Chapter 5 introduce z-Transform. In this chapter, you learned to: The z-transform, properties of the z-transform, causality and stability, inverse z-transform.
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Nội dung Text: Lecture Digital signal processing: Chapter 5 - Nguyen Thanh Tuan
- Chapter 5
z-Transform
Nguyen Thanh Tuan, Click
M.Eng.
to edit Master subtitle style
Department of Telecommunications (113B3)
Ho Chi Minh City University of Technology
Email: nttbk97@yahoo.com
- 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
Digital Signal Processing 2 z-Transform
- Content
1. z-transform
2. Properties of the z-transform
3. Causality and Stability
4. Inverse z-transform
Digital Signal Processing 3 z-Transform
- 1. The z-transform
The z-transform of a discrete-time signal x(n) is defined as the power
series:
X ( z)
n
x (n ) z n x( 2) z 2 x ( 1) z x (0) x(1) z 1 x(2) z 2
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 {z C | X ( z ) x ( n) z
n
n
}
The z-transform of impulse response h(n) is called the transform
function of the filter:
H ( z) h (
n
n ) z n
Digital Signal Processing 4 z-Transform
- 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
Digital Signal Processing 5 z-Transform
- Example 2
Determine the z-transform of the signal
a) x(n)=(0.5)nu(n)
b) x(n)=-(0.5)nu(-n-1)
Digital Signal Processing 6 z-Transform
- 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 a causal signal is the ROC of an anticausal signal
exterior of a circle. is the interior of a circle.
Digital Signal Processing 7 z-Transform
- Example 3
Determine the z-transform of the signal
x(n) a nu(n) b nu(n 1)
The ROC of two-sided signal is a ring (annular region).
Digital Signal Processing 8 z-Transform
- 2. Properties of the z-transform
Linearity:
if x1 (n)
z
X 1 ( z ) with ROC1
and x2 (n)
z
X 2 ( z ) with ROC2
then
x(n) x1 (n) x2 (n)
z
X ( z) X1 ( z) X 2 ( 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)
Digital Signal Processing 9 z-Transform
- 2. Properties of the z-transform
Time shifting:
if x(n)
z
X ( z)
then x(n D)
z
z D X ( z)
The ROC of z D X (z ) is the same as that of X(z) except for z=0 if
D>0 and z= if D
- 2. Properties of the z-transform
Time reversal:
if x(n)
z
X ( z ) ROC : r1 | z | r2
1 1
then x ( n)
z
X ( z 1 ) ROC : | z |
r2 r1
Example: Determine the z-transform of the signal x(n)=u(-n).
Scaling in the z-domain:
if x(n)
z
X ( z ) ROC : r1 | z | r2
then a n x(n)
z
X (a 1 z ) ROC : | a | r1 | z || a | r2
for any constant a, real or complex
Example: Determine the z-transform of the signal x(n)=ancos(w0n)u(n).
Digital Signal Processing 11 z-Transform
- 3. Causality and stability
A causal signal of the form
x(n) A1 p1nu(n) A2 p2nu(n)
will have z-transform
A1 A2
X ( z) 1
1
ROC| z | max | pi |
1 p1 z 1 p2 z i
the ROC of causal signals are outside of the circle.
A anticausal signal of the form
x(n) A1 p1nu(n 1) A2 p2nu(n 1)
A1 A2
X ( z) 1
1
ROC| z | min | pi |
1 p1 z 1 p2 z i
the ROC of causal signals are inside of the circle.
Digital Signal Processing 12 z-Transform
- 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.
Digital Signal Processing 13 z-Transform
- 4. Inverse z-transform
transform
x(n) z X ( z ), ROC
X ( z), ROC inverse
z- x(n)
transform
x(n)
z
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
Note that with X ( z ) -1 we have
1 - az
a nu (n) if ROC | z || a | (causal signals)
x ( n) n
a u (n 1) if ROC | z || a | (anticausal signals)
Digital Signal Processing 14 z-Transform
- Partial fraction expression method
In general, the z-transform is of the form
N ( z ) b0 b1 z 1 bN z N
X ( z)
D( z ) 1 a0 z 1 aM z M
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 p1 z 1 )(1 p2 z 1 )(1 pM z 1 )
If N < M and all M poles are single poles.
where
Digital Signal Processing 15 z-Transform
- 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
Digital Signal Processing 16 z-Transform
- Example 5od
Digital Signal Processing 17 z-Transform
- Partial fraction expression method
If N=M
Where and for i=1,…,M
If N> M
Digital Signal Processing 18 z-Transform
- 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
Digital Signal Processing 19 z-Transform
- Example 6 (cont.)
Digital Signal Processing 20 z-Transform
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