Slide010 Filter Theory

8/22/2019 Slide010 Filter Theory

1/90

Filter Theory

Instructor :Muhammad Riaz

1


2/90

ObectivesM.A.J.U.

y Introduction and filters specifications.

y Transfer functions.

y Filters approximations and prototype design.

y requency rans orma on an magn u e sca ng.

y Passive filter desi n and realization.

y Design examples

2 M Riaz, EE,


3/90

M.A.J.U.

Filters are frequency selective networks.

Filters allows certain freq. band to pass through them

.

An ideal filter will have an am litude res onse that is

unity (or at fixed gain) for the frequencies of interest

(pass band) and zero everywhere else (stop band).

The frequency at which the response changes from

- - .

3 M Riaz, EE,


4/90

M.A.J.U.

y These are used to

y Stabilize the amplifiers performance.

.

y

Separate the bands of frequencies (e. g. radio receiver).

y Eliminate the effect of aliases in A/D systems.

y Eliminate the effect of harmonics in the reconstruction (D/A

conversion) of the signals, etc,.

4 M Riaz, EE,


5/90

Filter

Types

M.A.J.U.

Idealcharacteristics5

M Riaz, EE,


6/90

Practical filter and filter parameters M.A.J.U.

6 M Riaz, EE,


7/90

Thetransferfunctions M.A.J.U.

Thefilterhasfrequencydependentresponsebecausethe

impedance(capacitor

and

inductor).

vc

cii

L

LLv

diL

dt

= c cidvC

dt

=

CZ

Cs=

LZ Ls=

s = +

eper reqradianfreq

7 M Riaz, EE,


8/90

Thetransferfunctions M.A.J.U.

Transferfunctionsaretheratiooftheoutputvoltage(current)totheinput

voltage(current).

11 1 01

( ) ...( )...

m mo m m

n n

V s a s a s a s aH sV s b s b s b s b

+ + + += =+ + + +

n n

H(s) is the rational function of s with real coefficients.The roots of the numerator polynomial are called zeros of H(s).The roots of denominator polynomial are called poles of H(s).The numerical values of as and bs determine the filter response (LP, HP,

, ,.The degree of the denominator polynomial determines the order of thefilter. Normally n m.

8 M Riaz, EE,


9/90

Low

ass

function

M.A.J.U.

0aH s =0

2( )

aH s

b s b s b=

+ +

1 0

s +Firstorder 20dB/decade

rolloffinstopband

Secondorder 40dB/decaderoll

offinstopband

Aninductorinseriesor/andcapacitorinparallelperformlowpassfunctions.

sRL

LR LR

CC

iV oViV oV

oVLRiV

( ) LR

H s = LRH s = 2( )LRH s =

Ls

s L s LR R Cs R R+ + L Ls s

9 M Riaz, EE,


10/90

Hi h

ass

function

M.A.J.U.

1a sH s =

22

2( )

a sH s =

1 0

s + 2 1 0

Firstorder 20dB/decade

Secondorder 40dB/decaderoll

offinsto band

Acapacitorinseriesor/andinductorinparallelperformhighpassfunctions.

( ) LR Cs

H s = ( )L

R LsH s =

2

2( ) L

R LCsH s

R LCs Ls R=

+ +L

s L s L s

10 M Riaz, EE,


11/90

Band

ass

function

M.A.J.U.

1a s= Secondorder 40dB/decaderoll2

2 1 0b s b s b+ + offinstopband

Aseriesor and arallelcombinationofaninductorandaca acitor erformband

passfunctions.

2( )

1L

R CsH s

LCs R Cs=

+ + 2( ) L

R LsH s =

s s s

11 M Riaz, EE,


12/90

Band

sto

function

M.A.J.U.

22 0a s aH s

+= Secondorder 40dB/decaderoll

2 1 0

s s+ + o nstop an

Aseriesor/andparallelcombinationofaninductorandacapacitorperformband

stop unctionsass ownin igure.

( )2 1LR LCs +=

( )2 1( )

LR LCsH s

+=

( ) 2s L s L L sR R LCs R R Cs R R+ + + + L Ls s+ +

12 M Riaz, EE,


13/90

GeneralBiquads M.A.J.U.

2

02 2

( ) z

p

H s H

=

2

02 2

( )p

sH s H

s s

=

+ +p

p

Q pQ

Lowpass highpass

02 2( )

z

z

p

sQ

H s H

=

2 2

0

2 2

( ) z

p p

sH s H

s s

+=

+ +

2 2zs s

+

p

pQ p

BandpassBandstop

02 2

( ) zp

p

p

QH s H

s sQ

=+ +

Delayequalizer

13 M Riaz, EE,


14/90

Band

ass

Exam le

M.A.J.U.

1a sH s = Secondorder 40dB/decaderoll2 1 0s s+ + offinstopband

For

100 , 0.1 , 1LL mH C F R k= = =

4

2 4 810( )Cs

H s =

2( )

1L

L

R CsH s

LCs R Cs=

+ +

1 ,

1

pLC

L

=

( )

( )2 2 2( )

1L z z

L p p p

R L s Q sH s

s R L s LC s Q s

= =

+ + + +

p p L LLR C

= = =

410 , 1p pQ = =

14 M Riaz, EE,

Magnitude response and phase plot is shown on thenext slides.


15/90

Gain Plot

-5

0

-10

ain(dB)

-20

-15

nitude

)

-30

-25Ma(dB

-35

15 Frequency in rad/sec

102

103

104

105

106

-

Frequency in rad/sec


16/90

Attenuation Plot40

30

n20

25

tio

n

ttenuatio

dB)

10

15

Attenu

(dB)

5


102

103

104

105

106

0



17/90

Phase Response

80

100

40

60

ha

sein

eg

rees

-20

0

20

se

(deg.)

-60

-40Pha

102

103

104

105

106

-100

-80



18/90

Second order Low-2

0( )zH s H

=

1 LCL

passp

p

s sQ

+ +L

LRCiV oV ( )

2 1 1L

ss R C s LC

=+ +

1 , LC

LC Q R = =

100 0.1 10 10

For

L mH C F R k= = = =

8

2 3 8

10

( ) 10 10H s s s=

+ +

100 , 0.1 , 2 , 2L p

For

L mH C F R k Q= = = =

8

2 4 8

10( )

0.5 10 10H s

s s=

+ +

100 , 0.1 , 707 , 0.707L p

For

L mH C F R Q= = = =

8

2 4 8

10( )

1.414 10 10H s

s s=

+ +

18 100 , 0.1 , 500 , 0.5L p

For

L mH C F R Q= = = =

8

2 4 8

10( ) 2 10 10H s s s= + +


19/90

Gain Response

40

Q=10

0

20

Q=0.707Q=0.5

-20

nitude

)

-60

-40

Ma(dB

-80

19

102

103

104

105

106

-100



20/90

Phase Response

-20

0

Q=10

Q=2

-60

-40

.

Q=0.5

-100

-80

se

(deg.)

-140

-120Ph

-160

20

102

103

104

105

106

-



21/90

Second order High-

( )

2

2( )

1 1L

sH s

s R C s LC =

+ +Cpass

LR

iV oVL 1 ,p p L

CLC Q R

L = =

100 , 0.1 , 10 , 10L p

For

L mH C F R k Q= = = =

2

2 3 8( )

10 10

sH s

s s

=+ +

100 , 0.1 , 2 , 2L p

For

L mH C F R k Q= = = =

2

2 4 8( )

0.5 10 10

sH s

s s=

+ +

100 , 0.1 , 707 , 0.707L por

L mH C F R Q= = = =2

2 4 8( )

1.414 10 10sH s

s s=

+ +

2121

100 , 0.1 , 500 , 0.5L pL mH C F R Q= = = =2

2 4 8( )

2 10 10

sH s

s s

=

+ +


22/90

40

20

Q=10

Q=2

Q=0.707

0

Q=0.5

-40

-

ni

tude

)

-60

Ma(dB

-100

-80

22

102

103

104

105

106



23/90

160

180

Q=10

Q=2

120

140

Q=0.707

Q=0.5

80

100

se

(deg.)

40

60Ph

20

23

102

103

104

105

106



24/90

Effect of Qp

( )

( )2 2 2( )

1L z z

L p p p

R L s Q sH s

s R L s LC s Q s

= =

+ + + +

41 10 / secp LC rad = =

For

p

L

Q

R C

=

42 10 s, . , , .L p

For

= = = =

2 4 82 10 10s s+ +

4

2 10 sH s =p

100 0.1 500 2

For

L mH C F R= = = =

2 4 82 10 10s s+ +

40.5 10 sH s

=

p

For

= = = =

.s s+ +

40.1 10 sH s

=

24

, . , ,L p .s s+ +


25/90

Gain Response

0

-20

-

-30

nitude

)

-40 Q=0.5

Q=0.707

Ma(Db

-50 Q=2Q=10

25

102

103

104

105

106

-60



26/90

Phase Response

80

100

Q=0.5

40

60

.

Q=2Q=10

0

20

se

(deg.)

-

-40

-20

Ph

2 3 4 5 6-100

-80

26Frequency in rad/sec


27/90

Band

sto

function

M.A.J.U.

sR

LR

CiV oV

L

LLRiV

oV

2 1s LC+ 2 1s LC+

( )2( )

|| 1L

L s s L

H sR R s R R L s LC

=+ + + ( )2

( )1 1L

H ss R C s LC

=+ +

1 ,p z p LCLC Q RL

= = =11 , ||p z p

s L

LLC QR R C

= = =

27 M Riaz, EE,


28/90

( )( )

2

2

1( )

1 1L

s LCH s

s R C s LC

+=

+ +

1 ,p z p L

CLC Q RL

= = =LLRiV

oV

100 0.1 10 10

For

L mH C F R k= = = =

2 8

2 3 8

10( )

10 10

sH s

s s

+=

+ +

100 , 0.1 , 2 , 2L p

For

L mH C F R k Q= = = =

2 8

2 4 8

10( )

0.5 10 10

sH s

s s

+=

+ +

100 , 0.1 , 707 , 0.707L p

For

L mH C F R Q= = = =

2 8

2 4 8

10( )

1.414 10 10

sH s

s s

+=

+ +

28 100 , 0.1 , 500 , 0.5L p

For

L mH C F R Q= = = =

2 8

2 4 8

10

( ) 2 10 10

s

H s s s

+

= + +


29/90

Phase Response

-10

0

-20

=

Q=2Q=0.707

Q=0.5

nitude

)-40

-30

Ma(Db

-50

-

-60

29


102

103

104

105

106


30/90

Phase Response

80

100

Q=10

40

60

=

Q=0.707Q=0.5

se

(deg.)

0

20

Ph

-40

-20

-80

-60

30

Frequency in rad/sec10

210

310

410

510

6-100


31/90

-

2

0

2 2

( ) zH s H

= M.A.J.U.

p pps sQ

31 M Riaz, EE,


32/90

2

0

2 2

( )s

H s H

= M.A.J.U.

p

p

ps sQ

32 M Riaz, EE,


33/90

and

Band

ass0

( )

z

z

sQ

H s H

=M.A.J.U.

2 2p

p

p

s sQ

+ +

cp

H L

fCentral freq.QBandwidth f f

= =

c H Lf f f=

33 M Riaz, EE,


34/90

2 2

0( )z

sH s H

+= M.A.J.U.

p

pp

p

s sQ

+ +

34 M Riaz, EE,


35/90

Second

order

low

ass

hase

Res onse

M.A.J.U.

35 M Riaz, EE,


36/90

M.A.J.U.

p

NotchFilterWidthversusFrequencyforVariousQValues

36 M Riaz, EE,


37/90

Notchfilter haseRes onseM.A.J.U.

37 M Riaz, EE,


38/90

M.A.J.U.

2 2zz

z

s sQ

H s H

+

=

2 2p

p

p

s s

Q

+ +

38 M Riaz, EE,


39/90

Assignment 31:

- -

M.A.J.U.

.plane. (c) Magnitude Plot. (d) Phase Plot on the semilog graph paper. (e) compute3 dB down frequency and show it on the plot. Also show brief detail of yourcalculations.

Important:You have to show the work in the next class (22-05-13)personally in original form otherwise you will be marked as absent.40 F40 F5mH

10

iV oV

5mH

10iV o

V

5mH 40 F

10iVoV

40 F10

iV oV

5mH

39 M Riaz, EE,


40/90

Filter Approximations

Filter specifications are approximated by using some well-described rational functions.

.

The most popular approximations are:

Butterworth,Chebyshev,

BesselElliptic (or Cauer) types.

-

filters.

High-pass, band-pass, and band-reject filters are realized

40

by using frequency transformation from low-pass function.


41/90

Butterworth Filters M.A.J.U.

Flat response in pass-band and monotonically increasing

in sto band.2

22

1( )

1n

H s

=

+ Gain function

p

2

210lo 1 dB

n

A

= +

Loss (Attenuation)

Functions.p

0.1 minA

( )0.1 maxlog 10 1An

=

n is order of thefilter.

ogp

41 M Riaz, EE, MAJ U.


42/90

Example:

Find order for Butterworth filter for the following low-pass specifications:

Wp=1000 rad/sec, Ws=1100 rad/sec, Amax=0.1 dBs, and Amin = 25 dBs

Solution:

n=log10((10 2.5-1)/(10 .01-1))/(2*log10(1100/1000))

= .n=50, very high order because transition band is very narrow.

42


43/90

Normalized Butterworth Filters M.A.J.U.

1p =

( ) 2 210log 1 dBnA = +

22 2 2

nn s

j

22

( ) 1

n

sThe solution of the loss function L s =0= +gives the poles of the normalised LP BW filter, which are

2

,n

ks = e 0,1, 2, , 2 1k n= "

are equal.

43 M Riaz, EE,


44/90

M.A.J.U.

80

Normalized Butterworth Response

60

70

on(dB)

n=1n=2n=3

30

40

50

Attenuati

10

20

10-1

100

101

102

0

Frequency rad/sec

44 M Riaz, EE,

Ch b h Filt


45/90

Chebyshev Filters M.A.J.U.

2

2 2

1( )H =

( ) ( )2 210log 1 ( ) dBA F = +

2 0.1 max10 1

A =

2 2 2=

( )12

1

cos cos 1( )

n for Let the loss function F

=

45 M Riaz, EE,


46/90

Chebyshev Filters order M.A.J.U.

120.10.11 maxmincosh 10 1 10 1AA

( )1

cosh s pn

=

0,1, 2, ,k n= "

11 1 (2 1)sinh sinh sin2

k

k

n n

+ =

11 1 (2 1)cosh sinh cos2

k

k

n n

+ = >> [z,p,k]=cheb1ap(2,3)

2 2

2 1 2 1

11 1 1 1

sinh sinh cosh sinh

k k

+ =

Which is Equation ofEllipse so poles are on

e pse.

46 M Riaz, EE,


47/90

M.A.J.U.

47 M Riaz, EE,


48/90

M.A.J.U.

48 M Riaz, EE,

Chebyshev Magnitude (Attenuation)


49/90

Chebyshev Magnitude (Attenuation)

70

80n =1

n =2

n =3

n =4

60

dB)

40

50

n

uation(

20

30Att

10

49

10-1

100

101

0

(rad/sec)

M Riaz, EE, MAJ U.

Pass band expanded view Chebyshev


50/90

Pass-band expanded view Chebyshev

8

n =1

n =2

4

6 n =3

n =4

)

2

a

tion(d

0

Atten

-4

-2

50

10-1

100

101

(rad/sec)

M Riaz, EE,

Chebyshev Magnitude (Gain)


51/90

y g ( )Response

-10

0n =1

n =2

n =3

n =4

-30

-20

-

-40

a

in(dB)

-60

10-1

100

101

-80

-70

51

Maximum pass-band attenuation of 3 dB at PB edge frequency = 1 rad/sec

M Riaz, EE, MAJ U.

Pass-band expanded view Chebyshev


52/90

p yResponse

2

n =1

n =2

n =3

n =4

0

-4

-2

n

(dB)

-6

Ga

10-1

100

101

-8

52

(rad/sec)

M Riaz, EE, MAJ U.


53/90

MATLAB Code for Fre uenc

>> w=.1:.001:5;*

Response

>> [z,p,k]=cheb1ap(1,3);

>> Hs1=k./(s-p(1));>> [z,p,k]=cheb1ap(2,3);

*. - . ->> [z,p,k]=cheb1ap(3,3);>> Hs3=k./((s-p(1)).*(s-p(2)).*(s-p(3)));>> [z,p,k]=cheb1ap(4,3);

* * *. - . - . - . ->> semilogx(w,-20*log10(abs(Hs1)),w,-20*log10(abs(Hs2)),w,-20*log10(abs(Hs3)),w,-20*log10(abs(Hs4)));grid minor>> figure

semilogx(w,20*log10(abs(Hs1)),w,20*log10(abs(Hs2)),w,20*log10(abs(Hs3)),w,20*log10(abs(Hs4)));grid minor>>

53 M Riaz, EE,

MA J U


54/90

M.A.J.U.

54 M Riaz, EE,


55/90

Example:

Find order for Chebyshev filter for the following low-pass specifications:

Wp=1000 rad/sec, Ws=1100 rad/sec, Amax=0.1 dBs, and Amin = 25 dBs

Solution:

n=acosh(((10 2.5-1)/(10 .01-1)) .5)/(acosh(1100/1000))

n = 12.2858

= ,

response in pass-band) but still high.


Invrse Chebyshev Filters (Equiripple pass band MA J U


56/90

Invrse Chebyshev Filters (Equiripple pass band M.A.J.U.

2

22

1( )H =

( ) ( )2210log 1 (1 ) dBA T = +

2

0.1 max

1A

=

56 M Riaz, EE,

Inverse Chebyshev Filters MA J U


57/90

Inverse Chebyshev Filters M.A.J.U.

Poles of H(s) are

0,1, 2, ,k n= "

1

1

sinh sinh sin2

1 1 (2 1)cosh sinh cos

kn n

k

=

=

2n n

1min0.110 1A

=

57 M Riaz, EE,

M.A.J.U.


58/90

M.A.J.U.

58 M Riaz, EE,

Elliptic Filter Approximation


59/90

Elliptic Filter Approximation

It has finite poles of attenuation (zeros of gain function) inthe stopband.

The poles are chosen to have equiripple response in stopband.

The pole closest to the stopband edge significantly

increases the slo e in sto band.

Further poles are needed to maintain the level of min..

For given requirement, elliptic approximation will require a

59

ower or er t an t e utterwort or e s ev.

Fifth d Elli ti Filt i ti f ti


60/90

Fifth order Elli tic Filter a roximation function

2 20.0046205 4.36495 10.56773s s+ + + +

( ) ( )( )2 2( )

0.19255 1.03402 0.58054 0.52500 0.392612T s

s s s s s=

+ + + + +

60

Elliptic Filter Magnitude Response


61/90

180

140

160

) 100

120

ation(d

60

80

Atten

40

0

61 Maximum PB attenuation of 3 dB, SB attenuation of 40

dB and PB frequency = 1 rad/sec M Riaz, EE, MAJ U.

(rad/sec)10-1

100

101

-

Pass-Band Enlarged view


62/90

g

5

6n =1

n =2

n =3

n =4

n =6

4

2

3

1

-1

0

62

M Riaz, EE,

10-0.9

10-0.8

10-0.7

10-0.6

10-0.5

10-0.4

10-0.3

10-0.2

10-0.1

100

Sto Band Enlar ed View


63/90

Sto -Band Enlar ed View

n =1

n =2

n =3

60

70 n =4

n =6

50

30

40

20

63

M Riaz, EE,

10 10

MATLAB Code


64/90

MATLAB Code

w=.1:0.00001:10;=*

[z,p,k]=ellipap(1,3,40)Hs1=k./(s-p(1));

[z,p,k]=ellipap(2,3,40)Hs2=k.*(s-z(1)).*(s-z(2))./((s-p(1)).*(s-p(2)));[z,p,k]=ellipap(3,3,40)Hs3=k*(s-z(1)).*(s-z(2))./((s-p(1)).*(s-p(2)).*(s-p(3)));[z,p,k]=ellipap(4,3,40)Hs4=k*(s-z(1)).*(s-z(2)).*(s-z(3)).*(s-z(4))./((s-p(1)).*(s-p(2)).*(s-p(3)).*(s-p(4)));[z,p,k]=ellipap(6,3,40)Hs6=k*(s-z(1)).*(s-z(2)).*(s-z(3)).*(s-z(4)).*(s-z(5)).*(s-z(6))./((s-p(1)).*(s-p(2)).*(s-p(3)).*(s-p(4)).*(s-p(5)).*(s-p(6)));

,- , ,- , ,- , ,- , ,-20*log10(abs(Hs6)));grid minor

64

M Riaz, EE,


65/90

Example:

Find order for Elliptic filter for the following low-pass specifications:

wp=1000 rad/sec, Ws=1100 rad/sec, Amax=0.1 dBs, and Amin = 25 dBs

Solution:

[N,wp]=ellipord(1000,1100,0.1,25,'s')

N = 6, reasonable order but at the expense of equiripple response both inpass and stop-bands.

65

M Riaz, EE, MAJ U.

50


66/90

50

30

40

on(dB)

20

Attenuati

101

102

103

104

0

rad sec

0.1

0.12

0.14

0

0.02

0.04

0.06

.

66

M Riaz, EE, MAJ U.101

102

103

-0.04

-0.02


67/90

0.05

0

0

-20

-10

-0.1

-0.05

-40

-30

10-1

100

-0.1510

-110

010

1-60

-50

67

M Riaz, EE, MAJ U.

Frequency Transformation LPP to LP


68/90

q y

( ) ( )p

sLP LPP sT s T s

=

=PP

L

PPsL PP PPsLsZ L s L s== = =

p

p p p

PP

p

C

PP

sC PP PPs

CsY C s C s

== = =

p p p

68

Frequency Transformation LPP to HP


69/90

q y

( ) ( ) pHP LPP ss

T s T s =

=1

1p pL PP PPsZ L s L s

== = =

PP pL

p

1

PP pC 1

p

p

C PP PPss

Y C s C s C s

== = =

69

Frequency Transformation LPP to BP


70/90

q y2 2

0( ) ( ) sBP LPP sBs

T s T s +==

2 2

0

2 2 20 0

2

1PPPP PPs

L PP PPsBs

s LL L

Z L s L s s BBs B Bs B sL

+=

+= = = + = +

( )20PP PPL B B L

( )2 2

0

2 2 20 0

20

1PPPP PPsC PP PPs

BsPP

s CC CY C s C s s

Bs B Bs B B C s

+

=

+= = = + = +

20

PP

PP

C B

B C

70

Frequency Transformation LPP to BS


71/90

2 20

( ) ( ) BsBS LPP ss

T s T s

=+

=20PPB L

2 22 2 11 1

BsL PP PPss

BsZ L s Ls

=+

= = =+

( )20PP PPL B B L s 1PP

L B

2 22 2

1

1 1BsC PP PPs

s

BsY C s C

s =

+

= = =+

1

C B

( )2

0PP PPC B B C s20PPB C 1 1

71

( )20C

PP PPC B B C s

Design procedure


72/90

1. specify the filter requirements.. e ec approx ma on ype accor ng o e

application.3. Determine the order of the filter.

4. Find transfer function for low-pass prototype (LPP)

filter.

. .6. Use frequency transformation to transform the LPP

filter according to the desired specifications.

. se magn u e sca ng o ac eve e mpe ance

matching.

72

Example: Design a complete Butterworth filter for the M.A.J.U.


73/90

Rs = RL=75 ohm:

A

35dB 35dB

0.5 dB

/rad sec

log scale

1000 1500 2000 3000

73

M Riaz, EE,

M.A.J.U.


74/90

Convert the specification into LP prototype Wp=1 Ws=(w2-w1)/(w4-w3)=(3000-1000)/(2000-1500)=2000/500=4 Determine the Butterworth order.

( ) ( )0.1 0.1 35min0.1 0.1 2

10 1 10 1log log

A

A

( )2log 42log

3.58s

p

n

= = =

The LP rotot e re uirements are shown in the next slide.

n =

74

M Riaz, EE,

M.A.J.U.


75/90

( ) dBA

35dB

0.5 dB

/rad sec41

log scale

75

M Riaz, EE,

Poles, transfer function and reflectioncoeff. of normalized LP rotot e filte

M.A.J.U.


76/90

2( 2 1) ,nj n kks = e+ + 0,1, 2, , 2 1k n= "8(5 2 )j k=

+ 0 1 2 7k= "

( )( )( )( )( )( )( )( )8 8 8 8 8 8 8 8

2

85 7 9 11 3 3

1 1( )

1j j j j j j j j

T s

ss e s e s e s e s e s e s e s e

= =

+

,k

( )( )( )( ) ( )( )( )( )8 8 8 8 8 8 8 85 5 7 7 3 3( ) ( )

j j j j j j j jT s T s

s e s e s e s e s e s e s e s e

=

1 1

= ( )( ) ( )( ) ( )( ) ( )( )2 2 2 28 8 8 82cos 5 1 2cos 7 1 2cos 1 2cos 3 1s s s s s s s s + + + +

2 2

1( )

0.76537 1 1.84776 1T s

s s s s=

+ + + +

8 42

8 2 2( ) ( )

s ss s = =

. .

76

M Riaz, EE, MAJ U.

The deriving point impedance and the circuit M.A.J.U.


77/90

3 21 ( ) 2.61313s +3.41422s +2.61313s+1sZ

= =

or pro o ype er s s own

1 ( ) 2 2.61313s +3.41422s +2.61313s+1s s+ +

4 3 22 2.61313s +3.41422s +2.61313s+1sY

+=

Realize the elements by continuous division method.

2.61313s +3.41422s +2.61313s+1

4 3 2

2.61313s +3.41422s +2.61313s+1 2 2.61313s +3.41422s +2.61313s+10 0.76537s

2 2.40434 2 0.76537 0.0

s

s s s s

+

+ + + +

-----------------------------------------------------

0.20879s 1.41422s 1.74676 10 2.61313s +3.41422s +2.6131s+ + + 3s+1

77

M Riaz, EE, MAJ U.

LP prototype filter is shown M.A.J.U.


78/90

4 0.7654 HL =2 1.8478 HL =

1 3 1.8478 FC

=1 0.7654 FC

=

78

M Riaz, EE, MAJ U.

Frequency transformationM.A.J.U.

ra sec=


79/90

ra sec=

2 6 2 20 3 10 /rad sec =

2 20( ) ( ) sBP LPP s

BsT s T s +==

( )( )

2 20 1

1 1

21 0

1

1 11.531

217.75

s CY C s C s mF s

BBs B H ssC

+= = = + = +

( )( )

2 20 2

2 2 2

2

1 13.6956

90.2

s LZ L s L s mH s

BBs B F ss

L

+= = = + = +

( )( )

2 20 3

3 3

2

3

1 13.6956

90.2

s CY C s C s mF s

BBs B H ss

+= = = + = +

( )2 2

0 44 4 4

1 11.531

217.75

s LZ L s L s mH s

BBs B F s

+= = = + = +

3 0

24 0L

79

M Riaz, EE, MAJ U.

The band ass filter with normalized terminations (one

M.A.J.U.


80/90

ohm) is shown

1.531 mF217.75 H90.2 F3.6956 mH1

90.2 F3.6956 mH1.531 mF217.75 H

80

M Riaz, EE, MAJ U.

M.A.J.U.


81/90

Increase the impedance of each element by multiplying the, .

20.4 F16.33 mH0.277 H75 1.2 F

1.2 F20.4 F16.33 mH 0.277 H

Final Circuit

81

M Riaz, EE, MAJ U.


82/90

omponen s va ues or norma zeButterworth filter with equal input and output

ermna ons

2sin , 1, 2, ......,2

k kL or C k n

n= =

82

Chebyshev Filter Design M.A.J.U.


83/90

Order for Cheb shev Filte

( ) ( )

120.10.11 maxmin

cosh 10 1 10 1AA

n

=

121 0.1 35 0.1 0.5

cos

cosh 10 1 10 1

s p

( )1cosh 4 .n = =

20.71570( )

0.62646 1.14245 0.62646T s

s s=

+ + +

83

M Riaz, EE, MAJ U.

Chebyshev Filter Design M.A.J.U.


84/90

Order for Cheb shev Filte

2 0.71570( )T s = . . .

( ( )2 6Denormalizing, by replacing s by 3 10 500 .s s+

3178,925( )

sT s =

. . . .s s s s s s+ + + + + +

84

M Riaz, EE, MAJ U.

Usin values from table, LP rotot e circuit is iven

M.A.J.U.


85/90

below

2 1.2804 HL =1

1.8636 FC =

1 1 .=

case of Butterworth filter design and complete the designyourselves.

85

M Riaz, EE, MAJ U.

Components values for normalized


86/90

11 1coshA =

terminations

( )2 1sin , 1, 2, ......,2

k

k

A = k nn

=

maxln coth17.37

A

=

2

sinh2

sin 1 2 ......2

Yn

kB =Y + k n

=

=

11

2 cosh

n

A AG =

Y

86

21

1 1

4 cosh, 2, 3, 4 ......,k k

k

k k

A A AG = k= n

B G

Components values for normalized 3rd order Chebyshev filter (usingformulae of previous slide with equal input and output terminations

0 1A


87/90

max0.1

1 1

10 1 0.3493

1 1 1 1cosh cosh 0.570978

A

A

= =

= = =

( )1 2 3

.

2 1sin sin 0.5, 1, 0.52 6

n

k

A = A An

= = = =

max 0.5ln coth ln coth 3.548167917.37 17.37

3.5481679

A

= = =

21 2

s n s n 0.62643635292 6

sin2

Yn

kB ,B =Y +

= = =

20.6264363529 sin 1.14242252= +

= n

1 32 0.5 cosh 0.570978, 1.86369

0.6264363529G G = =

87

2 21 2

2

1 1

4 cosh 4 0.5 1 cosh 0.5709781.28038

1.1424225 1.86369

A A AG = =

B G

=

- -

Assignment # 04 M.A.J.U.

filt i t


88/90

filter requirements:pass band = 3.742 to 3.778 GHz with Amax= 1 dBstop band = 0 to 3.74 and 3.78 to infinity with Amin = 15 dB

The terminations are Rs=RL=600 ohms.Design complete chebyshev filter. Give circuit diagram, locations of poles and

zeros. Give magnitude and phase plots using P-spice.

( ) dBA

15dB 15dB

Gf Hz3.74 3.742 3.778 3.78

log scale


Assignment # 04 (cont.) M.A.J.U.

P bl 02 Use cheb she appro imations to design a band reject filter


89/90

Problem 02: Use chebyshev approximations to design a band reject filterwith the following specifications. Rs=RL=1000 ohm

( ) dBA

35dB1dB 1dB

100 200 400 800

log scale

89

M Riaz, EE, MAJ U.

M.A.J.U.Assignment # 04 (cont.)


90/90

Problem 03:A high-pass Butterworth filter must have at least 45 dB of attenuation below 300Hz, and the attenuation must be no more than 0.5 dB above 3000 Hz. Find theapproximation function and design the complete filter with Rs=RL=1500 ohm.

A low-pass filter requirement is specified by Amax = 1 dB, Amin = 35 dB, Fp =1000 Hz, Fs = 3500 Hz. Find the Butterworth approximation function, needed and

desi n the com lete Filter.

90

M Riaz, EE, MAJ U.

Slide010 Filter Theory

Documents