chap7-iir filter design

Copyright © 2005. Shi Ping CUC

Chapter 7IIR 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.)( 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


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


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


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


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

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



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



j)()( sHsH aa

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

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



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)(



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



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



)()( 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.



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


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 ,


,

)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


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:


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


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


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


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



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



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


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)(


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)(


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


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


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


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


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:


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


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


432 9528.04539.17426.02756.02456.0)(

sssssHan

4102

sss

c

434291418

18

105.9866105.7398101.8420104.2954103.8278

)(

ssss

sHa



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



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


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



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



|| ),(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



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


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



An alternative method

|| ),()2

()(

1)(

)()(

11

TjHk

Tj

TjHeH

ze

TAzH

nTThnh

ak

aj

N

kTsk

a

k



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.


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

)(


N

kTs

k

ze

AzH

k1

11)(



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


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

)(



)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



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


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



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


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.



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.


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



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


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


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



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



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 , , , sp

highpass filter 21 , , , ps

bandpass filter 212211 , , , , , spps

bandstop filter 212211 , , , , , pssp



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



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.



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



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



Lowpass - Highpass

1

11

1

Z

Zz

]2/)cos[(

]2/)cos[(

cc

cc

: Cutoff frequency of new digital filterc



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



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



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

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



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



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


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


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


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



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


Solution

2.0p



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




3.7842)0.3450tan(2)2

tan(2

0.6498)1.0tan(2)2

tan(2

ss

pp

T

T


4656.07134.0

4149.0)(

2

sssHa


21

21

5157.01997.11

)1(0704.0)(

zz

zzH L


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



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


Solution

2.0p



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




6217.0)0.0959tan(2)2

tan(2

0.3168)1.0tan(2)2

tan(2

ss

pp

T

T


0.01561243.03131.0

0156.0)(

23

ssssHa


321

321

0.73352.36922.62251

0.00160.00490.00490.0016)(

zzz

zzzzH L


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


0 30 40 50 60

0.707

1

Magnitude Response

Analog frequency in rad/s

N=2N=4N=8N=16

return

)( jHa


0 2 4 6

0.707

1

Magnitude Response


Am

plit

ud

e

0 2 4 6

-3

-1

0

1

3

Phase Response


Ph

as

e in

ra

d

)( jHa

return


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


de

cib

els

0 2 3 5

x 104

-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 104

0

0.1778

0.89131

Magnitude Response


H

0 2 3 5

x 104

-30

-15

-10

Magnitude in dB


de

cib

els

0 2 3 5

x 104

-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

)(1 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


Am

plit

ud

e4N


0 2 3 5

0.8913

1

Magnitude Response


Am

plit

ud

e

return

5N


0 1 2 3 5

0.89131

Magnitude Response


Am

plit

ud

e

0 1 2 3 5-3

-2

-1

0

1Phase Response


Ph

as

e in

ra

d

return

)( jHa


0 2 3 5

x 104

0

0.1778

0.89131

Magnitude Response


H

0 2 3 5

x 104

-30

-15

-10

Magnitude in dB


de

cib

els

0 2 3 5

x 104

-1

-0.5

0

0.5

1

Phase Response


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


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


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


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


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


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


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



3

6

9

12

15Group Delay

Frequency in pi units

sa

mp

les

return


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


0 0.2 0.3 1

-15

-10


Frequency in pi units0 0.6 1

-15

-10



return


0 0.46 0.6 1

0.1778

0.89131

Magnitude Response


-1

-0.5

0

0.5

1

Phase Response

Frequency in pi

pi u

nit

s

0 0.46 0.6 1

-15

-10



2

4

6

8

10Group Delay


sa

mp

les

return


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


Frequency in pi units0 0.2 0.4 0.5 0.7 1

3

6

9

12

15Group Delay


sa

mp

les

return


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


Frequency in pi units0 0.250.35 0.650.75 1

2

4

6

8

10Group Delay


sa

mp

les

return

chap7-iir filter design

Documents

h j h j

h s acopyrig

zeros of h

properties of h j

iir filters h

passband ripple h

s ha s

a1 e h e j e