Now consider what happens when we take the factored form of the general transfer function, Eq.(8.2), and set to to get the frequency response in factored form:
As usual for the frequency response, we prefer the polar form for this expression. Consider first the amplitude response .
In the complex plane, the number is plotted at the coordinates . The difference of two vectors and is , as shown in Fig.8.1. Translating the origin of the vector to the tip of shows that is an arrow drawn from the tip of to the tip of . The length of a vector is unaffected by translation away from the origin. However, the angle of a translated vector must be measured relative to a translated copy of the real axis. Thus the term may be drawn as an arrow from the th zero to the point on the unit circle, and is an arrow from the th pole. Therefore, each term in Eq.(8.3) is the length of a vector drawn from a pole or zero to a single point on the unit circle, as shown in Fig.8.2 for two poles and two zeros. In summary:
For example, the dc gain is obtained by multiplying the lengths of the lines drawn from all poles and zeros to the point . The filter gain at half the sampling rate is the product of the lengths of these lines when drawn to the point . For an arbitrary frequency Hz, we draw arrows from the poles and zeros to the point . Thus, at the frequency where the arrows in Fig.8.2 join, (which is slightly less than one-eighth the sampling rate) the gain of this two-pole two-zero filter is . Figure 8.3 gives the complete amplitude response for the poles and zeros shown in Fig.8.2. Before looking at that, it is a good exercise to try sketching it by inspection of the pole-zero diagram. It is usually easy to sketch a qualitatively accurate amplitude-response directly from the poles and zeros (to within a scale factor).