Transcript Minimum-Phase Systems - Embedded Signal Processing Laboratory

Minimum-Phase Systems
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Content and Figures are from Discrete-Time Signal Processing, 2e by Oppenheim, Shafer, and Buck, ©1999-2000 Prentice Hall
Inc.
Minimum-Phase System
• A system with all poles and zeros inside the unit circle
• Both the system function and the inverse is causal and stable
• Name “minimum-phase” comes from the property of the phase
– Not obvious to see with the given definition
– Will look into it
• Given a magnitude square system function that is minimum
phase
– The original system is uniquely determined
• Minimum-phase and All-pass decomposition
– Any rational system function can be decomposed as
Hz  Hmin zHap z
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing
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Example 1: Minimum-Phase System
• Consider the following system
1  3z 1
H1 z  
1
1  z 1
2
• One pole inside the unit circle:
– Make part of minimum-phase system
• One zero outside the unit circle:
– Add an all-pass system to reflect this zero inside the unit circle
1 1
1 z
1
3
 
1 1
3
1 z
3
1  3z 1
1
1
 1 1 
 1
H1 z 
3
z    3
z
1 1
1 1 
1
3
1 z
1 z
1  z 1 
2
2
2
1 1  1 1 

1

z  z  

3
3   H zH z

H1 z   3
min
ap
 1  1 z1  1  1 z1 



2
3



Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing
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Example 2: Minimum-Phase System
• Consider the following system
3 j / 4 1 
3


z 1  e  j / 4z 1 
1  e
2
2


H2 z   
1
1  z 1
3
• One pole inside the unit circle:
• Complex conjugate zero pair outside the unit circle
3 j / 4 1 
3  j / 4 1 

1

e
z
1

e
z 


2
2


H2 z   
1 1
1 z
3
3 j / 4 3  j / 4  2  j / 4
1  2
j / 4
1 
e
e
e

z
e

z



2
2
3
3




1
1  z 1
3
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing
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Example 2 Cont’d
9  2  j / 4
 2

 z 1  e j / 4  z 1 
 e
4 3
 3

H2 z  
1
1  z 1
3

1 


1 

9
2  j / 4 1 
2

z 1  e j / 4z 1 
1  e
4
3
3


H2 z  
1
1  z 1
3
2  j / 4 1 
e
z 1 
3

2  j / 4 1 
e
z 1 
3

2 j / 4 1 
e z 
3

2 j / 4 1 
e z 
3

 2  j / 4
 2

 z 1  e j / 4  z 1 
 e
3
 3

 
2  j / 4 1 
2 j / 4 1 

1

e
z
1

e z 


3
3



H2 z  Hmin zHap z
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing
5
Frequency-Response Compensation
• In some applications a signal is distorted by an LTI system
• Could filter with inverse filter to recover input signal
– Would work only with minimum-phase systems
• Make use of minimum-phase all-pass decomposition
– Invert minimum phase part
• Assume a distorting system Hd(z)
• Decompose it into
Hd z  Hd,minzHd,ap z
• Define compensating system as
Hc z 
1
Hd,min z
• Cascade of the distorting system and compensating system
Gz  Hc zHd z  Hd,min zHd,ap z
Copyright (C) 2005 Güner Arslan
1
Hd,min z
351M Digital Signal Processing
 Hd,ap z
6
Properties of Minimum-Phase Systems
• Minimum Phase-Lag Property
– Continuous phase of a non-minimum-phase system
  
  
  
argHd ej  argHmin ej  argHap ej
– All-pass systems have negative phase between 0 and 
– So any non-minimum phase system will have a more negative
phase compared to the minimum-phase system
– The negative of the phase is called the phase-lag function
– The name minimum-phase comes from minimum phase-lag
• Minimum Group-Delay Property
  
  
  
grdHd ej  grdHmin ej  grdHap ej
– Group-delay of all-pass systems is positive
– Any non-minimum-phase system will always have greater group
delay
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing
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Properties of Minimum-Phase System
• Minimum Energy-Delay Property
n
 hk
k 0
2

n
 h k
k 0
2
min
– Minimum-phase system concentrates energy in the early part
• Consider a minimum-phase system Hmin(z)
• Any H(z) that has the same magnitude response as Hmin(z)
– has the same poles as Hmin(z)
– any number of zeros of Hmin(z) are flipped outside the unit-circle
• Decompose one of the zeros of Hmin(z)

Hmin z  Qz 1  zkz1

• Write H(z) that has the same magnitude response as

Hz  Qz z1  zk
• We can write these in time domain
hminn  qn  zkqn  1
Copyright (C) 2005 Güner Arslan

hn  qn  1  zk qn
351M Digital Signal Processing
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Derivation Cont’d
• Evaluate each sum
n
 hk
2

k 0
n
k 0
 h k
2
min
k 0
 qk  1
n

 qk
n
k 0
2
2
 z q k  1qk   zk qk  1q k   zk qk 

k
 hk
k 0
2
n
2

2

 z q k  1qk   zk qk  1q k   zk qk  1

k
• And the difference is
n



 
n
  hmink   zk  1  qk   qk  1
2
k 0
2

2
2
k 0
2
  z
k
2

2

 1 qn
2
• Since |qk|<1
n
 hk 
k 0
Copyright (C) 2005 Güner Arslan
2
n
 hmink
2
k 0
351M Digital Signal Processing
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