512 C H A P T E R 9: The Z-Transform
localization of poles and zeros determines the type of filter. However, in the discrete domain there is
a greater variety of filters than in the analog domain.
9.2 Laplace Transform of Sampled Signals...................................................
The Laplace transform of a sampled signal
x(t)=
∑
n
x(nTs)δ(t−nTs) (9.1)
is given by
X(s)=
∑
n
x(nTs)L[δ(t−nTs)]
=
∑
n
x(nTs)e−nsTs (9.2)
By lettingz=esTs, we can rewrite Equation (9.2) as
Z[x(nTs)]=L[xs(t)]
∣∣
z=esTs
=
∑
n
x(nTs)z−n (9.3)
which is called the Z-transform of the sampled signal.
RemarksThe function X(s)in Equation (9.2) is different from the Laplace transforms we considered
before:
n Letting s=j, X()is periodic of period 2 π/Ts(i.e., X(+ 2 π/Ts)=X()for an integer k). Indeed,
X(+ 2 π/Ts)=
∑
n
x(nTs)e−jn(+^2 π/Ts)Ts=
∑
n
x(nTs)e−jn(Ts+^2 π)=X()
n X(s)may have an infinite number of poles or zeros—complicating the partial fraction expansion when
finding its inverse. Fortunately, the presence of the{e−nsTs}terms suggests that the inverse should be done
using the time-shift property of the Laplace transform instead, giving Equation (9.1).
nExample 9.1
To see the possibility of an infinite number of poles and zeros in the Laplace transform of a sam-
pled signal, consider a pulsex(t)=u(t)−u(t−T 0 )sampled with a sampling periodTs=T 0 /N.
Find the Laplace transform of the sampled signal and determine its poles and zeros.
Solution
The sampled signal is
x(nTs)=
{
1 0≤nTs≤T 0 or 0≤n≤N
0 otherwise