filter designing complete

8/10/2019 Filter Designing Complete

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1

Chapter:01

Filter:Filter can be considered can be considered as frequency selective networks. A filter is

required to separate an unwanted signal from a mixture of wanted and unwanted signals.

The filter specification are generally given in terms of cutoff frequencies, pass

band (P.B) and stop band (s.b) regions. P. B is the frequency band of wanted signal and S.B

is the frequency band of unwanted signal. An ideal filter should pass the wanted signal with

no attenuation and provide infinite attenuation.

Depending upon the components used, filters can be classified as:1. passive filters: Filters which are the compotnet such as R,L,C are the passive filters. The

Gains of such filters are always less than or equal to unity (i.e GS1). It is to be noted the

L and C are filter components, but R is not.

2. Active filters: The filters which use the components such as transistors, op-amp etc arethe active filters. The Gains of such filters are always greater than or equal to unity. ( G 1)

Gain and Attenuation:

Filternetwork

i/pV1(t)

o/pV2(t)

Let us consider the filters network with i/p V1(t) having power P1 and o/p V2(t) having

power p2 as shown in fig1. Then the transfer function is given by T(s) = V2(s)/V1(s)

Where , V1(s) and V2(s) are the Laplace Transform of V1(t) .

Also, T(s) = T(jw) =)(

)(

1

2

jwv

jwv

Then the voltage gain in db is given by ,

Av = 20log10 )(jwT dB .(1)

Or in term of power , the power gain is given by,

Ap = 10 log102

1

p

p

Now, the voltage attenuation is given by ,

= 1/Av

= -20log )(jwT dB.(2)

From equation 1 and 2 ,we can write,


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2

)(jwT = 100.05Av..(3)

)(jwT = 10-0.05

.(4)

Types of filters: ( According to the function)

Filters are classified according to the functions they are to perform. The pattern

of PB and SB that give rise to the most common filters as defined below:

1.

Low pass filters: (LPF): A LPF characteristics is one in which the PB extend from = 0 to = cwhere cis know as cut off frequency.

A

w

Fig. 1(a)

SBPB

wc

2. High pass filter: A high pass filter is a compolement of a low pass filter in that the

frequency range form o to cis the SB and from cto infinity is the PB.A

w

Fig. 1(b)

SB PB

wc

3. Band pass filter ( BPF): A BPF is one in which the frequency extending form L(or1) to u(2) are passed while signals at all other frequencies are stopped.

A

wFig. 1(c)

SBPBSB

wc

4.

Band stop filter(BSF): A BSF is complement of BPF where signal components at

frequencies form 1 to 2 are stopped and all others are passed. These filters aresometimes known as Notch filters.

A

wFig. 1(d)

SBPB PB

Notch filter

5. All pass filters (APF): It is a filter which passes all range of frequencies , i.e , PB


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3

ranges from o to infinity.A

wPB

Fig. 1(e)

Non- ideal Characteristics:

Filter Gain curve Attenuation curve

1. LPF

2.HPF

1. From the attenuation curve it to be noted that in the pass band the attenuation isalways less then a maximum value. Designated as max

2. In the stop band the attenuation is always larger then a minimum value designated as min.

3. Band between PB and SB so defined are known as transition bands. (TB).Bilinear Transfer function and its poles and zeroes:

We know,

T(s) = P(s)/Q(s) = N(s)/D(s)

T(s) =01

1

1

01

1

1

............

...........

bsbsbsb

asasasan

n

n

n

m

m

m

m

++++

++++

When , m = n = 1, then the T(s) of equation (i) will be bilinear , i.e

+

+=

=

+

+

=

+

+==

)(

)()(

)(

)(

)/(

)/(

)(

)()(

2

1

1

1

101

11

01

01

zs

zsGsTor

ps

zsG

bbsb

aasa

bsb

asa

sQ

sPsT

o

0.707

Wc

TB1

W Ws

A

Wc

A

Wp Ws

WsWcWp

A

A


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4

If z1< p1 If p1< z1

Here, G = a1/b1 = Gain

Z = -a0/a1= a zero

P1 = -b0/b1= a pole

Date:2065/4/22

Realisation of filter with passive elements:Let us now see how the bilinear transfer function and its various special cases can be

realized with passive elements.+

-

+-

v1 c

Fig 1.

Plot the magnitude and phase response of the ckt shown in fig (1) and identify the filter.

Solution:

Applying kirchoffs law for fig 1

)..(....................1

)......(..........1

2

11

iiidtL

V

iidtL

RV

=

+=

Taking laplace transform of equation (i) and (ii)

+

==

+=

csRsI

sIcs

sV

sVsV

iiisIcs

sRIsV

1)(

)(1

)(

)()(

).........().........(1)()(

1

22

1


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5

0

0)(

/1

1

)/1(

1

1

1

WS

WsT

RCS

RC

RCsRC

cs

Rcs

cs

+=

+=

+=

+=

Where, W0 = 1/RC

Now , for magnitude plot,

T(s) = T(jw) = W0/(jw+W0)

22

0)(

oww

wjwT

+=

Now when

W = 0 )(jwT = 1W= wo )(jwT = 0.707

W = , )(jwT = 0

0.707

Wc

1

W

T(jw)

Fig. 2. Magnitude plot

For phase plot:

(jw) = tan-1(o/w0) tan-1

(w/wo)

(jw) = tan-1(w/w0)When,

W = 0 , (j0) = 0

W = wo, (jwo) = -45

W = , (j ) = - 90


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6

WoW

90

-90

-45

45

2.

+

-

+-

v1

R 1

R 1

R

R 1 c v2

1 2

34

Above figure can be modified as:

v1-

+

-

1 1

3

2

4 4 From figure the potential of node 2, is V1/2 and the potential at node 3 is VsR/(1+1/cs)

V2= V1/2 - VsR/(1+1/cs)

V1/V2= - RCS/RCS+1T(s) = R(S+1- 2RCS)/2(RCS+1) = -{(RCS+1)/2(RCS+1)}

= RC(S+1/RC)/2RC(s+1/RC)

Where Wo= 1/RCT(jw) = -1/2 {(jw-wo)/(jw+wo)}

For magnitude plot ,

)(jwT =2

0

2

2

0

2

)(21

wwww

++


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7

)(jwT = 2

1 W

T(jw)

Phase plot:

(jw) = tan-1(-w/wo) - tan-1

( w/wo)

(jw) = -2tan-1(w/wo)when,

w = 0, (jw) = 0

w = 0, (jw) = -90

w= , (jw) = -180

WoW

90

-180

-90

45

-45

-135

From the magnitude plot, we see that the networking is all pass filter.

Assignment:3.

v2v1-

+

+

-

4.


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8

v2v1-+

c1

R 1R 2

5.

v2v1-

+R 2

C2

+

-

R 1

Date: 2065/4/28

Example :04

v2v1-+

c1

R 1R 2

From fig (i)

Y1= c1s+1/R1=1

11 1

R

SCR +

Z1= 1/Y1 =111

1

+SCR

R

Now applying kirchoffs voltage law, for fig (i).

V1= z1i+R2i

V1(s) = (z1s+R2)I1(s)And ,

V2(s) = R2I(s)

T(s) =

2

11

1

2

21

2

1

2

1

)()(

)(

RSCR

R

R

RsZ

R

sV

sV

++

=+

=

=

++

+

=++

+

112

21121

11

121

21121

112

)1

()1(

CRR

RRSCRR

CRSCRR

RSCRRR

SCRR


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9

=

1112

11

11

1

CRCRS

CRS

++

+

Or, T(s) =)(

)(

02

01

02

01

=

+

+

S

S

S

S

And ,0102

0102

,

or

wo2-wo1

For Magnitude plot:

T(jw) =2

02

2

2

01

2

02

01

ww

ww

wjw

wjw

+

+=

+

+

Now at w= 0,21

2

02

01)0(RR

R

w

wjT

+==

At w = , 1)(02

01 ==w

wjT

R2

R1+R2

1

w=0W

T(jw)

For Phase plot,

T(jw) =02

01

wjw

wjw

+

+

Where, w01= 1/R1C1

W02= 1/R1C1+1/R2C2

Therefore, (jw) =

02

1

01

1 tantanw

w

w

w

(jw) = z p

Since direct phase plot of above expression is very complicated, we will go it by indirectmethod. First we will plot the zero phase and then the pole phase and finally find the net

pole zero phase.

Zero plot (z)


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10

)(tantan)( 111

01

1CwR

w

wz

=

=

Now at w = 0

(z) = (j0) = 0

(z) = (jw0)= 45 Now at w = (j ) = 90

Pole plot (p)(p) = tan-1(w/w01)

=

+

1211

1

11tan

CRCR

w

Now at, w = 0

p= (j0) = 0at w = w02

p= (w02) = 45

at w = , p= (j ) = 90

W

90

-90

w=0

-45

45

(jw)

wo1

wo2

zero plot

pole zero plot

Pole plot

Thus the magnitude response of the above network shown that it is a high pass filter with dcgain R2/(R1+R2) and phase plot signifies it is leading type.

Insertion Gain and insertion loss:


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11

T(jw)

wwo

1

Insertion gain

1

wwo

wwo

1

Insertion gain

1

wwo

1

T(jw)

One of the important factor that should be consider in design is that the minimum value of should be zero degree. But this is not true in practical case since we are using active element

, this need not be the case because the active element may provided the gain greater than

one (1). If it is necessary to meet the specification exactly then it will be necessary to provide

ck t to reduce the gain. We call this unwanted gain as the insertion gain. On the other hand

there is a loss in the components of passive filter so it provides access attenuation and we

call this loss as insertion loss. To overcome this problem additional compensation circuit is

required.

Chapter- 2

Normalization and Renormalization:

In most of the cases we consider the values of R, L S& C to be the order of unity. It is very

difficult to built the capacitor of 1 f and inductor of 1 H . Besides this the practical values of

capacitors available in the electronic circuit is of the order of microfarad or Pico farad. The

circuit considered so for have normalized elemental values but practically these values are

not realizable. So we perform scaling to get the realizable components.

There are mainly two reasons for resorting the normalized design.1. Numerical computation become simple and it is easier to manipulate the numbers of

the order of unity.

2. If we have the normalized design of the filter then it is easy to generate the


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13

V1V2

+ +

- -

R=1 ohm

C =1F

Solution:

Rold= 1 Cold= 1 F

Now , let us assume that,

Cnew= 10 F

Note: Generally we assume new value of capacitor 1F or 10 F.We know that

Cnew= Cold/Km

Km= Cold/Cnew= 1F/10 F = 105

Therefore, Rnew= Km.Rold

= 105* 1

Rnew= 100K

V1V2

+ +

- -

R=100k

C =10 uf

Fig(ii) scaled ckt.

The transfer function for fig. (i) ,

Told(s) = 1/(s+1)

And, Tnew=

newnew

newnew

CRS

CR

1

1

+

= 1/s+1

Thus we see that there is no change in the following transfer function while doing magnitudescaling.

Date: 2065/5/3

2. Frequency scaling:

In frequency scaling our objective is to scale the frequency without affecting the

magnitude of the impedance , i.e

ZL= ( = XL) = LS = jWL

WLZL = is a constant.

Similarly,

Zc( = Xc) = 1/cs = 1/ jwc


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14

wcZc

1= is constant.

To do so any change in w must be compensated by corresponding change in L and c

If, w = old corner frequency

= new corner frequency.

= KfwWhere,Kf= frequency scaling factor.

If Kf> 1, then it is called expansion scaling

If, Kf< 1 , then it is called compression scaling.

o = 10

o

Expansion

o = 103

Compression

o = 1o = 103

Also, if T(jw) is old Transfer function, then the new transfer fucnti is T(j )= T (jKfw)

The resistance is unaffected by frequency scaling , i.e

Rnew= Rold.(v)

For inductor,

Xl= Ls = jwL = jwkf. L/kf

Or, XL= j(wkf) ( Lold/kf) since, L = Lold

= j ( Lold/kf)

Lold= Lold/ Kf.(vi)

For capacitor,

Cnew= Cold/ kf(vii)


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15

3. Both magnitude and Frequency scaling:

It is not necessary that we scale magnitude and scale in frequency separately. We can do

both at once. Cobining all the above equations.

Rnew= KmRold.(Viii)

Lnew

= Km/k

f. L

old(ix)

Cnew= Cold/Km.kf.(x)

These three equations are know as element scaling equations.

Example 01:1

1F

Solution:

W0= 1 , = 1000Therefore, kf= o/wo= 1000Now we know that

Cnew= Cold/kf= 1F/ 1000 = 1 mF

And , Rnew= Rold= 1

1k

1mF

Fig (ii): after frequency scaling.

Now,

Told(S) =1

1

1

1

0

+=

+s

CR

s

CR

oldold

oldld

And, Tnew(s) =10

10

1

1

+=

+s

CRs

CR

newnew

newnew

Example 02:

1

1F

R=1/10


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16

Perform frequency scaling with o= 1

Example 03:

v1 v2

R1

R2C1

T(s) = (s+0.5)/(s+3)

Perform magnitude and frequency scaling separately with wo= 3 and 0= 300.

Solution:

The transfer function of the above figure is

T(s) =

111

11

2

11

1

CRCRs

CRs

++

+

.(i)

But given ,

T(s) = (s+0.5)/(s+3) .(ii)

Comparing equation (i) and (ii)

1/R1C1= 0.5R1C1= 2 ..(iii)

Again, ( 1/R1+ 1/R2)1/C1= 3..(iv)

Let , C1= 1 F

For equation (iii) R1 1 = 2

R1= 2 Therefore from equation (iv)

(1/2 + 1/R2) 1/2 = 3

Therefore, R2 = 2/5

In order to perform magnitude scalingR1old= 2 R2old= 2/5 = 0.4 Cold= 1 F

Say, C1new= 10 FThen, Km= Cold/Cnew= 1F/ 10 F

Km= 105

Therefore, Rnew= kmR2old= 105 0.4 = 40 k

The selected ckt will be :


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17

v1 v2

200k

40k40.4F

Again for frequency scaling,

Wo= 3 , 0= 3000Therefore , kf= o/ wo= 3000/ 3 = 1000Therefore, R1new= R1old= 2

R2new= R2old= 0.4 C1old= C1old/kf= 1F/ 1000 = 1 mF.

Example 04:

+

_

C1= 1 F

C2= 1/10 F

R1= 1

R2= 1/100

Perform magnitude scaling to the ckt given.

Note: Take Cnew as the new value of capacitor for Cold where Cold represents the largest

value in the circuit.

Solution:

Here, R1old= 1 R2old= 2 C1old= 1 F

C2old= 1/10 F.

Take, Cnew= 10 F.Then for, magnitude scaling,

Cnew= Cold/km

Km= C1old/ C1new = 1F/ 10 F = 105

Therefore, C2new= C2old/km = 0.1 F/ 105

C2new= 1 FSimilarly,

R1old= km. R1old= 105 1 = 100 k

R2new= km. R2old= (1/100). 105 = 1 k.

+

_

10 uF100k

1k


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18

Fig: Magnitude Scaling Ckt.


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19

Chapter: 3

One port and two port passive network:

Positive real function: The filter circuit is complex transfer function that may be realizable

depending upon weather the transfer function exhibits PRF properties. I the transfer function

is PRF only ckt is realizable. There are two types of passive network : [i] one port network

[ii] Two port network.

1- port n/w

I(s)

V(s)

2- port n/w

I1(s)

V1(s)

I2(s)

Fig. 1(a) one port n/w Fig. 2(b) two port n/w

One port network: Let us suppose of fig of 1(a),

Then, z(s) = V(s) / I(s)

If V(s) = 3s+2

I(s) = 1

Then, z(s) = 3s+2

= Ls +R

V(s) 2

3H

Thus , the function is realization but if, z(s) = 3s-2 , then it is not realizable.

Date: 2065/ 5/10

Why? ( )

(i) If F(s) denote the function in S-domain, the F(s) indicates either driving pointimpedance or driving admittance. Which ever is concern to us.

(ii) F(s) should be for real value of S.(iii) The value of F(s) must be greater than or equal to zero. i.e Re[f(s)] 0.Thus in brief a PRF must be real and +ve .If F(s) = LS = jWL L must be +ve.

F(s) = 1/CS = 1/jwc C must be +ve

F(s) = R R must be +ve.

Properties of Passive n/w.A passive network is one

(i) The element of which one are +ve and real.


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20

(ii) The average Power dissipated (APD) by the n/w. for a sinusoidal i/p must be +ve.For one port n/w APD = 1/2 Re[ z(s)][I(s)]

20

Properties of PRF:1.

If F(s) is +ve and real , then 1/F(s) is also +ve and real.2.

The sum of DRFS is always PRF but the difference may not be PRF.

Example: Z1(s) = 5s+ 3 (PRF)

Z2(s) = 2s+ 5 ( PRF)

Then, z1(s)+z2(s) = 7s+8 (PRF)But, Z1(s) Z2(s) = 3s-2 (not PRF)

3. The Poles and zeros of PRF cannot be in the right half of the S-Plain.4.

Only poles with real residues can exists on the jw axis.

Example: F(s) = 6s/(s2+ 2 )

In this case, S = j

Residue = real and +ve.5.

The poles and zeroes of PRF Occurs in pairs.6.

The highest power of numerator and denominator polynomial may differ atmost by

unity.

Example:KSSSS

SSSSS

3324

233342346

12345

++++

+++++

7. The lowest power of numerator and denominator polynomial may differ atmost by

unity.

Example:KSSSS

SSSSS

3324

33342346

2345

++++

++++

8. The real part of F(s) must be greater than or equal to zero. i.e Re[F(s)] 0But , if Re[F(s)] = 0 , then the ckt do not consist resistive components. Hence only

capacitive and inductive components are presents. Hence only capacitive and inductive

components are present . Such a n/w whose transfer function satisfies this condition is

known as lossless n/w.

Example: Determine weather the function is PRF.(i) z(s) = 2s2+5/s(s2+1)

Hence , z(s) = 2s2+5/s(s+1)

A/s + Bs/(s2+1) = A/s + B/ (s

2+1)/s

0.)1(

52 2

=+

+

= ssss

s

A

3)1(

5)1(2

)1(.

)1(

522

2

2

2

=

+=

+

+

+=

ss

s

ss

sB

Z(s) = 5/3 + -3s/(s2+1)

Here, (-3) , the residues ( s2= -1) is ve , therefore z(s) is not PRF.

(ii) z(s) =

)3(1)3(

)4(2)4(

)3)(1(

)4)(1(

+++

+++=

++

++

sss

sss

ss

ss


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21

)()(

)3)(1(

521

34

86

21

2

2

szsz

ss

s

ss

ss

+=

++

++=

++

++=

Where z2(s) =3

2/1

1

2/3

31)3)(1(

52

+

+

+

=

+

+

+

=

++

+

ssS

B

s

A

ss

s

Therefore, z(s) = 1 +3

2/1

1

2/3

++

+ ss

It is not PRF.

(iii) z(s) =ss

sss

38

13483

23

+

+++

(iv) Y(s) =)4(

822

+

++

ss

ss

Basic ckt Synthesis Techniques:

Any one port n/w each can be represented by either admittance function Y(s) orimpedance function z(s) . i.e

)(

)(

)(

)(

)(

)(

............

.............)(

01

2

2

1

1

01

2

2

1

1

sP

sZ

sD

sN

sQ

sP

bsbsbsbsb

asasasasasF

m

m

m

m

m

m

n

n

n

n

n

n

=

=

=

+++++

+++++=

Design of LC Ckt . (Loss less ckt):

Consider a impedance function as

Z(s) =)()(

)()(

sOsE

sOsE

mm

nn

+

+

Where En(s) and Om(s) denote the even parts of numerator and denominator respectivelyand On(s) and On(s) denote odd part.

Z(s) =)()(

11

23456

345

sQsN

ssssssssss =

++++++++++

=

)(

)1(

)(

)1(

)(

)5(

)(

)1(

35246

3524

sO

ss

sE

sss

sO

ss

sE

ss

mm

nn

+++

+++

+++

++

For the loss less function , it is to be noted that,

Re[z(s)] = 0 .(i)

Now, z(s) = )()(

)()(

)()(

)()(

sOsE

sOsE

sOsE

sOsE

mm

mm

mm

nn

+

+

=)()(

)()()()()()()()(22

sOsE

sOsOsOsEsEsOsEsE

mm

mnmnmnmn

+


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22

=)()(

)()()()(

)()(

)().()()(2222

sOsE

sOsEsEsO

sOsE

sOsOsEsE

mm

mnmn

mm

mnmn

+

+

+

= Re[ z(s)] =)()(

)().()()(22

sOsE

sOsOsEsE

mm

mnmn

(ii)

Therefore from equation (i) and (ii).

0)()(

)().()()(

22

=

sOsE

sOsOsEsE

mm

mnmn

)().()()(

0)().()()(

sOsOsEsE

sOsOsEsE

mnmn

mnmn

=

=

)(

)(

)(

)(

sE

sO

sO

sE

m

n

m

m = .(iii)

The above equation (iii) indicates that LC ckt is even to odd ( or odd ) to even function.

Properties of LC Ckt:

1.0

4

4

2

2

0

4

2

2

2

............

.............)( bsbsbsb

asasasasF m

m

m

m

m

m

n

n

n

n

n

n

++++

++++=

The coefficients anand bm must be real and +ve and F(s) must be even to odd or odd to even

function.

2. The highest power of numerator and denominator can differ atmost by unity ( in thiscase it is 2). So does the lowest power.

3. The succeeding power of s in numerator and denominator must differ by the order of

2 all the way through . Example:ss

sss

4

165173

024

+

++

4. The poles and zeros must be alternatively placed on the jw axis and lie only on theimaginary axis.

5. There must be either a pole or a zero at the origin.Example: Test whether the following function is LC.

(i) z(s) = K (s2+1)(s2+5)/(s2+2)(s2+10) k>0It is not LC ckt function because,

1. There is neither pole or zero at the origin though the pole zero are alternatively placed onthe imaginary axis.

2. It is not even to odd or odd to even function.

(ii) Z(s) = z(s2+1)(s2+9)/s(s2+4)(iii) Z(s) = k s(s2+4)/(s2+1)(s2+3) , k> 0(iv) Z(s) = s5+4s3+5/(4s4+s2)

Date: 2065/5/12

Design of LC ckt by Fosters Method:

In this case ,

F(s) = skws

sk

S

k

i

i

+++

+ ............222

0 .(i)

This equation may represent z(s) or Y(s)

Case I : ( i.e when F(s) = z(s))


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23

Then,

Z(s) = skws

sk

S

k

i

i

+++

+ ............2

22

0

Here,

- ko/s will represent a capacitive reactance of 1/koF.- 2ki(s)/(s

2+w

2) will represent LC parallel combination.

Having capacitor of value 1/2kiF and inductor of value 2ki/wi2

. Thus the final circuit willbe:

.......

2ki/wi2

1/2ki

k

1/k0

z(s)

This method of circuit synthesis is known as foster impedance or series or 1st

method for LCckt.

Case II

In this case , F(s) = Y(s) , then equation (i) becomes

Y(s) = skws

sk

S

k

i

i

+++

+ ............2

22

0

Here,- KO/s represents admittance of inductor having value of 1/koH.- Ks represent admittance of capacitor having value KF.- 2ki(s)/s

2+w

2 represents admittance of series LC combination having inductor of value

1/2kiH and capacitor value wi2/2ki

The ckt can be realize as :

k

1/2ki1/k

wi2/2ki

This method of circuit synthesis is known as foster admittance or parallel or 2nd

method for

LC ckt.

Example 01:Design a Foster series n/w for the following n/w.

)9)(1(2)4()( 22

2

+++=ss

sssF

Solution:

It is Fosters series n/w


24/125



24

)9)(1(2

)4()()(

22

2

++

+==

ss

ssszsF

Now,91)9)(1(2

)4()(

2222

2

++

+=

++

+=

s

Bs

s

As

ss

sssz

Where, A =1

)9(.

)9)(1(2

)4(2

2

22

2

=

+

++

+

ss

s

ss

ss

= 16/3)91(2

41=

+

+

Therefore, A = 3/16

And B =9

)9(.

)9)(1(2

)4(2

2

22

2

=

+

++

+

ss

s

ss

ss

=16

5

82

5

)19(2

49=

=

+

+

Therefore, B = 5/16

)()(9

)6/5(

1

)16/3()( 2122 szsz

s

s

s

ssz +=

++

+=

The ckt will be as follows.

L 1= 3/16 H

C1 =16/3 Fz(s) C2 = 16/5 F

L 2 = 5/144 H

The first part of z(s) ( i.e z1(s) ) represents parallel LC combination having inductor L1of value 3/16 H and capacitor of value 16/3 F.

The 2ndpart of z(s) (i.e z2(s) ) represents parallel LC combination having inductor L2ofvalue 5/144 H and capacitor C2of value 16/5 F.

Example 02: Design Foster parallel n/w for the function

)9)(1(2

)4()(

22

2

++

+=

ss

sssF

Solution:

It is Fosters parallel n/w

)9)(1(2

)4()()(

22

2

++

+==

ss

sssYsF

Now, 91)9)(1(2

)4()(

2222

2

++

+=

++

+=

s

Bs

s

As

ss

sssz

Where, A =1

)9(.

)9)(1(2

)4(2

2

22

2

=

+

++

+

ss

s

ss

ss


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25

= 16/3)91(2

41=

+

+

Therefore, A = 3/16

And B =9

)9(.

)9)(1(2

)4(2

2

22

2

=

+

++

+

ss

s

ss

ss

=16

5

82

5

)19(2

49=

=

+

+

Therefore, B = 5/16

)()(9

)6/5(

1

)16/3()( 2122 sYsY

s

s

s

ssY +=

++

+=

The ckt will be as follows:

Figure:

The first part of Y(s) ( i.e Y1(s) ) represents series LC combination having inductor L1ofvalue 16/3 H and capacitor of value 16/3 F.

The 2ndpart of Y(s) (i.e Y2(s) ) represents series LC combination having inductor L2ofvalue 16/5 H and capacitor C2of value 144/5 F.

Example 03: Design Foster parallel n/w for the function)4(

)9)(1(2)(

2

22

+

++=

ss

sssF

Solution:It is Foster Parallel ,

)4(

)9)(1(2)()(

2

22

+

++==

ss

sssYsF

ss

ss

4

182023

24

+

++=

Therefore, Y(s) = 2s +ss

s

4

18123

2

+

+

= 2s +)4(

18122

2

+

+

ss

s

Y(s) = Y1(s) + Y2(s)

Now Y2(s) = 2s +4

)2/15(2/9

4)4(

1812222

2

++=

++=

+

+

s

s

ss

Bs

s

A

ss

s

Y(s) = 2s + )()()(4

)2/15(2/93212

sYsYsYs

s

s++=

++

Here Y1(s) = 2s , so C1= 2 F

S +4s )2s +20s +18( 2s

2s4+8s

2

12s2+18


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26

Y2(s) =s

2/9 , So, L1= 2/9 H

And Y3(s) =4

).2/15(2 +s

s

L2= 2/15 H

C2= 8/15 F

Therefore, The final ckt will be

C1 = 2 F

L 2 =2/15 H

C2 = 8/15 F

L 1= 2/9H

Fig. Fosters parallel n/w of LC ckt.

Assignment:

1. z(s) =

)1(

)9)(1(22

22

+

++

ss

ss

2. Y(s) =

)1)(3(

)4)(2(222

22

++

++

ss

ss

Date: 2065/5/17

Continued Fraction method or cauer method for LC Ckt

1. case- I

It is removed by successive removal of pole at . The ckt will be as follows:

....L2L1

C1 C2 Cn

Fig. For F(s) = z(s)

....

C1 C2 CnC3V(s)

L1 L2

Fig. For F(s) = Y(s)

Example 01:Synthesis the following function in cauer form.


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27

Z(s) =34

1612224

35

++

++

ss

sss

Solution:

In cauer n/w we proceed as follows:

8/32H

1/4 3/4Z(s)

2/3

Fig. Cauer n/w for LC series ckt

Example: 02:Y(s) =34

1612224

35

++

++

ss

sss

Y(s) 2F8/3 F 2/3 F

Fig: Cauer n/w for LC parallel ckt.

Example:03:Synthesis the following ckt in cauer form.

(i) Y(s) =)3)(1()4)(2(

22

22

++++

sssss (ii) Z(s) =

)3)(1()4)(2(

24

22

++++

sssss

Cauer II:

This is the case of removal of pole at origin.

....

L1 L2 Ln

C1 C2

Fig. Caure II n/w for LC series ckt.

S4+ 10s

2/4

2s

2s5+8s

3+6s

4S3+8S

3s2/2

S +4s +3) 2s + 12s + 16s (2s z1(s)

4s3+10s) s4+ 4s2+3 (s/4 Y2(s)

3S2/2+3) 4S

3+10S (8s/3 Z3(s)

2S) 3S2/2 +3 (3s/4 Y4(s)

3) 2s (2s/3 Z5(s)


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28

Example:01:Synthesize the following function in cauer form.

Z(s) =sss

ss

163122

345

24

++

++

Solution:

Since Z(s) is the case of pole at origin (i.e s = 0 ) z(s) can be rewrite as:

Z(s) =53

42

21216

43

sss

ss

++

++

16/3 176/40 88/3

21/44.447/64Z(s)

Fig. Cauer II n/w for LC ckt

Example:02:Y(s) =sss

ss

16122

3435

24

++

++

7/64 21/1936

16/388/3176/49Y(s)

Fig. Caure II n/w for parallel LC ckt.

R-C one port n/w: (R-C impedance /R-L admittance)

1. Foster 1stmethod:

In this case,

F(s) = z(s) , gives R-C impedance n/w.

16s+40s3/7

3s2/44

3+9s /4+ 3s /8

7s2/4+44s

4/88

44s3/7

16s+12s +2s ) 3+4s + s (3/16s z1(s)

7s2/4+5s

4/8)16s+12s

3+2s

5(64/7s Y2(s)

44s3/7+2s

5) 7s

2/4+5s

4/8 (49/176s Z3(s)

3s4/44) 44s

3/7 +2s

5((44)

2/21s Y4(s)

2s ) 3s /44(3/88s Z5(s)


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29

.......

1/k1

k

1/k0

z(s)

Foster method defines F(s) as

F(s) = z(s) = ko/s + k1/(s+1)+ k2/(s+2) + +kHere,

- ko/s represent capacitive reactance having capacitor of value 1/koF.- k represent resistor of value k.- ki/(s+i) represents RC parallel in which the resister has a value of ki/i and a

capacitor has value of 1/kiF.

Properties of RC impedance N/w:

1. the poles of RC impedance n/w are on the ve real axis.2. As in LC ckt, residues of poles (kis) are real and +ve i. z(s ) must be PRF.

3. At two critical frequencies i.e when s = o , i.e = 0 when s = i.e = 4. z(0) = if C0is present

= Ri, if C0is missing5. z() = k , R is present

= 0, R is missing6. z(0) z() is always true.7. The critical frequency nearest to the origin must be a pole.

8. The poles and zeroes must be alternatively placed.

Example:01State giving reasons which of the following if not RC impedance.

(a) Z(s) =)5)(2(

)9)(4)(1(

++

+++

sss

sss

(b) Z(s) =)4)(2(

)8)(1(

++

++

ss

ss

(c) Z(s) =)1(

)4)(2(

+

++

s

ss

(d) Z(s) =)3(

)2)(1(

+

++

ss

ss

Example:02:Synthesis the following function in Foster series form: F(s) =)3(

)4)(2(6

+

++

ss

ss

Solution:

Since it is foster series function z(s) =)3(

)4)(2(6

+

++

ss

ss

This is the RC impedance n/w.

Now,

(i) z(0) = , C0 is present .(ii) z() = , Ris also present.


30/125



30

Z(s) = ko/s + k + k1/(s+3) = ko/s + k1/(s+3)+6

Ko=0

.)3(

)4)(2(6

=+

++

ss

ss

ss= (6.2.4)/3 = 16

K2= 2

Z(s) = 16/s + 2/(s+3) + 6

The component values are as follows:16/s 1/cos c0 = 1/16 FR R = 6 2/(s+3) R1= 2/3 and C1= F

The ckt will be:

.......

1/2

61/16

z(s)

2/3

Date: 2065/5/19

F(s) =)3(

)4)(2(6+

++ss

ss

F(s) = z(s) =)3(

)4)(2(6

+

++

ss

ss

= 6+ 16/s + 2/(s+3)

Forster parallel method for R-C one port n/w:

In this case,

F(s) = Y(s)

Y(s) = ko/s + k1/(s+1)+ k2/(s+2) + +k. . . . . .

R1 R2

L2L1

Lo

Fig. (i) R-L admittance n/w for foster 2

ndmethod in this case

- ko/s represents inductor of value 1/ko

- k represents inductor of value 1/ko- ki/(s+i) represents RL series ckt having inductor of value 1/kiH and resister of

value i/k .


31/125



31

Properties:

Same as RC- impedance.

Example: 01:Synthesis the following function in foster parallel.

F(s) =)3(

)4)(2(6

+

++

ss

ss

Solution:Since it is Foster parallel,

F(s) = Y(s) =)3(

)4)(2(6

+

++

ss

ss

= 6 + 16/s + 2/(s+3)

The ckt will be:

2/3

1/2

1/161/6

Fig. R-L admittance ckt from foster parallel

Continued Fraction method or cauer method for R-C impedance or R-L Admittance:

1. If F(s) = z(s) , then it yields cauer 1 n/w.2. If F(s) = Y(s) , then it yields cauer 2 n/w.

For cauer 1 n/w:In this case F(s) = z(s)

Example:01:Synthesize the following function cauer 1 form.

F(s) =)3(

)4)(2(6

+

++

ss

ss

Solution:

F(s) = z(s) =)3(

)4)(2(6

+

++

ss

ss=

ss

ss

3

483662

2

+

++

Now,

The ckt will be:

6 54

1/18 1/144

Fig. Caure 1 n/w

18s

18s+48) s2+ 3s (s/18

S+3s)6S +36s+48(6

S2+18s

S +8s/3

s/3) 18s+ 4s (54

48) s/3 (s/3.48

s/3

Z1(s)

Y2(s)

Z3(s)

Y4(s)


32/125



32

Cauer 2 n/w:

Example: 02:Realise the given function in cauer 2 n/w F(s) =)3(

)4)(2(6

+

++

ss

ss

Solution:

In this case,

F(s) = Y(s) = )3(

)4)(2(6

+

++

ss

ss

In this case circuit will be :

1/6

1/144

1/54

1/18

Fig. Caure 2 method

R-L one-Port n/w:(R-L impedance or R-C admittance n/w)

1. Foster Series method: It yields R-L impedance ckt for which

F(s) = (s) = ko+ kis/(s+ 1) + k2s/(s+2) + ..+ ks

....

k1

k1/1

kko

z(s)

k2

k2/2

In this case,

- k0represent resistor of value ko.- ks represent inductor of value kH.

- kis/(s+i) represent RL parallel ckt with resistor of value kiand inductor of value

ki/i.This method of synthesis is know as foster series (1

st) method for R-L one port n/w.

Properties of R-L impedance n/w:

1. Poles are on the ve real axis.2. The residue of pole must be real and +ve i.e F(s) must be PRF.3. z(0) = k0if R0 is present.

= 0 if R0is missing.

4. z() = if L is present.= Riif L is missing.

5. z() z(0)6. Zero is nearest to the origin.7. The pole and zero must be alternatively placed.


33/125



33

2. Foster parallel method:In this case,

F(s) = Y(s) = ko+k1s/(s+i) + k2s/(s+2) + .+ kThe ckt will be as follows:

k1/k

0Y(s)1/k

11/k2

k1/1k2/2

This method of synthesis is known as Foster parallel method which yields R-C admittancen/w.

Properties:

Some as that of R-L impedance except F(s) = Y(s)

Example:01:Given F(s) =)6)(2(

)3)(1(4

++

++

ss

ss. Realise the above function in (a) Foster series

(b) Foster parallel.

Solution:

Since zero is nearest to the origin , (i.e s = -1f) the function yields R-L one port n/w.

(a)

Foster series:In this case F(s) = z(s) = )6)(2(

)3)(1(4

++

++

ss

ss

Thus, it yields R-L impedance n/w. To check the availability of components, we use.

Z(0) = (413)/(26 ) = 1 = ko. i.e Ro is present .

Z() = 4 = Ri , L is missing.

z (s) /s =)6)(2(

)3)(1(4

++

++

ss

ss=

62

1 21

++

++

s

k

s

k

s

K1=2

)2.()6)(2(

)3)(1(4

=+

++

++

ss

sss

ss

= )62(2)32)(12(4

+ ++

=

K2=6

)6.()6)(2(

)3)(1(4

=+

++

++

ss

sss

ss

=)26(6

)36)(16(4

+

++

K2= 5/2

z(s)/s =6

).2/5(

2

).2/1(1

+

+

+

+

s

s

s

s

s


34/125



34

1/2

1/4

1

z(s)

5/2

5/12

Fig. Foster series n/w

(b)

Foster parallel:In this case,

F(s) = Y(s) =)6)(2(

)3)(1(4

++

++

ss

ss

Which yields R-C admittance n/w.

Y(s) =6

).2/5(

2

).2/1(1

++

++

s

s

s

s

1Y(s) 2 / 5

1 2 / 51

2

Fig. Foster Parallel ckt.

Cauer Method for R-L one port n/w:(1) If F(s) = z(s) , it is called cauer 1 method which yields R-L impedance ckt.(2) If F(s) = Y(s) , it is called caure 2 method which yields R-C admittance ckt.

Example: 01:Synthesize the following function in

(a) caure 1 n/w. (b) cauer 2 n/w.128

121642

2

++

++

ss

ss

Solution:

(a)cauer 1 n/w:In this case

F(s) = z(s) =)6)(2(

)3)(1(4

++

++

ss

ss=

128

121642

2

++

++

ss

ss

This way the ckt cannot be realize. Therefore z(s) is rewritten in form as:

Z(s) =2

2

812

41612

ss

ss

++

++

S +8s+12 ) 4s + 16s+12 ( 44s + 32s+4s-ve


35/125



35

10/49

16/71

2/3 5/7

Fig. cauer 1 n/w

(b)Cauer 2 n/w:In this case,

F(s) = Y(s) =12812164 2

2

++++

ssss = 2

2

81241612ssss

++++

k

1/2ki

1/k

wi2/2ki

Fig. Cauer 2 n/w

Assignment: 03

1. F(s) =)3)(1(

)4)(2(

++

++

ss

ssFind the n/w of the form (a) Foster series (b) Foster parallel.

2. Realize the n/w function F(s) =)4)(2(

)3)(1(

++

++

ss

ss (a) 1st Foster method. (b) 2

nd foster

method.

3. Realise the n/w function Y(s) =)3)(1(

)4)(2(

++

++

ss

ssas a cauer n/w.

4. z(s) =)2)(2(

)3)(1(

++

++

ss

ssRealise the function in foster and cauer n/w.

12+8s +s2

) 12+16s+4s2 ( 1 Z1(s)

12+ 8s+s2

8s +3s) 12+8s+s ( 3/2s Y2(s)

12+9s/2

7s/2 +s) 8s+3s ( 16/7 Z3(s)

8s+16s2/7

5s /7) 7s/2+s ( 49/10s Y4(s)

7s/2

s2

) 5s2/7 ( 5/7 Z5(s)

5s2/7


36/125



36

5. Realise the n/w Y(s) =)6)(1(

)4)(2(

++

++

ss

ss

Two port n/w:1.

Z-Parameter

2. Y Parameter3. ABCD Parameter

4.

Transformation of one parameter to other5. T and n/w6. Interconnection of two port n/w

a. Cascade b. series c. parallel.

Date: 2065/5/24

Chapter: 4

Low pass Filter Approximations:

w

SBPB

wo=1 Wp

1

W

T(jw)T(jw)

1

Ws Fig. (a) Ideal case (b) Non ideal case

The desirable feature of low pass approximation are

1. Minimum pass band attenuation, p2. Maximum stop band attenuation, s3. Low transition band ratio, ws/wp4. Simple network.

The approximation Method are:

1. Butterworth2. Chebyshev3. Inverse chebyshev4. Ellipse or Cauer

5. Bessel Thomson1. Butterworth low pass approximation: Generally signal become contaminated with

high frequency signal. It is evident that low pass filter are required to remove such

unwanted signals from the useful one. The desirable LPF response is shown in fig . 1(a)

Below the normalize frequency i.e w0 = 1, the amplitude )(jwT is

constant and above this frequency it is zero. Pass band and stop band are clearly separated at

wo= 1. But since the ideal response can not be achieve . We make the approximation basedon the ideal response.

We make the magnitude T(jw) nearly constant in PB. In the SB, we require sharp

roll off (n-pole roll off). Where n will be large no if abrupt transition from PB to SB isdesired.

Mathematically, we can write,

T(jw) = Re[ T(jw) ] + j Im[ T(jw)]


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37

Re[T(jw)] = Real part of T(jw)Im[T(jw)] = Imaginary part of T(jw).

Where it is to be noted that Re[T(jw)] indicates an even functions.

Where Im[T(jw)] indicates it is an odd function.

Agains,

T*(jw) = T(-jw) = Re[ T(jw)]+jIm [T(jw)] .(ii)The functions so obtained is called

conjugate of T(jw)

Thus (i) and (ii) gives

T(jw) T*(jw) =2

)(jwT = Re[T(jw)]2+ jIm[T(jw)]

2 (iii)

T(jw) T*(jw) = T(s) T*(s) =2

)(sT

The function2

)(sT (or2

)(jwT ) is called magnitude squared function.

Example 01:Find the magnitude square function for

T(s) = (s+2) / (s3+ 2s

2+ 2s+3)

T(s) = -s+2 / -s3+ 2s

2 2s +3

2

)(sT = T(s) . T(-s)

= (2+s)/(s3+2s

2+2s+3) (2-s)/(-s

3+2s

2 2s+3)

= ..

The magnitude square function is an even function which can be represented by using a

numerator and denominator polynomial that are both even, i.e

)(

)()(

2

22

wB

wAjwT =

n

n

n

n

wBwBwBBwAwAwAAjwT

2

2

4

4

2

20

2

2

4

4

2

202

...........

...........)(++++++++=

n

n wBwBwBB

AjwT

2

2

4

4

2

20

02

...........)(

++++=

Here A2= A4= A2n= 0 (assumption).

The choice has been made as per our inspection on the roll off that was directly dependent on

the number of poles. This means larger the difference between degree of A and B , we get

the larger roll-off . This will give us a direct n-pole roll off for Tn(jw) or Tn(s) which will be

know as All pole function.

Special case:

We assume ,

B2= B4 = 0

B2n= (1/w0)2n

. B0 and A0= B0

Now , putting these assumption in the equation (i) we get,

n

n

o

wBB

AjwT

2

20

2)(

+=

=

0

2

0

0

1B

wB

B

n

o

+


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38

=n

n

ww

2

2

0

11

1

+

2)(jwT =

n

w

w2

0

1

1

+

.(ii)

In generalize condition,

wo= 12

)(jwT =( ) nw 21

1

+ .(iii)

2)(jwT =

( ) nw 21

1

+ (iv)

From equation (iv) the following property can be written.

1. At w = 0 , i.e T(j0) = 1 for all values of n.

2.

At w = 1 (=w0), i.e T(j1) = 0.707 for all values of n.3. At w = , i.e T(j )= 0 for all value of n.4. For large values of w; Tn(jw) exhibits larger roll off.

5. Butterworth response , also known as, maximally flat response, is all pole functions.6.

Butterworth (BU) response can be expanded in Taylors series from as:2

)(jwT =( ) nw 21

1

+

= (1+w2n

)-1/2

= 1+ . w2n

+ (1/2)2. (w

2n)

2/2! - ..

1 . w2n

In Taylor series,

= nwjwT 2

2

11)( ..(v)

Again we know ,2

)(jwT =( ) nw 21

1

+

Putting jw = s2

)(sT =n

j

s2

1

1

+

=

n

n

j

s2

2

1

1

+

=

n

ns

)1(

1

12

+=

nns

2)1(1

1

+

2)(sT =

nns

2)1(1

1

+ (vi)

Which gives the butterworth response in s-domain

Evaluation of T(s) for BU Response:

(i) For n = 1 equation (vi) becomes


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39

2)(sT =

21

1

s

s2=1

s = 1

2

)(sT = 1/(1-s)(1+s)

= 1/(1+s). 1/(1-s)

= T(s) . T(-s)T(s) = 1/(s+1)

Date: 2065/5/29

Butterworth transfer function (continued )

(ii) For n = 2Equation (vi) becomes ;

42

2

)1(1

1)(

ssT

+=

=41

1

s+

jw

45135

225 315

To get the poles ,

1+s4 = 0

S4= -1

S = 1 (180 + k360 )/4 , k = 0, 1, 2, 3 [since n = 4]

S = 1 45 , 135, 225, 315

The poles that lie on the left half of s-plane are:

S = 1 135, 225

Or S = -0.0707j0.707 = s1, s2

T(s) =))((

1

21 ssss

=)707.0707.0)(707.0707.0(

1

jsjs +++

=12

12 ++ ss

(iii) For n = 3

NOTE:

(i) If sn= -1, then, s=1(180+k360)/n, k = 0, 1..(n-

1) in s domain.

(ii) If sn=-1, then, S = 1 k360/n, k =0,1, 2.(n-1)


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40

63)1(1

1)(

ssT

+=

=6

1

1

s

To get the pole

1-s6= 0

S

6

= 1S = 1 k360/n , k = 0,1,2 (2n-1)

S = 10, 60, 120, 180, 240, 300

The poles that lie on left half of s-plane are

S = 1 120, 180, 240

Or, = 1120, 1180, 1240

S1= -0.5 + j0.866S2= -1+j0

S3= -0.5 - 0.866j

))()((

1)(

321 sssssssT

=

=)866.05.0)(866.05.0)(1(

1

jsjss ++

=)1)(1(

12 +++ sss

jw

60120

180

320240

Order and cutoff frequency for Butterworth:

It is to noted that, at w =wp, = p= maxAnd at w = ws, = s= minWe know that

n

ow

wsT

2

2

1

1)(

+

=

Also the attenuation formula is given by ;

= -20log )(sT


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41

= -20log10

+

n

ow

w2

1

1

= -20log10

2

12

1

+

n

ow

w

= 10log10n

ow

w2

1

+ .(i)

/10 = log10n

ow

w2

1

+

10/10

=

n

ow

w2

1

+

n

ow

w2

= 10

/10-1

ow

w= (10

/10-1)

1/2n

w =n

w

2

1

10/ )110(

Now at w = wp, = max

wo=n

pw

2

1

10max/ )110( .(ii)

and at w = ws , = min

wo=n

sw

2

1

10min/ )110( ..(iii)

equating (i) and (ii) can be equated as:

n

pw

2

1

10max/

)110(

=

n

sw

2

1

10min/

)110(

=o

p

w

w

n

n

2

1

10min/

2

1

10max/

)110(

)110(

=

n

o

p

w

w2

)110(

)110(10min/

10max/

Taking log on both sides,

20 log =

n

o

p

w

w2

log)110(

)110(

10min/

10max/

n = log)110(

)110(10min/

10max/

/ 2 log

o

p

w

w


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42

Now let us find expression for transition band ratio , i.eTBR = ws/wp, where , TBR = Transition band ratio.

Ws/wp= [(10min/10

1)/(10max/10

-1)]1/2n

.(v)

Example 01: Consider a filter using a butterworth response to realize the following

specifications of LPF.

max

= 0.5 dB

min= 20 dBwp= 1000 rad/secws= 2000 rad/sec

Determine the order and cut off frequency for the filter.

Solution:

n = 4.83 5wo= 1234.12 rad/sec

Note: Always choose higher value of n ( i.e the order of filter )because it provides larger

roll off which decreases attenuation.

Date: 2065/6/2

2.

Chebyshev Approximation Method For LPF :

1

W

T(jw)

Wo

C- R

Wo

1

W

T(jw)

BU-R

Fig (i) (a) Chebyshev response (b) butterworth response

The generalize low pass filter can be represented by2

)(jwTn = 2)]([1

1

wFn

+ .(i)

For Butterworth

Fn(w) = (w/wo)n

With w0= 1

Fn(w) = wn

Similarly to butterworth we have to determine the function Fn(w) for chebyshev response

for which the concept of Lissagious figure is required.

Lissagious figure:


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43

sin

Horizontal plate

Line of axis

Vertical plate

Fig (ii) (a) CRO Lissagious figure.n=1

n=2

n=3

n=4

x

y

Fig(ii) (b) Lissagious figure for n = 1,2,3 and 4

When adjustable frequency multiple of fixed frequency is applied , stationary figures are

obtained which are know as Lissagious figures.

Analysis:

Let the deflection due to voltage on horizontal plates bex = coskT .(ii)

Where , k = 2 /TThe deflection due to voltage on vertical plates will be then,

y = cosnkT .(iii) Where n is integer and proves the multiple frequencies.

From (ii),

KT = cos-1

x

y = cosn cos-1

x ..(iv)

cn(x) = cosn cos-1

x which is the equation for Lissagious figures.

Example:If n = 4

Assume, = cos-1xx = cos

Then,

y = cos4 x 4 y

0 1 0 1

22.5 0.924 90 045 0.707 180 -1

67.5 0.383 270 0

90 0 360 1


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44

x=1

1

0.383

0.707-1

y

# Analyse the same for n = 3 and 5.

Chebyshev magnitude Response:

We know that ,2

)(jwTn = 2)]([1

1

wFn+

Where Fn(w) = cn(w) ; 1Where cn(w) = cosn cos

-1w

Therefore the magnitude square response will be2

)(jwTn =)(1

122

wcn+ .(vi)

This function (i.e cn(w)) is valid within the range w = 1. However , the function must

also be valid for longer value of w for which we should refine our assumption for cn(w).

w > 1,

Let,Cos

-1(w) = jz

w = cosjz

we know that ,

cosjz =2

)()( jzjjzjee

+=

2

zzee

+ = coshz

cosjz = coshz

w = coshz

Z = cosh

-1

w w = cosj cosh

-1w

cos-1

(w) = jcosh-1

w

cn(w) = cosn cos-1

w

= cosnj cosh-1

w

= cosj(ncosh-1

w)

= coshn cosh-1

w

cn(w) = cosh cosh-1

w , w> 1

Cn(w) = cosn cos-1

w, w = 1


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45

Properties of magnitude response for Chebyshev:

We know that,2

)(jwTn =)(1

122

wcn+

)(jwTn =

)(1

1

22

wcn+

Where, cn(w) = cosn cos-1

w w 1= coshn cosh

-1w w 1 and 1

1. At w = 0,

Cn(0) = cosn /2 ; 0,1,2.)(jwTn = 1 for n = odd

=21

1

+ for n = even

2. w = 1

cn(1) = 1 for all values of n.

)(jwTn =21

1

+

1

ww=1

1

w=1w

Fig (iii) (a) C-R for n = odd (b) C-R for n = even

Order of C-R filter:

We know , the attenuation formula is given by

= -20log )(jwTn dB

But, )(jwTn =

)(1

1

22

wcn+

=2

1

22

)(1

1

+ wcn

= - 20log2

1

22 )(1

1

+ wcn

= -10log)(1

122

wcn+

= 10 log )(1 22 wcn+ (vii)

= 10log 212 )cos(cos1 wn + w 1

for w > 1,

= 10 log 212 )cosh(cosh1 wn + ............(ix)

Now ,


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46

max occurs when , cn(w) = 1

equation (vii) reduces to ,

= max= 10 log ( 1+ 2 .1)(x)max/ 10 = log ( 1+ 2 .1)1+ 2 = 10

max/10

2

1

10max/ )110( = .(xi)

Date: 2065/6/7

Herewe know that

w = wnp, then, 1)(22 = wcn

)coshcosh(1

)( 1 npnpn wnwc =

= [since wnp>1]

Cosh-1

(ncosh-1

whp) =

1

Cosh-1

(ncosh-1

whp) =

1

Cosh-1

whp = 1/n. cosh-1

(

1)

wnp= cosh(1/n. cosh-1(1 )) (xii)

Wnp= cosh [1/n. cosh-1

({10max/10

-1}1/2

)]

Now = min when w = wsmin= 10 log10( )(1

22

sn wc+

)(22

sn wc = 10min/10

1

2(cosh ncos-1ws)2 = 10min/10-1Or, ( cosh ncosh

-1ws)

2= (10

min/10-1)/ (10

max/10-1)

n cosh-1

ws= cosh-1

[(10min/10

-1)/ (10max/10

-1)]1/2

n = {cosh-1

[(10min/10

-1)/ (10max/10

-1)]1/2

}/cosh-1

ws..(xiii)

Example: Given wp = 1 , ws = 2.33 , max = 0.5dB , min = 22 dB. Calculate n forButterworth and chebyshev filters which filter would you select.

Solution: For Butterworth filter , the order is given by

n = log10[(10max/10

-1)/(10min/10

-1)]/ 2 log (wp/ws)

= log[(100. 5/10

-1)/(1022/10

-1)]/2log (1/2.33)

= 4.234 5

n for BU = 5For Chebyshev the order is given by ,

n = cosh-1

[(10min/10

-1)/(10max/10

-1)]/cosh-1

(2.33)

= 2.89 3


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47

n for chebyshev = 3 .

Since the order of chebyshev filter (i.e n =3) is less then the order of butterworth filter (i.e n

= 5) and both filter provides the same roll- off for the specification, n would choose

chebyshev filter.

Chebyshev poles location and network function:

We know2

)(jwT =)(1

122 wc

n+..(i)

Substituting s = jw equation (i) becomes,2

)(sT =)/(1

122

jscn+..(ii)

To determine the poles,

0)/(122 =+ jscn

=

1)/( jjscn (iii)

Again,

Cn(s/j) = cosn cos-1

(s/j)

Let

Cos-1

(s/j) = x = u + jvThen, cn(s/j) = cosnx = cosn (u+jv)

= cosnu. Cosnjv sinnu. Sin njv

= cosnu coshnv jsin nu . sinh nv= 0

1j [ from equ. (iii)]

Thus, comparing , we get, [ cosjnv = coshnv

Cosnu . cosh nv = 0 [ sinjnv = jsinhv]-sinnu. Sinhnv = 0

The minimum value of

Coshnv = 1, coshnv not equal to 0

cosnu = 0

Or cosnuk= cos(2k+1). /2, k = 0,1,2.Uk= (2k+1) /2n .(v)Now ,

-sinnuk= sinhnvk=

1

But, sin nuk= +- 1

+-1 . sinhnvk=

1

Or sinhnvk=

1

Nvk= sinh-1

(

1)

Vk = 1/n. sinh-1

(

1)


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48

Again, we know thatCos

-1(s/j) = x = u +jv

s/j = cosx = cos(u+jv)

in general,

sk= jcos(uk+jv)

= j[cosuk.cosjv sinuk. sinjv]

= j[cosuk. coshv jsinuk. sinhv ]

Sk= sinuk. sinhv + jcosuk. coshv (vi) , k = 0,1,2.(2n-1)

Again,

Sk= sin[(2k +1) /2n] sinhv + jcos[(2k+1)/2n] coshvOr , sk= k + jwk ..(viii)Where,

k= sin[(2k+1) /2n] sinhv .(ix) Form euation (ix) ,

Now adding equation (xi) and (xii) we get,

Which is equation of ellipse . Therefore we can say that the poles of chebyshev filter lie on

the ellipse.

Date: 2065/6/9

Example:01Obtained the 4thorder network function of a low pass chebyshev filter with

max= 0.75 dBSolution: n = 4 max= 0.75 dBNow = ( 10max/10-1)1/2 whp= cosh (1/n. cosh

-1(1/))

= (100.75/10

-1)1/2

= 0.434And whp= cosh ( 1/n. cosh

-1(1/)) =

Pole location is given by


49/125



49

Sk= sinuksinhv + jcosuk coshvWhere, uk= (2k+1) ./2n ; k = 0, 1, 2n-1

V = 1/n. sinh-1

(1/)

uo= /8 u1= 3/8 , u2= 5/8, u3= 7/8 , u4= 9/8u5= 11/8 u6= 13/8 , u7 = 15//8

v = 0.393 (adjust calculator in radian)

s0= 0.154 + 0.996j

s1= 0.373+ 0.413j

s2= 0.373 0.413j

s3= 0.154-0.996j

s4= -0.154 0.996j

s5= -0.373 0.413j

s6= -0.373 + 0.413j

s7= -0.154 + 0.996j

The transfer function (or n/w function) for forth order chebyshev filter is given by ,

T(s) = 1/(s+s4)(s+s5)(s+s6)(s+s7)

jw

S0

S1

S2

S3S4

S5

S6

S7

Home Assignment:

Example:02: Determine the network function for 3rdorder chebyshev LPF with max= 0.75dB ( =p; pass band attenuation)

Date: 2065/6/14

Inverse chebyshev low pass approximation:


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50

W

T (jw)

BU-ResponseW

T (jw)

Ideal LPF

W

T (jw)

Chebyshev-respone

T (jw)

Winverse-Chebyshev-response

W

1- T (jw)2

Fig: intermediate stage to obtain inverse chebyshev response.

T (jw)ic

2

Fig: The reciprocal value of w of intermediate stage give the value of w in I-C response.

We know the response of chebyshev is given by2

)(jwT =)(1

122 wcn+

1-2

)(jwTc = 1-)(1

122 wc

n+

=)(1

)(22

22

wc

wc

n

n

+

Now replace w by 1/w


51/125



51

2)(jwTIC =

)/1(1

)/1(22

22

wc

wc

n

n

+

(i)

Where,2

)(jwTIC is the magnitude square response for I-C.

We know ,

cn(1/w) = cosncos-1(1/w)at for w = 1

cn(1) = 1 for all value of n

Thus equation (i) becomes

2)1.(jTIC = 2

2

1

1.

+

)1.(jTIC = 2

2

1

1.

+

. (ii)

We know that ,

min= -20log )1.(jTIC dB (iii)

Using equation (ii) on equation (iii) , we get,

= min= -20log )1.(jTIC dB

= - 20log

2/1

2

2

1

+

= 10log

+2

21

min= 10 log [ 1+ 21

}

Or , 10min/10

-1 =2

1

( ) 21

10min/ 1010 = . (iv)

Again in general, the attenuation formula can be written as:

= -10log

+

)/1(1

)/1(22

22

wc

wc

n

n

= 10 log

+

)/1(

11

22wcn

Now at w = wp = maxThen above equation becomes

= max= 10 log

+

)/1(

11

22

pn wc

(10max/10

1) =)/1(

122

pn

wc

)/1(2

pn wc =)110(

1.

110max/2


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52

)/1(2

pn wc =)110(

)110(10max/

10min/

)/1( pn wc =)110(

)110(10max/

10min/

.(v)

)/1( pn wc = coshn cosh-1

(1/wp) . (vi)

[

wp< 1, 1/wp> 1]Thus equating equation (v) and (vii)

Coshn cosh-1

(1/wp) =2

1

10max/

10min/

)110(

)110(

n =)/1(cosh

)110(

)110(cosh

1

2

1

10max/

10min/1

pw

(vii)

Which gives the required order for the inverse chebyshev filter.

Now , for half power frequency i.e at w = wp)1.(jTIC = 1/2

2)1.(jTIC =

Which means,

)/1(22

Pn wc = 1

)/1(2

npn wc = 21

)/1( wcn =

1

Coshn.cosh-1

(1/wnp) =cosh-1

1

n coshn.cosh-1

( )/1( npw = cosh-1

(

1)

cosh-1

( )/1( npw = 1/n. cosh-1

(

1)

1/wnp= cosh[1/n. cosh-1

(

1)]

= )

1(cosh

1cosh

1

1

n

wnp < 1 .(viii)

Which gives the desire half power frequency.

Example: 01Given, max= 0.5 dB

min = 22 dBwp= 0.9

n = ?

wnp= ?


53/125



53

Assignment:

Example:02 Differentiate between Butterworth , chevyshev and inverse chebysehev filters.

Pole zero location for inverse chebyshev:

We know that ,

2)(jwTIC =)/1(1

)/1(22

22

wcwc

n

n

+

T(s). T(-s) = z(s).z(-s)/[p(s).p(-s)]

Where, z(s) z(-s) |s = jw= )/1(22

wcn

P(s) P(-s)|s = jw= 1 + )/1(22

wcn

For zero location:

)/1(22

kn wc

0 0)/1(2 = kn wc

0)/1( =kn wc

Cosn cos-1

(1/wk) = cos(k/2) for k = 1,3,5 ..(i.e odd)ncos

-1(1/wk) = k/2

1/wk= cos(k/2n) which gives the zero for inverse chebyshev.Wk= sec(k/2n)

For poles:

1+ 0)/1(22

= kn wc The poles location are similar to chebyshev.

Simply replacing wkby 1/wk

i.e if chebyshev poles = pi

Then , inverse chbyshev poles = 1/pi

Fig. Pole locationFig. Zero location

Example:01Given,

min= 18 dB

max= 0.25 dBws= 1.4 rad/secwp= 1 rad/sec

Find out the pole and zero for inverse chbyshev response.


54/125



54

Chapter 5

Frequency transformation:

Frequency transformation is important because the prototype LPF with any type of

approximation can be converted into high pass band pass , band stops filters within the same

characteristics easily.

0.707

Wc

1

T(jw)

The effect of frequency transformation are:

1. Magnitude response 2. Network function

3. Location of poles and zeroes. 4. Network elements.

Types of transformation:

1. LP to LP transformation

WW0 0

Transformation

Old LPF New LPF

Replace s by wo/o.si.e

w0= 1 ( in normalized case)

s s/0

TLP(new)(s) = TLP(old)(s/0)For eamaple,

If

TLP(s) = 1/S+1

Then

TLP(old)(s) = 1/s+1

TLP(new)(s) = TLP(old)(s/o) = 1/(s/o)+1 = 0/(s+0)

1. For resistor:- No change.

2. For inductor:XL= LS

Putting0

s

s


55/125



55

XL = Lold0

s= SLs

Lnew

old .0

=

Lnew= Lold/0

3. For capacitor:

Xc= 1/cs

Putting0

s

s

Xc =

0

1

sCold

=sC

sC

newold .

1

.

1

0

=

Cnew = Cold/s

2 LP to HP Transformation:

WW0 0

LPF with W0 HPF with 0

Transformation

In this case we replacesw

s.0

0

Or ,s

s0 [Since w0= 1]

THP(s) = TLP(s) ( )SLPs

Ts

0

0

=

=

Example if TLP(s) = 1/(s+1)

Then, THP(s) =

1

1

0 +s

=s

s

+ 0

(1)For resistor:

No change

(2) For inductor:

XL= LS

Puttings

s 0


56/125



56

XL= L.

s

0 =

sL

.1

1

0

Comparing

sL

.1

1

0

with 1/CS

C =0

1

L

(3)For capacitor:Xc= 1/cs

Puttings

s 0

XL = LSSCc

s

cs

=

=

=

.

1

.

1

000

Comparing SC

.1

0

with LS

L =

0

1

C

Date: 2056/6/15

(3) LP to BP Transformation:

W

T (jw)LP

L U

T (j )BP

Transmission

WsWp

In this case,

Lu

sws

+

22

0 .

Here, u L= B

And w0= 1

Bs

ss

22

. +

Where 02= L. u


57/125



57

(1) For resistor

- no change

(2)

For inductor:

XL= LS

The new value of inductive reactance is given by:

XL= L.

+

Bs

s2

0

2

XL=

Bs

Ls

B

L2

0. + =

sL

Bs

B

L

.

1.

2

0

+

The new component are inductor and capacitor in series.

L B

Lo2B

(3)

For capacitor:

The new capacitive reactance form LP to BP is given by :

=

sc

Bs

B

c

Bs

cs

B

c

Bs

ccs

Bs

sc

.

1

111

.

1

2

0

2

0

2

0

22

0

2

+

=

+

=+

=+

The new components (i.e inductor and capacitor) are in parallel as shown in fig. below:

B

Co2

C

B

LP to BS Transformation:

W

T (jw)LP

L U

T (j )BS

Transmission

WsWp


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58

In this case s is replaced by 020

2.w

s

Bs

+

But w0= 1, 20

2 +

s

Bss

(1)

For resistor :

Resistor value remain same.

(2) For inductor:

XL= LS

XL= 20

2.

+s

BsL =

sLB

sLBLBsLBS

s

LBS

s

.

11

111.

2

0

2

022

0

2

+

=

+

=+

The new component (i.e inductor and capacitor ) are in parallel as in figure below:

LB

o2

1

LB

(3) For capacitor:

Xc= 1/cs

Xc=s

CBs

CBCBsCBs

s

CBS

s

s

Bsc .

1.

1

.

1

2

0

2

022

0

2

2

0

2

+=

+=+

=

+

CB

o2

1

CB

Example:01:If T(s) =1

1

+s, then change the above function from LP to BP. Given , L=

10 and u = 20.Solution:

Then, TLP(s) =1

1

+s , L= 10 , u = 20

We know ,

02

= L. u = 10. 20 = 200For Lp to BP we replace

s

s

s

s

B

ss

10

200

)1020(

2000 222

0

2+

=

+=

+


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59

Thus,

110

200

1)(

10

200 22

++

==+=

s

ssT

s

ss

T BPLP = 20010

102 ++ ss

s

TBP(s) =20010

)(102 ++ ss

s

Example:02:Obtain the transfer function of the 4thorder Butter worth HPF with 0= 2

104rad/sec.

TLP(s) =161313.241921.361313.2

1234 ++++ ssss

We know that ,

ss

0

=

161313.241921.361313.2

1

0

2

0

3

0

4

0 +

+

+

+

ssss

Example:03:The filter shown in the figure below is a 4thorder chebyshev low pass filter with

p= 1 dB and wp= 1. Obtain a bandpass filter from this low pass with o= 400 rad/sec andB = 150.

+

-

} } }BA C

D E

+

-

V1 V2

Solution:

For LP to BP conversion , we replace

Bs

ss

2

0

2 +

Where, o= 400 rad/sec , B = 150

Now for section A:L = 1 .2817

Which changes to series LC component as shown below:L B

Lo2B

The new inductor value is = L/B = 1.2817/150 = 8.54 mH

and the new value of capacitor is = B/L20

= 150/(1.28174002

) = 731.45 F.

For section B:


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60

C= 1. 9093

Which changes for LP to BP As:

B

Co2

C

B

New inductor value = B/C o2= 150/(1.9093400

2)= 491.01 F

New capacitor value = C/B = 1.9093/150 = 12.72 F

For section C:L = 1 .4126

B

Lo2

L

B

For section D:

B

Co2

C

B

For section E:R = 1 R = 1

+

-V

+

-

8. 54 mH731. 45uF

12.72 uF

6. 99 mF

663. 66uF

9. 41 mH

893. 71mH491uH 1

Date: 2065/6/16


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61

Doubly Terminated LC-Ladder ckt:

Loss less Ladder

1

R 2z i n

R 1

+

-

V s V 2

V 1

I 2I 1 +

-

Fig.1 Doubly Terminated LC ladder ckt.

From figure(i)

I1= Vs/(R1+Vin) (i)

Where,Zin= Rin+ jxin.(ii)

Since the ckt is loss less

Input power = output power

P1= zin|I1(jw)|2= |V2(jw)|

2/R2 (iii)

From equation (i) and (iii)zin|Vs(jw)|

2/(R1+zin) = |V2(jw)|

2/R2

or , |V2(jw)|2/|vs(jw)|

2= zinR2/(R1+zin)

2..(iv)

Now for matched source.R1= zin

Which means

V1= vs/2

P1max= |v1(jw)|2/R1= |vs(jw)|

2/4R1

Also it is to remember that ,

P2= |v2(jw)|2

/R2|(jw)|

2= p2/p1max= [|v2(jw)|

2/R2]/ |vs(jw)|

2/4R1= 4R1/R2. |v2(jw)/vs(jw)|

2..(vi)

Form equation (iv) and (vi)

|H(jw)|2= 4R1/R2. {zinR2/(R1+zin)}

= 4R1zin/ (R1+zin)2= 1- (R1-zin)

2/(R1+zin)

2

(R1-zin)2/(R1+zin)

2= |(jw)|

2

= reflection coefficient

2

1

2

1

)(

)()().(

in

in

zR

zRss

+

=

)(

)()(

1

1

in

in

zR

zRs

+

= .....................(vii)

From equation (vii) , we get

)(1

)(1.1

s

sRzin

+

= 1st zin ..(viii)

)(1

)(1.1

s

sRzin

+= -----------2

ndzin

Generally we take R1= 1. Both impedances in equation (viii) are reciprocal impedance.

Synthesis of Doubley Terminated LC ladder with equal terminal (All pass filter)


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62

For butterworth response:

)()(

)()(

1

1)((

2

22

sDsD

sNsN

wjwHjwT

n

=

+== [since w0= 1]

n

n

nw

ww

jwHsHs 2

2

2222

1111)(1)(1)(

+=

+===

)().(

).(

)().(1)().(

2

2

2

sDsD

ss

sDsD

w

w

wss

nnn

n

n

=

=

+= (ix)

Now,

For n = 1

D(s) = s+1 [since T(s) = H(s) = 1/S+1]

Form equation (ix)

(s) = sn/D(s)

= s1

/s+1 = s/s+1

zin1=)(1

)(1.1

s

sR

+

=

11

11

.1

++

+

s

s

s

s

=ss

ss

++

+

1

1.

Zin1=12

1.

+s.............(a)

Zin2= 2s+1 .(b)

Zin2= 2s+ 1 = Ls + R

i.e L = 2, and R = 1

The ckt will be

vs-

+

R 1

R 2

1

1

2

From equation (a) , zin1= 1/(2s+1) i.e c = 2, and R = 1

vs-

+

1

1

2

v2

+

-

For n = 2

D(s) = s2+2s + 1

12)(

)(2

++

==

ss

s

sD

ss

nn

zin1=)(1

)(1

s

s

+

=

)12/(1

)12/(122

22

+++

++

sss

sss=

)12(

)12(22

22

sss

sss

+++

++


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63

zin1=)122(

)12(2 ++

+

ss

s ..(a)

Similary,

Zin2=12

122 2

+

++

s

ss .(b)

Taking equation (b)

The ckt will be as follows:

vs-

+

1

1

1 . 4 1

v2

+

-

1 . 4 1

vs-

+

1

1

2

1 . 4 1

1 . 4 1

Home work : For n = 3 and n = 4

Date: 2065/6/17

Synthesis of Doubly Terminated LC - Ladder with unequal termination: ( R1R2) :For R1R2 the butter worth response is given by ,

2

2

22

)(1

)0()( jwT

w

HjwH

n=

+=

Generally we take,

R11 and R1R2

+

-R2

R1

Vs ZinV2

I2I1

LC

2s2+2 s

1

1 ) 2.s + 1( 2 s Y2(s)2 s

1) 1 (1 z3(s)

2 s+1) 2s +2 s +1 (2.s z1(s)


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64

From figure, the transform function , T(s) =sV

V2

From which we get ,

T(0) =12

2

RR

R

+

Now we know2

2

2

12

)(

)(.

4)(

sV

sV

R

RsH

s

=

2

2

12 )(.4

)( sTR

RsH =

)(.2)(2

1 sTR

RSH =

21

2

2

1

2

1 .2)0(.2)0(RR

RRRT

RRH

+==

21

12 .2)0(

RR

RRH

+=

Example:01:Realize the doubly terminated ladder filter with a Butter worth response for n

= 3, R1= 1, R2= 2 .

Solution:

We know, for unequal termination ( i.e R1R2) the Butterworth response is given by,

nw

HjwH

2

22

1

)0()(

+=

Here, n = 3, R1= 1 & R2= 2

H2(0) =

( ) ( ) 98

21

2.1.4.422

12

12 =+

=+RR

RR

n

wjwH

2

2

1

9/8)(

+=

The reflection coefficient function is22

)(1)( jwHjw =

=n

w21

9/81

+ =

n

n

n

n

w

w

w

w2

2

2

2

1

9/1

1

9/81

+

+=

+

+

6

6

32

322

1

)/(9/1

1

)/(9/1)(

w

js

w

jsjw

+

+=

+

=

Or,66

232

6

62

1

)3/1)(3/1(

1

)()3/1(

1

)(9/1)(

s

ss

s

s

s

ss

+=

=

=

)()3/1(.

)()3/1()().(

33

sDs

sDsss

+=

Where, D(s). D(-s) = 1- s6


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65

)(

3/1)(

3

sD

ss

+=

For n = 3,

D(s) = s3+2s

2+ 2s+1 (from table)

The first impedance is ,

Zin1=

122

3/11

122

3/1

1

)(1

)(1

23

3

23

3

+++

++

+++

+

=

sss

s

sss

s

s

s

Zin1=3/4222

3/22223

2

+++

++

sss

ss .(a)

Zin2=3/222

3/42222

23

++

+++

ss

sss ..(b)

Now using continued fraction method for equation (b)

+-

11 2

22/3

Home Assignment:

Try it for n = 1, 2, 3 and 4 , for unequal terminal i. e R1= 1 and R2= 2.[ for n = 4, D(s) = s

4+2.16s

3+3.14s

2+2.6s+1]

Review of ideal and non ideal properties of operational amplifiers, GBP, CMRR,Inverting and non inverting A/F.

2s2+2s+2/3 ) 2s

3+2s

2+2s + 4/3 ( s z1(s)

4/3.s+4/3 ) 2s +2s+ 2/3(3/2. s Y2(s)

2/3 ) 4/3.s +4/3(2s z3(s)

4/3 ) 2/3 (1/2 Y4(s)

2s +2s +2/3.s

2s2+2s

4/3.s

2/3


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66

Fundamental of Active filter circuit:-

Ideal & Non-ideal properties of op-amp.

Gain Bandwidth product( GBP)

CMRR & its importance.

The main advantage of Active filter:-

Small in size

Provide grater amplification

Cheaper than passive filter.

The limitation area:-

Extra Vccis required

Sensitive to temperature

Low gain at high temperature

Low gain at high frequencies

CMRR should be high

Certain important configuration of op-amp:-

-+

R

Rf

Vo

ViR

RV

f

o .=

(2) Non-investing:-

-+

R

Rf

Vo

Vi

ViR

RFVo

+= 1

(3)Integration:-


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67

c

-+ Vo

R

Vi

ViSRC

ViRCS

Vo

==

11.

1

If R1=1 & C = 1, then

SVi

Vo 1= I.e. Integrator always contributes polo.

(4) Differentiator:-

-

+c

Rf

Vo

Vi

R

OVo

CS

OVi =

1

( )ViCRSVo =

If Ro= 1& Co= 1, Then

SViVo =

(5) Summer:-

-+

Rf

Vo

R1

R1

V1

V2

( )21 VVRi

RFVo +=

(6) Subtract or (Difference A/F)


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68

-+

Rf

Vo

R1

R1

V1

V2

Rf

( )12 VVRi

RFVo =

Design of Active filters (op-amp based):-

(1)

Investing type:-

-

+ V2

V1

Z2

Z1

From fig.

R(S) =1

2

1

1

)(

)(

Z

Z

SV

SV =

(a)T(S) = -K/S

Since, the above T(S) contributes polo we can reduce the T(S) with T(S) of integrator

I.e. ( )S

K

RCSST

==

1

RCK

1=

If R=1, then,

C=1/K

If C=1, then, R=1/K

Thus the design will be

-+ V2

R =1

1/k

(b)T(S)= -KS (Do yourself)

(c) T(S) = -K(S+a1)

We can compare with the general T(S) of investing type ie.

T(S) =1

2

Z

Z


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69

( )11

2 aSkZ

Z+=

Or )(1

1

1

2

1

1

2 aSky

y

y

y+==

If y 2= 1, then,

Y1= KS+Ka1

Y21

Y1

1/ka1

1/ka1

1

-+ V2

Fig:- Design for R(s) = - (s + a1)

(d) T(S) =1PS

K+

Let we can write,

11

2

PS

K

Z

Z

+=

( )12

1

PS

K

y

y

+=

( )K

PSy

y

12

1 1

+= fig: Design for T(S)= -K/(S+P1)

y1=1, then

K

P

K

S

K

PSy 112 +=

+=

(e) T(S) =1ps

ks

+

KS

P

K

ZZ

11

2

11

+

=

If Z2=1, then,

k/p1

-+ V21

1/k

-+

1

V2V1

1/k k/p1


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70

KS

P

KZ 11

1+=

(f) ( )1

1

PS

qSKST

+

+=

+

+

=1

1

1

2

ps

qs

KZ

Z

+

+=

1

1

2

1

ps

qsk

y

y

Let y1= ks + kq1

Then, y2= s + p1

1/p1

-+ V2

1

1/ka1

k

V1

Fig: Design for T(S) =( )

( )11

ps

qsk

+

+

# 2nd

approach of above problem (Do YourselfDo YourselfDo YourselfDo Yourself)

(2)Non-investing type:-

-

+

V2V1

Z2

Z1

(a) T(S) = ( )( )1

1

ps

qsk

+

+ Where, q1>p1

Comparing,

++=+

1

1

1

21psqsk

zz

-+ V2

1/a-p1

a1-p1/p1

1


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71

11

1

1

2

+

+=

ps

qsk

z

z

=1

11

ps

pskqks

+

+

( ) ( )( )

1

11

1

2 1

ps

pkqks

z

z

+

+=

For, k = 1

1

11

1

2

ps

pq

z

z

+

= T(s) =

( )( )1

1

ps

qsk

+

+ for k = 1

11

1

11

2

1 1

pq

p

pq

sy

y

+

=

If y1= 1, then

y2=1111

1

pq

p

pq

s

+

For, k 1

(

filter designing complete

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