Transcript Z-transform
1
Lecture 7: Z-transform
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
[email protected]
Office Hours:
Room 2030
Class web site:
http://ee.lamar.edu/gleb/ds
p/index.htm
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Definitions
Z-transform converts a discrete-time signal into a complex
frequency-domain representation. It is similar to the Laplace
transform for continuous signals.
X ( z ) xn z
n
n is an integer time index;
When the magnitude r =1,
n
(7.2.1)
If (where) it exists!
z re j
is a complex number; - angular freq.
z e j
X ( z) X e j xne jn
n
If it exists!
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(7.2.2)
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Region of Convergence (ROC)
The Region of convergence (ROC) is the set of points z in the complex plane,
for which the summation is bounded (converges):
n
x
z
n
(7.3.1)
n
Since z is complex:
z re j
(7.3.2)
X ( z) xn r n e jn xn r n e jn
n
In general, z-transform exists for
Im
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r+
(7.3.3)
n
r r r
(7.3.4)
r z r
(7.3.5)
Re
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4
Region of Convergence (ROC)
Examples of ROCs
from Mitra
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Region of Convergence (ROC)
X ( z ) xn z n
n
xn ...,a 2 ,a 1 ,,a,a 2 ,...
Example 7.1: Let xn = an
There are no values of z satisfying:
n
xn z n
Example 7.2: Let xn = an un – a causal sequence
X ( z ) a un z
n
n
n
az 1
n
n 0
1
1 az 1
(7.5.1)
Im
for az 1 1 z a
ROC
a
We can modify (7.5.1) as
X ( z)
N ( z )0z 0zero( s)
D( z )0z a pole( s)
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z
N ( z)
z a D( z )
Re
(7.5.2)
roots of numerator: X(z) = 0
roots of denominator: X(z)
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x
6
Region of Convergence (ROC)
X ( z ) xn z n
n
Example 7.3: Let xn = -an u-n-1 – an anticausal sequence
X ( z ) a u n 1 z
n
n
n
az
m 1
1 m
1
n 1 0n 1 a n z 1
n
n
a z
1
m 1
m
a 1 z
z
1 a 1 z z a
(7.6.1)
Im
for a 1 z 1 z a
x
a
Conclusion 1: z-transform exists only within the ROC!
Conclusion 2: z-transform and ROC uniquely specify the signal.
Conclusion 3: poles cannot exist in the ROC; only on its boundary.
Note: if the ROC contains the unit circle (|z| = 1), the system is stable.
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Re
7
The transfer function
X ( z ) xn z n
n
Time shift:
xnm xnm z ( nm) z m {l n m} z m xl z l z m X ( z )
z
n
l
i
LTI:
a y
Consider an LCCDE:
and take z-transform
utilizing time shift
i
n i
b j xn- j
(7.7.2)
j
i
j
a
z
Y
(
z
)
b
z
i
j X ( z)
i
n
(7.7.3)
j
yn hn xn Y ( z ) hm xn m z n hm xn m z ( nm ) z m H ( z ) X ( z )
z
(7.7.1)
m
m
(7.7.4)
n
X (z)
M
The system transfer function
H ( z)
Y ( z)
X ( z)
b z
j 0
N
i
a
z
i
i 0
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j
j
hn z n
n
(7.7.5)
8
Rational z-transform
X ( z ) xn z n
n
Frequently, a z-transform can be described as a rational function, i.e. a
ratio of two polynomials in z-1:
P( z ) n0 n1 z 1 ... nM 1 z ( M 1) nM z M
(7.8.1)
H ( z)
1
( N 1)
N
D( z ) d0 d1 z ... d N 1 z
dN z
Here M and N are the degrees of the numerator’s and denominator’s
polynomials. An alternative representation is a ratio of two polynomials in z:
M
( M 1)
... nM 1 z nM
P( z )
N M n0 z n1 z
H ( z)
z
D( z )
d0 z N d1 z N 1 ... d N 1 z d N
(7.8.2)
Finally, a rational z-transform can be written in a factorized form:
M
H ( z)
n0 (1 z j z 1 )
j 1
N
d0 (1 pi z 1 )
M
z ( N M )
i 1
zeros: numerator = 0
n0 ( z z j )
j 1
N
d 0 ( z pi )
(7.8.3)
i 1
Poles: denominator = 0
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Notes on poles of a system function
Positions of poles of a transfer function are used to evaluate system
stability. Let assume a single real pole at z = . Therefore:
Y ( z)
1
H ( z)
zY z Y z X z
X ( z) z
(7.9.1)
The difference equation is:
yn1 yn xn yn yn1 xn1
(7.9.2)
Therefore, the impulse response is:
for
hn hn1 n1
(7.9.3)
n 0,1, 2,3,...hn 0,1, , 2 ,...
(7.9.4)
Iff || < 1, hn decays as n and the system is BIBO stable; otherwise, hn
grows without limits. Therefore, poles of a stable system (and signals
in fact) must be inside the unit circle.
Zeros may be placed anywhere. Zeros at the origin produce a time delay.
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The transfer function and the
Frequency response
BIBO:
j
h
z
1
RO
C
H
e
H ( z) ze j
n
10
X ( z ) xn z n
n
(7.10.1)
n
M
H ( z)
b z
j 0
N
b0 z
j
a z
i 0
j
i
i
M
M
(z z )
j 1
N
j
a0 z N ( z pi )
(7.10.2)
i 1
where zj are zeros and pi are poles of the transfer function.
M
e j z j
M
H e j b0 1 jN1
b0 e j ( M N ) (e j z j )
a0
j 1
e j pi
(7.10.3)
H e j
N
i 1
a0 (e j pi )
M
N
i 1
j
j
j
H e b0 a0 ( M N ) (e z j ) (e pi )
j 1
i 1
BIBO:
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The transfer function and the
Frequency response
11
X ( z ) xn z n
n
A good way to evaluate the system’s frequency response:
H e j
k
2
k
p
M
P DFT bm 0
N
P DFT ak 0
(7.11.1)
Zero-padded
When the frequency approaches a pole, the frequency response has a local
maximum, a zero forces the response to a local minimum.
For real systems, poles and zeros are symmetrical with respect to the real axis.
Magnitude of H()
1.5
1
0.5
0
0
0.1
pole
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0.2
0.3
Fractional frequency,
0.4
0.5
zero
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The transfer function and the SFG
zS ( z ) AS ( z ) bX ( z )
Sn 1 ASn bxn
T
T
z
y
c
S
dx
Y
(
z
)
c
S ( z ) dX ( z )
n
n
n
(7.12.1)
zI A S ( z ) bX ( z )S ( z ) zI A bX ( z )
1
Y ( z ) c T zI A bX ( z ) dX ( z )
(7.12.2)
H ( z ) c T zI A b d
(7.12.3)
1
1
Poles of H(z) correspond to the eigenvalues of the
system matrix.
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More on Transfer function
M
H ( z)
P( z )
D( z )
b
m0
N
m
z
m
k
a
z
k
k 0
b0 z
M
M
(z z )
j 1
N
j
a0 z N ( z pi )
zeros
M
b0 ( M N )
z
a0
i 1
(z z
m 1
N
m
)
(7.13.1)
(z p )
k 1
k
poles
1) N > M: zeros at z = 0 of multiplicity N-M
2) M > N: poles at z = 0 of multiplicity M-N
j ( M N )
e
1constmagnitude
j ( M N )
e
distortionless
j ( M N )
(M N )lin. phase
e
H e j
BIBO z | 1 ROCH ( z ) j H e j
j
z e
H
e
H ( zi ) 0Y ( zi ) 0;H ( pi ) Y ( pi )
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(7.13.2)
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Types of digital filters
1. FIR (“all-zero”) filter:
bn n M
H ( z ) bm z hn
m 0
0otherwise
M
m
All poles are at z = 0: a “nest of poles”
ROC: the entire z-plane except of the origin (z = 0).
FIR filters are stable.
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(7.14.1)
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Types of digital filters
2. IIR (“all-pole”) filter:
H ( z)
b0
N
a z
k 0
k
All zeros are at z = 0: a “nest of zeros”
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k
(7.15.1)
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Types of digital filters
3. General IIR (“zero-pole”) filter:
M
H ( z)
b
m0
N
k
a
z
k
k 0
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m
z
m
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(7.16.1)
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On test signals…
Y ( z ) H ( z ) X ( z ) yn hn xn
(7.17.1)
z
n hn
(7.17.2)
xn xk nk yn
(7.17.3)
k
Calculate
yn xk hnk
and compare to
yn
k
We don’t need any other that a delta function test signals since a unit-pulse
response is a complete system’s description.
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Types of sequences and convergence
1. Two-sided:
G( z )
N
n M
gn z n
Converges everywhere except of z = 0 and z =
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(7.18.1)
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Types of sequences and convergence
2. Right-sided:
G( z )
Blows up at z =
n M
gn z n
1
n M
gn z n gn z n
(7.19.1)
n 0
Assume: if converges at z = z0, converges for |z| > | z0|
ROC: r - < |z| < - exterior ROC
For a causal sequence: |z| > r - = max|pk| - a max pole of G(z)
To be causal, a sequence must be right-sided (necessary but not sufficient)
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Types of sequences and convergence
3. Left-sided:
G( z )
N
n
gn z n
1
n
N
gn z n gn z n
Converges at z0
n 0
Blows up at z = 0
ROC: 0 < |z| < r + - interior ROC
When encountering an interior ROC, we need to check convergence at
z = 0. If the sequence “blows up” at zero – it’s an anti-causal sequence
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(7.20.1)
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Properties
from Mitra
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Common pairs
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Inverse z-transform
xn
1
2 j
X ( z ) z n1dz
(7.23.1)
c
Where C is a counterclockwise closed path encircling the origin and is entirely in
the ROC. Contour C must encircle all the poles of X(z).
In general, there is no simple way to compute (7.23.1)
A special case:
C is the unit circle (can be used when the ROC includes the unit circle).
The inverse z-transform reduces to the IDTFT.
1
xn
2
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X e j e jn d
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(7.23.2)
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Inverse z-transform
X ( z ) xn z n
n
A. Via Cauchy residue theorem
xn i
(7.24.1)
i
For all poles of X(z)zn-1 inside C (contour of integration)
Where i are the residues of X(z)zn-1
for a pole of multiplicity k:
Residue function:
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1 d k 1i ( z )
i
(k 1)! dz k 1
i ( z ) X ( z ) z
Fall 2008
n 1
(7.24.2)
z pi
z pi
k
(7.24.3)
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Inverse z-transform: Example
X ( z ) xn z n
n
X ( z)
Example:
xn
1
2 j
C
z
; a z
za
z n1
1
z dz
za
2 j
C
Im
a n 0
zn
dz
z a
a 0 n 0
0 is a residue of X(z)z-n-1 at z=0 – involves pole of
x
a
C
Re
multiplicity –n wnen n < 0.
1
d k 1
zn
k
n
a
(
z
a
)
z
(k 1)! dz k 1
z a k 1
z a
an
multiplicity
1 d k 1 k ( z a)1
1
k
k
n
Let n k(n 0)0
z
(
1)(
2)...(
(
k
1))(
z
a
)
a
a
z 0
(k 1)! dz k 1
z k
(k 1)!
n
a
n 0
n
xn n
x
a
un
n
n
a a 0n 0
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Inverse z-transform
X ( z ) xn z n
n
B. Via recognition (table look-up)
Sometimes, the z-transform can be modified such way that it can be
found in a table…
Example:
X ( z) e
n
an n a
a 1
X ( z) e
z ; 1 z a
z
n 0 z n !
n 0 n !
a
z
Therefore:
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a
z
an
xn un
n!
Fall 2008
27
Inverse z-transform
X ( z ) xn z n
n
C. Via long division
1. Right-sided z-transform sequences can be expanded into a power series in
z-1. The coefficient multiplying z-n is the nth sample of the inverse z-transform.
Example:
z 2 2 z 1 2
X ( z)
; z 1
z 1 1
Lower powers first:
and long division:
2 2 z 1 z 2
X ( z)
; z 1
1 z 1
2 z2 z3 z4 ...
1 z 1| 2 2 z 1 z 2
2 2 z 1
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z 2
z 2 z 3
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xn {2,0,1,
1,1,...}
x0 x1 x2 x3 x4
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Inverse z-transform
X ( z ) xn z n
n
2. Left-sided z-transform sequences – into a power series in z1…
Example:
z 2 2 z 1 2
X ( z)
; z 1
z 1 1
Multiply both numerator and denominator by z2 …
z 2 z 2 2 z 1 2 1 2 z 2 z 2
X ( z) 2
z
z 1 1
z z2
Long division…
X ( z)z 1 1 z z 2 z3 z 4 ...
x1
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x0 x-1
x-2
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x-3
x-4
Non-causal
29
Inverse z-transform
X ( z ) xn z n
n
Example:
Example:
z 2 2 z 1 2
1
X ( z ) 1
;
z 1
1
2
z
1
z
2
z
X ( z) z 1
; z 1
z 1
z 1
1
xn n 1 n (1)n u n 1
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not suitable for
long division!
30
Inverse z-transform
X ( z ) xn z n
n
D. Via partial fraction expansion (PFE)
LetG ( z )
N ( z)
;r z r
D( z )
(7.30.1)
If the degree of the numerator is equal or greater than the degree of the
denominator: M N, G(z) is an improper polynomial. Then:
N ( z ) M N l N1 ( z )
G( z )
cl z
D( z ) l 0
D( z )
(7.30.2)
A proper fraction: M1 < N
Then:
G( z )
M N
N
cl z
l 0
l
l 1
l
1 pl z
1
Simple poles: multiplicity of 1.
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(7.30.3)
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Inverse z-transform
X ( z ) xn z n
n
Here l is a residue
l ( z pl )G ( z ) z n l
z pl
(7.31.1)
poles
Therefore:
gn
M N
c
l 0
l
n
p
u
if
external
ROC
:
p
r
l
l n
l n
p
u
if
internal
ROC
:
p
r
l 1
l
l n l
N
n l
This method is suitable for complex poles.
Problem: large polynomials are hard to manipulate…
??QUESTIONS??
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(7.31.2)