lectures 9 iir systems: first order system · 2008. 2. 29. · 1 ee3054 signals and systems...
TRANSCRIPT
-
1
EE3054
Signals and Systems
Lectures 9
IIR Systems: First Order System
Yao Wang
Polytechnic University
Some slides included are extracted from lecture presentations prepared by McClellan and Schafer
2/29/2008 © 2003, JH McClellan & RW Schafer 2
License Info for SPFirst Slides
� This work released under a Creative Commons Licensewith the following terms:
� Attribution� The licensor permits others to copy, distribute, display, and perform
the work. In return, licensees must give the original authors credit.
� Non-Commercial� The licensor permits others to copy, distribute, display, and perform
the work. In return, licensees may not use the work for commercial purposes—unless they get the licensor's permission.
� Share Alike� The licensor permits others to distribute derivative works only under
a license identical to the one that governs the licensor's work.
� Full Text of the License
� This (hidden) page should be kept with the presentation
-
2
FIR system: Review
� Described by a feedforwarddifference equation
� Impulse response is finite duration (finite impulse response or FIR)
� Characterized by impulse response h[n], system function H(z) (Z-transform of h[n]) and frequency response H(e^jw)
∑=
−==
=
M
k
n
knxkhnhnxny
bnh
0
][][][*][][
][
∑=
−=M
k
k knxbny
0
][][
)()(][ ω̂jeHzHnh ↔↔
IIR System: General
∑∑==
−+−=M
k
k
N
k
k knxbknyany
01
][][][
Weighted average of
input samples
Weighted average of
past output samples
(feedback)
Still a linear time-invariant system
Impulse response is infinitely long generally
Called Infinite Impulse Response (IIR) system
-
3
Roadmap
� First discuss first order system
� Time domain: output for given input, impulse response
� Z-domain: transfer function, characterization by poles, how to compute output using Z-domain
� Frequency response
� Next discuss second order system
� Finally to general IIR system
∑=
−+−=M
k
k knxbnyany
0
1 ][]1[][
∑=
−+−+−=M
k
k knxbnyanyany
0
21 ][]2[]1[][
2/29/2008 © 2003, JH McClellan & RW Schafer 6
ONE FEEDBACK TERM (First
Order System)
� CAUSALITY
� NOT USING FUTURE OUTPUTS or INPUTS
y[n] = a1y[n −1]+ b0x[n] +b1x[n −1]FIR PART of the FILTER
FEED-FORWARDPREVIOUS
FEEDBACK
� ADD PREVIOUS OUTPUTS
-
4
2/29/2008 © 2003, JH McClellan & RW Schafer 7
FILTER COEFFICIENTS
� ADD PREVIOUS OUTPUTS
� MATLAB
� yy = filter([3,-2],[1,-0.8],xx)
y[n] = 0.8y[n −1]+ 3x[n] −2x[n −1]
SIGN CHANGEFEEDBACK COEFFICIENT
2/29/2008 © 2003, JH McClellan & RW Schafer 8
COMPUTE OUTPUT
-
5
2/29/2008 © 2003, JH McClellan & RW Schafer 9
COMPUTE y[n]
� FEEDBACK DIFFERENCE EQUATION:
y[n] = 0.8y[n −1]+ 5x[n]
y[0] = 0.8y[−1] + 5x[0]
� NEED y[-1] to get started
2/29/2008 © 2003, JH McClellan & RW Schafer 10
AT REST CONDITION
� y[n] = 0, for n
-
6
2/29/2008 © 2003, JH McClellan & RW Schafer 11
COMPUTE y[0]
� THIS STARTS THE RECURSION:
� SAME with MORE FEEDBACK TERMS
y[n]= a1y[n −1]+ a2y[n − 2] + bk x[n − k]k=0
2
∑
2/29/2008 © 2003, JH McClellan & RW Schafer 12
COMPUTE MORE y[n]
� CONTINUE THE RECURSION:
-
7
2/29/2008 © 2003, JH McClellan & RW Schafer 13
PLOT y[n]y[n] has infinite duration!
Is IIR system LTI?
� If x[n]=0, y[n]=0 for n
-
8
Properties of LTI system:
Review
� Any LTI system can be characterized by
its impulse response h[n]=T(δ[n])
� Output to any input is related by
� y[n]=x[n]*h[n]
2/29/2008 © 2003, JH McClellan & RW Schafer 16
y[n]= a1y[n −1]+ b0x[n]
IMPULSE RESPONSE
u[n] =1, for n ≥ 0
h[n]= a1h[n −1]+ b0δ[n]
][)(][ 10 nuabnhn=
h[n] has infinite duration!
-
9
2/29/2008 © 2003, JH McClellan & RW Schafer 17
PLOT IMPULSE RESPONSE
h[n] = b0 (a1)nu[n] = 3(0.8)nu[n]
� Show that for the example system� y[n]=0.8 y[n-1] + 5 x[n]
y[n] = x[n]* h[n] yields same result as the direct computation using recursion
h[n]=5 * 0.8^n u[n]
x[n]=2δ[n]-3δ[n-1]+2 δ[n-3]
By linearity and time invariance:
y[n] = 2 h[n] – 3 h[n-1] + 2 h[n-3]
-
10
� When x[n] and h[n] are both infinite duration,
numerical computation of convolution is
generally infeasible.
� But we can still use the recursion based on the
difference equation, although this is tedious.
� Z-transform comes to rescue!
� Y(z)= X(z) H(z)
� Determine H(z), X(z), Y(z)
� From Y(z), determine y[n] (inverse Z-transform)
2/29/2008 © 2003, JH McClellan & RW Schafer 20
CONVOLUTION PROPERTY
� MULTIPLICATION of z-TRANSFORMS
� CONVOLUTION in TIME-DOMAIN
Y (z) = H(z)X(z)X(z)
h[n]y[n] = h[n]∗ x[n]x[n]
H(z)
IMPULSE
RESPONSE
-
11
2/29/2008 © 2003, JH McClellan & RW Schafer 21
System Function of First
Order System
� Impulse response:
� Infinite duration!
� Z-transform (System
Function) H(z) = h[n]z−n
n=−∞
∞
∑
∑∑∞
=
−−∞
−∞=
==0
1010 ][)()(n
nnn
n
n zabznuabzH
][)(][ 10 nuabnhn=
2/29/2008 © 2003, JH McClellan & RW Schafer 22
Derivation of H(z)
� Recall Sum of Geometric Sequence:
� Yields a COMPACT FORMrn
n=0
∞
∑ =1
1− r
11
1
0
0
1
10
0
10
if1
)()(
azza
b
zabzabzHn
n
n
nn
>−
=
==
−
∞
=
−∞
=
− ∑∑If |r|
-
12
2/29/2008 © 2003, JH McClellan & RW Schafer 23
Recap:
� FIRST-ORDER IIR FILTER:
y[n]= a1y[n −1]+ b0x[n]
H(z) =b0
1− a1z−1
][)(][ 10 nuabnhn=
Transform pair
2/29/2008 © 2003, JH McClellan & RW Schafer 24
Another first order system
y[n] = a1y[n −1]+ b0x[n] +b1x[n−1]
H(z)=b0
1− a1z−1 +
b1z−1
1− a1z−1 =
b0 + b1z−1
1 − a1z−1
]1[)(][)(][ 11110 −+=− nuabnuabnh nn
shifta is1−z
-
13
Can we determine H(z) more
easily
� Can we determine H(z) w/o determining
h[n] first?
� YES: by apply Z-transform to the
difference equation!
2/29/2008 © 2003, JH McClellan & RW Schafer 26
DELAY PROPERTY of X(z)
� DELAY in TIMEMultiply X(z) by z-1
x[n]↔ X(z)
x[n −1]↔ z −1X(z)
Proof: x[n −1]z −n
n= −∞
∞
∑ = x[ℓ]z− (ℓ+1)ℓ=−∞
∞
∑
= z−1 x[ℓ]z −ℓ
ℓ= −∞
∞
∑ = z−1X(z)
-
14
2/29/2008 © 2003, JH McClellan & RW Schafer 27
y[n] = a1y[n −1]+ b0x[n] +b1x[n −1]
Z-Transform of IIR Filter
� DERIVE the SYSTEM FUNCTION H(z)
� Use DELAYDELAY PROPERTY
� Apply transform on both sides
Y (z) = a1z−1Y(z) + b0X(z) + b1z
−1X(z)
2/29/2008 © 2003, JH McClellan & RW Schafer 28
H(z)=Y (z)
X(z)=b0 +b1z
−1
1− a1z−1 =
B(z)
A(z)
Y (z) − a1z−1Y(z) = b0X(z) + b1z
−1X(z)
(1 − a1z−1)Y (z) = (b0 + b1z
−1)X(z)
Y (z) = a1z−1Y(z) + b0X(z) + b1z
−1X(z)
-
15
2/29/2008 © 2003, JH McClellan & RW Schafer 29
Example
� DIFFERENCE EQUATION:
�� READREAD the FILTER COEFFS:
y[n] = 0.8y[n −1]+ 3x[n] −2x[n −1]
)(8.01
23)(
1
1
zXz
zzY
−
−=
−
−
H(z)
2/29/2008 © 2003, JH McClellan & RW Schafer 30
POLES & ZEROS
� ROOTS of Numerator & Denominator
POLE: H(z) ���� inf
ZERO:
H(z)=0
H(z) =b0 + b1z
−1
1 − a1z−1 → H (z) =
b0z + b1z − a1
b0z + b1 = 0 ⇒ z = −b1
b0
z − a1 = 0 ⇒ z = a1
-
16
2/29/2008 © 2003, JH McClellan & RW Schafer 31
EXAMPLE: Poles & Zeros
� VALUE of H(z) at POLES is INFINITEINFINITE
POLE at z=0.8
ZERO at z= -1
∞→=−
+=
=−−
−+=
−
+=
−
−
−
−
0)(8.01
)(22)(
0)1(8.01
)1(22)(
8.01
22)(
29
1
54
1
54
1
1
zH
zH
z
zzH
2/29/2008 © 2003, JH McClellan & RW Schafer 32
POLE-ZERO PLOT
2 + 2z −1
1−0.8z−1
ZERO at z = -1
POLE at
z = 0.8
-
17
Stability of the System
� FIRST-ORDER IIR FILTER:
y[n]= a1y[n −1]+ b0x[n]
H(z) =b0
1− a1z−1
][)(][ 10 nuabnhn=
Pole at z=a_1
When |a_1| < 1
h[n] = b0 (a1)nu[n] = 3(0.8)nu[n]
-
18
When |a_1| >1
� Show h[n]
� System produce unbounded output for
finite input!
� Unstable!
� BIBO stability
� BIBO: bounded input bounded output
Stability from Pole Location
� A causal LTI system with initial rest conditions is stable if all of its poles lie strictly inside the unit circle!� Our example is for 1st order system with 1 pole only. Above statement is true for systems of any order, which can be decomposed into sum of first order systems.
� Zero locations do not affect system stability
� FIR systems are always stable (poles at zeros only)
-
19
2/29/2008 © 2003, JH McClellan & RW Schafer 37
FREQUENCY RESPONSE
� SYSTEM FUNCTION: H(z)
� H(z) has DENOMINATOR
� FREQUENCY RESPONSE of IIR
� We have H(z)
� THREE-DOMAIN APPROACH
H(ej ?ω ) = H(z ) z = e j ?ω
h[n]↔ H (z)↔ H(e j?ω )
2/29/2008 © 2003, JH McClellan & RW Schafer 38
FREQUENCY RESPONSE
� EVALUATE on the UNIT CIRCLE
H(ej ?ω ) = H(z ) z = e j ?ω
-
20
2/29/2008 © 2003, JH McClellan & RW Schafer 39
FREQ. RESPONSE FORMULA
ω
ωω
ˆ
ˆˆ
1
1
8.01
22)(
8.01
22)(
j
jj
e
eeH
z
zzH
−
−
−
−
−
+=→
−
+=
=2ˆ)( ωjeH ω
ω
ω
ω
ω
ω
ˆ
ˆ
ˆ
ˆ2
ˆ
ˆ
8.01
22
8.01
22
8.01
22j
j
j
j
j
j
e
e
e
e
e
e
−
+⋅
−
+=
−
+−
−
−
−
=−−+
+++−
−
ωω
ωω
ˆˆ
ˆˆ
8.08.064.01
4444jj
jj
ee
ee
ω
ωˆcos6.164.1
ˆcos88
−
+
?ˆ@,40004.0
88)(,0ˆ@2ˆ πωω ω ==
+== jeH
2/29/2008 © 2003, JH McClellan & RW Schafer 40
Frequency Response Plot
ω
ωω
ˆ
ˆˆ
8.01
22)(
j
jj
e
eeH
−
−
−
+=
freqz(b,a)
b=[2,2]
a=[1, -0.8]
-
21
2/29/2008 © 2003, JH McClellan & RW Schafer 41
UNIT CIRCLE
� MAPPING BETWEEN
z = e j?ω
z = 1 ↔ ?ω = 0
z = −1 ↔ ?ω = ±π
z = ± j ↔ ?ω = ± 12π
z and ?ω
2/29/2008 © 2003, JH McClellan & RW Schafer 42
SINUSOIDAL RESPONSE
� x[n] = SINUSOID => y[n] is SINUSOID
� Get MAGNITUDE & PHASE from H(z)
ω
ω
ωω
ω
ˆ)()( where
)(][ then
][ if
ˆ
ˆˆ
ˆ
jez
j
njj
nj
zHeH
eeHny
enx
==
=
=
-
22
2/29/2008 © 2003, JH McClellan & RW Schafer 43
POP QUIZ
� Given:
� Find the Impulse Response, h[n]
� Find the output, y[n]
� Whenx[n] = cos(0.25πn)
H(z) =2 + 2z−1
1− 0.8z −1
2/29/2008 © 2003, JH McClellan & RW Schafer 44
Evaluate FREQ. RESPONSE
2 + 2z−1
1 − 0.8z −1at ?ω = 0.25π
zero at ω=πω=πω=πω=π?ω = 0.25π
0ˆ is 1 == ωz
-
23
2/29/2008 © 2003, JH McClellan & RW Schafer 45
POP QUIZ: Eval Freq. Resp.
� Given:
� Find output, y[n],
when
� Evaluate at
x[n] = cos(0.25πn)
H(z) =2 + 2z−1
1− 0.8z −1
z = e j0.25π
y[n]= 5.182cos(0.25πn − 0.417π )
309.1
25.0
22
22
182.58.01
)(22)(
j
je
e
jzH
−
−=
−
−+=
π
2/29/2008 © 2003, JH McClellan & RW Schafer 46
THREE DOMAINS
Use H(z) to get
Freq. Response
z = e j?ω
-
24
READING ASSIGNMENTS
� This lecture focuses on First Order
System
� Chapter 8, Sects. 8-1, 8-2, 8-3, 8-4, 8-5
� 8-3.2 Block diagram structure: study by
yourself
� 8-5: Frequency response of first order
system: study by yourself (Slides 36-45)