Signals and Systems - Electrical Engineering

518 C H A P T E R 9: The Z-Transform

nExample 9.3

Find the Z-transform of a discrete-time pulse

x[n]=

{

1 0≤n≤ 9

0 otherwise

Determine the region of convergence ofX(z).

Solution

The Z-transform ofx[n] is

X(z)=

∑^9

n= 0

1 z−n=

1 −z−^10

1 −z−^1

=

z^10 − 1

z^9 (z− 1 )

(9.8)

That this sum equals the term on the right can be shown by multiplying the left term by the

denominator 1−z−^1 and verifying the result is the same as the numerator in negative powers of

z. In fact,

( 1 −z−^1 )

∑^9

n= 0

1 z−n=

∑^9

n= 0

1 z−n−

∑^9

n= 0

1 z−n−^1

=( 1 +z−^1 +···+z−^9 )−(z−^1 +···+z−^9 +z−^10 )= 1 −z−^10

Sincex[n] is a finite sequence there is no problem with the convergence of the sum, althoughX(z)

in Equation (9.8) seems to indicate the need forz6=1 (z=1 makes the numerator and denom-

inator zero). From the sum, if we letz=1, thenX( 1 )=10, so there is no need to restrictzto be

different from 1. This is caused by the pole atz=1 being canceled by a zero. Indeed, the zeroszk

ofX(z)(see Eq. 9.8) are the roots ofz^10 − 1 =0, which arezk=ej^2 πk/^10 fork=0,..., 9. Therefore,

the zero whenk=0, orz 0 =1, cancels the pole at 1 so that

X(z)=

∏ 9

k= 1 (z−e

jπk/ (^5) )
z^9
That is,X(z)has nine poles at the origin and nine zeros around the unit circle except atz=1. Thus,
the wholez-plane excluding the origin is the region of convergence ofX(z). n
ROC of Infinite-Support Signals
Signals of infinite support are either causal, anti-causal, or a combination of these or noncausal. Now
for the Z-transform of a causal signalxc[n] (i.e.,xc[n]=0,n<0)
Xc(z)=

∑∞

n= 0

xc[n]z−n=

∑∞

n= 0

xc[n]r−ne−jnω

to converge we need to determine appropriate values ofr, the damping factor. The frequencyω

has no effect on the convergence. IfR 1 is the radius of the farthest-out pole ofXc(z), then there is