chap7-iir filter design
TRANSCRIPT
Copyright © 2005. Shi Ping CUC
Chapter 7IIR Filter Design
Content
Preliminaries
Characteristics of Prototype Analog Filters
Analog-to-Digital Filter Transformations
Frequency Transformations
Copyright © 2005. Shi Ping CUC
Preliminaries
How to design a digital filter
First: Specifications
The design of a digital filter is carried out in three steps:
Before we can design a filter, we must have some specifications. These specifications are determined by the applications. Second: Approximations
Once the specifications are defined, we use various concepts and mathematics to come up with a filter description that approximates the given set of specifications. This step is the topic of filter design.
Copyright © 2005. Shi Ping CUC
Preliminaries
Third: Implementation
The product of the above step is a filter description in the form of either a difference equation, or a system function, or an impulse response. From this description we implement the filter in hardware or software on a computer.
In this and the next chapter we will discuss in detail only the second step, which is the conversion of specification into a filter description.
Copyright © 2005. Shi Ping CUC
Preliminaries
In many applications, digital filters are used to implement frequency-selective operations;
Therefore, specifications are required in the frequency-domain in terms of the desired magnitude and phase response of the filter;
Generally a linear phase response in the passband is desirable;
An FIR filter is possible to have an exact linear phase;
An IIR filter is impossible to have linear phase in passband. Hence we will consider magnitude-only specifications.
The specifications
Copyright © 2005. Shi Ping CUC
Preliminaries
There are two ways to give the magnitude specifications
Absolute specifications
Provide a set of requirements on the magnitude response function and generally used for FIR filters.)( jeH
πωaeH
eHa
sj
pj
|| )(
|| 1)(1
2
1
],0[ p Passband
],[ s Stopband
],[ sp Transition band
The ending frequency of the passband. BandwidthpThe beginning frequency of the stopband. s
The tolerance (or ripple) in passband1a
The tolerance (or ripple) in stopband2a
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Preliminaries
Relative specifications (dB)
Provide requirements in decibels (dB). This approach is the most popular one in practice and used for both FIR and IIR filters
2
0
2
1
0
1
lg20)(lg20)(
)(lg20
)1lg(20)(lg20)(
)(lg20
aeHeH
eH
aeHeH
eH
s
s
p
p
j
j
j
j
j
j
The maximum tolerable passband ripple1pR
The minimum tolerable stopband attenuation2sA
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Preliminaries
Examples
In a certain filter’s specifications the passband ripple is 0.25dB, and the stopband attenuation is 50dB. Determine the a1 and a2.
0.003210 ,lg2050
0.0284101 ),1lg(2025.0
)2050
(
222
)2025.0
(
111
aa
aa
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Preliminaries
dB60001.0lg20lg20
0.1755dB)02.01lg(20)1lg(20
22
11
a
a
Given the passband tolerance a1=0.02 and the stopband tolerance a2=0.001, determine the passband ripple and the stopband attenuation1 2
Copyright © 2005. Shi Ping CUC
Preliminaries
The basic technique of IIR filter design
IIR filters have infinite-length impulse responses, hence they can be matched to analog filters.
Analog filter design is a mature and well developed field.
We can begin the design of a digital filter in the analog domain and then convert the design into the digital domain
Copyright © 2005. Shi Ping CUC
Preliminaries
There are two approaches to this basic technique
Approach 1
Design analog lowpass filter
Apply freq. band transformation
s → s
Apply filter transformation
s → z
Designed IIR filter
Approach 2
Design analog lowpass filter
Apply filter transformation
s → z
Apply freq. band transformation
z → z
Designed IIR filter
return
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Characteristics of Prototype Analog Filters
Magnitude-squared function
sa
pa
AjH
jH
|| ,1
)(0
|| ,1)(1
1
2
2
2
2
Let be the frequency response of an analog filter
)( jHa
is a passband ripple parameter
is the passband cutoff frequency in rad/secpis the stopband cutoff frequency in rad/secs
is a stopband attenuation parameterA
sa
pa
AjH
jH
at 1
)(
at 1
1)(
2
2
2
2
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Characteristics of Prototype Analog Filters
jsaa sHjH )()(
The properties of 2
)( jHa
jsaaaaaaa sHsHjHjHjHjHjH )()()()()()()(2
)(tha is a real function
The poles and zeros of are distributed in
a mirror-image symmetry with respect to the axis.
For real filters, poles and zeros occur in complex
conjugate pairs.
j)()( sHsH aa
22
2)()()(
saaa jHsHsH
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Characteristics of Prototype Analog Filters
j)()( sHsH aa
0
2
2
s-plane
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Characteristics of Prototype Analog Filters
How to construct )(sHa
)(sHa is the system function of the analog filter. It must be causal and stable. Then all poles of must lie within the left half-plane.
)(sHa
)()( sHsH aa All left-half poles of should be assigned to
)(sHa
)(sHa )( sHa Zeros are not uniquely determined. They can be halved between and . (Zeros in each half must occur in complex conjugate pairs)
If a minimum-phase filter is required, the left-half zeros should be assigned to )(sHa
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Examples
)36)(49(
)25(16)(
22
222
jHa
)36)(49()25(16
)()()(22
222
22 sss
jHsHsHs
aaa
poles 6 ,7 ss 2th order zeros 5js
We can assign left-half poles and a pair
of conjugate zeros to
6 ,7 ss5js )(sHa
)6)(7()25(
)(2
0
ss
sKsHa
4
) ()(
0
00
K
jHsH asa
4213
1004)6)(7(
)25(4)(
2
22
ss
sss
ssHa
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Characteristics of Prototype Analog Filters
Butterworth lowpass filters
This filter is characterized by the property that its magnitude response is flat in both passband and stopband. The magnitude-squared function of an Nth-order lowpass filter is given by
N
c
a jH2
2
1
1)(
Copyright © 2005. Shi Ping CUC
Characteristics of Prototype Analog Filters
The properties of Butterworth lowpass filters
At , for all N1)( jHa0
707.02
1)( jHa At , for all N, which
implies a 3dB attenuation at
c
c
)( jHa is a monotonically decreasing function of
)( jHa approaches an ideal lowpass filter as N
)( jHa 0 is maximally flat at since derivatives of
all orders exist and are equal to zero
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Characteristics of Prototype Analog Filters
The poles and zeros of )()( sHsH aa
Nc
N
Nc
N
c
jsaaa js
j
js
jHsHsH22
2
2/
2
) (
) (
1
1) ()()(
Nkejs Nkj
ccN
k 2,,2,1 ,)()1()
212
21(
21
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Characteristics of Prototype Analog Filters
)()( sHsH aa There are 2N poles of , which are
equally distributed on a circle of radius with angular
spacing of radians.c
N
If the N is odd, there are poles on real axis.
If the N is even, there are not poles on real axis.
The poles are symmetrically located with respect to the imaginary axis.
A pole never falls on the imaginary axis, and falls on the real axis only if N is odd.
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Characteristics of Prototype Analog Filters
N
kk
Nc
a
sssH
1
)()(
Nkes Nkj
ck ,,2,1 ,)
212
21(
In general, we consider and this results in a
normalized Butterworth analog prototype filter
rad/s 1c
)(sHan
)()(c
anasHsH
When designing an actual filter with , we
can simply do a replacement for s, that is
)(sHa rad/s 1c
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Designing equations
Given , two parameters are required to
determine a Butterworth lowpass filters : 21 ,,, sp
cN ,
2
2
1
2
1lg20
1lg20
N
c
ss
N
c
pp
at
at
Solving these two equations for cN ,
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,
)lg(2
)110/()110(lg
101021
NNN
s
p
pSince the actual N chosen is larger than required, specifications can be either met or exceeded at or s
pTo satisfy the specifications exactly at
sTo satisfy the specifications exactly at
N
sc
2 10 1102
N
pc
2 10 1101
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Example
Determine the system function of 3th-order Butterworth analog lowpass filter. Suppose rad/s 2c
62
2
21
1
1
1)(
N
c
a jH
6,,2,1 ,2)
612
21(
keskj
k
641
1)()(6s
sHsH aa
884
8
))()(()(
23321
3
sssssssss
sH ca
Solution:
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Design the above filter with normalized Butterworth analog prototype filter. See table 6-4 on page 261
1 ,)( 02210
0
NNN
an aasasasaa
dsH
in case of 1)0( jHa00 ad
3NFor 2 ,2 21 aaWe can find
3232
2
4888
)2
()2
(2)2
(21
1
)()(
ssssss
sHsH ssana
2sss
c
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Design a lowpass Butterworth filter to satisfy:
rad/s 1020for dB 1 41 Passband
Stopband rad/s 105.12for dB 15 42
Solution:
15105.12
1lg20
1102
1lg20
24
24
N
c
N
c
rad/s 105.12 dB, 15
rad/s 102 dB, 14
s2
4p1
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rad/s 1013.12 4 c
5.8858
105.12102
lg2
)110/()110(lg
)lg(2
)110/()110(lg
4
4
10
15
10
1
101021
s
pN
6N
4
12 1015
4
2 10
101.12792110
105.12
1102
N
sc
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Look for table 6-4 on page 261
864.3 ,464.7 ,142.9 ,464.7 ,864.3 54321 aaaaa
65432
655
44
33
221
864.3464.7142.9464.7864.31
1
1
1)(
ssssss
ssasasasasasHan
41013.12
)()()(
ssans
sana sHsHsHc
6554103152202429
29
102.74103.76103.27101.90106.97101.28
101.28
ssssss
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Look for table 6-6 on page 263
259.0966.0
707.0707.0
966.0259.0
4,3
5,2
6,1
js
js
js
To construct a cascade structure
)1.001.93)(1.001.41)(1.00 0.52(
1
))()()()()((
1)(
222
654321
ssssss
sssssssssssssHan
41013.12
ss
sc
)105.04101.37)(105.0410)(105.04 103.69(
1028.1)(
952952942
29
ssssss
sHa
Copyright © 2005. Shi Ping CUC
Characteristics of Prototype Analog Filters
Chebyshev lowpass filters There are two types of Chebyshev filters
Chebyshev-I: equiripple in the passband and monotonic in the stopband.
Chebyshev-II: monotonic in the passband and equiripple in the stopband.
Chebyshev filters can provide lower order than Butterworth filters for the same specifications.
cN
a
CjH
22
2
1
1)(
Chebyshev-I
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N is the order of the filter
is the Nth-order Chebyshev polynomial given by2NC
cN x
xxN
xxNxC where
1|| ),ch(ch
1|| ),coscos()(
-1
1
xxCxC
xCxxCxC NNN
)(,1)(
)()(2)(
10
11
is the passband ripple factor. 10
cN
a
CjH
22
2
1
1)(
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cN
a
C
jH221
1)(
The properties of Chebyshev lowpass filters
At :
0
even is for 1
1)0(
odd is for 1)0(
2NjH
NjH
a
a
At :
c NjH ca all for 21
1)(
For :
c021
1 ~ 1 between oscillates
)( jHa
c For :
0 tolly monotonica decreases )( jHa
s For : A
jH sa1)(
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Designing equations
Given , two parameters are required to
determine a Chebyshev-I filter: 21 ,,, sc
N,
110 11.02
c
sch
ch
N1
1.01 110 2
1101 21.0
1chN
chcs
11 1
3 chN
chcdB Note: this is only for dBc 3
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Determine system function
To determine a causal and stable , we must find
the poles of and select the left half-plane
poles for . The poles are obtained by finding the
roots of
)(sHa
)()( sHsH aa )(sHa
01 22
cN j
sC
It can be shown that if
are the (left half-plane) roots of the above polynomial,
then
Nkjs kkk ,,2,1
Nk
b
Nk
a
ck
ck
2)12(
cos)(
2)12(
sin)(
Nk ,,2,1
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1112
)(21),(
21 1111
NNNN ba
N
kk
a
ss
KsH
1
)()(
even is N ,
1
1odd is N ,1
)0(2
jHa
Where K is a normalizing factor chosen to make
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Determine poles by geometric method
The poles of fall on an ellipse with major axis
and minor axis .
)()( sHsH aa cb
ca
j
cbca
N
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Determine the system function of 2th-order Chebyshev-I lowpass filter. Suppose and rad/s 1c dB 11
2589.0110110 1.01.02 1 From table 6-5 on page 261
20
210
0
0977.11025.1)(
ss
d
ssaa
dsHa
0977.1
1025.1
1
0
a
a
0.89132589.11
1
1)0(2
jHa
9827.0 ,8913.01025.1
)( 00
0
d
dsH
sa
Examples
Solution:
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Design a lowpass Chebyshev-I filter to satisfy:
rad/s 102 4 cPassband cutoff:
dB 11 Passband ripple:
Stopsband cutoff: rad/s 105.12 4 s
Stopband attenuation: dB 152
Solution: 5088.0110110 1.01.0 1
3.1978)5.1(
(10.8761)
110
1
1
1
1.01
2
ch
ch
ch
ch
N
c
s
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4N 5088.0Look for table 6-5 on page 261
dB 11
9528.0 ,4539.1 ,7426.02756.0 3210 aaaa ,
4320
433
2210
0
9528.04539.17426.02756.0
)(
ssss
d
ssasasaa
dsHan
0.89132589.11
1
1)0(2
jHa
0.2456 ,0.89132756.0
)( 00
0
d
dsH
san
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432 9528.04539.17426.02756.02456.0)(
sssssHan
4102
sss
c
434291418
18
105.9866105.7398101.8420104.2954103.8278
)(
ssss
sHa
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Look for table 6-7 on page 264
4073.03369.0 ,9834.01395.0 3,24,1 jsjs
To construct a cascade structure
)0.27946738.0)(0.9865 2790.0(2456.0
))()()(()(
22
4321
0
ssss
ssssssssd
sHan
4102
sss
c
)101.1030104.2336)(103.8945 101.7530(108278.3)(
942942
18
sssssHa
return
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Analog-to-Digital Filter Transformations
Impulse invariance transformation Definition
To design an IIR filter having a unit sample response h(n) that is the sampled version of the impulse response of the analog filter. That is
)()( nThnh aT : Sampling interval
Tjj eeT or ,
Since this is a sampling operation, the analog and digital frequencies are related by
Copyright © 2005. Shi Ping CUC
The system function and are related by)(sHa)(zH
k
aezk
TjsH
TzH sT )
2(
1)(
This implies a mapping from the s-plane to the z-plane
T
T
j
0
T3
T3
]Re[z
]Im[zj
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Analog-to-Digital Filter Transformations
Properties
Using ]Re[s
UC) the of (outside 1|z| into maps 0
UC) the (on 1|z| into maps 0
UC) the of (inside 1|z| into maps 0
Since the entire left half of the s-plane maps into the unit circle, a causal and stable analog filter maps into a causal and stable digital filter.
All semi-infinite left strips of width map into . Thus this mapping is not unique but a many-to-one mapping
T/2 1|| z
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Analog-to-Digital Filter Transformations
|| ),(1
)(T
jHT
eH ajthen
There will be no aliasing.
Frequency response
ka
j
T
kjH
TeH )
2(
1)(
TTjHjH aa
||for 0)()(If
To minimize the effects of aliasing, the T should be selected sufficiently small.
If the filter specifications are given in digital frequency domain, we cannot reduce aliasing by selecting T.
Aliasing occurs if the filter is not exactly band-limited
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Analog-to-Digital Filter Transformations
Digitalizing of analog filters
N
k k
ka
ss
AsH
1
)(
Using partial fraction expansion, expand into)(sHa
The corresponding impulse response is
N
k
tskaa tueAsHLth k
1
1 )()]([)(
N
k
nTsk
N
k
nTska nueAnueAnThnh kk
11
)()()()()(
To sample the )(tha
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The z-transform of is)(nh
N
kTs
k
n
N
k
nTsk
n
n
ze
AzeAznhzH
k
k
11
0 1
1
1)()()(
Conclusions:
N
k k
ka
ss
AsH
1
)(Compared with
The pole in s-plane is mapped to the pole in z-planeks Tske
The partial fraction expansion coefficient of is the same as that of )(sHa
)(zH
The zeros in the two domains do not satisfy the same relationship
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Analog-to-Digital Filter Transformations
An alternative method
|| ),()2
()(
1)(
)()(
11
TjHk
Tj
TjHeH
ze
TAzH
nTThnh
ak
aj
N
kTsk
a
k
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Analog-to-Digital Filter Transformations
Advantages and disadvantages
The digital filter impulse response is similar to that of a analog filter. This means we can get a good approximations in time domain.
Due to the presence of aliasing, this method is useful only when the analog filter is essentially band-limited to a lowpass or bandpass filter in which there are no oscillations in the stopband.
It is a stable design and that the frequencies and are linearly related. So a linear phase analog filter can be mapped to a linear phase digital filter.
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Design procedure
Choose T and determine the analog frequencies
Transform analog poles into digital poles to obtain the digital filter
TTs
sp
p
,
Given the digital lowpass filter specifications 21 , , , sp
Design an analog filter using the specifications )(sHa
21 , , , sp
N
k k
ka
ss
AsH
1
)(
Using partial fraction expansion, expand into)(sHa
N
kTs
k
ze
AzH
k1
11)(
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Analog-to-Digital Filter Transformations
Examples
Transform3
1
1
1
34
2)(
2
sssssHa
into a digital filter using the impulse invariance method in which T=1
)(zH
21
1
4231
31
311
0183.04177.01
3181.0
)(1
)(
11)(
zz
z
ezeez
eeTz
ez
T
ez
TzH
TTT
TT
TT
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Design a lowpass digital filter using a Butterworth prototype to satisfy
dB15 ,3.0
dB1 ,2.0
2
1
s
p
Solution
Let T=1, and then
3.0 ,2.0 TT
ss
pp
Design an analog filter using the specifications )(sHa
21 , , , sp
65432 717.2691.3179.3825.1664.0121.0
121.0)(
sssssssHa
N
k k
ka
ss
AsH
1
)(
Using partial fraction expansion, expand into)(sHa
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)4970j 4970(
0711
49704970
0711
j0.679) 0.182(
j0.249 0.144
j0.679) 0.182(
j0.249 0.144
j0.182) 0.679(
j1.6070.928
j0.182) 0.679(
j1.6070.928)(
..s
.
). j .(s
.
s
ssssHa
Transform analog poles into digital poles to obtain the digital filter
N
kTs
k
ze
AzH
k1
11)(
21
1
21
1
21
1
645.0297.11
446.0287.0
370.0069.11
1450.1143.2
257.0997.01
630.0859.1)(
zz
z
zz
z
zz
zzH
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Analog-to-Digital Filter Transformations
Bilinear transformation Definition
This is a conformal mapping that transforms the -axis into the unit circle in the z-plane only once, thus avoiding aliasing of frequency components. This mapping is the best transformation method.
j
2
tan 1TcTj
Tj
Tj
Tj
Tj
Tj
e
ec
ee
eecj
1
1
11
11
1
1
22
22
1
1
1
1
1
11
1
z
zc
e
ecs
Ts
Ts
sc
scz
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Ts
Ts
e
ecs
1
1
1
1
Tsez 1
1
1
1
1
z
zcs
T
T
1j
10
s1-plane
]Re[z
]Im[zj
0
z-plane
0
j
s-plane
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Analog-to-Digital Filter Transformations
Parameter c
Tc
Tc
Tc
2 then ,
22tan 1
11
Keeping a good corresponding relationship between the analog filter and the digital filter in low frequencies. i.e. in low frequencies1
2
cot then 2
tanc2
tan 1 cc
ccc c
Tc
Keeping a good corresponding relationship between the analog filter and the digital filter in a specific frequency (for example, in the cutoff frequency, )Tcc 1
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Properties Using , we obtain js
22
22
)(
)(|| ,
)(
)(
c
cz
jc
jc
sc
scz
So 1|| 0 ,1|| 0 ,1|| 0 zzz Using , we obtainjez
jjce
ec
z
zcs
j
j
)2
tan(1
1
1
11
1
The imaginary axis maps onto the unit circle in a one-to-one fashion. Hence there is no aliasing in the frequency domain.
The entire left half-plane maps into the inside of the unit circle. Hence this is a stable transformation.
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Analog-to-Digital Filter Transformations
Advantages and disadvantages
It is a stable design;
There is no aliasing;
There is no restriction on the type of filter that can be transformed;.
The frequencies and are not linearly related. So a linear phase analog filter cannot be mapped to a linear phase digital filter.
Copyright © 2005. Shi Ping CUC
Design procedure
Choose a value for T. We may set T=1
)2
tan(2
),2
tan(2 s
sp
p TT
Given the digital lowpass filter specifications 21 , , , sp
Prewarp the cutoff frequencies and ; that isp s
Design an analog filter to meet the specifications
21 , , , sp )(sHa
Finally, set )1
12()(
1
1
z
z
THzH a
and simplify to obtain as a rational function in)(zH 1z
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Analog-to-Digital Filter Transformations
Examples
Transform into a digital filter using the bilinear transformation.
Choose T=1
34
2)(
2
sssHa
21
21
1
12
1
1
1
1
1
1
1
0.070.131
13.00.2713.0
311
2411
2
2
)1
12()
1
12()(
zz
zz
zz
zz
z
zH
z
z
THzH a
T
a
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Design the digital Chebyshev-I filter using bilinear transformation. The specifications are:
dB15 ,3.0
dB1 ,2.0
2
1
s
p
Solution
Let T=1
1.0191)15.0tan(2)2
tan(2
0.6498)1.0tan(2)2
tan(2
ss
pp
T
T
Prewarp the cutoff frequencies
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Design an analog Chebyshev-I filter to meet the specifications
21 , , , sp )(sHa
0.04920.20380.61406192.0
0438.0)(
234
sssssHa
)0.64931.55481)(0.84821.49961(
)1(0018.0
0.55072.29253.82903.05431
0.00180.00730.01100.00730018.0)(
2121
41
4321
4321
zzzz
z
zzzz
zzzzzH
Copyright © 2005. Shi Ping CUC
Analog-to-Digital Filter Transformations
Comparison of three filters
Using different prototype analog filters will give out different N and the minimum stopband attenuations.
dB15 ,3.0
dB1 ,2.0
2
1
s
pGiven the digital filter specifications:
prototype Order N Stopband Att.
Butterworth 6 15 dB
Chebyshev-I 4 25 dB
Elliptic 3 27 dB
return
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Introduction
The treatment in the preceding section is focused primarily on the design of digital lowpass IIR filters. If we wish to design a highpass or a bandpass or a bandstop filter, it is a simple matter to take a lowpass prototype filter and perform a frequency transformation.
Frequency transformations in the analog domain
Frequency transformations in the digital domain
There are two approaches to perform the frequency transformation
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Approach 1
Analog lowpass filter
Frequency transformation
s → s
Filter transformation
s → z
Designed IIR filter
Approach 2
Analog lowpass filter
Filter transformation
s → z
Frequency transformation
z → z
Designed IIR filter
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Frequency Transformations
Specifications of frequency-selective filters
Lowpass filter 21 , , , sp
highpass filter 21 , , , ps
bandpass filter 212211 , , , , , spps
bandstop filter 212211 , , , , , pssp
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Frequency Transformations
Frequency transformations in the digital domain
)(zH L the given prototype lowpass digital filter
)(ZHd the desired frequency-selective digital filter
)( 11 ZGzDefine a mapping of the form
Such that)( 11)()(
ZGzLd zHZH
To do this, we simply replace everywhere in by the function
1z )(zH L
)( 1ZG
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Given that is a stable and causal filter, we also want to be stable and causal. This imposes the following requirements:
)(zH L
)(ZHd
The unit circle of the z-plane must map onto the unit circle of the Z-plane
The inside of the unit circle of the z-plane must also map onto the inside of the unit circle of the Z-plane.
1Z)( 1ZG must be a rational function in so that )(ZHd
is implementable.
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Let and be the frequency variables of and , respectively. That is . Then
z Z jj eZez ,
)](arg[)()( jeGjjjj eeGeGe
)](arg[ ,1)( jj eGθeG
Hence the is an all-pass function)( 1ZG
1|| ,1
)(1
1
111
k
N
k k
k aZa
aZZGz
By choosing an appropriate order N and the coefficients , we can obtain a variety of mappings ka
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Frequency transformation formulae (P296)
Lowpass - Lowpass
1
11
1
Z
Zz
]2/)sin[(
]2/)sin[(
cc
cc
: Cutoff frequency of new digital filterc
c The cutoff frequency of prototype lowpass digital filter
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Lowpass - Highpass
1
11
1
Z
Zz
]2/)cos[(
]2/)cos[(
cc
cc
: Cutoff frequency of new digital filterc
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Lowpass - Bandpass
111
22
21
12
1
ZZ
ZZz
1
1 ,
1
2
2tan)
2cot( ,cos
]2/)cos[(
]2/)cos[(
21
120
12
12
k
k
k
k
k c
: lower cutoff frequency of bandpass digital filter1: upper cutoff frequency of bandpass digital filter2: center frequency of the passband0
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Lowpass - Bandstop
: lower cutoff frequency of bandstop digital filter1: upper cutoff frequency of bandstop digital filter2: center frequency of the stopband0
k
k
k
k c
1
1 ,
1
2
2tan)
2tan( ,cos
]2/)cos[(
]2/)cos[(
21
120
12
12
111
22
21
12
1
ZZ
ZZz
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Design procedure Determine the specifications of the digital prototype
lowpass filter;
Determine the specifications of the analog prototype lowpass filter;
Design the analog prototype lowpass filter;
Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation;
Perform the frequency transformation in digital domain to obtain the desired frequency-selective filters.
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Examples
Given the specifications of Chebyshev-I lowpass filter
dB15 ,3.0
dB1 ,2.0
2
1
s
p
Design a highpass filter with the above tolerances but with passband beginning at 6.0p
and its system function
)6493.05548.11)(8482.04996.11(
)1(01836.0)(
2121
41
zzzz
zzH L
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Solution
3820.0]2/)6.02.0cos[(
]2/)6.02.0cos[(
)4019.00416.11)(7647.05561.01(
)1(0243.0
)()(
2121
41
3820.01
3820.01
11
ZZZZ
Z
zHZHZ
ZzLd
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Frequency Transformations
Using the Chebyshev-I prototype to design a highpass digital filter to satisfy
dB15 ,46.0
dB1 ,6.0
ss
pp
A
R
Determine the specifications of the digital prototype lowpass filter
Solution
2.0p
Choose the passband frequency with a reasonable value:
Determine the stopband frequency by1
11
1
Z
Zz
)1
arg( 1
j
j
j
jj
e
e
e
ee
Copyright © 2005. Shi Ping CUC
3820.0]2/)6.02.0cos[(
]2/)6.02.0cos[(
]2/)cos[(
]2/)cos[(
pp
pp
3.0)3820.01
3820.0arg()
1arg(
46.0
46.0
j
j
j
j
s e
e
e
es
s
Determine the specifications of the analog prototype lowpass filter
Set T = 1 and prewarp the cutoff frequencies
1.0191)15.0tan(2)2
tan(2
0.6498)1.0tan(2)2
tan(2
ss
pp
T
T
Copyright © 2005. Shi Ping CUC
Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,, ,
0.04920.20380.61406192.0
0438.0)(
234
sssssHa
)0.64931.55481)(0.84821.49961(
)1(0018.0
0.55072.29253.82903.05431
0.00180.00730.01100.00730018.0)(
2121
41
4321
4321
zzzz
z
zzzz
zzzzzH L
Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation
Copyright © 2005. Shi Ping CUC
Perform the frequency transformation in digital domain to obtain the desired digital highpass filter
)4019.00416.11)(7647.05561.01(
)1(0243.0
)()(
2121
41
1 1
11
ZZZZ
Z
zHZHZ
ZzLh
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Using the Chebyshev-I prototype to design a bandpass digital filter to satisfy
dB15 ,7.0 ,2.0
dB1 ,5.0 ,4.0
21
21
sss
ppp
A
R
Determine the specifications of the digital prototype lowpass filter
Solution
2.0p
Choose the passband frequency with a reasonable value:
Copyright © 2005. Shi Ping CUC
Determine the stopband frequency by
69.0)1
arg(22
22
12
2
212
ss
ss
jj
jj
s ee
ee
1584.0]2/)4.05.0cos[(
]2/)4.05.0cos[(
]2/)cos[(
]2/)cos[(
12
12
pp
pp
0515.22
2.0tan)
2
4.05.0cot(
k
111
22
21
12
1
ZZ
ZZz
3446.01
1 ,2130.0
1
221
k
k
k
k
Copyright © 2005. Shi Ping CUC
Determine the specifications of the analog prototype lowpass filter
Set T = 1 and prewarp the cutoff frequencies
3.7842)0.3450tan(2)2
tan(2
0.6498)1.0tan(2)2
tan(2
ss
pp
T
T
Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,, ,
4656.07134.0
4149.0)(
2
sssHa
Copyright © 2005. Shi Ping CUC
21
21
5157.01997.11
)1(0704.0)(
zz
zzH L
Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation
Perform the frequency transformation in digital domain to obtain the desired digital bandpass filter
4321
42
1
0.71060.48147020.15731.01
0205.00410.00205.0
)()(1
12
2
21
12
1
ZZZZ
ZZ
zHZHZZ
ZZzLbp
Copyright © 2005. Shi Ping CUC
Frequency Transformations
Using the Chebyshev-I prototype to design a bandstop digital filter to satisfy
dB 20 ,65.0 ,35.0
dB 1 ,75.0 ,25.0
21
21
sss
ppp
A
R
Determine the specifications of the digital prototype lowpass filter
Solution
2.0p
Choose the passband frequency with a reasonable value:
Copyright © 2005. Shi Ping CUC
Determine the stopband frequency by
0.1919)1
arg(11
11
12
2
212
ss
ss
jj
jj
s ee
ee
0]2/)25.075.0cos[(
]2/)25.075.0cos[(
]2/)cos[(
]2/)cos[(
12
12
pp
pp
0.15842
2.0tan)
2
25.075.0tan(
k
111
22
21
12
1
ZZ
ZZz
0.72651
1 ,0
1
221
k
k
k
Copyright © 2005. Shi Ping CUC
Determine the specifications of the analog prototype lowpass filter
Set T = 1 and prewarp the cutoff frequencies
6217.0)0.0959tan(2)2
tan(2
0.3168)1.0tan(2)2
tan(2
ss
pp
T
T
Design an analog Chebyshev-I prototype lowpass filter to satisfy the specification: spsp AR ,, ,
0.01561243.03131.0
0156.0)(
23
ssssHa
Copyright © 2005. Shi Ping CUC
321
321
0.73352.36922.62251
0.00160.00490.00490.0016)(
zzz
zzzzH L
Transform the analog prototype lowpass filter into a digital prototype lowpass filter using bilinear transformation
Perform the frequency transformation in digital domain to obtain the desired digital bandstop filter
)0.3391)(0.7761.2481)(0.7761.2481(
)1(0.132
)()(
22121
32
111
22
21
12
1
ZZZZZ
Z
zHZHZZ
ZZzLbs
return
Copyright © 2005. Shi Ping CUC
0 30 40 50 60
0.707
1
Magnitude Response
Analog frequency in rad/s
N=2N=4N=8N=16
return
)( jHa
Copyright © 2005. Shi Ping CUC
0 2 4 6
0.707
1
Magnitude Response
Analog frequency in rad/s
Am
plit
ud
e
0 2 4 6
-3
-1
0
1
3
Phase Response
Analog frequency in rad/s
Ph
as
e in
ra
d
)( jHa
return
Copyright © 2005. Shi Ping CUC
0 2 3 5
x 104
0
0.1778
0.89131
Magnitude Response
Analog frequency in pi units
H
0 2 3 5
x 104
-30
-15
-10
Magnitude in dB
Analog frequency in pi units
de
cib
els
0 2 3 5
x 104
-1
-0.5
0
0.5
1
Phase Response
Analog frequency in pi units
P
0 0.5 1 1.5
x 10-4
0
10000
20000
Impulse Response
time in seconds
ha
(t)
return
Copyright © 2005. Shi Ping CUC
0 2 3 5
x 104
0
0.1778
0.89131
Magnitude Response
Analog frequency in pi units
H
0 2 3 5
x 104
-30
-15
-10
Magnitude in dB
Analog frequency in pi units
de
cib
els
0 2 3 5
x 104
-1
-0.5
0
0.5
1
Phase Response
Analog frequency in pi units
P
0 0.5 1 1.5
x 10-4
-5000
0
5000
10000
15000
20000
Impulse Response
time in seconds
ha
(t)
return
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(0 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(1 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(2 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(3 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(4 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(5 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
)(6 xC
Copyright © 2005. Shi Ping CUC
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-2
-1
0
1
2
3
return
Copyright © 2005. Shi Ping CUC
0 2 3 5
0.8913
1
Magnitude Response
Analog frequency in rad/s
Am
plit
ud
e4N
Copyright © 2005. Shi Ping CUC
0 2 3 5
0.8913
1
Magnitude Response
Analog frequency in rad/s
Am
plit
ud
e
return
5N
Copyright © 2005. Shi Ping CUC
0 1 2 3 5
0.89131
Magnitude Response
Analog frequency in rad/s
Am
plit
ud
e
0 1 2 3 5-3
-2
-1
0
1Phase Response
Analog frequency in rad/s
Ph
as
e in
ra
d
return
)( jHa
Copyright © 2005. Shi Ping CUC
0 2 3 5
x 104
0
0.1778
0.89131
Magnitude Response
Analog frequency in pi units
H
0 2 3 5
x 104
-30
-15
-10
Magnitude in dB
Analog frequency in pi units
de
cib
els
0 2 3 5
x 104
-1
-0.5
0
0.5
1
Phase Response
Analog frequency in pi units
Ph
as
e in
pi u
nit
s
0 1 2 3 4
x 10-4
-5000
0
5000
10000
15000
20000
Impulse Response
time in seconds
ha
(t)
return
Copyright © 2005. Shi Ping CUC
0 pi/T 2*pi/T
0.2
0.4
0.6
0 pi 2*pi
0.2
0.4
0.6
)( jHa
)( jeH
1T
Copyright © 2005. Shi Ping CUC
0 pi/T 2*pi/T
0.2
0.4
0.6
0 pi 2*pi
0.2
0.4
0.6
)( jHa
)( jeH
1.0T
return
Copyright © 2005. Shi Ping CUC
0 0.2 0.3 1
0.1778
0.89131
Magnitude Response
Frequency in pi0 0.2 0.3 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
0 0.2 0.3 1
-15
-10
Magnitude Response in dB
Frequency in pi0 0.2 0.3 1
2
4
6
8
10Group Delay
Frequency in pi
return
Copyright © 2005. Shi Ping CUC
0 pi/T 2*pi/T
0.2
0.4
0.6
Magnitude Response
frequency in rad/s
0 pi 2*pi
0.2
0.4
0.6
frequency in rad/sample
)( jHa
)( jeH
1T
Copyright © 2005. Shi Ping CUC
0 pi/T 2*pi/T
0.2
0.4
0.6
Magnitude Response
frequency in rad/s
0 pi 2*pi
0.2
0.4
0.6
frequency in rad/sample
)( jHa
)( jeH
1.0T
return
Copyright © 2005. Shi Ping CUC
0 0.2 0.3 1
0.1778
0.89131
Magnitude Response
Frequency in pi units0 0.2 0.3 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
pi u
nit
s
0 0.2 0.3 1
-15
-10
Magnitude Response in dB
Frequency in pi units0 0.2 0.3 1
3
6
9
12
15Group Delay
Frequency in pi units
sa
mp
les
return
Copyright © 2005. Shi Ping CUC
0 0.2 0.3 1
0.1778
0.89131
Magnitude Response
Frequency in pi units0 0.6 1
0.1778
0.89131
Magnitude Response
Frequency in pi units
0 0.2 0.3 1
-15
-10
Magnitude Response in dB
Frequency in pi units0 0.6 1
-15
-10
Magnitude Response in dB
Frequency in pi units
return
Copyright © 2005. Shi Ping CUC
0 0.46 0.6 1
0.1778
0.89131
Magnitude Response
Frequency in pi units0 0.46 0.6 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
pi u
nit
s
0 0.46 0.6 1
-15
-10
Magnitude Response in dB
Frequency in pi units0 0.46 0.6 1
2
4
6
8
10Group Delay
Frequency in pi units
sa
mp
les
return
Copyright © 2005. Shi Ping CUC
0 0.2 0.4 0.5 0.7 1
0.1778
0.89131
Magnitude Response
Frequency in pi units0 0.2 0.4 0.5 0.7 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
pi u
nit
s
0 0.2 0.4 0.5 0.7 1-100
-80
-60
-40
-20
0
Magnitude Response in dB
Frequency in pi units0 0.2 0.4 0.5 0.7 1
3
6
9
12
15Group Delay
Frequency in pi units
sa
mp
les
return
Copyright © 2005. Shi Ping CUC
0 0.250.35 0.650.75 1
0.1
0.89131
Magnitude Response
Frequency in pi units0 0.250.35 0.650.75 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
pi u
nit
s
0 0.250.35 0.650.75 1
-30
-20
-10
0
Magnitude Response in dB
Frequency in pi units0 0.250.35 0.650.75 1
2
4
6
8
10Group Delay
Frequency in pi units
sa
mp
les
return