Transcript Transform Analysis of LTI systems

Transform Analysis of
LTI systems
主講人：虞台文
Content

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
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The Frequency Response of LTI systems
Systems Characterized by ConstantCoefficient Difference Equations
Frequency Response for Rational System
Functions
Relationship btw Magnitude and Phase
Allpass Systems
Minimum-Phase Systems
Generalized Linear-Phase Systems
Transform Analysis of
LTI systems
Frequency Response
of LTI systems
Time-Invariant System
y(n)=x(n)*h(n)
x(n)
h(n)
X(z)
H(z)
Y(z)=X(z)H(z)
Frequency Response
Y ( z)  H ( z) X ( z)
j
j
j
Y (e )  H (e ) X (e )
j
j
Magnitude
j
| Y (e ) || H (e ) |  | X (e ) |
j
j
Phase
j
 | Y (e ) |  | H (e ) |  | X (e ) |
Ideal Frequency-Selective Filters
Hlp (e j )
Ideal Lowpass
Filter
Computationally
Unrealizable

 c
1 |  | c
H lp (e )  
0 c    
j
c

sin c n
hlp ( n) 
n

Ideal Frequency-Selective Filters
H hp (e j )
Ideal Highpass
Filter
Computationally
Unrealizable

 c
0 |  | c
j
H hp (e )  
1 c    
c


hhp (n)  (n)  hhp (n)
sin c n
 (n) 
n
Ideal Frequency-Selective Filters

Such filters are
–
–
–

Noncausal
Zero phase
Not Computationally realizable
Causal approximation of ideal frequencyselective filters must have nonzero phase
response.
Phase Distortion and Delay --Ideal Delay
j
 jnd
hid (n)  (n  nd ) Hid (e )  e
j
| Hid (e ) | 1
j
Hid (e )  nd
Delay Distortion
Linear Phase
|  | 
Delay Distortion would be considered a rather
mild form of phase distortion.
Phase Distortion and Delay --A Linear Phase Ideal Filter
 jn d

e
j
H lp (e )  
0
sin c (n  nd )
hlp (n) 
,
(n  nd )
|  | c
c    
  n  
Still a noncausal one.
Not computationally realizable.
Phase Distortion and Delay --Group Delay


A convenient measure of the linearity of phase.
Definition:
d
j
()  grd [ H (e )]  
{arg[ H (e )]}
d
j


Linear Phase  ()=constant
The deviation of () away from a constant
indicates the degree of nonlinearity of the
phase.
Transform Analysis of
LTI systems
Systems Characterized
by
Constant-Coefficient
Difference Equations
Nth-Order Difference Equation
N
a
k 0
M
k
y (n  k )   br x(n  r )
r 0
N
M
k 0
r 0
Y ( z ) ak z k  X ( z ) br z r
M
 br z
r
Y ( z ) r 0
H ( z) 
 N
X ( z )  a z k
k
k 0
Representation in Factored Form
Contributes poles at 0 and zeros at cr
M
(1  c z


 b0
H ( z )   
 a0 
r
1
)
r 1
N
1
(
1

d
z
)

r
k 1
Contributes zeros at 0 and poles at dr
Example
Two zeros at z = 1
(1  z 1 ) 2
H ( z) 
3 1
1 1
(1  2 z )(1  4 z )
poles at
z =1/2 and z =  3/4
1  2 z 1  z 2
Y ( z)
H ( z) 

3 2
1 1
1 4 z  8 z
X ( z)
(1  14 z 1  83 z 2 )Y ( z)  (1  2z 1  z 2 ) X ( z)
y(n)  14 y(n 1)  83 y(n  2)  x(n)  2x(n 1)  x(n  2)
Stability and Causality
For a given ration of polynomials,
different choice of ROC will lead to
different impulse response.
 We want to find the proper one to build a
causal and stable system.


How?
Stability and Causality

For Causality:
–

For Stability:
–

ROC of H(z) must be outside the outermost pole
ROC includes the unit circle
For both
–
All poles are inside the unit circle
Stability and Causality
5
y
(
n
)

 Example:
2 y(n  1)  y(n  2)  x(n)
Im
H ( z) 
1
1  52 z 1  z  2
1

(1  12 z 1 )(1  2 z 1 )
Discuss its stability and causality
1
Re
Inverse Systems
X(z)
Y(z)
H(z)
X(z)
Hi(z)
1
H i ( z) 
H ( z)
G(z)= H(z)Hi(z)=1
g(n) = h(n)* hi(n) = (n)
Does every system have an inverse system?
Inverse Systems
Give an example.
X(z)
Y(z)
H(z)
X(z)
Hi(z)
1
H i ( z) 
H ( z)
G(z)= H(z)Hi(z)=1
g(n) = h(n)* hi(n) = (n)
Inverse Systems
M
 b0 
H ( z )   
 a0 
1
(
1

c
z
 r )
r 1
N
1
(
1

d
z
)

r
k 1
N
 a0 
H i ( z )   
 b0 
1
(
1

d
z
 r )
r 1
M
1
(
1

c
z
 r )
k 1
Zeros
Zeros
Poles
Poles
Minimum-Phase Systems

A LTI system is stable and causal and also
has a stable and causal inverse iff both poles
and zeros of H(z) are inside the unit circle.
Such systems are referred to as minimum-phase
systems.
Impulse Response for Rational
System Functions

By partial fraction expansion:
M
H ( z) 
r
b
z
 r
r 0
N
 ak z
k
M N
Ak
H ( z )   Br z  
1
k

1
1

d
z
r 0
k
N
r
k 0
h(n) 
M N
N
r 0
k 1
n
B

(
n

r
)

A
d
 k k u (n)
 r
FIR and IIR
M N
Zero poles
Ak
H ( z )   Br z  
1
k 11  d k z
r 0
r
N
nonzero
poles
h(n) 
M N
 B (n  r )   A d
r 0
N
r
k 1
k
n
k
u (n)
FIR and IIR
M N
Zero poles
Ak
H ( z )   Br z  
1
k 11  d k z
r 0
h(n) 
r
N
FIR:
nonzero
The system contains
poles
only zero poles.
M N
 B (n  r )   A d
r 0
N
r
k 1
k
n
k
u (n)
FIR and IIR
M N
Zero poles
Ak
H ( z )   Br z  
1
k 11  d k z
r 0
r
N
IIR:
The system contains
nonzero poles (not
canceled by zeros).
h(n) 
M N
nonzero
poles
 B (n  r )   A d
r 0
N
r
k 1
k
n
k
u (n)
FIR
M
H ( z )   bk z
k 0
k
y(n)   bk x(n  k )
bn
h(n)   bk (n  k )  
k 0
0
M
M
k 0
0nM
otherwise
One pole is
canceled by
zero here.
Example:FIR
a
h( n)  
0
0nM
otherwise
n
7th-order
pole
M=7
Does this system have
nonzero pole?
M
H ( z)   a z
n 0
n n
1 M 1
1  (az )

1  az1
M 1  M 1
1 a z

1  az1
Example:FIR
a
h( n)  
0
0nM
otherwise
n
7th-order
pole
M=7
Write its system function.
M
y ( n)   a x ( n  k )
k
k 0
y(n)  ay(n 1)  x(n)  a M 1 x(n  M 1)
Example:IIR
y(n)  ay(n  1)  x(n)
1
H ( z) 
1
1  az
h(n)  a u(n)
n
Transform Analysis of
LTI systems
Frequency Response
of For Rational
System Functions
Rational Systems
M
H ( z) 
b z
k 0
N
M
k
k
H ( e j ) 
k
a
z
 k
k 0
 j k
b
e
k
k 0
N
 jk
a
e
 k
k 0
M
 b0 
H (e )   
 a0 
j
 j
(
1

c
e
 k )
k 1
N
 j
(
1

d
e
 k )
k 1
Log Magnitude of H(ej) --Decibels (dBs)
Gain in dB = 20log10|H(ej)|
M
b0
| H (e ) |
a0
j
 j
1

c
e
 k
k 1
N
 j
1

d
e
 k
k 1
N
b0 M
 j
20log10 | H (e ) | 20log10
  20log10 1  ck e  20log10 1  d k e  j
a0 k 1
k 1
j
Scaling
Contributed by zeros
Contributed by poles
Advantages of Representing the
magnitude in dB
j
j
j
j
j
Y (e )  H (e ) X (e )
j
| Y (e ) || H (e ) |  | X (e ) |
j
j
j
20log10 | Y (e ) | 20log10 | H (e ) | 20log10 | X (e ) |
The magnitude of
Output FT
The Magnitude
Of Impulse Response
The magnitude of
Input FT
Phase for Rational Systems
M
N


b
H (e j )   0    (1  ck e  j )  (1  d k e  j )
k 1
 a0  k 1
N
M
d
d
j
 j
grd[ H (e )]  
arg(1  d k e ) 
arg(1  ck e  j )
k 1 d
k 1 d
2
 j
2
 j
M
|
d
|

Re
{
d
e
}
|
c
|

Re
{
c
e
}
j
k
k
k
k
grd[ H (e )]  

2
 j
2
 j
1

|
d
|

2
Re
{
d
e
}
1

|
c
|

2
Re
{
c
e
}
k 1
k 1
k
k
k
k
N
Systems with a Single Zero or Pole
j 1
1  re z
1
j 1
1  re z
r

r

Frequency Response of a Single
Zero or Pole
j 1
1  re z
H (e )  1  re e
1
j 1
1  re z
1
H (e ) 
j  j
1  re e
j
j
j  j
Frequency Response of a Single Zero
j 1
1  re z
j
j
j  j
H (e )  1  re e
j  j
 j
j
| H (e ) |  (1  re e )(1  re e )
2
 1  r  2r cos(  )
2
j
20log10 | H (e ) | 10log10 [1  r  2r cos(  )]
2
|H(ej)|2:
maximum of
is ata =.
FrequencyIts
Response
Single Zero
max |H(ej)|2 =(1+r)2
Its minimum is at =0.
j 2
2
j 1 min |H(e )| j=(1r)
j  j
1  re z
j
H (e )  1  re e
j  j
 j
j
| H (e ) |  (1  re e )(1  re e )
2
 1  r  2r cos(  )
2
j
20log10 | H (e ) | 10log10 [1  r  2r cos(  )]
2
Frequency Response of a Single Zero
j 1
1  re z
j
j  j
H (e )  1  re e
 r sin(  ) 

H (e )  tan 
 1  r cos(  ) 
j
1
2
2
r

r
cos(



)
r
 r cos(  )
j
grd[ H (e )] 

2
1  r  2r cos(  )
| H (e j ) |2
Frequency Response of a Single Zero
10
10
0
0
0
-10
-10
-10
dB
10
Radians
-20
0
2
-20
-2
0
2
-20
2
2
2
0
0
0
-2
Samples
-2
-2
0
2
-2
-2
0
2
-2
5
5
5
0
0
0
-5
-5
-5
-10
-2
0
r = 0.9
=0
2
-10
-2
0
r = 0.9
 = /2
2
-10
-2
0
2
-2
0
2
-2
0
2
r = 0.9
=
於處有最大凹陷(1r)2
Frequency Response of a Single Zero
10
10
0
0
0
-10
-10
-10
dB
10
Radians
-20
0
2
-20
-2
0
2
-20
2
2
2
0
0
0
-2
Samples
-2
-2
0
2
-2
-2
0
2
-2
5
5
5
0
0
0
-5
-5
-5
-10
-2
0
r = 0.9
=0
2
-10
-2
0
r = 0.9
 = /2
2
-10
-2
0
2
-2
0
2
-2
0
2
r = 0.9
=
於||處最高(1+r)2
Frequency Response of a Single Zero
10
10
0
0
0
-10
-10
-10
dB
10
Radians
-20
0
2
-20
-2
0
2
-20
2
2
2
0
0
0
-2
Samples
-2
-2
0
2
-2
-2
0
2
-2
5
5
5
0
0
0
-5
-5
-5
-10
-2
0
r = 0.9
=0
2
-10
-2
0
r = 0.9
 = /2
2
-10
-2
0
2
-2
0
2
-2
0
2
r = 0.9
=
於處phase直轉急上
Frequency Response of a Single Zero
10
10
0
0
0
-10
-10
-10
dB
10
Radians
-20
0
2
-20
-2
0
2
-20
2
2
2
0
0
0
-2
Samples
-2
-2
0
2
-2
-2
0
2
-2
5
5
5
0
0
0
-5
-5
-5
-10
-2
0
r = 0.9
=0
2
-10
-2
0
r = 0.9
 = /2
2
-10
-2
0
2
-2
0
2
-2
0
2
r = 0.9
=
Frequency Response of a Single Zero
10
dB
0
-10
-20
-2
0
2
r=2
r=1.5
r=1.25
r=1/0.9
Radians
4
Zero outside
the unit circle
0
-2
Samples
-10
-20
-2
0
2
-2
0
2
-2
0
2
2
0
-2
-2
0
-4
2
10
10
Note that the group
delay is always
positive when r>1
5
0
0
4
2
-4
10
-2
0
=0
2
5
0
=
Frequency Response of a Single Zero
20
0
0
-20
-20
dB
20
Radians
-40
-2
0
4
4
2
2
0
-2
-4
-2
0
2
50
Samples
-40
2
Some zeros inside
the unit circle
And some outside
-4
-50
2
-2
0
2
-2
0
2
50
-50
0
2
-2
0
-2
0
0
0
-100
-2
-100
Frequency Response of a Single Pole
 The
converse of the single-zero case.
 Why?
 A stable system: r < 1
 Excise:
Use matlab to plot the frequency
responses for various cases.
Frequency Response of
Multiple Zero and Poles
 Using
additive method to compute
Magnitude
– Phase
– Group Delay
–
Example
Multiple Zero and Poles
0.05634(1  z 1 )(1  1.0166z 1  z 2 )
H ( z) 
(1  0.683z 1 )(1  1.4461z 1  0.7957z 2 )
zeros
Radius
1
1
Angle

1.0376 (59.45)
poles
Radius
0.683
0.892
Angle
0
0.6257 (35.85)
Example
Multiple Zero and Poles
dB
0
-50
-100
-3
-2
-1
0
1
2
3
Radians
4
2
0
-2
-4
-3
-2
-1
0
1
2
3
10
Samples
zeros
Angle

1.0376 (59.45)
poles
5
0
Radius
1
1
-3
-2
-1
0
1
2
3
Radius
0.683
0.892
Angle
0
0.6257 (35.85)
Transform Analysis of
LTI systems
Relationship btw
Magnitude and Phase
Magnitude and Phase
Know magnitude
Know Phase?
Know Phase
Know Magnitude?
In general, knowledge about the magnitude provides
no information about the phase, and vice versa.
Except when …
Magnitude
j
j
j
| H (e ) |  H (e ) H * (e )
2
 H ( z)H * (1/ z*) z e j
M
H ( z) 
b0  (1  ck z )
k 1
N
1
a0  (1  d k z 1 )
k 1
*
M


*
b
(1

c
z
*)
b
(1

c
k
0
k z)
 0

  kN1
H *(1/ z*)   kN1
a
 a
*
(1

d
z
*)
(1

d
k
0
k z)
 0

k 1
 k 1

M
Magnitude
C( z)  H ( z) H * ( z)
M
 b0 
  
 a0 
2
 (1  c z
1
 (1  d
1
k
k 1
N
k
)(1  c z )
*
k
z )(1  d z )
*
k
k 1
M
H ( z) 
b0  (1  ck z )
k 1
N
1
a0  (1  d k z 1 )
k 1
*
M


*
b
(1

c
z
*)
b
(1

c
k
0
k z)
 0

  kN1
H *(1/ z*)   kN1
a
 a
*
(1

d
z
*)
(1

d
k
0
k z)
 0

k 1
 k 1

M
Magnitude
M
 b0 
C ( z )   
 a0 
2
1
*
(
1

c
z
)(
1

c
 k
k z)
k 1
N
 (1  d
k 1
Zeros of H(z):
Poles of H(z):
ck
dk
1
k
z )(1  d z )
*
k
Conjugate reciprocal pairs
Zeros of C(z):
Poles of C(z):
*
k
ck and 1/ c
dk and 1/ d
*
k
Magnitude
M
b 
C ( z )   0 
 a0 
2
1
*
(
1

c
z
)(
1

c
 k
k z)
k 1
N
1
*
(
1

d
z
)(
1

d
 k
k z)
k 1
Given C(z), H(z)=?
How many choices if the
numbers of zeros and poles
are fixed?
Allpass Factors
1
z  a*
H ap ( z ) 
1  az1
1/a*
a
Pole at a
Zero at 1/a*
Allpass Factors
1
z  a*
H ap ( z ) 
1  az1
j
| Hap (e ) | 1
 j
j
e

a
*
1

a
*
e
 j
H ap (e j ) 

e
 j
 j
1  ae
1  ae
Allpass Factors
have the same frequency-response
There are infinite many systems to
magnitude?
H1(z)
| H ( z ) || H1 ( z ) |
H1(z)
Hap(z)
H ( z)  H1 ( z) H ap ( z)
Transform Analysis of
LTI systems
Allpass Systems
General Form
1
1
1
z  dk
( z  e )(z  ek )
H ap ( z )  
1 
* 1
1

d
z
(
1

e
z
)(
1

e
k 1
k 1
k
k
kz )
Mr
Mc
Real Poles
1 / ek*
1/ dk
dk
ek
*
k
1
Complex Poles
|Hap(ej)|=1
|Hap(ej)|=?
ek*
1 / ek
grd[Hap(ej)]=?
AllPass Factor
z 1  a *
H af ( z ) 
1
1  az
Consider a=rej
 j
 j
e

re
H af (e j ) 
j  j
1  re e
Always positive
for a stable and
causal system.
 r sin(  ) 
H af (e )    2 tan 

1

r
cos(



)


j
1
2
2
1

r
1

r
grd[ H af (e j )] 

2
1  r  2 cos(  ) | 1  re je  j |2
Example: AllPass Factor
Real poles
0.9
0.9
0
-1
1
dB
dB
1
-2
0
-1
2
0
-2
0
2
-2
0
2
-2
0
2
20
Samples
Samples
0
0
-5
2
20
10
0
-2
5
Radians
Radians
5
-5
0
-2
0
2
10
0
Example: AllPass Factor
Real Poles
0.9
0
-1
-2
0
Radians
Phase is
nonpositive
for 0<<.
0
-2
0
0
2
-2
0
2
-2
0
2
0
-5
2
20
Group delay is
positive
10
-2
0
2
Samples
Samples
-2
5
20
0
0
-1
2
5
-5
1
dB
dB
1
Radians
0.9
10
0
Example: AllPass Factor
Complex Poles
dB
1
/4
0
-1
-2
0
2
Radians
5
Continuous phase
is nonpositive
for 0<<.
0
-5
-2
0
2
-2
0
2
Group delay is
positive
Samples
20
10
0
Example: AllPass Factor
Complex Poles
dB
1
/4
1/2
-2
0
2
-2
0
2
-2
0
2
5
Continuous phase
is nonpositive
for 0<<.
Radians
3/4
-1
0
-5
15
Group delay is
positive
Samples
4/3
0
10
5
0
Transform Analysis of
LTI systems
Minimum-Phase
Systems
Properties of Minimum-Phase Systems




To have a stable and causal inverse systems
Minimum phase delay
Minimum group delay
Minimum energy delay
Rational Systems vs.
Minimum-Phase Systems
H(z)
How?
Hmin(z)
H ( z)  H min ( z) H ap ( z)
Hap(z)
Rational Systems vs.
Minimum-Phase Systems
Hmin(z)
H(z)
Hap(z)
Rational Systems vs.
Minimum-Phase Systems
Hmin(z)
H(z)
Pole/zero
Canceled
Hap(z)
Frequency-Response Compensation
s(n)
Distorting
System
Hd(z)
sd(n)
Compensatiing
System
s(n)
Hc(z)
The inverse system of Hd(z) iff
it is a minimum-phase system.
Frequency-Response Compensation
s(n)
Distorting
System
sd(n)
Hd(z)
s(n)
s(n)
Compensatiing
System
Hc(z)
Distorting
System
Allpass
System
Hdmin(z)
Hap(z)
sd(n)
Compensatiing
System
1
Hdmin(z)
sˆ(n)
Frequency-Response Compensation
G( z)  H d ( z) Hc ( z)
 H dmin ( z ) H ap ( z )
1
H dmin ( z )
Hd(z)
s(n)
 H ap ( z )
Hc(z)
Distorting
System
Allpass
System
Hdmin(z)
Hap(z)
sd(n)
Compensatiing
System
1
Hdmin(z)
sˆ(n)
Example:
Frequency-Response Compensation
H ( z )  (1  0.9e
j 0.6  1
 (1  1.25e
4th order
pole
z )(1  0.9e
j 0.8  1
 j 0.6  1
z )(1  1.25e
z )
 j 0.8  1
z )
Example:
Frequency-Response Compensation
40
dB
20
0
-20
-2
0
2
-2
0
2
-2
0
2
Radians
5
0
-5
4th order
pole
Samples
10
0
-10
Example:
Frequency-Response Compensation
H ( z )  (1.25) 2 (1  0.9e j 0.6  z 1 )(1  0.9e  j 0.6  z 1 )
 (1  0.8e j 0.8 z 1 )(1  0.8e  j 0.8 z 1 )
4th order
pole
Example:
Frequency-Response Compensation
40
dB
20
0
-20
-2
0
2
-2
0
2
-2
0
2
Radians
5
0
-5
4th order
pole
Samples
5
0
-5
-10
Example:
Frequency-Response Compensation
20
20
dB
40
dB
40
0
-20
0
-2
0
2
-20
0
-5
-2
0
2
-2
0
2
-2
0
2
5
Samples
Samples
0
0
-5
2
10
0
-10
-2
5
Radians
Radians
5
-2
0
2
Nonminimum Phase
0
-5
-10
Minimum Phase
Minimum Phase-Lag
j
j
j
H (e )  Hmin (e )Hap (e )
j
j
j
H (e )  Hmin (e )  Hap (e )
more
negative than
Nonpositive
For 0
Minimum Group-Delay
j
j
j
H (e )  Hmin (e )Hap (e )
grd[H (e j )]  grd[Hmin (e j )]  grd[Hap (e j )]
more
positive than
Nonnegative
For 0
Minimum-Energy Delay
H ( z)  H min ( z) H ap ( z)
| h(0) || hmin (0) |
Apply initial value theorem
Transform Analysis of
LTI systems
Generalized
Linear-Phase Systems
Linear Phase


Linear phase with integer (negative slope) --simple delay
Generalization: constant group delay
Example: Ideal Delay
j
Hid (e )  e
j
| Hid (e ) | 1
 j
,
|  | 
j
Hid (e )  
j
grd[Hid (e )]  
sin (n  )
h ( n) 
,
(n  )
  n  
Example: Ideal Delay
j
Hid (e )  e
j
| Hid (e ) |
1
 j|H(e
j)|
|  | 
,
1
j

Hid (e )  
j
grd[Hid (e )]  
H(ej)


sin (n  )
h ( n) 
,
(n  )

  n  
Impulse response is symmetric about
n = nd , i.e., h(2nd n)=h(n).
Example: Ideal Delay
1
0.5
0
-0.5
-5
0
5
10
15
If =nd (e.g., =5) is an integer, h(n)=(nnd).
sin (n  )
h ( n) 
,
(n  )
  n  
h(2n)=h(n).
Example: Ideal Delay
1
0.5
0
-0.5
-5
0
5
10
15
The case for 2 (e.g., =4.5) is an integer.
sin (n  )
h ( n) 
,
(n  )
  n  
Asymmetry
Example: Ideal Delay
1
0.5
0
-0.5
-5
0
5
10
 as an arbitrary number (e.g., =4.3).
sin (n  )
h ( n) 
,
(n  )
  n  
More General Frequency Response
with Linear Phase
j
j
H (e ) | H (e ) | e
 j
,
|  | 
j
|H(e )|

j
e
Zero-phase
filter
Ideal delay
More General Frequency Response
with Linear Phase
j
j
H (e ) | H (e ) | e
 j
,
|  | 
j
|H(e )|

j
e
Zero-phase
filter
Ideal delay
Example:
Ideal Lowpass Filter
e
H lp (e )  
0
j
 j
|  | c
1
|  | c  
 c
|H(ej)|
c


H(ej)

sin c (n  )
hlp (n) 
,
(n  )

  n  

Example:
Ideal Lowpass Filter
Show that
If 2 is an interger, h(2 n)=h(n).
That is, it has the same symmetric
property as an ideal delay.
sin c (n  )
hlp (n) 
,
(n  )
  n  
Generalized Linear Phase Systems
j
j
H (e )  A(e )e
 j  j
Real function.
and 
Possibly bipolar. are constants
j
H (e )    
j
grd[ H (e )]  
h(n) vs.  and 
j
j
H (e )  A(e )e
 j  j
 A(e j ) cos(  )  jA(e j ) sin(  )
j
H (e ) 

 h(n)e
 jn
n  



n  
n  
 h(n) cosn  j  h(n) sin n
h(n) vs.  and 
H (e j )  A(e j ) cos(  )  jA(e j ) sin(  )
H (e j ) 


n  
n  
 h(n) cosn  j  h(n) sin n
sin(  )

tan(  ) 
cos(  )


 h(n) sin n
n  

 h(n) cosn
n  
h(n) vs.  and 
H (e j )  A(e j ) cos(  )  jA(e j ) sin(  )
H (e j ) 


n  
n  
 h(n) cosn  j  h(n) sin n
sin(  )

tan(  ) 
cos(  )


 h(n) sin n
n  

 h(n) cosn
n  
h(n) vs.  and 


n  
n  
 h(n) cosn sin(  )   h(n) sin n cos(  )  0

 h(n) sin((n  )  )  0
n  
sin(  )

tan(  ) 
cos(  )


 h(n) sin n
n  

 h(n) cosn
n  
Necessary Condition for
Generalized Linear Phase Systems
j
j
H (e )  A(e )e
 j  j

 h(n) sin((n  )  )  0
n  
Let’s consider special cases.
Necessary Condition for
Generalized Linear Phase Systems
=0 or 
h(n) sin((n  )  )  0

2 = M = an integer
n  

=0 or 
Such a condition must
hold for all  and 
h(2  n)  h(n)

 h(n) sin (n  )  0
n  

 h(2  n) sin (  n)  0
n  

 h(2  n) sin (n  )  0
n  
Necessary Condition for
Generalized Linear Phase Systems
=0 or 
h(n) sin((n  )  )  0

2 = M = an integer
n  

 is an integer

2 is an integer
h(2  n)  h(n)

Necessary Condition for
Generalized Linear Phase Systems
=/2 or 3/2
h(n) sin((n  )  )  0

2 = M = an integer
n  

=/2 or 3/2
Such a condition must
hold for all  and 
h(2  n)  h(n) 

 h(n) cos(n  )  0
n  

 h(2  n) cos(  n)  0
n  

 h(2  n) cos(n  )  0
n  
Necessary Condition for
Generalized Linear Phase Systems
=/2 or 3/2
h(n) sin((n  )  )  0

2 = M = an integer
n  

 is an integer

2 is an integer
h(2  n)  h(n)

Causal
Generalized Linear Phase Systems
Generalized Linear
Phase System
Causal Generalized
Linear Phase System

 h(n) sin((n  )  )  0
n  

 h(n) sin((n  )  )  0
n 0
h(n)  0,
n  0 and n  M
Causal
Generalized Linear Phase Systems
h(M  n) 0  n  M
h(n)  
otherwise
0
H (e j )  Ae (e j )e jM / 2
Type I FIR linear phase system
0
1
2
3 …
M
M is even
Type II FIR linear phase system
0
1
2
3 …
M
M is odd
Causal
Generalized Linear Phase Systems
 h(M  n) 0  n  M H (e j )  jAo (e j )e  jM / 2
h(n)  
otherwise
0
 Ao (e j )e  jM / 2 j / 2
0
1
Type III FIR linear phase system
2…
…M
0
1
M is even
Type IV FIR linear phase system
2…
…M
M is odd
Type I FIR Linear Phase Systems
h(M  n) 0  n  M
h(n)  
otherwise
0
H (e j )  Ae (e j )e jM / 2
Type I FIR linear phase system
M is even
0
1
2
3 …
M
M
M /2
n 0
k 0
H (e j )   h(n)e  jn  e  jM / 2  a(k ) cosk
k 0
h(M / 2)
a(k )  
2h(M / 2  k ) k  1,2,, M / 2
Example:
Type I FIR Linear Phase Systems
1
1
2
3
4
H ( z)   z
n 0
4
n
1 z

1  z 1
 j5
4
2
0
5
-2
0
2
4
Radians
0
|H(ej)|
6
2
0
1 e
H (e ) 
1  e  j
e  j5 / 2 e j5 / 2  e  j5 / 2
 j 2 sin(5 / 2)
  j / 2 j / 2  j / 2  e
e
e
e
sin( / 2)
j
-2
-4
-2
0
2
Example:
Type I FIR Linear Phase Systems
1
1
2
3
4
H ( z)   z
n 0
4
n
1 z

1  z 1
 j5
4
2
0
5
-2
0
2
4
Radians
0
|H(ej)|
6
2
0
1 e
H (e ) 
1  e  j
e  j5 / 2 e j5 / 2  e  j5 / 2
 j 2 sin(5 / 2)
  j / 2 j / 2  j / 2  e
e
e
e
sin( / 2)
j
-2
-4
-2
0
2
Example:
Type II FIR Linear Phase Systems
1
1
2
3
5
H ( z )   z n
n 0
4
5
1 z

1  z 1
 j6
1

e
H (e j ) 
1  e  j
e  j3 e j3  e  j3
  j / 2 j / 2  j / 2
e
e
e
4
2
0
6
-2
0
2
-2
0
2
4
Radians
0
|H(ej)|
6
2
0
-2
-4
e
 j5 / 2
sin(3)
sin( / 2)
Example:
Type III FIR Linear Phase Systems
1
2
2
1
|H(ej)|
0
1.5
1
1
0.5
H ( z)  1  z 2
0
 j 2 sin()e
 j
 2 sin()e j j / 2
0
2
-2
0
2
2
Radians
H (e j )  1  e j2
 e j (e j  e j )
-2
1
0
-1
-2
Example:
Type IV FIR Linear Phase Systems
1
1
2
1.5
|H(ej)|
0
1
1
0.5
H ( z)  1  z 1
H (e )  1  e
 e j/ 2 (e j/ 2  e j/ 2 )
 j 2 sin( / 2)e
 j / 2
 2 sin( / 2)e j/ 2 j / 2
-2
0
2
-2
0
2
2
 j
Radians
j
0
1
0
-1
-2
Zeros Locations for
FIR Linear Phase Systems (Type I and II)
M
H ( z )   h( n) z  n
n 0
M
  h( M  n) z  n
n 0
M
  h(n) z ( M n )
n 0
H ( z)  z
M
1
H (z )
Let z0 be a zero of H(z)
H (1/ z0 )  z0 M H ( z0 )  0
1/z0 is a zero
M
 z  M  h( n) z n
n 0
z
M
1
H (z )
If h(n) is real
z0* and 1/ z0* are zeros
Zeros Locations for
FIR Linear Phase Systems (Type I and II)
( z0* )1
z21
z4
z1
z2
z0
1
H (z )
Let z0 be a zero of H(z)
z3
z11
H ( z)  z
M
*
0
z
z01
H (1/ z0 )  z0 M H ( z0 )  0
1/z0 is a zero
If h(n) is real
z0* and 1/ z0* are zeros
Zeros Locations for
FIR Linear Phase Systems (Type I and II)
( z0* )1
z21
z4
z1
z2
1
H (z )
Consider z = 1
z0
z3
z11
H ( z)  z
M
H (1)  (1)
M
H (1)
z0*
z01
if M is odd,
z = 1 must be a zero.
Zeros Locations for
FIR Linear Phase Systems (Type III and IV)
H ( z)   z
M
1
H (z )
( z0* )1
Let z0 be a zero of H(z)
M
0
H (1/ z0 )   z
H ( z0 )  0
1/z0 is a zero
If h(n) is real
z0* and 1/ z0* are zeros
z21
z4
z1
z2
z0
z3
z11
z0*
z01
z = 1 must be a zero.
Zeros Locations for
FIR Linear Phase Systems (Type III and IV)
H ( z)   z
M
1
H (z )
( z0* )1
Consider z = 1 H (1)   H (1)
Consider z = 1
( M 1)
H (1)  (1)
H (1)
z21
z4
z1
z2
z3
z11
if M is even,
z = 1 must be a zero.
z0
z0*
z01