Chapt 5 Design of Digital Filters_IIR
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7/31/2019 Chapt 5 Design of Digital Filters_IIR
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2/2/2012Watumull Institute of Technology, Worli
Prepared by Chandrashekhar Padole, DSP-BE Computer, Mumbai Uni 1
Chapt 05
Design of Digital Filters ( IIR)B.E. Comps, Mumbai Uni
PrePrepared by Chandrashekhar Padole
Lecturer
Watumull Institute of Tech , Worli
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Chapt 05 Design of Digital Filters
Design of FIR filters
Design of IIR filters from analog filters
Frequency transformation
Design of digital filters based on least-squares method
Digital filters from analog filters
Properties of FIR filters
Design of FIR filters using windows
Comparison of IIR and FIR filters
Linear phase filters
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Need/Applications of filters
In wireless transmission such as FM transmission or mobilecommunication frequency selective filters are used to separate outthe one channel signal from another.
For noise removal
Speech Processing- Speech recognition, TTS
Biomedical signal processing to diagnosis
Signal analysis and synthesis
Data compression
DTMF( Dual Tone Multi-frequency Receivers)
Mobile receivers
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Common Filter Types
1
0.707
1 2
Low Pass Filter
Pass-band Transitionband
Stop-band
3-dB or cut-offfrequency
H(w)
1
0.707
12
High Pass Filter
Pass-bandTransitionband
Stop-band
3-dB or cut-offfrequency
H(w)
1
0.707
1 2
Band-pass Filter
P.B.T.B. S.B.
3-dB or cut-offfrequency
H(w)
S.B. T.B.
1
0.707
1 2
Band-stop Filter
P.B. T.B.
3-dB or cut-offfrequency
H(w)
S.B. T.B. P.B.
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Design of Digital Filter
1. Designing an analog prototype filter (s-domain) from a givenspecification
2. Transforming analog filters ( s-domain) to a digital filter( z-
domain)
3. Realization of z-domain transfer function
Step 1 is not only dependent on the problem specificationbut also on the transformation technique to be used in step2 to convert the analog filter into digital filter. Intransformation , analog frequencies are not mapped as it isin digital frequencies. This leads to a need of very important
step of modification of given specification while designingthe analog filter.
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Contd..
Filter
Specification
Modificationof
frequencies
Analog filter
Design
Transformation: analog to
digital
Digital Filter
z=f(s)s=f(z)
iableFrequencyDigital
iableFrequencyAna
var__
var_log_
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Designing Analog Filter
Depending on application requirements and availability , one has todecide which kind of response is suitable. ( Butterworth or Chebyshev orElliptic or any other type)Normally factors on which particular type of filter is selected , are
Order of filter ( trade-off between sharp cut-off and computations)
Allowed ripples in pass-band and/or stop-band
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Contd..
% order of the filtersN = 5;% cut-off normalized frequency
band = 0.5;% ripple in the pass-band (dB)
Rpass = 0.5;% ripple in the stop-band (dB)Rstop = 20; w = 0:pi/255:pi;[num, den] = butter(N, band, 'low');
butterfilter = abs(freqz(num, den ,w));
[num, den] = cheby1(N, Rpass, band);cheby1filter = abs(freqz(num, den ,w));[num, den] = cheby2(N, Rstop, band);cheby2filter = abs(freqz(num, den ,w));
[num, den] = ellip(N, Rpass, Rstop, band);
ellipfilter = abs(freqz(num, den ,w));F = [w/pi ; butterfilter; cheby1filter; cheby2filter; ellipfilter];F = F';
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Design of Butterworth type filter
The magnitude function of the Butterworth low pass filter is given by
( )[ ]
2/121
1)(
N
c
jH
+
=
where N is the order of filterc is cut-off frequency (3dB frequency)
At < c , |H(jw)| 1At > c , |H(jw)| decreases rapidlyAt = c , |H(jw)| =0.707 ( -3dB)
1
0.707
c
H(jw)
As order of filter increases , transition region becomes steeperand approaches ideal filter response as N , which is
practically impossible
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Contd..
Designing Butterworth filter has following steps
1. Determine the order of filter
2. Cut-off frequency
3. Determine transfer function in s-domain
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To determine the order of filter
In problem specification/requirements, two gains ( orattenuation) at two frequencies are given , which can be used
to determine the order of filter
1
Ap
As
1 2
H(w)
Since designed analog filter would undergo
through transformation by one of the technique( BLT, IIT etc) , modification of frequencies are
required.
For bilinear transformation
For impulse invariant method
2tan
2
T=
T=
Thus, now we have analog frequencyspecification1 and 2 ,
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Contd..
Now calculate the order of filter using this equation
p
s
A
A
p
s
N
log
1
1log
2
1 2
2
1
1
General formula
This equation is validonly if frequency
response is specified in
gains
p
s
P
S
N
log
1
1log
2
1
Normalized gains( A)
Gains in dB ()
Attenuation in dB()
where2
1
sAS = 2
1
pAP =
sS1.0
10
=
sS1.0
10=
pP1.0
10
=
pP1.0
10=
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Derivation for order of filter ( N )
For designing Butterworth filter , we will have analog frequencyspecificationp and s with corresponding gains Ap and As resp.
We have ,
( )[ ]2/121
1)(
N
c
jH
+
=
( ) 111
2
2
+
Np
c
pA
At p ,
( )[ ]2
2
11 sN A
c
s
+
At s ,
( ) 11 22
p
N
Acp ( ) 11
2
2
s
N
Acs
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Contd..
Considering equality from both equation and taking
ratio, we get
( )
=
11
11
2
22
p
sN
A
A
p
s
Taking log and solving we get,
p
s
A
A
p
s
N
log
1
1log
21
2
2
1
1Log can be of any
base but take same
base for num anddenum
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To determine cut-off frequency
Cut-off frequency,
N
p
p
c
A
21
112
=
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Derivation for cut-off frequency
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To determine the analog transfer function ( s-domain)
).().........)()((
1)(
1210 =
N
assssssss
sH
Where, each pole isgiven by
++
=
N
kj
ck es2
12
2
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Derivation for pole
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Butterworth polynomials for different orders N
Order ,n Factors of Polynomial Bn(s)
1 (s + 1)
2 s2 + 1.4142s + 1
3 (s + 1)(s2
+ s + 1)4 (s2 + 0.7654s + 1)(s2 + 1.8478s + 1)
5 (s + 1)(s2 + 0.6180s + 1)(s2 + 1.6180s + 1)
6 (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9319s + 1)
7 (s + 1)(s2 + 0.4450s + 1)(s2 + 1.2470s + 1)(s2 + 1.8019s + 1)
8 (s2
+ 0.3902s + 1)(s2
+ 1.1111s + 1)(s2
+ 1.6629s + 1)(s2
+ 1.9616s + 1)
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Design steps for Butterworth Filter ( summary)
Calculate the order of filter usingthis equation
p
s
A
A
p
s
N
log
1
1log
2
1 2
2
1
1 Cut-off
frequency, N
p
p
c
A
21
11
2
=
).().........)()((
1)(
1210 =
N
assssssss
sH
Where, each pole isgiven by
++
=
N
kj
ck es2
12
2
Transfer function
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Problem
Determine analog filter H(s) to design digital filter by IIT methodsatisfying following constraints
1)(707.0 jweH
2.0)( jweH
2/0
4/3
with T=1 s
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Solution
4/32/
2
1
=
=Given
2.0707.0
2
1
=
=
AA
Transformation technique is IIT
T
=
1
0.707
1 2
H(w)
1 2
2/11 ==T
with T=1 s 4/322 ==T
Order of filter
89.3
2/
4/3log
1
1log
2
1
log
11log
2
1 2
2
2
2
707.0
1
2.01
1
1
=
=
p
s
A
A
p
s
N
89.3N
4=NAs order filter should be integer
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Contd..
2/
1707.0
1
2/
11
81
21
22
=
=
=N
p
p
c
A
Cut-off frequency
Transfer function).().........)()((
1)(
1210 =
N
assssssss
sH
++
=
N
kj
ck es2
12
2
Ha(s) poles
++
=
Nkj
k es2
122
We use normalized poles ( with c =1 ) as indigital transformation , actual c would beconsidered
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Contd..
k ( 0N-1) = kth pole
0 -0.382+j0.92
1 -0.92+j0.382
2 -0.92-j0.382
3 -0.382-j0.92
++
=
N
kj
k es2
12
2
+
=82
0
j
es
+
=8
3
2
1
j
es
+
=8
5
2
2
jes
+
=8
7
2
3
j
es
=8
5
0
j
es
=8
7
1
j
es
=8
9
2
j
es
=8
11
3
j
es
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Contd..
))()()((1)(
3210 sssssssssHa
=
))92.0382.0())(382.092.0())(382.092.0())(92.0382.0((
1
jsjsjsjs ++=
)184.1)(1765.0(
1)(
22++
=ssss
sHa
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Problem
Design a low pass filter ( Butterworth) with following specificationsPass-band 0-500HzStop-band 2-4kHzPass-band ripple 3dB
Stop-band attenuation 20dBSampling Frequency 8kHzBilinear transformation to be used
Ans- Frequency warping2
tan2
T=
Order of filter N=2
Cut-off frequency c = ???
Transfer function12
1)(
2+
=ss
sHa
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Design of Chebyshev type filter
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Contd..
Chebyshev low-pass filter has magnitude response , given by
2/1
221
)(
+
=
c
NC
AjwH
where A is filter gain 11
2
1
=A
and c is cut-off frequency and
The chebyshev polynomial of Nth order CN(x) is given by
=
)coshcosh(
)coscos()(
1
1
xN
xNxCN
1||
1||
xfor
xfor
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Design Steps
Determine the order of filter
1
21
2
2
1
cosh
111coshA
N
Transfer function
11
2
1
=
A
Poles are calculated as :
N/12 11
++=
2
12
11
+=r
2
12
12
=r
Now, the kth normalized pole is given by
kkk jrrs sincos 12 += 1,......1,0.. = Nkfor
whereN
kk
2
)12(
2
++=
Actual poles can be given assk=sk c
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Problem
Determine analog chebyshev filter for following specifications ofdigital filter by using bilinear transformation technique
1)(8.0 jweH
2.0)( jweH
2.00
6.0
with T=1 s
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Solution
6.02.0
2
1
==Given
2.08.0
2
1
==
AA
Transformation technique is bilinear
1
0.707
1 2
H(w)
1 2
with T=1 s
Order of filter
20.1N 2=NAs order filter should be integer
2tan
2
T=
65.02
tan2 1
1 ==T
75.22
tan2 2
2 ==T
1
21
2
2
1
cosh
111
coshA
N
75.018.0111 22
1
===A
( ) 20.112.2
563.2
23.4cosh
)53.6(cosh
65.0
75.2cosh
12.0
1
75.0
1cosh
1
1
1
2
1
===
N
C d
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Contd..
2nd order transfer function
Now, the kth normalized pole is given by
kkk jrrs sincos 12+=
1,......1,0..=
Nkfor
whereN
kk
2
)12(
2
++=
2
12
11
+
=r
2
12
12
=rand
where
75.018.0
11
122
1
===A
N/12 11
++=
where
C td
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Contd..
73.175.0
175.01
2/1
2=
++=
75.0
73.1*2
173.165.0
2
1 22
11 =+
=+
=
r 375.0
73.1*2
173.165.0
2
1 22
12 =
=+
=
r
4
3
420 =+=
N
kk
2
)12(
2
++=
4
5
4
3
21
=+=
kkk jrrs sincos 12 +=
53.0265.04
3sin75.0
4
3cos375.00 jjs +=
+
=
53.0265.04
5sin75.0
4
5cos375.01 jjs =
+
=
C td
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Contd..
))((1)(
10 sssssHa
=
))53.0265.0())(53.0265.0((
1)(
jsjssHa
+=
35.053.0
1)(
2++
=ss
sHa
R t P bl f h b h
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Repeat Problem for chebyshev
Determine analog filter H(s)( chebyshev) to design digital filter byIIT method satisfying following constraints
1)(707.0 jweH
2.0)( jweH
2/0
4/3
with T=1 s
Solution
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Solution
4/32/
2
1
=
=Given
2.0707.0
2
1
=
=
AA
Transformation technique is IIT
1
0.707
1 2
H(w)
1 2
with T=1 s
Order of filter
358.2N 3=NAs order filter should be integer
1
21
2
2
1
cosh
111
coshA
N
11707.0111 22
1
===A
358.296.0
27.2
2/
4/3cosh
12.0
1cosh
1
2
1
==
N
2/11 ==T
4/322 ==T
T=
Compare with Butterworth .
Order of chebyshev is
smaller but at the cost ofripples in pass band
Contd
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Contd..
Transfer function for 3rd
order chebyshev
Analog to digital Transformation( Step II)
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Analog to digital Transformation( Step II)
Transforming analog filter in to digital filter ( s-domain z-domain)
From first step , we get the analog filter transfer function in s-domain with a low pass characteristics always.
If requirement is of other than low pass filter, transfer functionwould be transformed for desired characteristic filter ( high pass, band
pass ,band reject etc).
This transformation can be done either in analog domain itselfor in digital domain.
Cont
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Cont
Normalized LP-Analog filter( s-domain)
Transformation in s-
domain for HP/BP/BR
Analog to digitaltransformation
s z
Analog to digitaltransformation sz
( BLT, IIT etc)
Digital FilterLP
Transformation ins-domain forHP/BP/BR
Digital FilterHP/BP/BR
Digital FilterHP/BP/BR
De-normalization
Ss/
Frequency transformation in s-domain
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Frequency transformation in s-domain
Normalized LPanalog filter in s-
domain
Frequencytransformation in
s-domain
Desired Filter in S-domain
Desired Filter Frequency transformation
Low pass filter
High pass filter
Band-pass filter
Band-stop filter
c
sS
sS c
s
sQS
0
2
0
2 )(
+ 21
2
0 =
12
0
=Q
)(2
0
2
0
+
sQ
sS
Problem
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Problem
From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
cs
BcsHa
+=)(
Solution
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Solution
A ) High Pass filter
cs
BcsH
LP
a+
=)(
ss
LP
a
HP
acsHsH /)()(
=
sc
Bcs
c
Bc
csc +
=+
=
c
HP
acs
BssH
+
=)(
Contd
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Contd..
B) Band pass filter
cs
sQ
BcsH
BP
a
+
+=
0
2
0
2 )()(
csQQs
sBc
.02
0
2
0
++
=
( )( ) 20
20
0
++=
ss
sBc
Qc
Q
Contd..
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Contd..
C) Band reject filter
csQ
s
BcsH
BR
a
++
=
)(
)(
2
0
2
0
)(
)(.2
0
2
0
2
0
2
++
+=
sQcs
sQBc
2
002
2
0
2 )(
+
+
+=
sQ
s
sB
c
20
02
2
0
20
02
2
++
+
++
=
sQ
s
B
sQ
s
Bs
cc
Analog to Digital Transformation
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Analog to Digital Transformation
Mapping of s-plane to z-plane is done such that
1. Points in left-hand plane of s-plane mapped into correspondingpoints
2. Points in right-hand plane of s-plane mapped into corresponding
points outside unit circle of z-plane3. j axis is mapped along the perimeter of unit circle of z-plane
j
s-planez-plane
1
Different Approaches
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pp
Following are different methods for obtaining digital filter from ananalog filter ( mapping of s-plane to z-plane)
1. Approximation of derivatives2. Impulse-invariant transformation
3. Bilinear Transformation4. Matched z-transformation
Approximation of Derivatives
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pp
Differentiation is given by
nT-T nT
y(t)
T
TnTynTy
dt
tdy
nTt
)()()( =
=
)1()( = nyny
where T is sampling interval
Differentiation in s-domain,
Differentiatory(t)
dt
tdy )(
ssH = )(
Differentiation in z-domain,
Differentiatory(n)
TzzH
1
1)(
=
T
nyny )1()(
Contd..
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Thus , if we replace s by 1-z
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
Problem
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From a normalized low pass filters transfer function given below,obtain a) HP b) BP and c) BR filters
End of Chapter 05
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Queries ???