1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

9.1 • THE Z-TRANSFORM 343

•EXAMPLE 9.6 Find the inverse z-transform Xn = x [n] = 3 -^112 ;:_ 1 ]. Use (a)
series, (b) the table of z.-transforms, and ( c) residues.

Solut ion

(a) Method of series.

Expand X(z} = 2 ; :_ 1 in a series involving powers of~·

Z 1

00

( 1 )n

00

X(z) = - = = ~ -z-^1 = ~ -^1 z-n

z -! 1 - lz-1 L.J 2 L.J zn

2 2 n=O n=O

The sequence of coefficients in the Laurent series is what we desire,

and we see that

1

Xn = x[nj = zn.

(b) Writing X(z) = z_:!, we identify b = 4 and use line 3 in Table 9.1 to

obtain

1

Xn = x[nj = b" = zn.

(c) Writing X(z} = ,.:; , we see that X(z) has a simple pole at zo = !·

Using Corollary 9.1 for finding the inverse z-transform we obtain

Xn = x[n] = Res [X(z)zn-^1 ,zo]

= Res [ z ~! zn-t, ~] = Res [ z ~ 4' ~].

Using the function f(z) = X(z}z"-^1 and value zo = 4 in Corollary

9.2 we get

1 z" 1

Xn = x[nj = lim (z-zo)f(z) = Jim (z - 2)-- 1 = lim Zn= zn.

z-zo z-+4 Z - 2 z-!