significantly reduces notation. In fact, in digital signal processing we will always evaluate the Z transform on the unit circle – with A = 1 – which means that we will only really use the discrete-time Fourier transform and we will never use its generalized form – the Z transform. We do however use the Z transform notation. It is much easier to write z^{-k} rather than e^{-j ω k}.

## Some useful properties of the Z transform

One useful property of the Z transform is that it is linear, much like the Fourier transforms.

$$Z(a_1 x_1(k) + a_2 x_2(k))=a_1 Z(x_1(k))+a_2 Z(x_2(k))$$

Second, time-shifting of the signal x(k) results in the following.

$$Z(x(k-m))=\sum_{k=-\infty}^{\infty} x(k-m) \, z^{-k}$$ $$=\sum_{k=-\infty}^{\infty} x(k-m) z^{-(k-m)} z^{-m}=Z(x(k)) \, z^{-m}$$

This means that, if the output of a system a(k) on the input x(k) is given by

$$y(k)=\sum_{n=0}^{N-1} a(n) \, x(k-n)$$

where a(k) is a finite impulse response filter, then

$$Z(y(k))=Z(\sum_{n=0}^{N-1} a(n) \, x(k-n))=(\sum_{n=0}^{N-1} a(n) \, z^{-n}) Z(x(k))$$

and

$$H(z)=\frac{Z(y(k))}{Z(x(k))}=\sum_{n=0}^{N-1} a(n) \, z^{-n}$$

which is the transfer function of the system a(k). This transfer function has nice properties. In linear time-invariant systems, the magnitude |H(z)| of H(z), z = A e^{-j ω}, when evaluated at A = 1 (on the unit circle), produces the magnitude response of the system

$$|H(e^{-j \, \omega})|=\sqrt{Re(H(e^{-j \, \omega}))^2+Im(H(e^{-j \, \omega}))^2}$$

The phase response of the system is

$$\Phi(e^{-j \, \omega})=atan2(\frac{Im(H(e^{-j \, \omega}))}{Re(H(e^{-j \, \omega}))})$$

The notation |H(e^{-j ω})| and Φ(e^{-j ω}) is used to show that the magnitude response and phase response are evaluated with e^{-j ω} and not with z = A e^{-j ω}.

## Example transfer functions of a comb filter with the Z transform

A feedforward comb filter is just a simple delay. It has the form

$$y(k)=x(k)+g \, x(k-m)$$

where g is some gain, 1 – g is the decay of the delayed signal, and m is the delay in number of samples. The Z transform of both sides of the above equation produces

$$Z(y(k))=Z(x(k)+g \, x(k-m))=Z(x(k))+g \, z^{-m} Z(x(k))=Z(x(k))(1+g \, z^{-m})$$

The transfer function of this comb filter then is

$$H(z)=\frac{Z(y(k))}{Z(x(k))}=1+g \, z^{-m}$$

The magnitude response of the comb filter at a particular frequency ω is

$$H(e^{-j \, \omega})=|1+g \, z^{-m}|=|1+g \, e^{-j\, \omega \, m}| = |1+g \, \cos(\omega \, m)-g \, \sin(\omega \, m)| $$ $$= \sqrt{(1+g \, \cos(\omega \, m))^2+(g \, \sin(\omega \, m))^2}= \sqrt{(1+g)^2+2 \, g \, \cos(\omega \, m)}$$

This function of ω, when plotted, produces the characteristic comb like magnitude response of the comb filter shown below. This example feedforward comb filter was computed with the sampling frequency f_{s} = 2000 Hz, a delay of m = 50 samples, and with g = 0.8.

## Other examples

For other examples of using the Z transform, see the topics on Comb filter, Transfer function, Gibbs phenomenon, Butterworth filter, All pass filter, and Shroeder-Moore filter.

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