Bài giảng Digital signal processing - Chapter 5: Z-Transform - Nguyễn Thanh Tuấn
Chapter 5
z-Transform
Nguyen Thanh Tuan,M.Eng.
Department of Telecommunications (113B3)
Ho Chi Minh City University of Technology
Email: [email protected]om
The z-transform is a tool for analysis, design and implementation of
discrete-time signals and LTI systems.
Convolution in time-domain multiplication in the z-domain
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Content
1. z-transform
2. Properties of the z-transform
3. Causality and Stability
4. Inverse z-transform
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1. The z-transform
The z-transform of a discrete-time signal x(n) is defined as the power
series:
X(z) x(n)zn
n
The region of convergence (ROC) of X(z) is the set of all values of
z for which X(z) attains a finite value.
ROC {zC| X (z) x(n)zn }
n
The z-transform of impulse response h(n) is called the transform
function of the filter:
H(z) h(n)zn
n
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Example 1
Determine the z-transform of the following finite-duration signals
a) x1(n)=[1, 2, 5, 7, 0, 1]
b) x2(n)=x1(n-2)
c) x3(n)=x1(n+2)
d) x4(n)=(n)
e) x5(n)=(n-k), k>0
f) x6(n)=(n+k), k>0
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Example 2
Determine the z-transform of the signal
a) x(n)=(0.5)nu(n)
b) x(n)=-(0.5)nu(-n-1)
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z-transform and ROC
It is possible for two different signal x(n) to have the same z-
transform. Such signals can be distinguished in the z-domain by their
region of convergence.
z-transforms:
and their ROCs:
ROC of an anticausal signal
is the interior of a circle.
ROC of a causal signal is the
exterior of a circle.
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Example 3
Determine the z-transform of the signal
x(n) anu(n)bnu(n1)
The ROC of two-sided signal is a ring (annular region).
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2. Properties of the z-transform
Linearity:
z
x1(n)X1(z) with ROC1
if
z
x2 (n)X2 (z) with ROC2
and
then
z
x(n) x1(n) x2 (n)X(z) X1(z) X2 (z) with ROC ROC1 ROC2
Example: Determine the z-transform and ROC of the signals
a) x(n)=[3(2)n-4(3)n]u(n)
b) x(n)=cos(w0 n)u(n)
c) x(n)=sin(w0 n)u(n)
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2. Properties of the z-transform
Time shifting:
z
x(n) X(z)
if
z
x(n D)zD X(z)
then
zD X(z)
The ROC of
is the same as that of X(z) except for z=0 if
D>0 and z= if D<0.
Example: Determine the z-transform of the signal x(n)=2nu(n-1).
Convolution of two sequence:
z
z
x1(n) X1(z)
x2 (n) X2 (z)
if
and
z
then
x(n) x1(n)x2 (n)X(z) X1(z)X2 (z)
the ROC is, at least, the intersection of that for X1(z) and X2(z).
Example: Compute the convolution of x=[1 1 3 0 2 1] and h=[1, -2, 1] ?
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2. Properties of the z-transform
Time reversal:
z
x(n)X(z) ROC:r | z | r
if
1
2
1
1
z
x(n) X (z1) ROC: | z |
then
r2
r
1
Example: Determine the z-transform of the signal x(n)=u(-n).
Scaling in the z-domain:
z
x(n)X(z) ROC:r | z | r
1
2
if
then
z
an x(n)X(a1z) ROC:| a | r | z || a | r
1
2
for any constant a, real or complex
Example: Determine the z-transform of the signal x(n)=ancos(w0n)u(n).
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3. Causality and stability
A causal signal of the form
x(n) A pnu(n) A pnu(n)
1 1
2 2
will have z-transform
A
A2
1
X (z)
ROC| z | max | pi |
1 p1z1 1 p2z1
i
the ROC of causal signals are outside of the circle.
A anticausal signal of the form
x(n) A pnu(n1) A pnu(n1)
1 1
2 2
A
A2
1
X (z)
ROC| z | min | pi |
1 p1z1 1 p2z1
i
the ROC of causal signals are inside of the circle.
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3. Causality and stability
Mixed signals have ROCs that are the annular region between two
circles.
It can be shown that a necessary and sufficient condition for the
stability of a signal x(n) is that its ROC contains the unit circle.
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4. Inverse z-transform
ztransform
x(n) X(z), ROC
inversez-transform
X(z), ROCx(n)
z
x(n) X(z), ROC
In inverting a z-transform, it is convenient to break it into its partial
fraction (PF) expression form, i.e., into a sum of individual pole
terms whose inverse z transforms are known.
1
X (z)
Note that with
we have
1-az-1
anu(n) if ROC| z || a |(causal signals)
anu(n 1) if ROC| z || a |(anticausalsignals)
x(n)
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Partial fraction expression method
In general, the z-transform is of the form
N(z) b0 b z1 bN zN
1
X (z)
D(z) 1 a0 z1 aM zM
The poles are defined as the solutions of D(z)=0. There will be M
poles, say at p1, p2,…,pM . Then, we can write
D(z) (1 p1z1)(1 p2z1)(1 pM z1)
If N < M and all M poles are single poles.
where
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Example 4d
Compute all possible inverse z-transform of
Solution:
- Find the poles: 1-0.25z-2 =0 p1=0.5, p2=-0.5
- We have N=1 and M=2, i.e., N < M. Thus, we can write
where
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Example 5od
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Partial fraction expression method
If N=M
Where
and
for i=1,…,M
If N> M
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Example 6
Compute all possible inverse z-transform of
Solution:
- Find the poles: 1-0.25z-2 =0 p1=0.5, p2=-0.5
- We have N=2 and M=2, i.e., N = M. Thus, we can write
where
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Example 6 (cont.)
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