One of the primary motivating factors for utilizing the z-transform and analyzing the pole/zero plots is due to their relationship to the frequency response of a discrete-time system. Based on the position of the poles and zeros, one can quickly determine the frequency response. This is a result of the correspondence between the frequency response and the transfer function evaluated on the unit circle in the pole/zero plots. The frequency response, or DTFT, of the system is defined as:

$$\begin{array}{rcl}\hfill H\left(w\right)& \hfill =\hfill & H\left(z\right){|}_{z,z={e}^{iw}}\hfill \\ \hfill & \hfill =\hfill & \frac{\sum _{k=0}^{M}{b}_{k}{e}^{-(iwk)}}{\sum _{k=0}^{N}{a}_{k}{e}^{-(iwk)}}\hfill \end{array}$$

1

$$H\left(w\right)=\left|\frac{{b}_{0}}{{a}_{0}}\right|\frac{\prod _{k=1}^{M}\left|{e}^{iw}-{c}_{k}\right|}{\prod _{k=1}^{N}\left|{e}^{iw}-{d}_{k}\right|}$$

2

$$\left|H\left(w\right)\right|=\left|\frac{{b}_{0}}{{a}_{0}}\right|\frac{\prod \mathrm{"distances\; from\; zeros"}}{\prod \mathrm{"distances\; from\; poles"}}$$

3

Let us now look at several examples of determining the magnitude of the frequency response from the pole/zero plot of a z-transform. If you have forgotten or are unfamiliar with pole/zero plots, please refer back to the Pole/Zero Plots module.

Example 1

In this first example we will take a look at the very simple z-transform shown below: $$H\left(z\right)=z+1=1+{z}^{-1}$$ $$H\left(w\right)=1+{e}^{-(iw)}$$ For this example, some of the vectors represented by $\left|{e}^{iw}-h\right|$, for random values of $w$, are explicitly drawn onto the complex plane shown in the figure below. These vectors show how the amplitude of the frequency response changes as $w$ goes from $0$ to $2\pi $, and also show the physical meaning of the terms in Equation 2 above. One can see that when $w=0$, the vector is the longest and thus the frequency response will have its largest amplitude here. As $w$ approaches $\pi $, the length of the vectors decrease as does the amplitude of $\left|H\left(w\right)\right|$. Since there are no poles in the transform, there is only this one vector term rather than a ratio as seen in Equation 2.

Example 2

For this example, a more complex transfer function is analyzed in order to represent the system's frequency response. $$H\left(z\right)=\frac{z}{z-\frac{1}{2}}=\frac{1}{1-\frac{1}{2}{z}^{-1}}$$ $$H\left(w\right)=\frac{1}{1-\frac{1}{2}{e}^{-(iw)}}$$

Below we can see the two figures described by the above equations. The Link represents the basic pole/zero plot of the z-transform, $H\left(w\right)$. Link shows the magnitude of the frequency response. From the formulas and statements in the previous section, we can see that when $w=0$ the frequency will peak since it is at this value of $w$ that the pole is closest to the unit circle. The ratio from Equation 2 helps us see the mathematics behind this conclusion and the relationship between the distances from the unit circle and the poles and zeros. As $w$ moves from $0$ to $\pi $, we see how the zero begins to mask the effects of the pole and thus force the frequency response closer to $0$.

In conclusion, using the distances from the unit circle to the poles and zeros, we can plot the frequency response of the system. As $w$ goes from $0$ to $2\pi $, the following two properties, taken from the above equations, specify how one should draw $\left|H\left(w\right)\right|$.

While moving around the unit circle...

- if close to a zero, then the magnitude is small. If a zero is on the unit circle, then the frequency response is zero at that point.
- if close to a pole, then the magnitude is large. If a pole is on the unit circle, then the frequency response goes to infinity at that point.