This is called a ``narrow-band filter" because in the Fourier domain
the function is large only in a narrow band of frequencies. Setting to half its peak value of ,we find a half-bandwidth of .The damping time constant of the damped sinusoid bt is shown in the exercises following this section to be .
Naturally we want a real-time function, so we multiply the filter 1/(1-Z/Zp) times .The resulting time function is real because conjugate poles are like the conjugate roots. The spectrum of the conjugate factor is like (38), except that is replaced by .Multiplying the response (38) by itself with yields the symmetric function of displayed on the right in Figure 9.
You might be disappointed if you intend to apply the filter of Figure 9 as a narrow-band filter. Notice that the passband is asymmetric and that it passes the zero frequency. Equation (38) is symmetric about ,but taking the product with its image about has spoiled the symmetry. Should we be concerned about this ``edge effect''? The answer is yes, whenever we handle real data. For real data, is usually small enough. Recall that .Consider a pole at a particular :decreasing pushes towards zero, which is where a pole and its mate at negative frequency create the asymmetrical response shown in Figure 9.
So in practice we might like to add a zero at zero frequency and at the Nyquist frequency, i.e., (1-Z)(1+Z), as shown in Figure 10.
Compare Figure 10 and 9. If the time functions were interchanged, could you tell the difference between the figures? There are two ways to distinguish them. The most obvious is that the zero-frequency component is made evident in the time domain by the sum of the filter coefficients (theoretically, F(Z=1)). A more subtle clue is that the first half-cycle of the wave in Figure 10 is shorter than in Figure 9; hence, it contains extra high frequency energy, which we can see in the spectrum.