1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.3 • DIGITAL SIGNAL FILTERS 387

lmz

7

Figure 9.10 Amplitude response A(9) and zero-pole plot for y(n] = ~ E x(n - k].

k=O

Solution

(a) We use the property (ii) boosting up filter. The conjugate pair of

poles ielf a.nd ie-if lie on the circle lzl = ~· Then we calculate

There are no zeros, so the transfer function is H(z) = 1+a, z b1\ 112 z

and we see that bo = 1, a1 = -iv'2 and a2 = ~· The filter is

y[n] = box[n] - a 1 y [n - l ] -a2y[n - 2)

2 4

= x[n) +

3

J2y[n-1] - gy[n-2).

This is the same filter that was investigated in Example 9.22.

Remark 9. (^16 2)
The transfer function can be written H (z) = z (^2) - &\Ji ~ z + ~ and has a zero of order
two at the origin, and two poles inside the unit circle. The arguments of the
poles ~e±if correspond to frequencies that are boosted up by the filter. The
situation is illustrated in Figure 9.11. •