1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

388 CHAPTER 9 • z- TRANSFORMS AND APPLICATIONS

lmz
A(8')
2.5

1.5

0.5

Figure 9.11 Amplit ude response A(11) and zero-pole plot for the boosting up filter

y[n] = x[n] + ~ ./2y(n - l] -~y(n - 2].

Solution

(b) Use the additional pole at ~eiO = ~ and calculate

(1-~J2z-^1 + ~z-^2 ) (1-~z-1)

= 1 - ~ ( 3 + 4J2) z-^1 + ~ ( 4 + 3J2) z-^2 - ~z-3.

The transfer function is H(z) = 1+a, z t+a~z "+a 3 z s where bo = 1,

a1 = -~(3 + 4J2), a2 = b(4 + 3J2), and aa = -~. The desired

filter is

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

= x[n) + ~ ( 3 + 4J2) y[n - 1) - ~ ( 4 + 3J2) y[n - 2) + ~y[n - 3).

R e ma r k 9. 17

The transfer function can be written H(z) = z•-H3+ 4 v'2)z~:H4+av'2)•-a and

has a zero of order three at the orig!,n, and three poles inside the unit circle. The

arguments of the poles ~e'*^0 , ~e::l:'f correspond to frequencies that a.re boosted

up by the filter. The situation is illustrated in Figure 9.12. •