Lecture 20

Digital Signal Processing

Analog Filter Design

1Copyright © 2010, S. K. Mitra

Analog Analog LowpassLowpass Filter Filter SpecificationsSpecifications

• Typical magnitude response of an analog lowpass filter may be given as indicated below

)( ΩjHa

1



• In the passband, defined by , we require

i.e., approximates unity within an error of

)( ΩjHa

pΩ≤Ω≤0

ppap jH Ω≤Ωδ+≤Ω≤δ− ,1)(1

)( ΩjHa

∞≤Ω≤Ωs

∞≤Ω≤Ωδ≤Ω ssa jH ,)(

• In the stopband, defined by , we require

i.e., approximates zero within an error of

pδ±

sδ2



• - passband edge frequency• - stopband edge frequency• - peak ripple value in the passband• - peak ripple value in the stopband• Peak passband ripple

dB• Minimum stopband attenuation

dB

pΩ

sΩ

sδpδ

)1(log20 10 pp δ−−=α

)(log20 10 ss δ−=α

3



• Magnitude specifications may alternately be given in a normalized form as indicated below

4



• Here, the maximum value of the magnitude in the passband assumed to be unity

• - Maximum passband deviation, given by the minimum value of the magnitude in the passband

• - Maximum stopband magnitude

21/1 ε+

A1

5


Analog Analog LowpassLowpass Filter DesignFilter Design• Two additional parameters are defined -

(1) Transition ratio

For a lowpass filter

(2) Discrimination parameterUsually

s

pkΩΩ

=

121 −=

Ak ε

1<k

11 <<k

6


Butterworth ApproximationButterworth Approximation

• The magnitude-square response of an N-thorder analog lowpass Butterworth filteris given by

• First derivatives of atare equal to zero

• The Butterworth lowpass filter thus is said to have a maximally-flat magnitude at

12 −N 2)( ΩjHa

Nc

a jH 22

)/(11)(ΩΩ+

=Ω

0=Ω

0=Ω7



• Gain in dB is

• As and dB

is called the 3-dB cutoff frequency

210 )(log10)( Ω=Ω jHaG

0)0( =G30103.3)5.0(log10)( 10 −≅−==ΩcG

cΩ

8



• Typical magnitude responses with 1=Ωc

0 1 2 30

0.2

0.4

0.6

0.8

1

Ω

Mag

nitu

de

Butterworth Filter

N = 2N = 4N = 10

9



• Two parameters completely characterizing a Butterworth lowpass filter are and N

• These are determined from the specified bandedges and , and minimum passband magnitude , and maximum stopband ripple

cΩ

pΩ sΩ21/1 ε+A/1

10


Butterworth ApproximationButterworth Approximation• and N are thus determined from

• Solving the above we get

cΩ

222 1

)/(11)(

AjH N

cssa =

ΩΩ+=Ω

222

11

)/(11)(

ε+=

ΩΩ+=Ω N

cppa jH

)/1(log)/1(log

)/(log]/)1[(log

21

10

110

10

2210

kkAN

ps=

ΩΩ−⋅= ε

11


Butterworth ApproximationButterworth Approximation• Since order N must be an integer, value

obtained is rounded up to the next highest integer

• This value of N is used next to determine by satisfying either the stopband edge or the passband edge specification exactly

• If the stopband edge specification is satisfied, then the passband edge specification is exceeded providing a safety margin

cΩ

12


Butterworth ApproximationButterworth Approximation• Transfer function of an analog Butterworth

lowpass filter is given by

where

• Denominator is known as the Butterworth polynomial of order N

∏ −Ω=

∑+Ω==

=−=

N

Nc

NN

Nc

Na pssdssD

CsH1

10 )()()(

l lll

l

Nep NNjc ≤≤Ω= −+ ll

l 1,]2/)12([π

)(sDN

13


Butterworth ApproximationButterworth Approximation• Example - Determine the lowest order of a

Butterworth lowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40dB at 5 kHz

• Now

which yieldsand

which yields

11

1log10 210 −=⎟⎠⎞

⎜⎝⎛+ ε

25895.02 =ε

401log10 210 −=⎟⎠⎞

⎜⎝⎛

A000,102 =A

14


Butterworth ApproximationButterworth Approximation• Therefore

and

• Hence

• We choose N = 4

51334.19611 2

1=−= ε

Ak

51 =ΩΩ=

p

sk

2811.3)/1(log)/1(log

10

110 ==kkN

15


ChebyshevChebyshev ApproximationApproximation• The magnitude-square response of an N-th

order analog lowpass Type 1 Chebyshev filteris given by

where is the Chebyshev polynomialof order N:

)/(11)( 22

2

pNa T

sHΩΩ+

=ε

⎩⎨⎧

>ΩΩ≤ΩΩ

=Ω −

−

1),coshcosh(1),coscos(

)( 1

1

NN

TN

)(ΩNT

16


ChebyshevChebyshev ApproximationApproximation

• Typical magnitude response plots of the analog lowpass Type 1 Chebyshev filter are shown below

0 1 2 30

0.2

0.4

0.6

0.8

1

Ω

Mag

nitu

de

Type 1 Chebyshev Filter

N = 2N = 3N = 8

17


ChebyshevChebyshev ApproximationApproximation• If at the magnitude is equal to 1/A,

then

• Solving the above we get

• Order N is chosen as the nearest integer greater than or equal to the above value

sΩ=Ω

2222 1

)/(11)(

ATjH

psNsa =

ΩΩ+=Ω

ε

)/1(cosh)/1(cosh

)/(cosh)/1(cosh

11

1

1

21

kkAN

ps−

−

−

−

=ΩΩ−= ε

18


ChebyshevChebyshev ApproximationApproximation• The magnitude-square response of an N-th

order analog lowpass Type 2 Chebyshev(also called inverse Chebyshev) filter is given by

where is the Chebyshev polynomialof order N

22

2

)/()/(

1

1)(

⎥⎦

⎤⎢⎣

⎡ΩΩΩΩ

+

=Ω

sN

psNa

TT

jH

ε

)(ΩNT

19


ChebyshevChebyshev ApproximationApproximation• Typical magnitude response plots of the

analog lowpass Type 2 Chebyshev filter are shown below

0 1 2 30

0.2

0.4

0.6

0.8

1

Ω

Mag

nitu

de

Type 2 Chebyshev Filter

N = 3N = 5N = 7

20


ChebyshevChebyshev ApproximationApproximation• The order N of the Type 2 Chebyshev filter

is determined from given , , and Ausing

• Example - Determine the lowest order of a Chebyshev lowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz -

)/1(cosh)/1(cosh

)/(cosh)/1(cosh

11

1

1

21

kkAN

ps−

−

−

−

=ΩΩ−= ε

ε sΩ

6059.2)/1(cosh)/1(cosh

11

1== −

−

kkN

21


Elliptic ApproximationElliptic Approximation• The square-magnitude response of an

elliptic lowpass filter is given by

where is a rational function of order N satisfying , with the roots of its numerator lying in the interval

and the roots of its denominator lying in the interval

)/(11)( 22

2

pNa R

jHΩΩ+

=Ωε

)(ΩNR)(/1)/1( Ω=Ω NN RR

10 <Ω<∞<Ω<1

22


Elliptic ApproximationElliptic Approximation

• Typical magnitude response plots with are shown below

1=Ω p

0 1 2 30

0.2

0.4

0.6

0.8

1

Ω

Mag

nitu

de

Elliptic Filter

N = 3N = 4

23

22

22


Analog Analog LowpassLowpass Filter DesignFilter Design• Example - Design an elliptic lowpass filter

of lowest order with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHz

• Code fragments used[N, Wn] = ellipord\(Wp, Ws, Rp, Rs, ‘s’\);[b, a] = ellip\(N, Rp, Rs, Wn, ‘s’\);with Wp = 2*pi*1000;

Ws = 2*pi*5000;Rp = 1;Rs = 40;

24


Analog Analog LowpassLowpass Filter DesignFilter Design

• Gain plot

0 2000 4000 6000-60

-40

-20

0

Frequency, Hz

Gai

n, d

B

Lowpass Elliptic Filter

25


Design of Analog Design of Analog HighpassHighpass, , BandpassBandpass and and BandstopBandstop FiltersFilters

• Steps involved in the design process:Step 1 - Develop of specifications of a prototype analog lowpass filter from specifications of desired analog filter

using a frequency transformationStep 2 - Design the prototype analog lowpass filterStep 3 - Determine the transfer function of desired analog filter by applying the inverse frequency transformation to

)( sH LP

)(sHLP

)(sHD

)(sHD

)(sHLP26


Design of Analog Design of Analog HighpassHighpass, , BandpassBandpass and and BandstopBandstop FiltersFilters• Let s denote the Laplace transform variable

of prototype analog lowpass filter and denote the Laplace transform variable of desired analog filter

• The mapping from s-domain to -domain is given by the invertible transformation

• Then

s

s)ˆ(sHD

)ˆ()()ˆ( sFsLPD sHsH ==

)(ˆ 1)ˆ()( sFsDLP sHsH −==

)ˆ(sFs =

)(sHLP

27


Analog Analog HighpassHighpass Filter DesignFilter Design• Spectral Transformation:

where is the passband edge frequency of and is the passband edge

frequency of• On the imaginary axis the transformation is

ss ppˆΩΩ

=

pΩpΩ

ΩΩΩ

−=Ω ˆˆ pp

)(sHLP)ˆ(sHHP

28


Analog Analog HighpassHighpass Filter DesignFilter Design

ΩΩΩ

−=Ω ˆˆ pp

Ω

sΩ− ˆ pΩsΩpΩ− ˆHighpass

PassbandPassband Stopband

sΩ− pΩ sΩpΩ−Lowpass

Passband

Stopband StopbandΩ

Ω

0

0

29

Lowpass to Highpass spectral Transformation:

30


Analog Analog HighpassHighpass Filter DesignFilter Design• Example - Design an analog Butterworth

highpass filter with the specifications: kHz, kHz, dB, dB

• Choose• Then

• Analog lowpass filter specifications: ,, dB, dB

4ˆ =pF 1ˆ =sF40=αs

1.0=α p

1=Ω p

1=Ω p

410004000

ˆˆ

ˆ2

ˆ2===

ππ

=Ωs

p

s

ps F

FFF

1.0=α p 40=αs4=Ωs31


Analog Analog HighpassHighpass Filter DesignFilter Design• Code fragments used

[N, Wn] = buttord\(1, 4, 0.1, 40, ‘s’\);[B, A] = butter\(N, Wn, ‘s’\);[num, den] = lp2hp(B, A, 2*pi*4000);

• Gain plots

0 2 4 6 8 10-80

-60

-40

-20

0

Ω

Gai

n, d

B

Prototype Lowpass Filter

0 2 4 6 8 10-80

-60

-40

-20

0

Frequency, kHz

Gai

n, d

B

Highpass Filter

32


Analog Analog BandpassBandpass Filter Filter DesignDesign

• Spectral Transformation

where is the passband edge frequency of , and and are the lower and upper passband edge frequencies of desired bandpass filter

)ˆˆ(ˆˆˆ

12

22

pp

op s

ssΩ−ΩΩ+Ω=

pΩ1ˆ pΩ 2ˆ pΩ

)ˆ(sHBP

)(sHLP

33



• On the imaginary axis the transformation is

where is the width of passband and is the passband center frequency of the bandpass filter

• Passband edge frequency is mapped into and , lower and upper passband edge frequencies

w

op BΩ

Ω−ΩΩ−=Ω ˆ

ˆˆ 22

12 ˆˆ ppwB Ω−Ω=oΩ

1ˆ pΩm 2ˆ pΩ±pΩ±

34



w

op BΩ

Ω−ΩΩ−=Ω ˆ

ˆˆ 22

1ˆ sΩ− oΩ1ˆ sΩ1ˆ pΩ−

BandpassPassbandPassband Stopband


Passband

Stopband StopbandΩ

ΩStopband Stopband

0

02ˆ sΩ

2ˆ pΩ− 1ˆ pΩ 2ˆ pΩoΩ− ˆ↓ ↓ ↓ ↓2ˆ sΩ−

35

Lowpass to Bandpass spectral Transformation:

36



• Stopband edge frequency is mapped into and , lower and upper stopband edge frequencies

• Also,

• If bandedge frequencies do not satisfy the above condition, then one of the frequencies needs to be changed to a new value so that the condition is satisfied

sΩ±1ˆ sΩm 2ˆ sΩ±

21212 ˆˆˆˆˆ ssppo ΩΩ=ΩΩ=Ω

37



• Case 1:To make we can either increase any one of the stopband edges or decrease any one of the passband edges as shown below

2121 ˆˆˆˆ sspp ΩΩ>ΩΩ

1ˆ pΩ 2ˆ pΩ1ˆ sΩ 2ˆ sΩ

→ →

←←

Ω

2121 ˆˆˆˆ sspp ΩΩ=ΩΩ

Passband

Stopband Stopband

38



(1) Decrease tolarger passband and shorter

leftmost transition band(2) Increase to

No change in passband and shorter leftmost transition band

1ˆ pΩ

1ˆ sΩ

221 ˆ/ˆˆ pss ΩΩΩ

221 ˆ/ˆˆ spp ΩΩΩ

39



• Note: The condition can also be satisfied by decreasing which is not acceptable as the passband is reduced from the desired value

• Alternately, the condition can be satisfied by increasing which is not acceptable as the upper stop band is reduced from the desired value


2ˆ pΩ

2ˆ sΩ

40



• Case 2:To make we can either decrease any one of the stopband edges or increase any one of the passband edges as shown below

2121 ˆˆˆˆ sspp ΩΩ<ΩΩ

1ˆ pΩ 2ˆ pΩ1ˆ sΩ 2ˆ sΩ

→ →

←←

Ω


Passband

Stopband Stopband

41



(1) Increase tolarger passband and shorter

rightmost transition band(2) Decrease to

No change in passband and shorter rightmost transition band

2ˆ pΩ

2ˆ sΩ



42



• Note: The condition can also be satisfied by increasing which is not acceptable as the passband is reduced from the desired value

• Alternately, the condition can be satisfied by decreasing which is not acceptable as the lower stopband is reduced from the desired value


1ˆ pΩ

1ˆ sΩ

43



• Example - Design an analog elliptic bandpass filter with the specifications:

kHz, kHz, kHzkHz, dB, dB

• Now and• Since we choose

4ˆ 1 =pF 7ˆ 2 =pF8ˆ 2 =sF

3ˆ 1 =sF1=pα 22=sα

621 1028ˆˆ ×=pp FF 6

21 1024ˆˆ ×=ss FF

2121 ˆˆˆˆ sspp FFFF >

428573724

2

211 .

FFFFp

ssp === kHz

44



• Bandwidth kHz• We choose

• Analog lowpass filter specifications:, , dB, dB

1=Ω p

413725

924)(1

2121 .

)/(BFFFFws

ssss =

×−

=⋅−

=Ω

1=Ω p 4.1=Ωs 1=pα22=sα

725

7247 =−=wB

45



• Code fragments used[N, Wn] = ellipord\(1, 1.4, 1, 22, ‘s’\);[B, A] = ellip\(N, 1, 22, Wn, ‘s’\);[num, den]

= lp2bp(B, A, 2*pi*4.8989795, 2*pi*25/7);• Gain plot

0 2 4 6 8-60

-40

-20

0

Ω

Gai

n, d

B

Prototype Lowpass Filter

0 5 10 15-60

-40

-20

0

Frequency, kHz

Gai

n, d

B

Bandpass Filter

46


Analog Analog BandstopBandstop Filter DesignFilter Design

• Spectral Transformation

where is the stopband edge frequency of , and and are the lower and upper stopband edge frequencies of the desired bandstop filter

sΩ)(sHLP

2212

ˆˆ

)ˆˆ(ˆ

os

ss ss

sΩ+Ω−ΩΩ=

)ˆ(sHBS

1ˆ sΩ 2ˆ sΩ

47


Analog Analog BandstopBandstop Filter DesignFilter Design• On the imaginary axis the transformation is

where is the width of stopband and is the stopband center frequency of the bandstop filter

• Stopband edge frequency is mapped into and , lower and upper stopband edge frequencies

22 ˆˆˆ

Ω−ΩΩΩ=Ωo

ws

B

12 ˆˆ sswB Ω−Ω=oΩ

1ˆ sΩm 2ˆ sΩ±sΩ±

48


Analog Analog BandstopBandstop Filter DesignFilter Design• Passband edge frequency is mapped

into and , lower and upper passband edge frequencies

pΩ±1ˆ pΩm 2ˆ pΩ±

1ˆ sΩ−oΩ

1ˆ sΩ1ˆ pΩ−

BandpassPassbandPassband


Passband

Stopband StopbandΩ

ΩStopband Stopband

0

0

2ˆ sΩ2ˆ pΩ− 1ˆ pΩ 2ˆ pΩoΩ− ˆ↓ ↓ ↓ ↓

2ˆ sΩ−

Passband

49

Lowpass to Bandstop spectral Transformation:

50


Analog Analog BandstopBandstop Filter DesignFilter Design• Also,

• If bandedge frequencies do not satisfy the above condition, then one of the frequencies needs to be changed to a new value so that the condition is satisfied


51


Analog Analog BandstopBandstop Filter DesignFilter Design• Case 1:• To make we can either

increase any one of the stopband edges or decrease any one of the passband edges as shown below

2121 ˆˆˆˆ sspp ΩΩ>ΩΩ

1ˆ pΩ 2ˆ pΩ1ˆ sΩ 2ˆ sΩ→ →

←←

Ω


Stopband

PassbandPassband

52



(1) Decrease tolarger high-frequency passband

and shorter rightmost transition band(2) Increase to

No change in passbands andshorter rightmost transition band

2ˆ pΩ

2ˆ sΩ



53


Analog Analog BandstopBandstop Filter DesignFilter Design• Note: The condition

can also be satisfied by decreasing which is not acceptable as the low-frequency passband is reduced from the desired value

• Alternately, the condition can be satisfied by increasing which is not acceptable as the stopband is reduced from the desired value


1ˆ pΩ

1ˆ sΩ

54


Analog Analog BandstopBandstop Filter DesignFilter Design• Case 2:• To make we can either

decrease any one of the stopband edges or increase any one of the passband edges as shown below

2121 ˆˆˆˆ sspp ΩΩ<ΩΩ


1ˆ pΩ 2ˆ pΩ1ˆ sΩ 2ˆ sΩ→ →

← ←

ΩStopband

PassbandPassband

55



(1) Increase tolarger passband and shorter

leftmost transition band(2) Decrease to

No change in passbands andshorter leftmost transition band

1ˆ pΩ

1ˆ sΩ



56



• Note: The condition can also be satisfied by increasing which is not acceptable as the high-frequency passband is decreased from the desired value

• Alternately, the condition can be satisfied by decreasing which is not acceptable as the stopband is decreased


2ˆ pΩ

2ˆ sΩ

57

Lecture 20

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Transcript of Lecture 20