IIR filter design, Digital signal processing

Chapter 7
IIR Filter Design
Content
Preliminaries
Characteristics of Prototype Analog Filters
Analog-to-Digital Filter Transformations
Frequency Transformations

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.

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.

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

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.)( ωj
eH
πωaeH
eHa
s
j
p
j
≤≤≤
≤≤≤−
||)(
||1)(1
2
1
ω
ωω
ω
ω ],0[ pω Passband
],[ πωs Stopband
],[ sp ωω Transition band
The ending frequency of the passband. Bandwidthpω
The beginning frequency of the stopband.sω
The tolerance (or ripple) in passband1a
The tolerance (or ripple) in stopband2a

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
aeH
eH
eH
aeH
eH
eH
s
s
p
p
j
j
j
j
j
j
−=−==
−−=−==
ω
ω
ω
ω
δ
δ
The maximum tolerable passband ripple1δ pR
The minimum tolerable stopband attenuation2δ sA

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
)
20
50
(
222
)
20
25.0
(
111
==−==
=−=−−==
−
−
aa
aa
δ
δ

Preliminaries
dB60001.0lg20lg20
0.1755dB)02.01lg(20)1lg(20
22
11
=−=−=
=−−=−−=
a
a
δ
δ
 Given the passband tolerance a1=0.02 and the
stopband tolerance a2=0.001, determine the
passband ripple and the stopband attenuation1δ 2δ

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

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

 Magnitude-squared function
sa
pa
A
jH
jH
Ω≥Ω≤Ω≤
Ω≤Ω≤Ω≤
+
||,
1
)(0
||,1)(
1
1
2
2
2
2
ε
 Let be the frequency response of an analog
filter
)( ΩjHa
is a passband ripple parameterε
is the passband cutoff frequency in rad/secpΩ
is the stopband cutoff frequency in rad/secsΩ
is a stopband attenuation parameterA
sa
pa
A
jH
jH
Ω=Ω=Ω
Ω=Ω
+
=Ω
at
1
)(
at
1
1
)(
2
2
2
2
ε

Ω=
=Ω jsaa sHjH )()(
 The properties of
2
)( ΩjHa
Ω=
∗
−=Ω−Ω=ΩΩ=Ω 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.
Ωj
)()( sHsH aa −
22
2
)()()(
s
aaa jHsHsH
−=Ω
Ω=−

Ωj)()( sHsH aa −
σ
0
2
2
s-plane

 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

 Examples
)36)(49(
)25(16
)( 22
22
2
Ω+Ω+
Ω−
=ΩjHa
)36)(49(
)25(16
)()()( 22
22
2
22
ss
s
jHsHsH
s
aaa
−−
+
=Ω=−
−=Ω
poles 6,7 ±=±= ss 2th
order zeros 5js ±=
We can assign left-half poles and a pair
of conjugate zeros to
6,7 −=−= ss
5js ±= )(sHa
)6)(7(
)25(
)(
2
0
++
+
=
ss
sK
sHa
4
)()(
0
00
=∴
Ω= =Ω=
K
jHsH asa
4213
1004
)6)(7(
)25(4
)( 2
22
++
+
=
++
+
=
ss
s
ss
s
sHa

 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 jH 2
2
1
1
)(






Ω
Ω
+
=Ω

 The properties of Butterworth lowpass filters
 At , for all N1)( =ΩjHa0=Ω
707.0
2
1
)( ==ΩjHa At , for all N, which
implies a 3dB attenuation at
cΩ=Ω
cΩ=Ω
)( ΩjHa is a monotonically decreasing function of Ω
)( ΩjHa approaches an ideal lowpass filter as ∞→N
)( ΩjHa 0=Ω is maximally flat at since derivatives of
all orders exist and are equal to zero

 The poles and zeros of )()( sHsH aa −
N
c
N
N
c
N
c
js
aaa
js
j
j
s
jHsHsH 22
2
2
/
2
)(
)(
1
1)()()(
Ω+
Ω
=






Ω
+
=Ω=−
=Ω
Nkejs N
kj
cc
N
k 2,,2,1,)()1(
)
2
12
2
1(
2
1
=Ω=Ω−=
−+π

)()( sHsH aa − There are 2N poles of , which are
equally distributed on a circle of radius with angular
spacing of radians.
cΩ
N
π
 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.

∏=
−
Ω
= N
k
k
N
c
a
ss
sH
1
)(
)(
Nkes N
kj
ck ,,2,1,
)
2
12
2
1(
=Ω=
−+π
In general, we consider and this results in a
normalized Butterworth analog prototype filter
rad/s1=Ωc
)(sHan
)()(
c
ana
sHsH
Ω
=
When designing an actual filter with , we
can simply do a replacement for s, that is
)(sHa rad/s1≠Ωc

 Designing equations
Given , two parameters are required to
determine a Butterworth lowpass filters :
21 ,,, δδsp ΩΩ
cN Ω,
2
2
1
2
1lg20
1lg20
δ
δ
=














Ω
Ω
+Ω=Ω
=














Ω
Ω
+Ω=Ω
N
c
s
s
N
c
p
p
at
at
Solving these two equations for cN Ω,

 ,
)lg(2
)110/()110(lg 1010
21
NNN
s
p
′=
Ω
Ω






−−
=′
δδ
pΩ
Since the actual N chosen is larger than required,
specifications can be either met or exceeded at or sΩ
pΩTo satisfy the specifications exactly at
sΩTo satisfy the specifications exactly at
N
s
c
2 10
110
2
−
Ω
=Ω
δN
p
c
2 10
110
1
−
Ω
=Ω
δ

 Example
 Determine the system function of 3th
-order Butterworth
analog lowpass filter. Suppose rad/s2=Ωc
62
2
2
1
1
1
1)(





 Ω+
=






Ω
Ω+
=Ω N
c
a jH
6,,2,1,2
)
6
12
2
1(
==
−+
kes
kj
k
π
64
1
1)()( 6
s
sHsH aa
−
=−
884
8
))()((
)( 23
321
3
+++
=
−−−
Ω
=
sssssssss
sH c
a
Solution:

Design the above filter with normalized Butterworth
analog prototype filter. See table 6-4 on page 261
1,)( 02
210
0
==
++++
= NN
N
an aa
sasasaa
d
sH

in case of 1)0( =jHa00 ad =
3=NFor 2,2 21 == aaWe can find
32
32
2
488
8
)
2
()
2
(2)
2
(21
1
)()(
ssssss
sHsH ssana
+++
=
+++
=
= =
2
sss
c
=
Ω
→

 Design a lowpass Butterworth filter to satisfy:
rad/s1020fordB1 4
1 ×≤Ω≤≤ πδPassband
Stopband rad/s105.12fordB15 4
2 ××≥Ω≥ πδ
Solution:
15
105.12
1lg20
1
102
1lg20
2
4
2
4
=














Ω
××
+
=














Ω
×
+
N
c
N
c
π
π
rad/s105.12dB,15
rad/s102dB,1
4
s2
4
p1
××=Ω=
×=Ω=
πδ
πδ

rad/s1013.12 4
××=Ω πc
5.8858
105.12
102
lg2
)110/()110(lg
)lg(2
)110/()110(lg
4
4
10
15
10
1
1010
21
=






××
×






−−
=
Ω
Ω






−−
=′
π
π
δδ
s
p
N
6=N
4
12 10
15
4
2 10
101.12792
110
105.12
110
2
××=
−
××
=
−
Ω
=Ω π
π
δN
s
c

Look for table 6-4 on page 261
864.3,464.7,142.9,464.7,864.3 54321 ===== aaaaa
65432
65
5
4
4
3
3
2
21
864.3464.7142.9464.7864.31
1
1
1
)(
ssssss
ssasasasasa
sHan
++++++
=
++++++
=
===
××
=
Ω
= 4
1013.12
)()()(
π
s
sans
sana sHsHsH
c
6554103152202429
29
102.74103.76103.27101.90106.97101.28
101.28
ssssss +×+×+×+×+×+×
×

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)(1.000.52(
1
))()()()()((
1
)(
222
654321
++++++
=
−−−−−−
=
ssssss
ssssssssssss
sHan
4
1013.12 ××
=
Ω
→
π
ss
s
c
)105.04101.37)(105.0410)(105.04103.69(
1028.1
)( 952952942
29
×+×+×++×+×+
×
=
ssssss
sHa

 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.






Ω
Ω+
=Ω
c
N
a
C
jH
22
2
1
1)(
ε
Chebyshev-I

N is the order of the filter
is the Nth
-order Chebyshev polynomial given by
2
NC



Ω
Ω=
>
≤
=
−
c
N x
xxN
xxN
xC where
1||),ch(ch
1||),coscos(
)( -1
1
xxCxC
xCxxCxC NNN
==
−= −+
)(,1)(
)()(2)(
10
11
is the passband ripple factor.ε 10 << ε






Ω
Ω+
=Ω
c
N
a
C
jH
22
2
1
1)(
ε







Ω
Ω+
=Ω
c
N
a
C
jH
22
1
1)(
ε
 The properties of Chebyshev lowpass filters
 At
:
0=Ω
evenisfor
1
1)0(
oddisfor1)0(
2
NjH
NjH
a
a
ε+
=
=
 At
:
cΩ=Ω NjH ca allfor
2
1
1)(
ε+
=Ω
 For
:
cΩ≤Ω≤0
2
1
1~1betweenoscillates
ε+
Ω)( jHa
cΩ>Ω For
:
0tollymonotonicadecreases)( ΩjHa
sΩ=Ω For
: A
jH sa
1)( =Ω

 Designing equations
Given , two parameters are required to
determine a Chebyshev-I filter:
pAAssc ,,,ΩΩ
N,ε
110
1.02
−= pA
ε






Ω
Ω







 −
≥
−
−
c
s
As
ch
ch
N
1
1.0
1 110
ε















 −
Ω=Ω −
ε
1101 1.0
1
sA
cs ch
N
ch



Ω=Ω −
ε
11 1
3 ch
N
chcdB
Note: this is only for dBc 3Ω<Ω

 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
=





Ω
+
c
N
j
sCε
It can be shown that if
are the (left half-plane) roots of the above polynomial,
then
Nkjs kkk ,,2,1 =Ω+= σ



 −
Ω=Ω



 −
Ω−=
N
k
b
N
k
a
ck
ck
2
)12(
cos)(
2
)12(
sin)(
π
π
σ
Nk ,,2,1 =

111
2
++=
εε
γ
)(
2
1),(
2
1
1111
NNNN
ba
−−
+=−= γγγγ
∏=
−
= N
k
k
a
ss
KsH
1
)(
)(




+
= evenisN,
1
1
oddisN,1
)0(
2
ε
jHa
Where K is a normalizing factor chosen to make

 Determine poles by geometric method
The poles of fall on an ellipse with major axis
and minor axis .
)()( sHsH aa − cbΩ
caΩ
σ
Ωj
cbΩcaΩ
N
π

 Determine the system function of 2th
-order Chebyshev-I
lowpass filter. Suppose andrad/s1=Ωc
dB1=pA
2589.0110110 1.01.02
=−=−= Ap
ε
2
0
2
10
0
0977.11025.1
)(
ss
d
ssaa
d
sHa
++
=
++
=
0977.1
1025.1
1
0
=
=
a
a
0.8913
2589.1
1
1
1)0(
2
==
+
=
ε
jHa
9827.0,8913.0
1025.1
)( 0
0
0
=∴===
d
d
sH sa
 Examples
Solution:

 Design a lowpass Chebyshev-I filter to satisfy:
rad/s102 4
×=Ω πc
Passband cutoff:
dB1=pAPassband ripple:
Stopsband cutoff: rad/s105.12 4
××=Ω πs
Stopband attenuation: dB15=sA
Solution:
5088.0110110 1.01.0
=−=−= pA
ε
3.1978
)5.1(
(10.8761)
110
1
1
1
1.0
1
==






Ω
Ω







 −
≥ −
−
−
−
ch
ch
ch
ch
N
c
s
As
ε

4=N 5088.0=εdB1=pA
9528.0,4539.1,7426.02756.0 3210 ==== aaaa ，
432
0
43
3
2
210
0
9528.04539.17426.02756.0
)(
ssss
d
ssasasaa
d
sHan
++++
=
++++
=
0.8913
2589.1
1
1
1)0(
2
==
+
=
ε
jHa
0.2456,0.8913
2756.0
)( 0
0
0
=∴===
d
d
sH san

432
9528.04539.17426.02756.0
2456.0)(
ssss
sHan
++++
=
4
102 ×
=
Ω
→
π
sss
c
434291418
18
105.9866105.7398101.8420104.2954
103.8278
)(
ssss
sHa
+×+×+×+×
×
=

 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 a=
T : Sampling interval
Tjj
eeT Ω
=Ω= ω
ω or,
Since this is a sampling operation, the analog and digital
frequencies are related by

The system function and are related by)(sHa)(zH
∑
∞
−∞=
=
−=
k
aez
k
T
jsH
T
zH sT )
2
(
1
)(
π
This implies a mapping from the s-plane to the z-plane
T
π
T
π−
σ
Ωj
0
T
π3
T
π3−
]Re[z
]Im[ zj

 Properties
 Using ]Re[s=σ
UC)theof(outside1|z|intomaps0
UC)the(on1|z|intomaps0
UC)theof(inside1|z|intomaps0
>>
==
<<
σ
σ
σ
 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/2π 1|| <z

πω
ωω
<= ||),(
1
)(
T
jH
T
eH a
j
then
There will be no aliasing.
Frequency response ∑
∞
−∞=
−
=
k
a
j
T
k
jH
T
eH )
2
(
1
)(
πωω
TT
jHjH aa
πω
≥Ω==Ω ||for0)()(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

 Digitalizing of analog filters
∑= −
=
N
k k
k
a
ss
A
sH
1
)(
Using partial fraction expansion, expand into)(sHa
The corresponding impulse response is
∑=
−
==
N
k
ts
kaa tueAsHLth k
1
1
)()]([)(
∑∑ ==
===
N
k
nTs
k
N
k
nTs
ka nueAnueAnThnh kk
11
)()()()()(
To sample the )(tha

The z-transform of is)(nh
∑∑∑∑ =
−
∞
= =
−
∞
−∞=
−
−
===
N
k
Ts
k
n
N
k
nTs
k
n
n
ze
A
zeAznhzH k
k
1
1
0 1
1
1
)()()(
Conclusions:
∑= −
=
N
k k
k
a
ss
A
sH
1
)(Compared with
 The pole in s-plane is mapped to the pole in z-planeks Tsk
e
 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

 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.
Ω ω

 Design procedure
 Choose T and determine the analog frequencies
 Transform analog poles into digital poles to obtain the
digital filter
TT
s
s
p
p
ωω
=Ω=Ω ,
Given the digital lowpass filter specifications 21 ,,, δδωω sp
 Design an analog filter using the specifications)(sHa
∑= −
=
N
k k
k
a
ss
A
sH
1
)(
 Using partial fraction expansion, expand into)(sHa
∑=
−
−
=
N
k
Ts
k
ze
A
zH k
1
1
1
)(

 Examples
 Transform
3
1
1
1
34
2
)( 2
+
−
+
=
++
=
ssss
sHa
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
T
zH TTT
TT
TT

 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.
Ωj





 Ω
⋅=Ω
2
tan 1T
c Tj
Tj
T
j
T
j
T
j
T
j
e
e
c
ee
ee
cj 1
1
11
11
1
1
22
22
Ω−
Ω−
Ω
−
Ω
Ω
−
Ω
+
−
⋅=
+
−
⋅=Ω
1
1
1
1
1
1
1
1
−
−
−
−
+
−
⋅=
+
−
⋅=
z
z
c
e
e
cs Ts
Ts
sc
sc
z
−
+
=

Ts
Ts
e
e
cs 1
1
1
1
−
−
+
−
⋅= Ts
ez 1
=
1
1
1
1
−
−
+
−
⋅=
z
z
cs
T
π
T
π−
1Ωj
1σ0
s1-plane
]Re[z
]Im[ zj
0
z-plane
0
Ωj
σ
s-plane

 Parameter c
T
c
T
c
T
c
2
then,
22
tan 1
11
=Ω≈Ω
Ω
⋅≈




 Ω
⋅=Ω
 Keeping a good corresponding relationship between
the analog filter and the digital filter in low
frequencies. i.e. in low frequencies1Ω≈Ω






Ω=





⋅=




 Ω
⋅=Ω
2
cotthen
2
tanc
2
tan 1 c
c
cc
c c
T
c
ωω
 Keeping a good corresponding relationship between the
analog filter and the digital filter in a specific frequency
(for example, in the cutoff frequency, )Tcc 1Ω≈ω

 Properties
 Using , we obtainΩ+= js σ
22
22
)(
)(
||,
)(
)(
Ω+−
Ω++
=
Ω−−
Ω++
=
−
+
=
σ
σ
σ
σ
c
c
z
jc
jc
sc
sc
z
So 1||0,1||0,1||0 >→>=→=<→< zzz σσσ
 Using , we obtainωj
ez =
Ω=⋅=
+
−
⋅=
+
−
⋅= −
−
−
−
jjc
e
e
c
z
z
cs j
j
)
2
tan(
1
1
1
1
1
1
ω
ω
ω
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.

 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;.
Ω ω The frequencies and are not linearly related.
So a linear phase analog filter cannot be mapped
to a linear phase digital filter.

 Design procedure
 Choose a value for T. We may set T=1
)
2
tan(
2
),
2
tan(
2 s
s
p
p
TT
ωω
=Ω=Ω
Given the digital lowpass filter specifications 21 ,,, δδωω sp
 Prewarp the cutoff frequencies and ; that ispω sω
 Design an analog filter to meet the specifications
)(sHa
 Finally, set )
1
12
()( 1
1
−
−
+
−
=
z
z
T
HzH a
and simplify to obtain as a rational function in)(zH 1−
z

 Examples
 Transform
into a digital filter using the bilinear transformation.
Choose T=1
34
2
)( 2
++
=
ss
sHa
21
21
1
1
2
1
1
1
1
1
1
1
0.070.131
13.00.2713.0
3
1
1
24
1
1
2
2
)
1
1
2()
1
12
()(
−−
−−
−
−
−
−
−
−
=
−
−
−−
++
=
+





+
−
+





+
−
=
+
−
=
+
−
=
zz
zz
z
z
z
z
z
z
H
z
z
T
HzH a
T
a

 Design the digital Chebyshev-I filter using bilinear
transformation. The specifications are:
dB15,3.0
dB1,2.0
2
1
==
==
δπω
δπω
s
p
Solution
 Let T=1
1.0191)15.0tan(2)
2
tan(
2
0.6498)1.0tan(2)
2
tan(
2
===Ω
===Ω
π
ω
π
ω
s
s
p
p
T
T
 Prewarp the cutoff frequencies

 Design an analog Chebyshev-I filter to meet the
specifications 21 ,,, δδsp ΩΩ
)(sHa
0.04920.20380.61406192.0
0438.0
)( 234
++++
=
ssss
sHa
)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
zzzz
zH

 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
==
==
δπω
δπω
s
p
Given the digital filter specifications:
prototype Order N Stopband Att.
Butterworth 6 15 dB
Chebyshev-I 4 25 dB
Elliptic 3 27 dB
return

 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

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

 Specifications of frequency-selective filters
 Lowpass filter 21 ,,, δδωω sp
 highpass filter 21 ,,, δδωω ps
 bandpass filter 212211 ,,,,, δδωωωω spps
 bandstop filter 212211 ,,,,, δδωωωω pssp

 Frequency transformations in the digital domain
)(zHL 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
1−
z )(zHL
)( 1−
ZG

Given that is a stable and causal filter, we also
want to be stable and causal. This imposes the
following requirements:
)(zHL
)(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.
1−
Z)( 1−
ZG must be a rational function in so that )(ZHd
is implementable.

Let and be the frequency variables of and ,
respectively. That is . Then
θ ω z Z
ωθ jj
eZez == ,
)](arg[
)()(
ω
ωωθ j
eGjjjj
eeGeGe
−
−−−
==
)](arg[,1)( ωω jj
eGθeG −−
−==
Hence the is an all-pass function)( 1−
ZG
1||,
1
)(
1
1
1
11
<
−
−
== ∏=
−
∗−
−−
k
N
k k
k
a
Za
aZ
ZGz
By choosing an appropriate order N and the
coefficients , we can obtain a variety of mappingska

 Frequency transformation formulae
 Lowpass - Lowpass
1
1
1
1 −
−
−
−
−
=
Z
Z
z
α
α
]2/)sin[(
]2/)sin[(
cc
cc
ωθ
ωθ
α
+
−
=
: Cutoff frequency of new digital filtercω
cθ The cutoff frequency of prototype lowpass digital filter

 Lowpass - Highpass
1
1
1
1 −
−
−
+
+
−=
Z
Z
z
α
α
]2/)cos[(
]2/)cos[(
cc
cc
ωθ
ωθ
α
−
+
−=
: Cutoff frequency of new digital filtercω

 Lowpass - Bandpass
11
1
2
2
2
1
1
2
1
+−
+−
−= −−
−−
−
ZZ
ZZ
z
αα
αα
1
1
,
1
2
2
tan)
2
cot(,cos
]2/)cos[(
]2/)cos[(
21
12
0
12
12
+
−
=
+
=
−
==
−
+
=
k
k
k
k
k c
α
β
α
θωω
ω
ωω
ωω
β
: lower cutoff frequency of bandpass digital filter1ω
: upper cutoff frequency of bandpass digital filter2ω
: center frequency of the passband0ω

 Lowpass - Bandstop
: lower cutoff frequency of bandstop digital filter1ω
: upper cutoff frequency of bandstop digital filter2ω
: center frequency of the stopband0ω
k
k
k
k c
+
−
=
+
=
−
==
−
+
=
1
1
,
1
2
2
tan)
2
tan(,cos
]2/)cos[(
]2/)cos[(
21
12
0
12
12
α
β
α
θωω
ω
ωω
ωω
β
11
1
2
2
2
1
1
2
1
+−
+−
= −−
−−
−
ZZ
ZZ
z
αα
αα

 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.

 Examples
 Given the specifications of Chebyshev-I lowpass filter
dB15,3.0
dB1,2.0
2
1
==
==
δπθ
δπθ
s
p
Design a highpass filter with the above tolerances but
with passband beginning at πω 6.0=p
and its system function
)6493.05548.11)(8482.04996.11(
)1(01836.0
)( 2121
41
−−−−
−
+−+−
+
=
zzzz
z
zHL

Solution
3820.0
]2/)6.02.0cos[(
]2/)6.02.0cos[(
−=
−
+
−=
ππ
ππ
α
)4019.00416.11)(7647.05561.01(
)1(0243.0
)()(
2121
41
3820.01
3820.0
1
1
1
−−−−
−
−
−
−=
++++
−
=
=
−
−
−
ZZZZ
Z
zHZH
Z
Z
zLd

 Using the Chebyshev-I prototype to design a
highpass digital filter to satisfy
dB15,46.0
dB1,6.0
==
==
ss
pp
A
R
πω
πω
 Determine the specifications of the digital prototype
lowpass filter
Solution
πθ 2.0=p
Choose the passband frequency with a reasonable value:
Determine the stopband frequency by
1
1
1
1 −
−
−
+
+
−=
Z
Z
z
α
α
)
1
arg(
1 ω
ω
ω
ω
θ
α
α
θ
α
α
j
j
j
j
j
e
e
e
e
e −
−
−
−
−
+
+
−−=→
+
+
−=

3820.0
]2/)6.02.0cos[(
]2/)6.02.0cos[(
]2/)cos[(
]2/)cos[(
−=
−
+
−=
−
+
−=
ππ
ππ
ωθ
ωθ
α
pp
pp
π
α
α
θ ω
ω
3.0)
3820.01
3820.0
arg()
1
arg( 46.0
46.0
=
−
−
−=
+
+
−= −
−
−
−
j
j
j
j
s
e
e
e
e
s
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
===Ω
===Ω
π
θ
π
θ
s
s
p
p
T
T

 Design an analog Chebyshev-I prototype lowpass filter
to satisfy the specification: spsp AR ,,, ΩΩ
0.04920.20380.61406192.0
0438.0
)( 234
++++
=
ssss
sHa
)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
zzzz
zHL
 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 highpass filter
)4019.00416.11)(7647.05561.01(
)1(0243.0
)()(
2121
41
1 1
1
1
−−−−
−
+
+
−=
++++
−
=
=
−
−
−
ZZZZ
Z
zHZH
Z
Z
zLh
α
α

 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
πωπω
πωπω
lowpass filter
Solution
πθ 2.0=p

π
αα
αα
θ ωω
ωω
69.0)
1
arg( 22
22
1
2
2
21
2
=
+−
+−
−−= −−
−−
ss
ss
jj
jj
s
ee
ee
1584.0
]2/)4.05.0cos[(
]2/)4.05.0cos[(
]2/)cos[(
]2/)cos[(
12
12
=
−
+
=
−
+
=
ππ
ππ
ωω
ωω
β
pp
pp
0515.2
2
2.0
tan)
2
4.05.0
cot( =
−
=
πππ
k
11
1
2
2
2
1
1
2
1
+−
+−
−= −−
−−
−
ZZ
ZZ
z
αα
αα
3446.0
1
1
,2130.0
1
2
21 =
+
−
==
+
=
k
k
k
k
α
β
α

lowpass filter
3.7842)0.3450tan(2)
2
tan(
2
0.6498)1.0tan(2)
2
tan(
2
===Ω
===Ω
π
θ
π
θ
s
s
p
p
T
T
4656.07134.0
4149.0
)( 2
++
=
ss
sHa

21
21
5157.01997.11
)1(0704.0
)( −−
−
+−
+×
=
zz
z
zHL
to obtain the desired digital bandpass filter
4321
42
1
0.71060.48147020.15731.01
0205.00410.00205.0
)()(
1
1
2
2
2
1
1
2
1
−−−−
−−
+−
+−
−=
+−+−
+−
=
=
−−
−−
−
ZZZZ
ZZ
zHZH
ZZ
ZZ
zLbp
αα
αα

 Using the Chebyshev-I prototype to design a bandstop
digital filter to satisfy
dB20,65.0,35.0
dB1,75.0,25.0
21
21
===
===
sss
ppp
A
R
πωπω
πωπω
lowpass filter
Solution
πθ 2.0=p

π
αα
αα
θ ωω
ωω
0.1919)
1
arg( 11
11
1
2
2
21
2
=
+−
+−
−= −−
−−
ss
ss
jj
jj
s
ee
ee
0
]2/)25.075.0cos[(
]2/)25.075.0cos[(
]2/)cos[(
]2/)cos[(
12
12
=
−
+
=
−
+
=
ππ
ππ
ωω
ωω
β
pp
pp
0.1584
2
2.0
tan)
2
25.075.0
tan( =
−
=
πππ
k
11
1
2
2
2
1
1
2
1
+−
+−
= −−
−−
−
ZZ
ZZ
z
αα
αα
0.7265
1
1
,0
1
2
21 =
+
−
==
+
=
k
k
k
α
β
α

lowpass filter
6217.0)0.0959tan(2)
2
tan(
2
0.3168)1.0tan(2)
2
tan(
2
===Ω
===Ω
π
θ
π
θ
s
s
p
p
T
T
0.01561243.03131.0
0156.0
)( 23
+++
=
sss
sHa

321
321
0.73352.36922.62251
0.00160.00490.00490.0016
)( −−−
−−−
++−
+++
=
zzz
zzz
zHL
to obtain the desired digital bandstop filter
)0.3391)(0.7761.2481)(0.7761.2481(
)1(0.132
)()(
22121
32
11
1
2
2
2
1
1
2
1
−−−−−
−
+−
+−
=
−+−++
+×
=
=
−−
−−
−
ZZZZZ
Z
zHZH
ZZ
ZZ
zLbs
αα
αα
return

0 30 40 50 60
0.707
1
Magnitude Response
Analog frequency in rad/s
N=2
N=4
N=8
N=16
return
)( ΩjHa

0 2 4 6
0.707
1
Magnitude Response
Amplitude
0 2 4 6
-3
-1
0
1
3
Phase Response
Phaseinrad
)( ΩjHa
return

0 2 3 5
x 10
4
0
0.1778
0.8913
1
Magnitude Response
Analog frequency in pi units
H
0 2 3 5
x 10
4
-30
-15
-10
Magnitude in dB
decibels
0 2 3 5
x 10
4
-1
-0.5
0
0.5
1
Phase Response
P
0 0.5 1 1.5
x 10
-4
0
10000
20000
Impulse Response
time in seconds
ha(t)
return

0 2 3 5
x 10
4
0
0.1778
0.8913
1
Magnitude Response
H
0 2 3 5
x 10
4
-30
-15
-10
Magnitude in dB
decibels
0 2 3 5
x 10
4
-1
-0.5
0
0.5
1
Phase Response
P
0 0.5 1 1.5
x 10
-4
-5000
0
5000
10000
15000
20000
Impulse Response
time in seconds
ha(t)
return

-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

-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

-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

-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

-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

-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

-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

0 2 3 5
0.8913
1
Magnitude Response
Amplitude
4=N

0 2 3 5
0.8913
1
Magnitude Response
Amplitude
return
5=N

0 1 2 3 5
0.8913
1
Magnitude Response
Amplitude
0 1 2 3 5
-3
-2
-1
0
1
Phase Response
Phaseinrad
return
)( ΩjHa

0 2 3 5
x 10
4
0
0.1778
0.8913
1
Magnitude Response
H
0 2 3 5
x 10
4
-30
-15
-10
Magnitude in dB
decibels
0 2 3 5
x 10
4
-1
-0.5
0
0.5
1
Phase Response
Phaseinpiunits
0 1 2 3 4
x 10
-4
-5000
0
5000
10000
15000
20000
Impulse Response
time in seconds
ha(t)
return

0 pi/T 2*pi/T
0.2
0.4
0.6
0 pi 2*pi
0.2
0.4
0.6
)( ΩjHa
)( ωj
eH
1=T
Ω
ω

0 pi/T 2*pi/T
0.2
0.4
0.6
0 pi 2*pi
0.2
0.4
0.6
)( ΩjHa
)( ωj
eH
1.0=T
Ω
ω
return

0 0.2 0.3 1
0.1778
0.8913
1
Magnitude Response
Frequency in pi
0 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 pi
0 0.2 0.3 1
2
4
6
8
10
Group Delay
Frequency in pi
return

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
)( ΩjHa
)( ωj
eH
1=T
Ω
ω

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
)( ΩjHa
)( ωj
eH
1.0=T
Ω
ω
return

0 0.2 0.3 1
0.1778
0.8913
1
Magnitude Response
Frequency in pi units
0 0.2 0.3 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
piunits
0 0.2 0.3 1
-15
-10
0 0.2 0.3 1
3
6
9
12
15
Group Delay
samples
return

0 0.2 0.3 1
0.1778
0.8913
1
Magnitude Response
0 0.6 1
0.1778
0.8913
1
Magnitude Response
0 0.2 0.3 1
-15
-10
0 0.6 1
-15
-10
return

0 0.46 0.6 1
0.1778
0.8913
1
Magnitude Response
0 0.46 0.6 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
piunits
0 0.46 0.6 1
-15
-10
0 0.46 0.6 1
2
4
6
8
10
Group Delay
samples
return

0 0.2 0.4 0.5 0.7 1
0.1778
0.8913
1
Magnitude Response
0 0.2 0.4 0.5 0.7 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
piunits
0 0.2 0.4 0.5 0.7 1
-100
-80
-60
-40
-20
0
0 0.2 0.4 0.5 0.7 1
3
6
9
12
15
Group Delay
samples
return

0 0.250.35 0.650.75 1
0.1
0.8913
1
Magnitude Response
0 0.250.35 0.650.75 1
-1
-0.5
0
0.5
1
Phase Response
Frequency in pi
piunits
0 0.250.35 0.650.75 1
-30
-20
-10
0
0 0.250.35 0.650.75 1
2
4
6
8
10
Group Delay
samples
return

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