3 50 CHAPTER 9 • Z-TRANSFORMS AND APPLICATIONS
Solution(a) Take the z-transform of both sides
z
z(Y(z) - 2) - 2Y(z) = -
3.
z -Solve for Y(z) and get Y(z) = (z:-;>(;~ 3 ). Then expand and obtain2 3
Y(z) = 2 + --
2+ --
3.
z- z-Find the inverse z-transform of each term:y[n] = 3- 1[2] + 3- t[z: 21+3- 1[z: 3l
= 2o(n] + 2no(n - l ] + 3"o[n - l].
When n = 0 we get y[O] = 2 + 0 + 0 = 2, and when n :?: 1 the
expression for y (n ] simplifies to be(b) Start with the formula Y(z) = (z:z;)(;:_ 3 ) in part (a). Then use Corol-
laries 9.1 and 9.2 and residues to find 3-^1 [Y(z)].
y (n ] = 3-^1 (Y(z)] = Res(Y(z)zn- i, 2] + Res[Y(z)z"'-l, 3]
2z^2 - 5z n-1 2z^2 - 5z n-1
= Res[(z- 2)(z-3)z , 2] +Res((z-2)(z-3)z , 3]I. ( ) 2z2
-5z n-1 1· ( 3) 2z2
-5z n - 1
= z-2 im z - 2 ( Z - 2 )( Z - 3 )z + zim -3 z - ( Z - 2 )( Z - 3 )z1. 2z^2 - 5z n - l
1
. 2z^2 - 5z n-1
= lm Z + lffi Z
z - 2 Z - 3 z - 3 Z - 2
= - 2 2 ,,- 1 + ~ 3 n-l
-1 1
= 2" + 3"
- EXAMPLE 9. 10 Solve the difference equation y[n+ 1J-2y[n] = n with initial
condition y[O] = 1.
(a) Use the z-transform and Tables 9.1 and 9.2 to find the solution.(b) Use residues to find the solution.