数字信号处理chp5_第2讲

5.2 System functions for system characterized
by difference equations
Difference equation :
System function :
00
[][]
N
M
k
k
k k a
y n k b
x n k ==−=
−∑∑0
()()
N
M
k
k k k k k a z
Y z b z X z −−===∑∑00()
()()
M
k k k N
k
k k b z Y z H z X z a z
−=−===
∑∑ZT
101101
(1)
(1)
M
k k N
k
k c z b a d z −=−=−⎛⎞=⎜⎟⎝⎠−∏∏
straightforward
relationship
Example 5.2 Find the difference equation of the
given system function
1211(1)()13(1)(1)
24
z H z z z −−−+=
−+121212()
()13
()148
z z Y z H z X z z z −−−−++==
+−121213
(1)()(12)()48
z z Y z z z X z −−−−+−=++13
[][1][2][]2[1][2]
48
y n y n y n x n x n x n +−−−=+−+−IZT
Solution:
5.2.1 Stability and Causality
same algebraic
expression for H (z )different choices for ROC
different h [n ]
Causal:
Stable:ROC includes the unit circle (|z |= 1).
Causal and stable:
All poles are inside the unit circle.
different system
same difference equation
max{}k d z <≤∞
ROC:max{}k d z <≤∞
ROC:h [n ]: right-sided
h [n ]: absolutely summable
The difference equation does not uniquely specify a LTI system.
Example 5.3 Determining the ROC
5
[][1][2][]
2
y n y n y n x n −−+−=12
1111
()5
1
1(1)(12)
2
2
H z z z z z −−−−=
=
−+−−2
z >12
z <1
22
z <<ROC
——causal, not stable
——stable, not causal ——neither causal nor stable
Solution:
()()()1
i G z H z H z ==1(),()i H z H z =
5.2.2 Inverse systems
101101(1)()(1)
N
k
k M
k
k c z b H z a d
z −=−=−⎛⎞
=⎜⎟
⎝⎠
−∏∏101101
(1)()(1)
N
k k i M k k d z a H z b c z −=−=−⎛⎞
=⎜⎟
⎝⎠
−∏
∏
1()()
j i j H e H e ωω=[][]*[][]
i g n h n h n n δ==Question: What about the the ROC of H i (z ) ?Answer: The ROCs of H i (z ) and H (z ) must overlap.
The poles of H i (z ) are the zeros of H (z ), and vice versa
5.2.2 Inverse systems
1) Not all systems have an inverse,
e.g. Ideal lowpass filter.
2) Multi inverse system may exist for a system, example 5.5
Note:
Example 5.5 Find the inverse of the given system
110.5
(),ROC: 0.9
10.9z H z z z −−−=>−11
11
10.92 1.8()0.512i z z H z z z −−−−−−+=
=
−−——both causal and stable
2) ROC: 2
z >1) ROC: 2
z <1
1[]2(2)[1] 1.8(2)[]
n
n i h n u n u n −=−−−−12[]2(2)[] 1.8(2)[1]
n n i h n u n u n −=−+−——causal and unstable
——stable and noncausal
5.2.3 Impulse response for rational
system functions
101
101
(1)
()(1)
M
k k N
k
k c z b H z a d z −=−=−⎛⎞=⎜⎟⎝⎠−∏∏
-1
1(),1M N
N
r
k
r
r k k A H z B z
M N d z
−−===
+≥−∑∑
(assumed to be causal)
infinite impulse response (IIR) SY 0
()M
k
k
k H z b
z −==
∑0
,0[][]0,
otherwise
M
n k
k b n M h n b n k δ=≤≤⎧=
−=⎨⎩
∑finite impulse response (FIR) SY
1
[][][]
M N
N
n r k k r k h n B n r A d u n δ−===
−+∑
∑IZT N ≠0:
N = 0:Example 5.7 A simple FIR system
[],0[]0,
otherwise n a u n n M h n ⎧≤≤=⎨
⎩111
1()1M M M
n n n a z H z a z az +−−−−=−==
−∑zeros:2/(1)
,0,1,,j k M k z ae
k M
π+=="poles:,0,1,,k z a z k M
==="0
[][]
M
k k y n a x n k ==−∑1[][1][][1]
M y n ay n x n a x n M +−−=−−−The pole at z =a is canceled by zero at z 1=a
11
()M M M
z a z z a ++−=−M =7
0()M
n n
n H z a z −==∑11
1
1()1M M a z H z az +−−−−=
− 5.3 Frequency response for rational
system functions
101101(1)()(1)
M
k k N
k
k c z b H z a d
z −=−=−⎛⎞
=⎜⎟
⎝⎠
−∏∏0101
(1)()(1)
M
j k
j k N
j k
k c e b H e
a d
e ωω
ω−=−=−⎛⎞=⎜⎟
⎝⎠
−∏
∏j z e
ω
=
Magnitude response:
010
11()1M
j k
j k N
j k
k c e b H e a d
e ωωω
−=−=−=
−∏∏z
log magnitude (dB):
010********
020log ()20log 20log 120log 1M N
j j j k k k k b
H e c e d e a ωωω
−−===+−−−∑∑z
magnitude-squared function:
*
2
2
*
01*01(1)(1)
()()()(1)(1)
M
j j k
k
j j j k N
j j k
k k c e c e b H e H e H e a d e
d e ωω
ω
ω
ω
ω
ω
−=−=−−⎛⎞
==⎜⎟
⎝⎠
−−∏∏z Gain in dB 10= 20log ()
j H e ωz
Attenuation in dB 10=20log ()j H e ω−= -Gain in dB
Three parameters of frequency response
Phase response:
01
10()11M N
j j j k k k k b H e c e d e a ωω
ω−−==⎡⎤⎡⎤⎡⎤=+−−−⎢⎥⎣⎦⎣⎦⎣⎦∑∑))))Three parameters of frequency response
Group delay:
11grd ()arg ()arg 1arg 1N M
j j j j k k k k d d d H e H e d e c e d d d ω
ωω
ωωωω
−−==⎡⎤⎡⎤⎡⎤⎡⎤=−
=−−−⎣⎦⎣⎦⎣⎦⎣⎦∑∑[]
arg where represents the continuous phase 1011
01(1)()(1)
M
k k N
k
k c z b H z a d
z −=−=−⎛⎞
=⎜⎟
⎝⎠
−∏∏0101(1)()(1)
M
j k
j k N
j k
k c e b H e
a d
e ωω
ω−=−=−⎛⎞=⎜⎟
⎝⎠
−∏∏j z e
ω
=
Three parameters of frequency response
()j H e ω
ππ
⎡⎤−<≤⎣⎦ARG
The principal value of phase:
()ARG ()2()
j j H e H e r ωωπω⎡⎤=+⎣⎦)z Computation of principal value:
()ARG ()arctan ()j j I j R H e H e H e ωωω⎡⎤
⎡⎤=⎢⎥⎣
⎦⎣⎦
0101ARG ()ARG 2()ARG 1ARG 1M
j j k k N
j k k b H e c e a d r e ω
ωω
πω−=−=⎡⎤⎡⎤⎡⎤=+−⎢⎥⎣⎦⎣⎦
⎣⎦
⎡⎤−−+⎣⎦∑∑1ω2ω3ω4ω5ω6
ω0
Sketch frequency response by Geometric method
()0101
()()
M
j k j N M k N j
k k e c b e a e d
ωω
ω−==−⎛⎞=⎜⎟⎝⎠−∏∏k j j k k k e c e αωρ=−=G
c zero vector:k
j j k k k e d l e βω=−=G
d pol
e vector:j e z ω
=k
c k
d ω
k
G
c ()0101
k
k
M
j k
j N M k N
j k k e b
e a l e α
ω
βρ−==⎛⎞=⎜⎟⎝⎠
∏
∏01
01
(1)
()(1)M j k j k N
j k k c e b H e a d e ωωω−=−=−⎛⎞
=⎜⎟
⎝⎠
−∏∏0
10
1
()M
k j k N
k
k b H e
a l ω
ρ===∏
∏
k
G d 0
1
10
()()
M N
j k k k k b
H e N M a ωαβω==⎛⎞=+−+−⎜⎟⎝⎠∑∑((0
1010101011
020log ()20log 20log 20log M N
j k k k k b H e l a ω
ρ===+−∑∑j z e ω
=2
ππ
32
π
2π0
ω
()
j H e ω1
1
()11j H z cz re z
θ−−=−=−j z e ω=()1j j j j j j e re H e re e
e ωθωθω
ω
−−=−=
31
()j v H e v ω=
3
v =z Magnitude:
3
12min
:
1v v v r
ωθ==−=−312max
:
1v v v r
ωθπ=+=+=+0
θ=2
πθ=
θπ
=
r < 1(r = 0.9)
First-order FIR (a single zero) system
23(,):12cos()
v r r ωθθπωθ∈+=+−−2
3(,2):12cos()
v r r ωθπθπωθ∈++=+−−2()1 12cos()
j j j H e re e r r ωθω
ωθ−=−=+−−frequency shift
z
Phase:30:0
ωφω=−=0
θ=θπ
=sin()()arctan 1cos()j r ARG H e r ω
ωθωθ⎡⎤−⎡⎤=⎢⎥
⎣⎦−−⎣⎦3()j H e ωφω
=−)0
θ=3(0~):0ωπφω∈−>3(~2):0ωππφω∈−<2
π
π
32
π2π0
ω
()j ARG H e ω
⎡⎤⎣⎦
2
π
θ=
j z e ω
=3
φj e z ω
=1
v 3
v ω
j e z ω
=1
v 3
v (2)
πω−3(2)
πφ−()1j j j H e re e ωθω−=−frequency shift
1cos()sin()
r jr ωθωθ=−−+−32:0
ωπφω=−=3:0
ωπφω=−=dips more sharply as r becomes closer to 1
(the zero is more closer to the unit circle).()j H e ωWhen r =1,10()0,20log ()j j H e H e ω
ω
==−∞
0.5
r =0.8
r =1r =π2π
ω
()
j H e ω3
31
()j v H e v v ω=
=z
Magnitude:()
θπ=
r = 1312min :1v v v r ωθ==−=−312max :1v v v r
ωθπ=+=+=+3(,):v ωθθπ∈+3
(,2):v ωθπθπ∈++,ωπεπ=−→33/2,φπ=,ωπεπ=+→3/2,φπ=3/2φωπ−=−3/2
φωπ−=discontinuity of π
π2π0
ω
()j ARG H e ω
⎡⎤⎣⎦
/2
π/2
π−z
Phase:3()j H e ωφω
=−)0.5r =0.8
r =1
r =θπ
=r = 1
30,2:0ωπφω=−=3(0~):0
ωπφω∈−<3(~2):0ωππφω∈−>r < 13:0
ωπφω=−=
3
31
()j v H e v v ω
== 1.2r =1
r = 1.5
r =π
2π0
ω
()
j H e ω
z Magnitude:2
r =()
θπ=
r > 1dips more sharply as
r becomes closer to 1 (the zero is more closer to the unit circle).
()j H e ω
312min :1v v v r ωθ==−=−312max :1v v v r
ωθπ=+=+=+3(,):v ωθθπ∈+3
(,2):v ωθπθπ∈++,ωπεπ=−→32,φπ=,ωπεπ=+→30,φ=3φωπ
−=−3φωπ
−=π
2π0
ω
()j ARG H e ω
⎡⎤⎣⎦
π
−π 1.5
r =2
r =1.2
r =3()j H e ωφω
=−)z
Phase:
discontinuity of 2π
θπ
=r > 130,2:0ωπφω=−=3(0~):0ωπφω∈−<3(~2):0
ωππφω∈−>First-order IIR (a single pole) system
1
11()()1i H z H z cz −=
=
−j z e ω
=11
()()1j i j j j H e H e re e ωω
θω
−=
=−1
()()
j i j H e H e ωω=
max 1
:()1j i H e r
ωωθ==
−min 1:()1j i H e r
ωωθπ=+=
+π
2π0
ω
()
j i H e ω1
0.531.9
=1
100.1
=(0.9,0)r θ==z Magnitude (,2):()
j i H e ωωθπθπ∈++(,):()j i H e ω
ωθθπ∈+π2π
ω
()j H e ω
0.1
1.9
11
()()1j i j j j H e H e re e ωω
θω
−=
=−()()j j i ARG H e ARG H e ωω
⎡⎤⎡⎤=−⎣⎦⎣⎦
π2π0
ω
()j ARG H e ω⎡⎤⎣⎦
π
2π0
ω
()j i ARG H e ω
⎡⎤⎣⎦
z
Phase (0.9,0)
r θ==First-order IIR (a single pole) system
4
ππ
2π
ω
74
π34
π
54
π01020log ()
j H e ωExample 5.8
1
11
()(1)(1)
j j H z re z re z θθ−−−=
−−0.9,/4
r θπ==121
111
()()()(1)(1)j j H z H z H z re z re z θθ−−−=⋅=⋅−−101011022
1021020log ()20log ()20log ()
10log 12cos()10log 12cos()j j j H e H e H e r r r r ωωωωθωθ=+⎡⎤=−+−−⎣⎦⎡⎤−+−+⎣⎦
1) magnitude:
4
π
θ=
4
π
θ=−
1020log ()
j H e ω()j ARG H e ω
⎡⎤⎣⎦
4
π054π34π74
π2π
ω
4
π
θ=
4
π
θ=−
()j ARG H e ω
⎡⎤⎣⎦
2) phase:
12()()()sin()arctan 1cos()sin()arctan 1cos()j j j ARG H e ARG H e ARG H e r r r r ωωω
ωθωθωθωθ⎡⎤⎡⎤⎡⎤=+⎣⎦⎣⎦⎣⎦
⎡⎤
−=⎢⎥
−−⎣⎦
⎡⎤++⎢⎥
−+⎣⎦
Example 5.8
1
11
()(1)(1)
j j H z re z re z θθ−−−=
−−
Summary
1Contribution of a factor (1)or (1)to ()
j j j cz re e H e θωω−−−−As varies, the peak of is shifted in frequency
As varies, the dip of is shifted in frequency Poles position Zeros position peak at dip at dip at peak at 1,
dip of () is more sharp
j r H e ω→1,()0
j r H e ω==1,
peak of () is more sharp j r H e ω→1,()j r H e ω==∞
θ
()j H e ωθ()j H e ωωθ
=ωθπ
=+ωθ=ωθπ
=+Summary
单位圆附近的零点位置对幅度响应凹谷的位置和深度有明显的影响，零点在单位圆上，则幅度响应为0。

在单位圆内且靠近单位圆附近的极点对幅度响应的凸峰的位置和深度有明显影响；极点在单位圆上，则幅度响应为无穷大；极点在单位圆外，则系统不稳定。

适当控制零点、极点的分布，就能改变数字滤波器的频率响应特性，以达到预期的设计要求。

Homework
5.4, 5.6, 5.8, 5.36。