copyright © 2005. shi ping cuc chapter 6 digital filter structures content introduction iir filter...
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Copyright © 2005. Shi Ping CUC
Chapter 6Digital Filter Structures
Content
Introduction
IIR Filter Structures
FIR Filter Structures
Copyright © 2005. Shi Ping CUC
Introduction
What is a digital filter A filter is a system that is designed to remove some
component or modify some characteristic of a signal
A digital filter is a discrete-time LTI system which can process the discrete-time signal.
There are various structures for the implementation of digital filters
The actual implementation of an LTI digital filter can be either in software or hardware form, depending on applications
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Introduction
Basic elements of digital filter structures
Adder has two inputs and one output. Multiplier (gain) has single-input, single-output. Delay element delays the signal passing through it by
one sample. It is implemented by using a shift register.
z-1
a z-1
a
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a1
z-1
z-1a2
b0)(nx )(ny
)(nx )(ny
a1
a2
b0
z-1
z-1
1 2
3
4
5
)2()1()()()()()2()1()()()(
)2()1()()1()1()(
)()(
210501
2142315
34
23
2
nyanyanxbnwnxbnwnyanyanwanwanw
nynwnwnynwnw
nynw)2()1()()( 210 nyanyanxbny
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Introduction
Computational complexity refers to the number of arithmetic operations (multiplications, divi
sions, and additions) required to compute an output value y(n) for the system.
Memory requirements refers to the number of memory locations required to store the sy
stem parameters, past inputs, past outputs, and any intermediate computed values.
Finite-word-length effects in the computations refers to the quantization effects that are inherent in any digital i
mplementation of the system, either in hardware or in software.
The major factors that influence the choice of a specific structure
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IIR Filter Structures
The characteristics of the IIR filter IIR filters have Infinite-duration Impulse Responses The system function H(z) has poles in An IIR filter is a recursive system
||0 z
)(11)()(
)(1
1
110
1
0N
N
MM
N
k
kk
M
k
kk
zaza
zbzbb
za
zb
zXzY
zH
The order of such an IIR filter is called N if aN≠0
M
kk
N
kk knxbknyany
01
)()()(
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IIR Filter Structures
In this form the difference equation is implemented directly as given. There are two parts to this filter, namely the moving average part and the recursive part (or the numerator and denominator parts). Therefore this implementation leads to two versions: direct form I and direct form II structures
Direct form
N
kk
M
kk knyaknxbny
10
)()()(
N
k
kk
M
k
kk
za
zbzH
1
0
1)(
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Direct form I
)(2 ny)(nx
b1
b2
b0
z-1
z-1
bM-1
z-1bM
)1( nx
)2( nx
)1( Mnx
)( Mnx
)(ny
a1
a2
z-1
z-1
aN-1
z-1aN
)1( ny
)2( ny
)1( Nny
)( Nny
)(1 ny
N
kk
M
kk knyaknxbny
10
)()()(
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Direct form II
b1
b2
b0
z-1
z-1
bM-1
z-1bM
)(nx )(nyz-1
z-1
z-1
a1
a2
aN-1
aN
For an LTI cascade system, we can change the order of the systems without changing the overall system response
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IIR Filter Structures
In this form the system function H(z) is written as a product of second-order sections with real coefficients
Cascade form
21
21
1
11
1
1
1
11
1
1
1
0
)1)(1()1(
)1)(1()1(
1)(
N
kkk
N
kk
M
kkk
M
kk
N
k
kk
M
k
kk
zdzdzc
zqzqzpA
za
zbzH
21
21
1
22
11
1
1
1
22
11
1
1
)1()1(
)1()1()(
N
kkk
N
kk
M
kkk
M
kk
zazazc
zbzbzpAzH 21
21
22NNNMMM
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IIR Filter Structures
kk
k kk
kk zHAzaza
zbzbAzH )(
1
1)(
22
11
22
11
Each second-order section (called biquads) is implemented in a direct form ,and the entire system function is implemented as a cascade of biquads. If N=M, there are totally biquads.
21N
If N>M, some of the biquads have numerator coefficients that are zero, that is, either b2k = 0 or b1k = 0 or both b2k
= b1k = 0 for some k.
if N>M and N is odd, one of the biquads must have ak2 = 0, so that this biquad become a first-order section.
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22
11
22
11
1
1)(
zaza
zbzbzH
kk
kkkb1k
b2k
z-1a1k
a2k
biquad
z-1
First-order section
b1kz-1a1kb1kz-1a1k
a2kz-1
z-1a1k
a2kz-1
z-1a1k
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Example 321
321
81
43
451
21148)(
zzz
zzzzH
)5.01)(25.01()2644.52402.14)(3799.02(
)(211
211
zzz
zzzzH
-1.2402
5.2644
z-1
-0.5 z-1
z-10.25
4 2
-0.3799
)(nx )(ny
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IIR Filter Structures
Parallel form
In this form the system function H(z) is written as a sum of sections using partial fraction expansion. Each section is implemented in a direct form. The entire system function is implemented as a parallel of every section.
21
12
21
1
110
110
1
0
111
)(N
k kk
kkN
k k
kN
k
kk
M
k
kk
zaza
zbb
zc
AG
za
zbzH
21 2NNN
Suppose M=N
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IIR Filter Structures
2
1
11
21
1
110
0 1)(
N
k kk
kk
zaza
zbbGzH
if N is odd, the system has one first-order
section and second-order sections.
if N is even, the system has second-order
sections.
21N
2N
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Example
1111
111
)21
21(1)
21
21(1)
811)(
431(
)21)(321)(
211(10
)(zjzjzz
zzzzH
1
4
1
3
1
2
1
1
)21
21(1)
21
21(1)
811()
431(
)(
zj
A
zj
A
z
A
z
AzH
57.1425.12 ,57.1425.12 ,68.17 ,93.2 4321 jAjAAA
21
1
21
1
211
82.2650.24
323
871
90.1275.14)(
zz
z
zz
zzH
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21
1
21
1
211
82.2650.24
323
871
90.1275.14)(
zz
z
zz
zzH
-12.9z-17/8
-3/32 z-1
-14.75
26.82z-11
-1/2 z-1
24. 5
)(nx )(ny
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IIR Filter Structures
Transposition theorem
If we reverse the directions of all branch transmittances and interchange the input and output in the flow graph, the system function remains unchanged.
The resulting structure is called a transposed structure or a transposed form.
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FIR Filter Structures
The characteristics of the FIR filter FIR filters have Finite-duration Impulse Responses,
thus they can be realized by means of DFT The system function H(z) has the ROC of , t
hus it is a causal system An FIR filter is a nonrecursive system FIR filters can be designed to have a linear-phase re
sponse
0|| z
1
0
)()(N
n
nznhzHIt has N-1 order poles at
and N-1 zeros in 0|| z0z
The order of such an FIR filter is N-1
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FIR Filter Structures
In this form the difference equation is implemented directly as given.
Direct form
1
0
)()()(N
m
mnxmhny
z-1
h(1)
z-1z-1
h(2)
)(nx
)(nyh(0) h(N-2) h(N-1)
It requires N multiplications
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FIR Filter Structures
In this form the system function H(z) is converted into products of second-order sections with real coefficients
Cascade form
2
1
22
110
1
0
)()()(
N
kkkk
N
n
n zbzbbznhzH
It requires (3N/2) multiplications
z-1
z-1
z-1
z-1
z-1
z-1
b11
b21
b01
b12
b22
b02
b1x
b2x
b0x)(nx )(ny
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FIR Filter Structures
Linear-phase form
When an FIR filter has a linear phase response, its impulse response exhibits the above symmetry conditions. In this form we exploit these symmetry relations to reduce multiplications by about half.
Linear phase:The phase response is a linear function of frequency
Linear-phase condition
)1()()1()(nNhnhnNhnh
Symmetric impulse response
Antisymmetric impulse response
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If N is odd
211
21
0
)1(
12
1
0
)1(211
21
0
1
12
1
211
21
0
1
0
)2
1(])[(
)1()2
1()(
)()2
1()(
)()(
NN
n
nNn
N
m
mNN
N
n
n
N
Nn
nN
N
n
n
N
n
n
zNhzznh
zmNhzNhznh
znhzNhznh
znhzH
mNnlet 1
mnlet
)1()( nNhnh
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If N is odd
211
21
0
)1( )2
1(])[()(
N
N
n
nNn zNhzznhzH
1 for symmetric -1 for antisymmetric
z-1
±1z-1
z-1
±1z-1
±1
z-1
±1z-1
)12
1( Nh )2
1( Nh)2(h)1(h)0(h
)(nx
)(ny
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If N is even
12
0
)1(
12
0
)1(
12
0
1
2
12
0
1
0
])[(
)1()(
)()()()(
N
n
nNn
N
m
mN
N
n
n
N
Nn
n
N
n
nN
n
n
zznh
zmNhznh
znhznhznhzH
mNnlet 1
mnlet
)1()( nNhnh
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If N is even
1
2
0
)1( ])[( )(
N
n
nNn zznhzH
1 for symmetric -1 for antisymmetric
z-1
±1z-1
z-1
±1z-1
±1
z-1
±1z-1
)22
( Nh )12
( Nh)2(h)1(h)0(h
)(nx
)(ny
z-1±1
The linear-phase filter structure requires 50% fewer multiplications than the direct form.
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FIR Filter Structures
Frequency sampling form
This structure is based on the DFT of the impulse response h(n) and leads to a parallel structure. It is also suitable for a design technique based on the sampling of frequency response H(z)
1
0
1
01
)()(11
)(1)1()(N
kkc
N
kk
N
N zHzHNzW
kHN
zzH
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Nc zzH 1)(
It has N equally spaced zeros on the unit circle 1,,1,0 ,
2
Nkezk
Nj
k
)2
sin(21)( 2 NjeeeHNjNjj
c
)2
sin(2)( NeH jc
Nz
N2
N4
)( jc eH
0
2
All-zero filter or comb filter
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It has N equally spaced poles on the unit circle
1
01
1
0 1)(
)(N
kk
N
N
kk zW
kHzH
1,,1,0 ,2
Nkezk
Nj
k
resonant filter
The pole locations are identical to the zero locations and that
both occur at , which are the frequencies at which
the designed frequency response is specified. The gains of
the filter at these frequencies are simply the complex-valued
parameters
kN 2
)(kH
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1
0
1
01
)()(11
)(1)1()(N
kkc
N
kk
N
N zHzHNzW
kHN
zzH
Nz
z-10NW
)0(H
z-11NW
)1(H
z-1)1( NNW
)1( NH
)(nx )(ny
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FIR Filter Structures
Problems
It requires a complex arithmetic implementation
It is possible to obtain an alternate realization in which only real arithmetic is used. This realization is derived using the symmetric properties of the DFT and the factor.k
NW
It has poles on the unit circle, which makes this filter critically unstable
We can avoid this problem by sampling H(z) on a circle |z|=r where the radius r is very close to one but is less than one.
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1
011
)(1)(N
kk
N
rNN
zrW
kHNzrzH
r
jIm[z]
Re[z]
unit circle
)()( kHkH r
)(kH
)(kH r
1
011
)(1)(N
kk
N
NN
zrWkH
NzrzH
1
011
)(1)(N
kk
N
N
zWkH
NzzH
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1
011
)(1)(N
kk
N
NN
zrWkH
NzrzH
By using the symmetric properties of the DFT and the factor, a pair of single-pole filters can be combined to form a single two-pole filter with real-valued parameters.
kNW
1,,1,0 ,2
Nkezk
Nj
k
For poles
kkN zz
kN
kN
kNjkN
NjkN
N rWWrerrerW )()(2)(2
)(
h(n) is a real-valued sequence, so
)())(()( kRkNHkH NN
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So the kth and (N-k)th resonant filters can be combined to form a second-order section with real-valued coefficients)(zH k
221
110
11
1)(1
)2cos(211)(
1)(
1)(
1)(
)(
zrkN
rz
zbb
zrWkH
zrWkH
zrWkNH
zrWkH
zH
kkkN
kN
kNN
kN
k
even is ,12
,,2,1
odd is ,2
1,,2,1
NNk
NNk
])(Re[2
)](Re[2
1
0kNk
k
WkHrb
kHb
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If N is even, the filter has a pair of real-valued poles rz
If N is odd, the filter has a single real-valued pole rz
1
)2()(
1)0(
)(1
210
rz
NHzH
rzH
zH N
1
)0()(
10
rzH
zH
jIm[z]
Re[z]
N is even
|z|=r
0k2Nk
jIm[z]
Re[z]
N is odd
|z|=r
0k2Nk
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second-order section
b1kz-1
z-1
b0k
)2cos(2 kN
r
2r
first-order sections
z-1 r
)0(H
)(0 zH
z-1 r
)2(NH
)(2zHN
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N is even
12
120 )()()(1)(
N
kkN
NN
zHzHzHNzrzH
N is odd
2)1(
10 )()(1)(
N
kk
NN
zHzHNzrzH
NN zr
)(0 zH
)(1 zH
)(zH k
)(2zHN
N1
)(nx )(ny
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FIR Filter Structures
Fast convolution form
x(n): N1-point sequence
h(n): N2-point sequence
L ≥ N1 + N2 - 1
return
)(kXL-point
DFT
L-pointDFT
L-pointIDFT
)(nx
)(ny
)(kH
)(kY
)(nh
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return
0 1 2 3 4 5 6 7 8 9 100
2
4
6
h(n)=h(N-1-n)
n
0 1 2 3 4 5 6 7 8 90
2
4
6
h(n)=h(N-1-n)
n
0 1 2 3 4 5 6 7 8 9 10-4
-2
0
2
4h(n)=-h(N-1-n)
n
0 1 2 3 4 5 6 7 8 9-4
-2
0
2
4h(n)=-h(N-1-n)
n
11N
10N
11N
10N
return