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Transform Analysis ofLTI systems
主講人：虞台文
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
Transform Analysis ofLTI systems
Frequency Response of LTI systems
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)
Frequency Response
)()()( zXzHzY
)()()( jjj eXeHeY
|)(||)(||)(| jjj eXeHeY
( ) ( ) ( )j j jY e H e X e
MagnitudeMagnitude
PhasePhase
Ideal Frequency-Selective Filters
cc
)( jlp eH
c
cjlp eH
0
||1)(
n
nnh c
lp
sin
)(
Ideal Lowpass Filter
ComputationallyUnrealizable
ComputationallyUnrealizable
Ideal Frequency-Selective Filters
c
cjhp eH
1
||0)(
n
nn
nhnnh
c
hphp
sin)(
)()()(
Ideal Highpass Filter
ComputationallyUnrealizable
ComputationallyUnrealizable
cc
)( jhp eH
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
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.
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)(
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.
Transform Analysis ofLTI systems
Systems Characterized by
Constant-Coefficient Difference Equations
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
)(
)()(
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
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
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:– 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
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
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
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.
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
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:
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
)()()(
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 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.
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).
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
)()(
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.
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
Example:IIR
)()1()( nxnayny )()1()( nxnayny
11
1)(
azzH 11
1)(
azzH
)()( nuanh n
Transform Analysis ofLTI systems
Frequency Response of For Rational
System Functions
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()(
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
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
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
Systems with a Single Zero or Pole
11 zre j
11
1 zre j
r
r
Frequency Response of a Single Zero or Pole
11 zre j
11
1 zre j
jjj ereeH 1)(
jjj
ereeH
1
1)(
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
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
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
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
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
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
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
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
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
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.
Frequency Response of Multiple Zeros and Poles
Using additive method to compute– Magnitude– Phase– Group Delay
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)
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)
Transform Analysis ofLTI 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 …
In general, knowledge about the magnitude provides no information about the phase, and vice versa. Except when …
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
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
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
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?
Allpass Factors
1
1
1
*)(
az
azzH ap 1
1
1
*)(
az
azzH ap
a
1/a*
Pole at a
Zero at 1/a*
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
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?
Transform Analysis ofLTI systems
Allpass Systems
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)]=?
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.
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
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
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
-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
Transform Analysis ofLTI 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)H(z)
Hmin(z)Hmin(z) Hap(z)Hap(z)
)()()( zHzHzH apminHow?
Rational Systems vs. Minimum-Phase Systems
H(z)
Hmin(z)
Hap(z)
Rational Systems vs. Minimum-Phase Systems
H(z)
Hmin(z)
Hap(z)
Pole/zeroCanceled
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.
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
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
Example:Frequency-Response Compensation
)25.11)(25.11(
)9.01)(9.01()(18.018.0
16.016.0
zeze
zezezHjj
jj
4th orderpole
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
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
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
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
Minimum Phase-Lag
)()()( jap
jmin
j eHeHeH
)()()( jap
jmin
j eHeHeH
NonpositiveFor 0 more
negative than
Minimum Group-Delay
)()()( jap
jmin
j eHeHeH
)]([)]([)]([ jap
jmin
j eHgrdeHgrdeHgrd
NonnegativeFor 0 more
positive than
Minimum-Energy Delay
)()()( zHzHzH apmin
|)0(||)0(| minhh
Apply initial value theoremApply initial value theorem
Transform Analysis ofLTI systems
Generalized
Linear-Phase Systems
Linear Phase
Linear phase with integer (negative slope) simple delay
Generalization: constant group delay
Example: Ideal Delay
|| ,)( jjid eeH || ,)( jjid eeH
1|)(| jid eH )( j
id eH
)]([ jid eHgrd
n
n
nnh ,
)(
)(sin)(
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 =
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).
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).
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
More General Frequency Response with Linear Phase
|| ,|)(|)( jjj eeHeH || ,|)(|)( jjj eeHeH
|H(ej)||H(ej)| ejej
Zero-phasefilter
Ideal delay
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)(
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.
Generalized Linear Phase Systems
jjjj eeAeH )()(Real function.
Possibly bipolar. and
are constants
)( jeH
)]([ jeHgrdconstant group delay
constant group delay
h(n) vs. and jjjj eeAeH )()(
)sin()()cos()( jj ejAeA
nj
n
j enheH
)()(
nnhjnnhnn
sin)(cos)(
h(n) vs. and )sin()()cos()()( jjj ejAeAeH
nnhjnnheHnn
j
sin)(cos)()(
)cos(
)sin()tan(
nnh
nnh
n
n
cos)(
sin)(
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
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
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
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
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
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
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)(
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
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
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
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
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
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
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
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
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
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.
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
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.
