z-Transform Chapter 6 Chapter 6 D I Dr. Iyad Jafar d J f - - PowerPoint PPT Presentation
z-Transform Chapter 6 Chapter 6 D I Dr. Iyad Jafar d J f Outline Outline Definition Relation Between z Transform and DTFT Relation Between z-Transform and DTFT Region of Convergence g g Common z-Transform Pairs The
Chapter 6 Chapter 6 D I d J f
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The DTFT does not exist for all sequences! Can we still
analyze such sequences in the frequency domain? Addi i ll h DTFT i l f i f hi h Additionally, the DTFT is a complex function of ω, which makes it hard to manipulate. O d h h f !
One way around these issues is to use the z-transform! The z-transform
A generalization of the DTFT and it may exist for
sequences for which the DTFT does not exist. F l h f i l f ti f h
For a real sequence, the z-transform is a real function of the
complex variable z.This makes provides easier manipulation.
For a sequence g[n] the z transform is defined by For a sequence g[n], the z-transform is defined by
where z = Re(z) + j Im(z)
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n
The z-transform is defined in terms of the complex
j
z = Re(z) j Im(z) = r e
So, the z-transform can written as
( ) j ( )
j
G( ) [ ]
Z=rejω Im(z)
j
r =1 ω
j z re n
G(z) g[n] re
Re(z)
1
n j n n
g[n] r e
j
for r = 1,
j
j n j
G(z) g[n] e G(e )
So, the DTFT is the z-transform evaluated at the unit
j
z re n
G(z) g[n] e G(e )
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Thus, G(ejω) can be computed from the clockwise or
Im(z)
j ω
Re(z) ( ) Z=rejω
j r =1
Re(z)
1
j
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6 4 (z)| 2 |G(
DTFT
2 2
DTFT
2 2
I ( )
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Re(z) Im(z)
The z-transform exists if the summation
G(z) g[n] z
n
G(z) g[n] z
In other words, the summation should be absolutely
j n n
g[n] z < g[n] re <
⇒
j
j
n n n
g[n] re g[n] r e = g[n] r <
This implies that the z-transform may exist even if g[n] is
n n n
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The set of values of r for which the z-transform exists is
called the region of convergence (ROC)
In general, the ROC for a sequence g[n] is represented as
g g
R z < R
g g g g
where 0 R < R
in the complex plane (inside a circle/outside a circle / rings)
|z| > α |z| < α α < |z| < β α
α
α
β
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|z| > α α
ROC for causal sequences is the region outside |z| = α
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α
ROC for anti causal sequences is the region inside |z| = α ROC for anti-causal sequences is the region inside |z| = α To completely define the z-transform, we must define the ROC
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Most LTI systems have rational z-transforms of the form
where P(z) and D(z) are two polynomials in z-1 of degrees M where P(z) and D(z) are two polynomials in z of degrees M and N, respectively.
The rational z-transform can be written as
The rational z transform can be written as
M M 1 k k (N M) k 1 k 1 N N
p (1 z ) p (z ) G(z) z
N N 1 k k k 1 k 1
G(z) z d (1 z ) d (z )
ζl, G(z) 0 {ζl} are called the zeros of G(z)
For z = λl, G(z) = ∞ {λ l} are called the poles of G(z) If N > M dditi
l (N M) fi it t = 0
If N > M additional (N – M) finite zeros at z = 0 If N < M additional (M – N) finite poles t z = 0
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Example 6.3:
2 2
z z H(z)
2
( ) z 3z 2
0 and z 1 (circles in the z plane o ) H(z) has poles at z = -1 and z = -2 (crosses in the z-plane‘x’) H(z) has no additional zeros or poles at z = ∞ since M = N H(z) has no additional zeros or poles at z = ∞ since M = N
Im(z) R ( )
1
l
Re(z)
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z-plane
The z-transform is not defined at the poles. Thus, the ROC for any z-transform should not include
The rational z-transform can be completely specified by The rational z transform can be completely specified by
1
1 H(z) 1 z z H(z) z 1 ⇒
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1 1 X(z)
1 1
X(z) 1 0.5z 1 0.3z
ROC: ( Z > 0.5) ( Z > 0.3)
ROC: Z > 0.5 ⇒
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2
2 3 2
z 2z 3 X(z) z 3z z 5
All possible regions of convergence are
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The general form for computing the inverse z-transform
n 1
1 g[n] G(z)z dz 2 j
2 j ∫
Alternatively, we consider two approaches to compute
Power series in z (Long division)
Manipulate G(z) into recognizable pieces (partial
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Partial fraction expansion rearrange G(z) as a sum of
P( ) P(z) G(z) D(z)
1 1 1 2 3
P(z) (1 z )(1 z )(1 z )...
3 1 2 1 1 1 1 2 3
... (1 z ) (1 z ) (1 z )
N
N k 1 k k 1
P(z) G(z) D(z) 1 z
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k
1 k k z
(1 z )G(z)
z(z 2) H(z) (z 0.2)(z 0.6)
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L k m m 1 1
P(z) G(z) D(z) 1 z 1
k k 1 m 1
D(z) 1 z 1 z
L m
1 d
m
L m 1 L m L m 1 L m z
1 d (1 z ) G(z) (L m)!( ) d(z )
⎤
⎦
1 m L
2
z G(z) (z 0.5)(z 1)
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1 11
3 2 1 2 1
1 11 z z 3z 1 3 6 G(z) 1 5 z z 1
z 1 6 6
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For a LTI system, the input-output relation is given by
Computing the z-transform of both sides
k
y[n] x[n k] h[k]
Computing the z transform of both sides
n
Y(z) x[n k] h[k] z ⎛ ⎞
⎟ ⎜ ⎟ ⎝ ⎠
Interchanging the order of summation
n k
⎜ ⎟ ⎝ ⎠
⎛ ⎞
n k n
Y(z) h[k] x[n k] z
⎞
⎟ ⎜ ⎟ ⎝ ⎠
k k
h[k] X(z) z ∑
k k
Y(z) X(z) h[k] z ⇒
X(z) H(z)
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Consider a
2
N2
1
N n N
H(z) h[n] z ∑
If h[n] is causal, 0 ≤ N1 < N2 , then all the poles of H(z)
If h[n] is anticausal, N1 < N2 < 0, then all the poles of
1 2
If h[n] is two-sided, then the poles of H(z) will be at z =
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M 1
1
k 0
1 y[n] x[n k] M
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Consider an IIR LTI system described by the difference
N M
d [ k] [ k]
The transfer function H(z) is
k k k 0 k 0
d y[n k] p x[n k]
M
N M k k k k
d Y(z)z p X(z)z
M k k k 0 N
p z Y(z) H(z)
k k k 0 k 0
( ) p ( )
N k k k 0
( ) X(z) d z
In case h[n] is causal IIR, the ROC is the exterior of the
In case h[n] is an anti-causal IIR, the ROC is the interior
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2
z 0.2z 0.15 X( )
2
X(z) z 0.4z 0.36z 0.144
We showed earlier that the DTFT is basically the z-
j
j
H(e ) H(z)
For a rational transfer function, the frequency response
j
z e
H(e ) H(z)
M M k j k 1
p z (z ) H( )
k 1 N N k
H(e ) d z (z )
k 1 z e
j k
(e )
j (N M) k 1 N j k
( ) p e d (e )
k k 1
(e )
Accordingly, the magnitude response is
M j
e
k j k 1 N j
e p H(e ) d
j k k 1
e
N j j k k
p ( ) ( ) + (N-M) + (e )- (e ) d
k k k 1 k 1
( ) ( ) ( ) ( ) ( ) d
Effectively, the computation of H(ejω) can be done by
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z 1
X(z) z(z j)(z j)
A LTI is said to be BIBO stable if its impulse response is
absolutely summable, i.e.
h[n]
h
S
Can we determine the stability of a LTI system based on its
transfer function?
n
transfer function?
n n n
H(z) h[n]z h[n]z h[n] z
n n n
( ) [ ] [ ] [ ]
j
j z e n
H(z) H(e ) h[n]
This implies that a BIBO stable LTI system has |H(ejω)|<∞ .
In other words, the DTFT should exist. So, the ROC of A BIBO stable LTI system should include the unit circle.
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Inverse System
x[n] y[n]
2
1
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1 2
1 H (z) H (z)
0.2 z 0.6 H( ) 0 5
, z 0.5 (z 0.3)(z 0.5)
4 z 5 H(z) , z 0.5 (z 0.5)(z 0.3)
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Deconvolution Given that the relation between the input and output of a LTI p p system is convolution, y[n] = x[n] * h[n], can we determine x[n] if we know h[n] and y[n]? E l 6 22
Y(z) Y(z) = X(z) H(z) X(z) = H(z) ⇒
Example 6.22: y[n] = {6, 10, 3, -2, 5, -6} h[n] = {2, 4, 1, -3}
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Recall the rational z-transform The frequency response (magnitude and phase) at specific
f b d i ll b d h l i f frequency can be computed geometrically based on the location of the poles and zeros of the z-transform. Specifically the magnitude response is
M M
response is
M M j k j k 1 k 1 N N j
e Dis tan ce to zeros p H(e ) K d
j k k 1 k 1
d e Dis tan ce to poles
g p zeros
Given some filter type and specifications, can we design
yp p g the filter by proper placement of the poles and zeros?
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Example 6.23: Design a single-pole lowpass filter with unity magnitude at ω = 0 and 0.5 at ω = π/2 1) place the pole at ω = 0 2) place on zero at ω = π 3) the transfer function will be
r
z 1 H(z) K
H(z) K z r
j0 j0 z e
e 1 H(z) K 1 2K = 1 r e r
j /2
j /2
e 1 H(z) K 0 5
K = 3/8 and r = 1/4
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j /2
j /2 z e
H(z) K 0.5 e r
Example 6.23: continued
1
3 z 1 3 1 z H( )
1
1
H(z) 8 z 0.25 8 1 0.25z
1
3 Y( ) 1 0 25 X( ) 1
0.4 0.6 H(ej)
1
3 Y(z) 1 0.25z X(z) 1 z 8
y[n] 0 25y[n 1] x[n] x[n 1] ⇒
0.4
0.25y[n 1] x[n] x[n 1] 8 ⇒
2 4
x[n] y[n]
50 5 10 15 20
n
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