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Transform Analysis ofLTI systems
主講人:虞台文
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Content The Frequency Response of LTI systems Systems Characterized by Constant-Coefficien
t Difference Equations Frequency Response for Rational System Fun
ctions Relationship btw Magnitude and Phase Allpass Systems Minimum-Phase Systems Generalized Linear-Phase Systems
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Transform Analysis ofLTI systems
Frequency Response of LTI systems
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Time-Invariant System
h(n)h(n)
x(n) y(n)=x(n)*h(n)
X(z) Y(z)=X(z)H(z)H(z)
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Frequency Response
)()()( zXzHzY
)()()( jjj eXeHeY
|)(||)(||)(| jjj eXeHeY
( ) ( ) ( )j j jY e H e X e
MagnitudeMagnitude
PhasePhase
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Ideal Frequency-Selective Filters
cc
)( jlp eH
c
cjlp eH
0
||1)(
n
nnh c
lp
sin
)(
Ideal Lowpass Filter
ComputationallyUnrealizable
ComputationallyUnrealizable
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Ideal Frequency-Selective Filters
c
cjhp eH
1
||0)(
n
nn
nhnnh
c
hphp
sin)(
)()()(
Ideal Highpass Filter
ComputationallyUnrealizable
ComputationallyUnrealizable
cc
)( jhp eH
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Such filters are– Noncausal– Zero phase– Not Computationally realizable
Causal approximation of ideal frequency-selective filters must have nonzero phase response.
Ideal Frequency-Selective Filters
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Phase Distortion and Delay ---Ideal Delay
)()( did nnnh )()( did nnnh ( ) dj njidH e e ( ) dj njidH e e
1|)(| jid eH
|| )( dj
id neH
Delay DistortionLinear Phase
Delay Distortion would be considered a rather mild form of phase distortion.
Delay Distortion would be considered a rather mild form of phase distortion.
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Phase Distortion and Delay ---A Linear Phase Ideal Filter
Still a noncausal one.Not computationally realizable.
Still a noncausal one.Not computationally realizable.
c
cnj
jlp
deeH
0
||)(
n
nn
nnnh
d
dclp ,
)(
)(sin)(
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A convenient measure of the linearity of phase. Definition:
Phase Distortion and Delay ---Group Delay
)]}({arg[)]([)(
jj eH
d
deHgrd )]}({arg[)]([)(
jj eH
d
deHgrd
Linear Phase ()=constant The deviation of () away from a constant
indicates the degree of nonlinearity of the phase.
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Transform Analysis ofLTI systems
Systems Characterized by
Constant-Coefficient Difference Equations
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Nth-Order Difference Equation
M
rr
N
kk rnxbknya
00
)()(
M
rr
N
kk rnxbknya
00
)()(
M
r
rr
N
k
kk zbzXzazY
00
)()(
N
k
kk
M
r
rr
za
zb
zX
zYzH
0
0
)(
)()(
N
k
kk
M
r
rr
za
zb
zX
zYzH
0
0
)(
)()(
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Representation in Factored Form
N
kr
M
rr
zd
zc
a
bzH
1
1
1
1
0
0
)1(
)1()(
N
kr
M
rr
zd
zc
a
bzH
1
1
1
1
0
0
)1(
)1()(
Contributes poles at 0 and zeros at crContributes poles at 0 and zeros at cr
Contributes zeros at 0 and poles at drContributes zeros at 0 and poles at dr
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Example
)1)(1(
)1()(
1431
21
21
zz
zzH
)1)(1(
)1()(
1431
21
21
zz
zzH
)(
)(
1
21)(
2831
41
21
zX
zY
zz
zzzH
)()21()()1( 212831
41 zXzzzYzz
)2()1(2)()2()1()( 83
41 nxnxnxnynyny )2()1(2)()2()1()( 8
341 nxnxnxnynyny
Two zeros at z = 1Two zeros at z = 1
poles at z =1/2 and z = 3/4
poles at z =1/2 and z = 3/4
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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
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For Causality:– ROC of H(z) must be outside the outermost pole
For Stability:– ROC includes the unit circle
For both– All poles are inside the unit circle
Stability and Causality
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Example:
Stability and Causality
)21)(1(
1
1
1)(
1121
2125
zz
zzzH
)21)(1(
1
1
1)(
1121
2125
zz
zzzH
)()2()1()( 25 nxnynyny )()2()1()( 2
5 nxnynyny
Re
Im
1
Discuss its stability and causalityDiscuss its stability and causality
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Inverse Systems
H(z)X(z) Y(z)
Hi(z)X(z)
G(z)= H(z)H i(z)=1
g(n) = h(n)* hi(n) = (n)
)(
1)(
zHzH i
)(
1)(
zHzH i
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Inverse Systems
H(z)X(z) Y(z)
Hi(z)X(z)
G(z)= H(z)H i(z)=1
g(n) = h(n)* hi(n) = (n)
)(
1)(
zHzH i
)(
1)(
zHzH i
Does every system have an inverse system?
Does every system have an inverse system?
Give an example.Give an example.
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Inverse Systems
N
kr
M
rr
zd
zc
a
bzH
1
1
1
1
0
0
)1(
)1()(
N
kr
M
rr
zd
zc
a
bzH
1
1
1
1
0
0
)1(
)1()(
M
kr
N
rr
i
zc
zd
b
azH
1
1
1
1
0
0
)1(
)1()(
M
kr
N
rr
i
zc
zd
b
azH
1
1
1
1
0
0
)1(
)1()(
Zeros
Poles
Zeros
Poles
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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.
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Impulse Response for Rational System Functions
By partial fraction expansion:
N
k
kk
M
r
rr
za
zbzH
0
0)(
N
k
kk
M
r
rr
za
zbzH
0
0)(
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
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FIR and IIR
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
Zero polesZero poles
nonzero poles
nonzero poles
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FIR and IIR
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
Zero polesZero poles
nonzero poles
nonzero poles
FIR: The system contains only zero poles.
FIR: The system contains only zero poles.
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FIR and IIR
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k k
kNM
r
rr zd
AzBzH
11
0 1)(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
N
k
nkk
NM
rr nudArnBnh
10
)()()(
Zero polesZero poles
nonzero poles
nonzero poles
IIR: The system contains nonzero poles (not canceled by zeros).
IIR: The system contains nonzero poles (not canceled by zeros).
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FIR
M
k
kk zbzH
0
)(
M
k
kk zbzH
0
)(
M
k
nk otherwise
Mnbknbnh
0 0
0)()(
M
k
nk otherwise
Mnbknbnh
0 0
0)()(
M
kk knxbny
0
)()(
M
kk knxbny
0
)()(
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Example:FIR
otherwise
Mnanh
n
0
0)(
otherwise
Mnanh
n
0
0)(
1
11
1
11
0 1
1
1
)(1)(
az
za
az
azzazH
MMMM
n
nn1
11
1
11
0 1
1
1
)(1)(
az
za
az
azzazH
MMMM
n
nn
Does this system have nonzero pole?
7th-orderpole M=7
One pole is canceled by zero here.
One pole is canceled by zero here.
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Example:FIR
otherwise
Mnanh
n
0
0)(
otherwise
Mnanh
n
0
0)(
Write its system function.
7th-orderpole M=7
)()(0
knxanyM
k
k
)1()()1()( 1 Mnxanxnayny M
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Example:IIR
)()1()( nxnayny )()1()( nxnayny
11
1)(
azzH 11
1)(
azzH
)()( nuanh n
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Transform Analysis ofLTI systems
Frequency Response of For Rational
System Functions
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Rational Systems
0
0
( )
Mk
kkN
kk
k
b zH z
a z
0
0
( )
Mk
kkN
kk
k
b zH z
a z
N
k
kjk
M
k
kjk
j
ea
ebeH
0
0)(
N
k
kjk
M
k
kjk
j
ea
ebeH
0
0)(
N
k
jk
M
k
jk
j
ed
ec
a
beH
1
1
0
0
)1(
)1()(
N
k
jk
M
k
jk
j
ed
ec
a
beH
1
1
0
0
)1(
)1()(
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Log Magnitude of H(ej) ---Decibels (dBs)
N
k
jk
M
k
jk
j
ed
ec
a
beH
1
1
0
0
1
1|)(|
N
k
jk
M
k
jk
j
ed
ec
a
beH
1
1
0
0
1
1|)(|
Gain in dB = 20log10|H(ej)|
N
k
jk
M
k
jk
j edeca
beH
110
110
0
01010 1log201log20log20|)(|log20
N
k
jk
M
k
jk
j edeca
beH
110
110
0
01010 1log201log20log20|)(|log20
Scaling Contributed by zeros Contributed by poles
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Advantages of Representing the magnitude in dB
)()()( jjj eXeHeY
|)(||)(||)(| jjj eXeHeY
|)(|log20|)(|log20|)(|log20 101010 jjj eXeHeY
The magnitude ofOutput FT
The MagnitudeOf Impulse Response
The magnitude ofInput FT
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Phase for Rational Systems
N
k
jk
M
k
jk
j edeca
beH
110
0 )1()1()(
N
k
jk
M
k
jk
j edeca
beH
110
0 )1()1()(
M
k
jk
N
k
jk
j ecd
ded
d
deHgrd
11
)1arg()1arg()]([
M
k
jk
N
k
jk
j ecd
ded
d
deHgrd
11
)1arg()1arg()]([
M
kj
kk
jkk
N
kj
kk
jkkj
ecc
ecc
edd
eddeHgrd
12
2
12
2
}{2||1
}{||
}{2||1
}{||)]([
ReRe
ReRe
M
kj
kk
jkk
N
kj
kk
jkkj
ecc
ecc
edd
eddeHgrd
12
2
12
2
}{2||1
}{||
}{2||1
}{||)]([
ReRe
ReRe
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Systems with a Single Zero or Pole
11 zre j
11
1 zre j
r
r
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Frequency Response of a Single Zero or Pole
11 zre j
11
1 zre j
jjj ereeH 1)(
jjj
ereeH
1
1)(
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Frequency Response of a Single Zero
11 zre j jjj ereeH 1)(
)1)(1(|)(| 2 jjjjj ereereeH
)cos(21 2 rr
)]cos(21[log10|)(|log20 21010 rreH j )]cos(21[log10|)(|log20 2
1010 rreH j
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Frequency Response of a Single Zero
11 zre j jjj ereeH 1)(
)1)(1(|)(| 2 jjjjj ereereeH
)cos(21 2 rr
)]cos(21[log10|)(|log20 21010 rreH j )]cos(21[log10|)(|log20 2
1010 rreH j
|H(ej)|2: Its maximum is at =.
max |H(ej)|2 =(1+r)2
Its minimum is at =0.min |H(ej)|2 =(1r)2
|H(ej)|2: Its maximum is at =.
max |H(ej)|2 =(1+r)2
Its minimum is at =0.min |H(ej)|2 =(1r)2
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Frequency Response of a Single Zero
11 zre j jjj ereeH 1)(
)cos(1
)sin(tan)( 1
r
reH j
2
2
2
2
|)(|
)cos(
)cos(21
)cos()]([
j
j
eH
rr
rr
rreHgrd 2
2
2
2
|)(|
)cos(
)cos(21
)cos()]([
j
j
eH
rr
rr
rreHgrd
![Page 41: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/41.jpg)
Frequency Response of a Single Zero
r = 0.9 = 0
r = 0.9 = /2
r = 0.9 =
-2 0 2-20
-10
0
10
dB
-2 0 2-2
0
2
Ra
dia
ns
-2 0 2-10
-5
0
5
Sa
mp
les
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
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r = 0.9 = 0
r = 0.9 = /2
r = 0.9 =
-2 0 2-20
-10
0
10
dB
-2 0 2-2
0
2
Ra
dia
ns
-2 0 2-10
-5
0
5
Sa
mp
les
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
Frequency Response of a Single Zero
於處有最大凹陷 (1r)2於處有最大凹陷 (1r)2
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r = 0.9 = 0
r = 0.9 = /2
r = 0.9 =
-2 0 2-20
-10
0
10
dB
-2 0 2-2
0
2
Ra
dia
ns
-2 0 2-10
-5
0
5
Sa
mp
les
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
Frequency Response of a Single Zero
於 ||處最高 (1+r)2於 ||處最高 (1+r)2
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r = 0.9 = 0
r = 0.9 = /2
r = 0.9 =
-2 0 2-20
-10
0
10
dB
-2 0 2-2
0
2
Ra
dia
ns
-2 0 2-10
-5
0
5
Sa
mp
les
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
-2 0 2-20
-10
0
10
-2 0 2-2
0
2
-2 0 2-10
-5
0
5
Frequency Response of a Single Zero
於處 phase直轉急上於處 phase直轉急上
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Frequency Response of a Single Zero
-2 0 2-20
-10
0
10
-2 0 2-4
-2
0
2
4
-2 0 20
5
10
-2 0 2-20
-10
0
10
dB
-2 0 2-4
-2
0
2
4
Rad
ians
-2 0 20
5
10
Sam
ples
r=1/0.9r=1.25r=1.5r=2
= 0 =
Zero outsidethe unit circle
Note that the group delay is always positive when r>1
Note that the group delay is always positive when r>1
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Frequency Response of a Single Zero
-2 0 2-40
-20
0
20
dB
-2 0 2-4
-2
0
2
4
Ra
dian
s
-2 0 2-100
-50
0
50
Sa
mpl
es
-2 0 2-40
-20
0
20
-2 0 2-4
-2
0
2
4
-2 0 2-100
-50
0
50
Some zeros insidethe unit circle
And some outside
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Frequency Response of a Single Pole
The converse of the single-zero case.Why? A stable system: r < 1
Exercise: Use matlab to plot the frequency responses for various cases.
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Frequency Response of Multiple Zeros and Poles
Using additive method to compute– Magnitude– Phase– Group Delay
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Example Multiple Zeros and Poles
)7957.04461.11)(683.01(
)0166.11)(1(05634.0)(
211
211
zzz
zzzzH
)7957.04461.11)(683.01(
)0166.11)(1(05634.0)(
211
211
zzz
zzzzH
zerosRadius Angle
1 1 1.0376 (59.45)
polesRadius Angle
0.683 00.892 0.6257 (35.85)
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Example Multiple Zeros and Poles
-3 -2 -1 0 1 2 3-100
-50
0
dB
-3 -2 -1 0 1 2 3-4
-2
0
2
4
Rad
ians
-3 -2 -1 0 1 2 30
5
10
Sam
ples
zerosRadius Angle
1 1 1.0376 (59.45)
polesRadius Angle
0.683 00.892 0.6257 (35.85)
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Transform Analysis ofLTI systems
Relationship btw
Magnitude and Phase
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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 …
In general, knowledge about the magnitude provides no information about the phase, and vice versa. Except when …
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Magnitude
)(*)(|)(| 2 jjj eHeHeH
jez
zHzH *)/1(*)(
N
kk
M
kk
zda
zcbzH
1
10
1
10
)1(
)1()(
N
kk
M
kk
zda
zcbzH
1
10
1
10
)1(
)1()(
*
*0 0
1 1
*0 0
1 1
(1 *) (1 )*(1/ *)
(1 *) (1 )
M M
k kk kN N
k kk k
b c z b c zH z
a d z a d z
*
*0 0
1 1
*0 0
1 1
(1 *) (1 )*(1/ *)
(1 *) (1 )
M M
k kk kN N
k kk k
b c z b c zH z
a d z a d z
![Page 54: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/54.jpg)
Magnitude)(*)()( zHzHzC
N
kkk
M
kkk
zdzd
zczc
a
b
1
*1
1
*12
0
0
)1)(1(
)1)(1(
N
kk
M
kk
zda
zcbzH
1
10
1
10
)1(
)1()(
N
kk
M
kk
zda
zcbzH
1
10
1
10
)1(
)1()(
*
*0 0
1 1
*0 0
1 1
(1 *) (1 )*(1/ *)
(1 *) (1 )
M M
k kk kN N
k kk k
b c z b c zH z
a d z a d z
*
*0 0
1 1
*0 0
1 1
(1 *) (1 )*(1/ *)
(1 *) (1 )
M M
k kk kN N
k kk k
b c z b c zH z
a d z a d z
![Page 55: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/55.jpg)
Magnitude
N
kkk
M
kkk
zdzd
zczc
a
bzC
1
*1
1
*12
0
0
)1)(1(
)1)(1()(
Zeros of H(z):
Poles of H(z):
kc
kd
Zeros of C(z):
Poles of C(z):
*/1 and kk cc
*/1 and kk dd
Conjugate reciprocal pairsConjugate reciprocal pairs
![Page 56: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/56.jpg)
Magnitude
N
kkk
M
kkk
zdzd
zczc
a
bzC
1
*1
1
*12
0
0
)1)(1(
)1)(1()(
Given C(z), H(z)=?
How many choices if the numbers of zeros and poles are fixed?
How many choices if the numbers of zeros and poles are fixed?
![Page 57: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/57.jpg)
Allpass Factors
1
1
1
*)(
az
azzH ap 1
1
1
*)(
az
azzH ap
a
1/a*
Pole at a
Zero at 1/a*
![Page 58: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/58.jpg)
Allpass Factors
1
1
1
*)(
az
azzH ap 1
1
1
*)(
az
azzH ap
j
jj
ap ae
aeeH
1
*)(
j
jj
ae
eae
1
*1
1|)(| jap eH
![Page 59: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/59.jpg)
Allpass Factors
H1(z)H1(z)
H1(z)H1(z) Hap(z)Hap(z)
)()()( 1 zHzHzH ap
|)(||)(| 1 zHzH
There are infinite many systems to have the same frequency-response magnitude?
There are infinite many systems to have the same frequency-response magnitude?
![Page 60: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/60.jpg)
Transform Analysis ofLTI systems
Allpass Systems
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General Form
cr M
k kk
kkM
k k
kap zeze
ezez
zd
dzzH
11*1
1*1
11
1
)1)(1(
))((
1)(
cr M
k kk
kkM
k k
kap zeze
ezez
zd
dzzH
11*1
1*1
11
1
)1)(1(
))((
1)(
Real Poles Complex Poles
kdkd/1ke
*ke
ke/1
*/1 ke |Hap(ej)|=1
|Hap(ej)|=?
grd[Hap(ej)]=?
![Page 62: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/62.jpg)
AllPass Factor
1
1
1
*)(
az
azzH af 1
1
1
*)(
az
azzH af
Consider a=rej
jj
jjj
af ere
reeeH
1)(
jj
jjj
af ere
reeeH
1)(
)cos(1
)sin(tan2)( 1
r
reH j
af
2
2
2
2
|1|
1
)cos(21
1)]([
jjj
af ere
r
r
reHgrd
Always positive for a stable and causal system.
Always positive for a stable and causal system.
![Page 63: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/63.jpg)
Example: AllPass FactorReal Poles
-2 0 2-1
0
1
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 20
10
20
Sam
ple
s
-2 0 2-1
0
1
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 20
10
20
Sa
mp
les
0.9 0.9
![Page 64: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/64.jpg)
Example: AllPass FactorReal Poles
-2 0 2-1
0
1
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 20
10
20
Sam
ple
s
-2 0 2-1
0
1
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 20
10
20
Sa
mp
les
Phase is nonpositivefor 0<<.
Group delay is positive
0.9 0.9
![Page 65: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/65.jpg)
Example: AllPass FactorComplex Poles
-2 0 2-1
0
1
dB
-2 0 2-5
0
5
Rad
ians
-2 0 20
10
20
Sam
ples
/40.9
Continuous phaseis nonpositive
for 0<<.
Group delay is positive
![Page 66: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/66.jpg)
-2 0 2-1
0
1
dB
-2 0 2-5
0
5
Rad
ians
-2 0 20
5
10
15
Sam
ples
Example: AllPass FactorComplex Poles
/40.8
1/23/44/3
Continuous phaseis nonpositive
for 0<<.
Group delay is positive
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Transform Analysis ofLTI systems
Minimum-Phase Systems
![Page 68: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/68.jpg)
Properties of Minimum-Phase Systems
To have a stable and causal inverse systems Minimum phase delay Minimum group delay Minimum energy delay
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Rational Systems vs. Minimum-Phase Systems
H(z)H(z)
Hmin(z)Hmin(z) Hap(z)Hap(z)
)()()( zHzHzH apminHow?
![Page 70: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/70.jpg)
Rational Systems vs. Minimum-Phase Systems
H(z)
Hmin(z)
Hap(z)
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Rational Systems vs. Minimum-Phase Systems
H(z)
Hmin(z)
Hap(z)
Pole/zeroCanceled
![Page 72: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/72.jpg)
Frequency-Response Compensation
s(n) DistortingSystem
Hd(z)
DistortingSystem
Hd(z)
sd(n) CompensatiingSystem
Hc(z)
CompensatiingSystem
Hc(z)
s(n)
The system of Hd(z) is invertible iff it is a minimum-phase system.
The system of Hd(z) is invertible iff it is a minimum-phase system.
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Frequency-Response Compensation
s(n) DistortingSystem
Hd(z)
DistortingSystem
Hd(z)
sd(n) CompensatiingSystem
Hc(z)
CompensatiingSystem
Hc(z)
s(n)
DistortingSystem
Hdmin(z)
DistortingSystem
Hdmin(z)
AllpassSystem
Hap(z)
AllpassSystem
Hap(z)
s(n) sd(n) CompensatiingSystem
1Hdmin(z)
CompensatiingSystem
1Hdmin(z)
)(ˆ ns
![Page 74: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/74.jpg)
Frequency-Response Compensation
DistortingSystem
Hdmin(z)
DistortingSystem
Hdmin(z)
AllpassSystem
Hap(z)
AllpassSystem
Hap(z)
s(n) sd(n) CompensatiingSystem
1Hdmin(z)
CompensatiingSystem
1Hdmin(z)
)(ˆ ns
H d(z) H c(z)
)()(
1)()( zH
zHzHzH ap
dminapdmin
)()()( zHzHzG cd
![Page 75: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/75.jpg)
Example:Frequency-Response Compensation
)25.11)(25.11(
)9.01)(9.01()(18.018.0
16.016.0
zeze
zezezHjj
jj
4th orderpole
![Page 76: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/76.jpg)
Example:Frequency-Response Compensation
4th orderpole
-2 0 2-20
0
20
40
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 2-10
0
10S
am
ple
s
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Example:Frequency-Response Compensation
)8.01)(8.01(
)9.01)(9.01()25.1()(18.018.0
16.016.02
zeze
zezezHjj
jj
4th orderpole
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Example:Frequency-Response Compensation
4th orderpole
-2 0 2-20
0
20
40
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 2-10
-5
0
5S
am
ple
s
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Example:Frequency-Response Compensation
-2 0 2-20
0
20
40
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 2-10
-5
0
5S
am
ple
s
-2 0 2-20
0
20
40
dB
-2 0 2-5
0
5
Ra
dian
s
-2 0 2-10
0
10
Sa
mp
les
Minimum PhaseNonminimum Phase
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Minimum Phase-Lag
)()()( jap
jmin
j eHeHeH
)()()( jap
jmin
j eHeHeH
NonpositiveFor 0 more
negative than
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Minimum Group-Delay
)()()( jap
jmin
j eHeHeH
)]([)]([)]([ jap
jmin
j eHgrdeHgrdeHgrd
NonnegativeFor 0 more
positive than
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Minimum-Energy Delay
)()()( zHzHzH apmin
|)0(||)0(| minhh
Apply initial value theoremApply initial value theorem
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Transform Analysis ofLTI systems
Generalized
Linear-Phase Systems
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Linear Phase
Linear phase with integer (negative slope) simple delay
Generalization: constant group delay
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Example: Ideal Delay
|| ,)( jjid eeH || ,)( jjid eeH
1|)(| jid eH )( j
id eH
)]([ jid eHgrd
n
n
nnh ,
)(
)(sin)(
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Example: Ideal Delay
|| ,)( jjid eeH || ,)( jjid eeH
1|)(| jid eH )( j
id eH
)]([ jid eHgrd
n
n
nnh ,
)(
)(sin)(
1
|H(ej)|
H(ej)
slope =
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Example: Ideal Delay
-5 0 5 10 15-0.5
0
0.5
1
n
n
nnh ,
)(
)(sin)(
If =nd (e.g., =5) is an integer, h(n)=(nnd).
Impulse response is symmetric about n = nd , i.e., h(2nd n)=h(n).
Impulse response is symmetric about n = nd , i.e., h(2nd n)=h(n).
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Example: Ideal Delay
n
n
nnh ,
)(
)(sin)(
The case for 2 (e.g., =4.5) is an integer.
-5 0 5 10 15-0.5
0
0.5
1
h(2n)=h(n).h(2n)=h(n).
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Example: Ideal Delay
n
n
nnh ,
)(
)(sin)(
as an arbitrary number (e.g., =4.3).
-5 0 5 10-0.5
0
0.5
1
AsymmetryAsymmetry
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More General Frequency Response with Linear Phase
|| ,|)(|)( jjj eeHeH || ,|)(|)( jjj eeHeH
|H(ej)||H(ej)| ejej
Zero-phasefilter
Ideal delay
![Page 91: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/91.jpg)
1
|H(ej)|
H(ej)
slope =
cc
Example: Ideal Lowpass Filter
c
cj
jlp
eeH
||0
||)(
c
cj
jlp
eeH
||0
||)(
n
n
nnh c
lp ,)(
)(sin)(
![Page 92: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/92.jpg)
Example: Ideal Lowpass Filter
n
n
nnh c
lp ,)(
)(sin)(
Show that
If 2 is an interger, h(2 n)=h(n).
It has the same symmetric property as an ideal delay.
It has the same symmetric property as an ideal delay.
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Generalized Linear Phase Systems
jjjj eeAeH )()(Real function.
Possibly bipolar. and
are constants
)( jeH
)]([ jeHgrdconstant group delay
constant group delay
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h(n) vs. and jjjj eeAeH )()(
)sin()()cos()( jj ejAeA
nj
n
j enheH
)()(
nnhjnnhnn
sin)(cos)(
![Page 95: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/95.jpg)
h(n) vs. and )sin()()cos()()( jjj ejAeAeH
nnhjnnheHnn
j
sin)(cos)()(
)cos(
)sin()tan(
nnh
nnh
n
n
cos)(
sin)(
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h(n) vs. and
)cos(
)sin()tan(
nnh
nnh
n
n
cos)(
sin)(
0)cos(sin)()sin(cos)(
nnhnnhnn
0))(sin()(
nnhn
0))(sin()(
nnhn
sin( ) sin cos cos sin sin( ) sin cos cos sin
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Necessary Condition for Generalized Linear Phase Systems
jjjj eeAeH )()(
Let’s consider special cases.Let’s consider special cases.
0))(sin()(
nnhn
0))(sin()(
nnhn
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Necessary Condition for Generalized Linear Phase Systems
0))(sin()(
nnhn
0))(sin()(
nnhn
=0 or 2 = M = an integer
0)(sin)(
nnhn
=0 or =0 or
0)(sin)2(
nnhn
0)(sin)2(
nnhn
Such a condition must hold for all and
)()2( nhnh )()2( nhnh
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Necessary Condition for Generalized Linear Phase Systems
0))(sin()(
nnhn
0))(sin()(
nnhn
=0 or 2 = M = an integer
)()2( nhnh )()2( nhnh
is an integer
2 is an integer
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Necessary Condition for Generalized Linear Phase Systems
0))(sin()(
nnhn
0))(sin()(
nnhn
=/2 or 3/2
2 = M = an integer
0)(cos)(
nnhn
=/2 or 3/2=/2 or 3/2
0)(cos)2(
nnhn
0)(cos)2(
nnhn
Such a condition must hold for all and
)()2( nhnh )()2( nhnh
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Necessary Condition for Generalized Linear Phase Systems
0))(sin()(
nnhn
0))(sin()(
nnhn
=/2 or 3/2
2 = M = an integer
)()2( nhnh )()2( nhnh
is an integer
2 is an integer
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CausalGeneralized Linear Phase Systems
0))(sin()(
nnhn
0))(sin()(
nnhn
Generalized Linear Phase System
Causal Generalized Linear Phase System 0))(sin()(
0
nnhn
0))(sin()(0
nnhn
Mnnnh and 0 ,0)( Mnnnh and 0 ,0)(
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CausalGeneralized Linear Phase Systems
0 1 2 3 … M
0 1 2 3 … M
otherwise
MnnMhnh
0
0)()(
otherwise
MnnMhnh
0
0)()( 2/)()( Mjj
ej eeAeH
2/)()( Mjje
j eeAeH
Type I FIR linear phase system
M is even
Type II FIR linear phase system
M is odd
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CausalGeneralized Linear Phase Systems
otherwise
MnnMhnh
0
0)()(
otherwise
MnnMhnh
0
0)()(
2/2/
2/
)(
)()(
jMjj
o
Mjjo
j
eeA
eejAeH2/2/
2/
)(
)()(
jMjj
o
Mjjo
j
eeA
eejAeH
Type III FIR linear phase system
M is even
Type IV FIR linear phase system
M is odd
0 1 2 …
… M
0 1 2 …
… M
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Type I FIR Linear Phase Systems
0 1 2 3 … M
otherwise
MnnMhnh
0
0)()(
otherwise
MnnMhnh
0
0)()( 2/)()( Mjj
ej eeAeH
2/)()( Mjje
j eeAeH
Type I FIR linear phase system
M is even
M
n
njj enheH0
)()(
2/
0
2/ cos)(M
k
Mj kkae
2/,,2,1)2/(2
0)2/()(
MkkMh
kMhka
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Example:Type I FIR Linear Phase Systems
1
54
0 1
1)(
z
zzzH
n
n
j
jj
e
eeH
1
1)(
5
2/2/
2/52/5
2/
2/5
jj
jj
j
j
ee
ee
e
e
)2/sin(
)2/5sin(2
je
-2 0 20
2
4
6
|H(e
j)|
-2 0 2-4
-2
0
2
4
Rad
ian
s
0 1 2 3 4
1
![Page 107: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/107.jpg)
Example:Type II FIR Linear Phase Systems
1
65
0 1
1)(
z
zzzH
n
n
j
jj
e
eeH
1
1)(
6
2/2/
33
2/
3
jj
jj
j
j
ee
ee
e
e
)2/sin(
)3sin(2/5
je
0 1 2 3 4 5
1
-2 0 20
2
4
6
|H(e
j)|
-2 0 2-4
-2
0
2
4
Ra
dian
s
![Page 108: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/108.jpg)
Example:Type III FIR Linear Phase Systems
21)( zzH
21)( jj eeH)( jjj eee
jej )sin(2
0 1
1
2
1
2/)sin(2 jje
-2 0 20
0.5
1
1.5
2
|H(e
j)|
-2 0 2-2
-1
0
1
2
Rad
ians
![Page 109: Transform Analysis of LTI systems 主講人:虞台文. Content The Frequency Response of LTI systems Systems Characterized by Constant- Coefficient Difference Equations](https://reader033.vdocuments.pub/reader033/viewer/2022061602/56649f4a5503460f94c6c4a5/html5/thumbnails/109.jpg)
Example:Type IV FIR Linear Phase Systems
11)( zzH
jj eeH 1)()( 2/2/2/ jjj eee
2/)2/sin(2 jej
0
1
1
1
2/2/)2/sin(2 jje
-2 0 20
0.5
1
1.5
2
|H(e
j)|
-2 0 2-2
-1
0
1
2
Rad
ians
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Zeros Locations for FIR Linear Phase Systems (Type I and II)
M
n
nznhzH0
)()(
M
n
nznMh0
)(
M
n
nMznh0
)()(
M
n
nM znhz0
)(
)( 1 zHz M
)()( 1 zHzzH M )()( 1 zHzzH M
Let z0 be a zero of H(z)
0)()/1( 000 zHzzH M
1/z0 is a zero
If h(n) is realz0* and 1/ z0* are zeros
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Zeros Locations for FIR Linear Phase Systems (Type I and II)
)()( 1 zHzzH M )()( 1 zHzzH M
Let z0 be a zero of H(z)
0)()/1( 000 zHzzH M
1/z0 is a zero
If h(n) is realz0* and 1/ z0* are zeros
0z
10z
*0z
1*0 )( z
1z
11z
2z12z
3z4z
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Zeros Locations for FIR Linear Phase Systems (Type I and II)
)()( 1 zHzzH M )()( 1 zHzzH M
Consider z = 10z
10z
*0z
1*0 )( z
1z
11z
2z12z
3z4z)1()1()1( HH M
if M is odd,z = 1 must be a
zero.
if M is odd,z = 1 must be a
zero.
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Zeros Locations for FIR Linear Phase Systems (Type III and IV)
)()( 1 zHzzH M )()( 1 zHzzH M
Let z0 be a zero of H(z)
0)()/1( 000 zHzzH M
1/z0 is a zero
If h(n) is realz0* and 1/ z0* are zeros
0z
10z
*0z
1*0 )( z
1z
11z
2z12z
3z4z
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Zeros Locations for FIR Linear Phase Systems (Type III and IV)
)()( 1 zHzzH M )()( 1 zHzzH M
0z
10z
*0z
1*0 )( z
1z
11z
2z12z
3z4z
Consider z = 1 )1()1( HH
z = 1 must be a zero.z = 1 must be a zero.
Consider z = 1)1()1()1( )1( HH M
if M is even,z = 1 must be a
zero.
if M is even,z = 1 must be a
zero.