Analog Filters: The Approximation Franco Maloberti.

Analog Filters: The Approximation

Franco Maloberti

Franco Maloberti Analog Filters: The Approximation

2

Introduction

The first step of the normalized filter design is to find a magnitude characteristics, |H(j)|, that fulfills the specifications

Handling |H(j)| it is difficult. It is better to deal with |H(j)|2.

|H(j)|2 is a rationale function in 2

€

H( jω)2

= H(s)H(−s)s= jω

= H( jω)H*( jω)


3

Example

Consider the transfer function

€

H(s) =s+ 2

s3 + 2s2 + 2s+ 3

H(s)H(−s) =(s+ 2)(−s+ 2)

(s3 + 2s2 + 2s+ 3)(−s3 + 2s2 − 2s+ 3)=

−s2 + 4

−s6 + 8s2 + 9

H( jω)2

=ω2 + 4

ω6 − 8ω2 + 9

€

H( jω)2

=ω2 + 4

ω6 − 8ω2 + 9

Much easier to deal than


4

Methodology

The approximation phase is based on the following: Find a real rationale function in 2 capable to

met the magnitude specifications (pass-band and stop-band)

Live with the resulting phase response Correct (if necessary) the phase response (or

delay) with a cascaded network Work on low-pass characteristics (we will se

shortly how to achieve high-pass, band-pass, … from a low-pass TF)


5

Generic Low-Pass TF

A generic form of a low-pass TF is:

€

H( jω)2

=A0

1+ F(ω2)

F(ω2) <<1 for 0 ≤ω ≤ωp

F(ω2) >>1 for ω >>ωp

In the pass-band H()=A0

p

A0


6

Butterworth Characteristics

A very simple form for F(2) is

n is the order of the Butterworth filter Properties:€

F(ω2) =ω2n

H( jω)2

=A0

1+ω2n

€

H( jω)Max

= H( j0) =1

H( j1) =1

2 (−3 dB or half − power)


7

Maximally Flat

The magnitude response of the Butterworth filter is said to be maximally flat.

If we take the first 2n-1 derivatives of H(j), they are all zero at =0

€

H( jω) = (1+ω2n )−1/2

0 10 20 30 40 50 60 70 80 90 1000

0.1

0.2

0.3

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0.5

0.6

0.7

0.8

0.9

1

n=8

=0,…,2


8

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

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1

Butterworth Plots; |H(j)|2

clear all;omega=linspace(0,2,100);for k=1:6n=k;for i=1:100Butter(i,k)=1/(1+omega(i)^(2*n));endendplot(omega,Butter);

All curves cross 0.5For =1


9

Butterworth Plots

For >>1 |H(i)|2 ≈ 1/(2n) H(j)|dB = -20 n log

clear all;omega=linspace(0,100,100);for k=1:4n=k;for i=1:100Butter(i,k)=1/(1+omega(i)^(2*n));ButdB(i,k) = 20* log10(Butter(i,k));endendsemilogx(omega,ButdB);


10

Design of Butterworth filters

Specifications

€

α s =10log(1+ωs2n )

α p =10log(1+ωp2n )

ωs2n =10α s /10 −1

ωp2n =10α p /10 −1

ωsωp

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2n

=10α s /10 −1

10α p /10 −1

The values of αs, αp and the ratio s/p determine the ordern of the filter.

dB|H(j)|2pαssαp


11

Chebyshev Characteristics

The Chebishev filters are based on the Chebishev polynomials

n is the order It is a polynomial!

€

Cn (ω) = cos(ncos−1ω)

€

cosθ =ω θ = cos−1ω

Cn (ω) = cos(nθ)


12

Chebyshev Polynomials

€

Cn (ω) = cos(nθ)

Cn+1(ω) = cos(nθ +θ) = cosnθ cosθ − sinnθ sinθ

Cn−1(ω) = cos(nθ −θ) = cosnθ cosθ + sinnθ sinθ

Cn+1(ω) +Cn−1(ω) = 2cosnθ cosθ

Cn (ω) ω

Cn+1(ω) = 2ωCn (ω) −Cn−1(ω)


13

Chebyshev Polynomials

€

C0(ω) = cos0 =1

C1(ω) = cos(cos−1ω) =ω

C2(ω) = 2ω2 −1

C3(ω) = 2(2ω2 −1)ω −ω = 4ω3 − 3ω

C4 (ω) = 2(4ω3 − 3ω)ω − (2ω2 −1) = 8ω4 − 8ω2 +1

€

Cn (ω) = 2n−1ωn +K


14

-1.5 -1 -0.5 0 0.5 1 1.5-15

-10

-5

0

5

10

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Chebyshev Polynomials Features

If varies in the range -1 +1 Cn()=cos(n+) limited to the range -1 +1

If | | is larger than 1 Cn()=cosh(n cosh-1 ) increases monotonically

with n.

clear all;n=7xmax=1+5/n^2;w=linspace(-xmax,xmax,200);for i=1:200C(i)=cos(n*acos(w(i)));endplot(w,C);


15

Chebyshev low-pass

Choose a small number and let

the low-pass transfer function is

Normalize p = 1 Maximum attenuation in the pass-band

€

F(ω2) = ε 2Cn2(ω)

€

H( jω)2

=1

1+ ε 2Cn2(ω)

€

H( jω)min

2=

1

1+ ε 2

α s =10log(1+ ε 2)


16

Chebyshev low-pass Plots

clear all;n = input('what is the order? ');eps = input('what is the value of epsilon? ');xmax=2;w=linspace(0,xmax,200);for i=1:200

C(i)=cos(n*acos(w(i)));H(i)=1/(1+(eps*C(i))^2);

endplot(w,H);

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

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Order 4Eps=0.4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

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Order 7Eps=0.2


17

Start from specifications

Determine

The order of the filter must be such that

Remember that

Chebyshev Filter DesigndB|H(j)|2pαssαp

€

α p =10log(1+ ε 2)

€

= 10α p /10 −1

€

α s =10log[1+ ε 2Cn2(ωs /ωp ))]

€

Cn (ωs /ωp ) = cosh(cosh−1(ωs /ωp ))

n =cosh−1[Cn (ωs /ωp )]

cosh−1(ωs /ωp )


18

Other Chebyshev Responses

Define another function

Use in the Matlab program

€

Ha ( jω)2

=1− H( jω)2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

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H(i)=1-1/(1+(eps*C(i))^2);


19

Inverse Chebyshev High-pass

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

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1

€

Hb ( jω)2

=1

1+ ε 2Cn2(1 /ω)

clear all;n = input('what is the order? ');eps = input('what is epsilon? ');xmax=2;w=linspace(0,xmax,200);for i=1:200

C(i)=cos(n*acos((1/w(i))));H(i)=1/(1+(eps*C(i))^2);

endplot(w,H);


20

Inverse Chebyshev Low-pass

€

Hc ( jω)2

=1− Hb ( jω)2

=1−1

1+ ε 2Cn2(1 /ω)

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

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0.9

1


21

Elliptic-Function

Butterworth and Chebyshev don’t have finite zeros All-pole functions

Finite zeros help in the pass-band stop-band transition (see inverse Chebyshev)

Elliptic functions are a special category of zero-pole transfer functions

€

Hc ( jω)2

=P( jω)

2

Q( jω)2


22

Elliptic-Function (ii)

Define RN()

or

Poles and zeros are reciprocal of one another If 1, 2,… are in the range 0 ≤ ≤1 |R()|2 is

zero at 1, 2,… and infinite at 1/1, 1/2,…

€

RN (ω) =(ω1

2 −ω2)(ω22 −ω2)L (ωN

2 −ω2)

(1−ω12ω2)(1−ω2

2ω2)L (1−ωN2ω2)

€

RN (ω) =ω(ω1

2 −ω2)(ω22 −ω2)L (ωN

2 −ω2)

(1−ω12ω2)(1−ω2

2ω2)L (1−ωN2ω2)


23

Elliptic-Function (iii)

The elliptic characteristics is

€

H( jω)2

=1

1+ ε 2RN2 (ω)


24

Plot of the Elliptic Function

clear all;n=5;eps=8;xmax=0.91;om1sq=0.1;om2sq=0.4;om3sq=0.65;om4sq=0.8;omega=linspace(0,xmax,200);for i=1:200w=omega(i);Num=w*(om1sq-w*w)*(om2sq-w*w)*(om3sq-w*w)*(om4sq-w*w);Den=(1-om1sq*w*w)*(1-om2sq*w*w)*(1-om3sq*w*w)*(1-om4sq*w*w);R(i)=Num/Den;Ellip(i)=1/(1+(eps*R(i))^2);endplot(omega,Ellip);

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90.996

0.9965

0.997

0.9975

0.998

0.9985

0.999

0.9995

1

1.0005

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 20

0.5

1

1.5

2

2.5

3x 10-6

Pass

Stop


25

Determine the order of a Filter

BUTTORD Butterworth filter order selection. [N, Wn] = BUTTORD(Wp, Ws, Rp, Rs) returns the order N of the lowest order digital Butterworth filter that loses no more than Rp dB in the passband and has at least Rs dB of attenuation in the stopband. Wp and Ws are the passband and stopband edge frequencies, normalized from 0 to 1 (where 1 corresponds to pi radians). For example, Lowpass: Wp = .1, Ws = .2 Highpass: Wp = .2, Ws = .1 Bandpass: Wp = [.1 .8], Ws = [.2 .7] Bandstop: Wp = [.2 .7], Ws = [.1 .8] BUTTORD also returns Wn, the Butterworth natural frequency (or, the "3 dB frequency") to use with BUTTER to achieve the specifications. [N, Wn] = BUTTORD(Wp, Ws, Rp, Rs, 's') does the computation for an analog filter, in which case Wp and Ws are in radians/second.

»help buttord


26

Verification

»ButterorderN = 12Wn = 166.7598

0 0.2 0.4 0.6 0.8 1 1.2 1.40

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1clear all;omega=linspace(0,1.2,200);n=12;for i=1:200Butterw(i)=1/(1+omega(i)^(2*n));endplot(omega,Butterw);Wp=80;Rp=10*log10(Butterw(Wp+1));Ws=180;Rs=10*log10(Butterw(Ws));[N, Wn] = buttord(Wp, Ws, Rp, Rs, 's')

Butterorder


27

Order of Chebishev or Elliptic filters

[N, Wn] = CHEB1ORD(Wp, Ws, Rp, Rs, 's') does the computation for an analog filter, in which case Wp and Ws are in radians/second.

[N, Wn] = ELLIPORD(Wp, Ws, Rp, Rs, 's') does the computation for an analog filter, in which case Wp and Ws are in radians/second.

clear all;Wp=2*pi;k=input('normalized stop frequency ')Ws=k*Wp;Rp=input('attenuation passband ')Rs=input('attenuation stopband ')[Nbutter, Wn] = BUTTORD (Wp, Ws, Rp, Rs, 's')[NCheb1, Wn] = CHEB1ORD(Wp, Ws, Rp, Rs, 's')[Nellip, Wn] = ELLIPORD(Wp, Ws, Rp, Rs, 's')

Comparing the orders »OrderFilternormalized stop frequency 1.2attenuation passband 0.2attenuation stopband 25

Nbutter = 25Wn = 6.7203NCheb1 = 9Wn = 6.2832Nellip = 5Wn = 6.2832


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Butter and Cheb with same specs

stop frequency 1.2attenuation PB 0.2 dBattenuation SB 25 dB

Nbutter = 25Wn = 6.7203NCheb1 = 9Wn = 6.2832

0 0.5 1 1.5 2 2.50

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1


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Butter and Cheb with same specs (ii)

clear all;point=2000;range=2;omega=linspace(0,range,point)+range/2/point;n=25;m=9;eps=0.186;k=6.7203/6.2832;for i=1:pointButter(i)=1/(1+(omega(i)/k)^(2*n));C(i)=cos(m*acos(omega(i)));Cheb(i)=1/(1+(eps*C(i))^2);endplot(omega,Butter,omega,Cheb);Butter(1000)Cheb(1000)Butter(1200)Cheb(1200)

Analog Filters: The Approximation Franco Maloberti.

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Transcript of Analog Filters: The Approximation Franco Maloberti.