1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.1 • THE Z-T RANSFO.RM 351

Solution

(a) Take the z..transform of both sides

z

z(Y(z ) - 1) - 2Y(z) = (z _ l ) 2.

Solve for Y(z) and get Y(z) = (~~'"'i~;(zt_:~)' Then expand and obtain

1 2 4

Y(z)=l- - -+-.

(z-1)^2 z-1 z-2

Find the inverse z..transform of each term

y(n] =;;-t(l]-3-1[ 1 ]-3-1[_ 2_1+3 -1 [_ 4_ 1

(z - 1)^2 z - 1 z - 2

= o[n) - (n - l)o[n - 11- 28[n - 1) + 4 * 2n-^1 o[n - 11.

When n = 0 we get y [O) = 1 + 0 + 0 + 0 = 1, and when n :2: 1 the

expression for y[nl simplifies to be

y(n] = - 1 - n + 2"+1.

(b) Start with the formula Y (z ) = (~~],~;;::~) in part (a). Then use

Corollaries 9.1 and 9.2 and residues to find 3-^1 [Y(z)].

y(n] = 3-^1 [Y(z)) = Res[Y(z)z"-^1 , 1] + Res[Y(z)zn-^1 ,2 1

z^3 - 2z^2 +2z n-I z^3 - 2z^2 + 2z n - I

= Res[(z- 1)2(z- 2)z , 1) + Res((z- 1)2(z-2)z ,2]

. d 2 z^3 - 2z^2 + 2z n-l). z^3 - 2z^2 + 2z n-l

= !~dz((z-l) (z-1)2(z-2)z +J~(z-^2 )(z-1) 2 (z- 2 )z

Ii d (z

(^3) - 2z^2 + 2z n-l)

1. z
   (^3) - 2z^2 + 2z n - l
   = m- z + 1m z
   z - l dz (z - 2) z-2 (z - 1)2
   I. ((n+l)z
   (^3) - 4(n+ l )z (^2) +(6n+2)z- 4n n-l) 4
   = 1m z + -^2 n-l
   •-1 (z- 2)^2 l

= (n+ 1)- 4{n+ 1) + (6n+ 2)-4nl'•- 1 + i 2 n-l

1 1

= -1-n+ 2n+1

• EXAMPLE 9.11 Given the repeated dosage drug level model y [n + 1] =
  ay[nl + b with the initial condition y[OI =yo:

(a) Use the trial solution method.

(b) Use z-transforms to find the solution.