1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

362 CHAPTER 9 • Z·TRANSFORMS AND APPLI CATIONS

Step (ii) Solve equation (9-20) for Y(z).

Step (iii) Use partial fractions to expand Y(z) in a sum of terms, and look up
the inverse z-transform(s) us ing Table 9.1, to get the solution

y[n) = 3-^1 [Y(z)].

Step (iv) Alternate calculation using residues. Perform steps (i) and (ii), then

find y[n] using residues
k
y[n) = 3-^1 [Y(z)) = L Res[Y(z)zn-^1 ,z;J,
i=l
where Z!J z2, ... , Zk are the poles of f(z) = Y(z)zn-^1.

Remark 9.8

The function f ( z) = Y ( z )zn- l has real coefficients. Hence, if z; and z; are poles,

then we can use the computational fact:

Res[f(z),z;] = Res[f(z),z;). (9-2~

We now show how to obtain answers to Examples 9.12- 9.14 using z-transform

methods.

• EXAMPLE 9. 15

(a) Use z-transform methods to solve y[n+ 2] - 4y(n + 1) +3y(n) = 0 with

y[O) =Yo= 1 and y[l] =Yi = 5.

(b) Use z-transform methods to solve y [n + 2] -4y(n + 1) + 3y[n) = zn+^2

with y[O) =Yo = 1 and y(l) =Yi = 3.

Solution

(a) Take t he z-transforms of each term

z^2 {Y(z) - 1 - 5z-^1 ) - 4(z(Y(z) - 1)) + 3(Y(z)) = O.

Solve for Y(z) and get Y(z) = ez-"i»t:- 3 > •

Calculate the residues for f(z) = Y(z)zn- l = {z-zi'i{;_ 3 >zn- I at the

poles

z^2 + z

Res[f(z), 1) =Lim --

3

• zn-I = -l · l " -^1 = -1, and
  z~l z -
  2
  Res[f(z),3) =Lim z + z z"-^1 =6· 3n- l = 2 · 3".
  ,,,_3 z - 1