1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

346 CHAPTER 9 • Z-TRANSFORMS AND APPLICATIONS

Applying the time delay properties to the z..transforms of ea.ch term in {9-4),
we obtain X(z) + aX(z)z-^1 + bX(z)z-^2 = Y(z). Factoring, we get

X{z){l + az-^1 + bz-^2 ) = X(z)H(z), where

H(z) = 1 + az-^1 + bz-^2

(9-6}

(9-7)

represents the filter transfer function. Now, in order for the filter to suppress

the inputs cos(w7rn) and sin(w7rn), the zeros of H (z) must cancel the poles of

the inputs, namely eiw" and e-iww. Therefore, we must have

1 + az-^1 + bz-^2 = (1 -eiw" z-^1 )(1-e-iww z -^1 ) ,

and an easy calculation reveals that

a = -(eiw" + e-iw") = -2cos(w7r)

and

b = eiw1r e-iw1r = 1.

A complete discussion of this process is given in Section 9.3.

• EXAMPLE 9.8 (FIR filter design) Use residues to find the inverse z-transform

x(n] of X ( z) = 1 _ "' 2 z

1

1 + 2 = ( .ur. ) (

1

v~ .t l-e4z-^1 l -e""'"T"z,. -^1 ).

Then write down the FIR filter equation that suppresses x{n].

Solution WritingX(z)=. )( ~) weseethatX(z)hassimplepoles

( z-~'t z-~ 4

at z 1 =elf, and z2 = e=f', respectively. Using Corollaries 9.1and9.2 we obtain

= ,. eif 1• ( e u)" " = (1 - - -i) e 1.!Ut '.

e• -e-• 2 2

The residue at the pole z 2 = e-if is computed similarly.

[

zn+l '"]

Res[X(z)zn-

1

,z2) =Res (. ) ( -ill)'e-•

z -e'f z -e •

e-1f ( _i:l!)n (1 i) _iu

e • = -+- e •

e-1I- eT 2 2