3.9 Discrete time filter design

In this first example we will take a look at the very simple z-transform shown below: $$H(z)=z+1=1+z^{-1}$$ $$H(w)=1+e^{-(iw)}$$ 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 [link] above. One can see that when $w=0$ , the vector is the longest and thus the frequency responsewill have its largest amplitude here. As $w$ approaches $\pi$ , the length of the vectors decrease as does the amplitude of $\left|H(w)\right|$ . Since there are no poles in the transform, there is only this onevector term rather than a ratio as seen in [link] .

Pole/zero plot

Frequency response: |h(w)|

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

Below we can see the two figures described by the above equations. The [link] represents the basic pole/zero plot of the z-transform, $H(w)$ . [link] shows the magnitude of the frequency response. From the formulas and statements in the previoussection, 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 [link] helps us see the mathematics behindthis 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 poleand thus force the frequency response closer to $0$ .

Pole/zero plot

Frequency response: |h(w)|