Upload
lenhi
View
225
Download
3
Embed Size (px)
Citation preview
EE 477Digital Signal Processing
8aIIR Systems
EE 477 DSP Spring 2007 Maher 2
General Difference Equation
•FIR: output depends on current andpast inputs only
•IIR: output depends on current andpast inputs and past outputs
[ ] [ ] [ ]∑∑==
−+−=N
ll
M
kk lnyaknxbny
10
EE 477 DSP Spring 2007 Maher 3
IIR Block Diagram
z-1
z-1
z-1
Σb0
b1
b2
b3
x[n]
x[n-1]
x[n-3]
x[n-2]
z-1
z-1
z-1
Σ
a1
a2
a3
y[n-1]
y[n-3]
y[n-2]
y[n]
EE 477 DSP Spring 2007 Maher 4
IIR Recursion
•Since current output depends on pastoutputs, we must either know orassume the prior state
•Generally use initial rest assumptions:–Input is a right-handed sequence (zero
before some n=n0)–Output is zero prior to n0
EE 477 DSP Spring 2007 Maher 5
IIR Recursion (cont.)
•Example: compute response of systemdescribed by:
with input:]1[5.0]1[][][ −+−−= nynxnxny
]2[5.0]1[2.0][][ −+−−= nnnnx δδδ
1
-0.2
0.5 ΣZ-1 -1
ΣZ-1 -1 Z-10.5
EE 477 DSP Spring 2007 Maher 6
IIR Recursion (cont.)
•Assume initial restn x[n] y[n]-2 0 0-1 0 00 1 11 -0.2 -0.72 0.5 0.353 0 -0.3254 0 -0.16255 0 -0.081256 0 -0.040637 0 -0.020318 0 -0.010169 0 -0.0050810 0 -0.0025411 0 -0.00127
y[n]
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-2 0 2 4 6 8 10 12 14 16 18 20 22 24 26
n
EE 477 DSP Spring 2007 Maher 7
Impulse Response
•The general IIR system expression islinear and time invariant (assuminginitial rest)
•Output is given by the convolution sum:
•Due to recursion, unit sample responsecan be of infinite extent
∑∞
−∞=
−=k
knhkxny ][][][
EE 477 DSP Spring 2007 Maher 8
Response Example
•Example:•Unbounded unit sample response
]1[2][][ −+= nynxny
0
200
400
600
800
1000
1200
-2 -1 0 1 2 3 4 5 6 7 8 9 10
n
][2][ nunh n=
EE 477 DSP Spring 2007 Maher 9
Closed Form Summation
•Some responses can be described inclosed form. One useful example:
( )
=+−
≠−
−=
+
=∑
1,1
1,1
12
121
2
1
rLL
rr
rrr
LL
L
Lk
k
EE 477 DSP Spring 2007 Maher 10
Response Behavior•Consider unit step response of simple
IIR system: ][]1[][ 1 nxnyany +−=
1n
…1…
1+a1+a12+ a1
3 + a14 + a1
515
1+a1+a12+ a1
3 + a1414
1+a1+a12+ a1
313
1+a1+a1212
1+a111
110
00-1
y[n]x[n]n
∑=
n
k
ka0
1 1
1
1
11
11
11][
,nlimit thein
aa
aany
n
−−
→−
−=
∞→∞+
][][
isresponsesampleunit thatnote
1 nuanh n=
EE 477 DSP Spring 2007 Maher 11
Response Behavior (cont.)
•Now consider what happens if:|a1|<1: sum converges to
|a1|>1: sum diverges
|a1|=1: sum diverges if a1=1, or alternates[1,0,1,0,… ] if a1= -1
111a−
EE 477 DSP Spring 2007 Maher 12
Response Behavior (cont.)•In general, we can use convolution sum:
•In this example with unit step input:∑
∞
−∞=
−=k
knhkxny ][][][
1
11
1
1
11
01
1
11
11
11
0,
0,0
][][][
aa
a
aa
na
n
knuakuny
n
n
n
n
k
kn
k
kn
−−
=
−
−
=
≥
<=
−=
+
+
=
−
∞
−∞=
−
∑
∑
EE 477 DSP Spring 2007 Maher 13
IIR System Function
•General first-order IIR:
•Take z-transform of both sides:
•Collect terms
]1[][]1[][ 101 −++−= nxbnxbnyany
)()()()( 110
11 zXzbzXbzYzazY −− ++=
)()()()1()()()()(
110
11
110
11
zXzbbzYzazXzbzXbzYzazY
−−
−−
+=−
+=−
EE 477 DSP Spring 2007 Maher 14
IIR System Function (cont.)
•Since Y(z)=H(z)X(z), we can writeH(z)=Y(z)/X(z)
•Note form of numerator anddenominator: feed forward on top,1 with negative of feed back on bottom
)1()(
)()()( 1
1
110
−
−
−+
==zazbb
zXzYzH
EE 477 DSP Spring 2007 Maher 15
Poles and Zeros
•The numerator polynomial has rootsthat are the zeros of the system
•The denominator polynomial has rootsthat are the poles of the system
•If the order of the numerator anddenominator polynomials differ, therewill be zeros and/or poles at 0 or infinity
EE 477 DSP Spring 2007 Maher 16
Structures: Direct Form I•Direct Form I: feed-forward calculation
followed by feedback calculation
z-1
z-1
z-1
Σb0
b1
b2
b3
x[n]
x[n-1]
x[n-3]
x[n-2]
z-1
z-1
z-1
Σ
a1
a2
a3
y[n-1]
y[n-3]
y[n-2]
y[n]
EE 477 DSP Spring 2007 Maher 17
Direct Form II•Direct Form II: feedback first, then
feed-forward (OK because LTI system)
x[n]
z-1
z-1
z-1
Σb0
b1
b2
b3
w[n-1]
w[n-3]
w[n-2]
z-1
z-1
z-1
Σ
a1
a2
a3
w[n-1]
w[n-3]
w[n-2]
w[n]y[n]
EE 477 DSP Spring 2007 Maher 18
Direct Form II (cont.)•Note that delay lines contain the same
values, so they can be merged.
x[n] Σb0
b1
b2
b3
z-1
z-1
z-1
Σ
a1
a2
a3
w[n-1]
w[n-3]
w[n-2]
w[n]y[n]
EE 477 DSP Spring 2007 Maher 19
Transpose Forms
•Oddly enough, the delay/branch/sumblock diagram can be “run in reverse”tocreate an equivalent transpose form:–Reverse all arrows in block diagram–Branch points become sum points; sum
points become branch points–Input and Output are exchanged
EE 477 DSP Spring 2007 Maher 20
Transpose Example•Original: first-order direct form II
•Transposed version
x[n] Σb0
b1
z-1
Σ
a1
w[n-1]
w[n]y[n]
x[n]
Σ
b0
b1
z-1
Σ
a1
y[n]
EE 477 DSP Spring 2007 Maher 21
Region of Convergence
•The z-transform of the unit sampleresponse is the system function, H(z)
•The z-transform is computed from
•This was OK for FIR, but for IIR wemust be sure that the infinite sumactually converges!
∑∞
−∞=
−=n
nznhzH ][)(
EE 477 DSP Spring 2007 Maher 22
Region of Convergence (cont.)
•Example: h[n]=anu[n]
•If |az-1|<1, then sum converges
x[n]
z-1
Σ
a
y[n]
( )∑∑∞
=
−∞
=
− ==0
1
0)(
n
n
n
nn azzazH
( )1
0
1
11
−
∞
=
−
−=∑ az
azn
n
EE 477 DSP Spring 2007 Maher 23
v
Region of Convergence (cont.)
•Since |az-1|<1, this implies |z| > |a|.•This means that the ROC will be all
values of |z| outside a radius r = |a|
|a|
ROC
EE 477 DSP Spring 2007 Maher 24
Region of Convergence (cont.)
• In order for a causal system to be stable, theROC must include the unit circle.
• The poles of the system define the edge ofthe ROC.
• This means: a stable, causal system willhave all its poles inside the unit circle, so thatthe ROC extends outward (including the unitcircle) from the largest pole.