Filter Desing Complete

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Chapter:01

Filter: Filter can be considered can be considered as frequency selective networks. A filter isrequired to separate an unwanted signal from a mixture of wanted and unwanted signals.

The filter specification are generally given in terms of cutoff frequencies, passband (P.B) and stop band (s.b) regions. P. B is the frequency band of wanted signal and S.Bis 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. TheGains of such filters are always less than or equal to unity (i.e GS1). It is to be noted theL 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:

i/p o/p

V1(t) Filter V2(t)network

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) = v2(jw)v1(jw)

Then the voltage gain in db is given by ,

Av = 20log10 T(jw) dB .(1)

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

p1Ap = 10 log10

p2

Now, the voltage attenuation is given by ,

= 1/Av

= -20log T(jw) dB.(2)

From equation 1 and 2 ,we can write,

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T(jw)= 100.05Av..(3)

T (jw)= 10-0.05

.(4)

Types of filters: ( According to the function)

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

PB SB

wwcFig. 1(a)

2. High pass filter: A high pass filter is a compolement of a low pass filter in that thefrequency range form o to cis the SB and from cto infinity is the PB.

A

SB PB

wwcFig. 1(b)

3. Band pass filter ( BPF): A BPF is one in which the frequency extending formL(or

1) to u(2) are passed while signals at all other frequencies are stopped.A

SB PB SB

wwc

Fig. 1(c)

4. Band stop filter(BSF): A BSF is complement of BPF where signal components atfrequencies form 1 to 2 are stopped and all others are passed. These filters aresometimes known as Notch filters.

A

PB SB PB

wFig. 1(d) Notch filter

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



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ranges from o to infinity.A

PBw

Fig. 1(e)

Non- ideal Characteristics:

Filter Gain curve Attenuation curve

1. LPF

2.HPF

A

1 TB0.707

Wp

Wc

Ws

A

A

WpWc Ws

A

Ws WcWp

1. From the attenuation curve it to be noted that in the pass band the attenuation is always

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)

amsm +am1s

m1 +........... +a1s +a0

T(s) =b s

n +b sn 1 + +b s +b

n

n1

10

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

T(s)=P(s)

=a

1

s

+

a0

Q(s) b1s +b0

=a1(s +ao/a1)+ b0/b1)b1(s

G(s z ) G(s +z )

1 1= or T (s)= (sp1) (s+z2)



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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 berealized 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

1V =R + idt.......... ......(i)1 1 L

1 ............V2= idt.......... (ii)L

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

1V (s)=RI (s)+ I (s)..................(iii)

1 cs

1I (s)V

2

(s) csV2(s) = =V1(s) 1

I (s)R +cs



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1

cs 1

=Rcs +1=RC(s +1 /RC)cs

1

RC=

S +1 /RC

W0

T (s)=S+W0

Where, W0= 1/RCNow , for magnitude plot,

T(s) = T(jw) = W0/(jw+W0)w0T ( jw) =

w2 +wo 2

Now when

W = 0 T(jw)= 1

W= wo T (jw)= 0.707

W = , T(jw) = 0T(jw)

10.707

WWc

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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90

45

WoW

-45

-90

2.R 1

1 R 1 2 +

+v1 R 1 c v2-

-4 R 3

Above figure can be modified as:1 1

3

+ 2

v1 -

-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 - Vs R/(1+1/cs)

V1/V2= - RCS/RCS+1

T(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 ,

T(jw)=1w

2 +(w

0)2

2w

2 +w0

2

6



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T(jw)

1T (jw) = W

2

Phase plot:

(jw) = tan-1

(-w/wo) - tan-1

(

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

(w/wo)when,w = 0, (jw) = 0

w = 0, (jw) = -90w= , (jw) = -180

90

45

WoW

-45 -90

-135 -180

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

Assignment:

3.+

v1+ v2-

-

4.

7



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c1

R 1v1 + R 2 v2

-

5.R 1

+

R 2v1

+ v2

-C2

-

Date: 2065/4/28Example :04

c1

R 1v1 + R 2 v2

-

From fig (i)Y1= c1s+1/R1=

R1C1S

+

1

R1

Z1= 1/Y1 =R1

R1C1S +1Now applying kirchoffs voltage law, for fig (i).

V1= z1i+R2i

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

V2(s) = R2I(s)

V2(s) R2T(s) = = =V1(s) Z1(s)+R2

R2(R1C1S +1)=

R1 +R2R1C1S +R2

R2R1 +R2

R1C1S + 11R R C(S+ )

1 2 1 R1C1=

R1 +R2R1R2C1 S +

R2R1C1



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S +1

=R

1

C1

S +1

+1

R2C1 R1C1

S +01 S (01)Or, T(s) = =

S +02 S (02)

-wo1wo2

02

> 01And ,

or, 02 < 01

For Magnitude plot:

jw +w w2 +w 201

01T(jw) = =

jw +w02 w2 +w022w

01 R2Now at w= 0,

T (

j0) = =w02 R1+R2

At w = , T(j)=w01 =1

w02

T(jw)

1

R2R1+R2

w=0

For Phase plot,

T(jw) =jw

+

w01

jw +w02

Where, w01= 1/R1C1

W02= 1/R1C1+1/R2C2w w

1 1

Therefore, (jw) = tan tanw

01w

02

(jw) = z pSince 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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w(z) =tan 1 =tan 1(wR C )

1 1w

01Now at w = 0(z) = (j0) = 0(z) = (jw0)= 45

Now at w =

(j ) = 90

Pole plot ( p)

(p) = tan-1

(w/w01)

1 w= tan1 1

+

R1C1R

2

C1

Now at, w = 0p= (j0) =

0 at w = w02p= (w02) = 45

at w = , p= (j ) = 90

(jw)

zero plot90

45 pole zero plot

wo2w=0 wo1

-45

Pole plot-90

Thus the magnitude response of the above network shown that it is a high pass filter with dc

gain R2/(R1+R2) and phase plot signifies it is leading type.

Insertion Gain and insertion loss:

10



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T(jw)

1 1

w wwo wo

1

wwo

Insertion gain

T(jw)

11

wwo

Insertion gain

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 tto 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 thisloss 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 verydifficult to built the capacitor of 1 f and inductor of 1 H . Besides this the practical values ofcapacitors available in the electronic circuit is of the order of microfarad or Pico farad. Thecircuit considered so for have normalized elemental values but practically these values arenot 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 ofthe order of unity.

2. If we have the normalized design of the filter then it is easy to generate the11



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filter of similar characteristics of varying center frequency and impedance levelwithout redesigning the whole circuit.

The actual or the required elemental values of the

Filter ckt which is obtained after scaling is called demoralized values of the circuit.

Scaling: While designing the ckt sometimes the value of components may not be available

so we change them with the available one, which is called scaling. To obtained the elemental

values of the required filter we amplitude and frequency scale the normalized design.

Types of scaling:

1. Impedance (Magnitude or amplitude) scaling: In this scaling, the magnitude of theimpedance is increased or decreased. To scale in magnitude , z(s) (the impedance) is

multiplied by a constant factor Km.

Z(s)

If Km. 1, then it is called scale up.

If Km< 1, then it is called scale down.

Let, Rold= old value of Resistor.

Lold= old value of inductor

Cold

= old value of capacitor.

The new values of R, L and C are given by

Rnew= KmRold.(i)Also,

XLKm = LoldS Km = (KmLold)S = LnewS

Lnew= KmLold.(ii)Again,

1 1

XcKm= 1/colds . Km =C =Cnew.Solds

Km

C0ld

Cnew = .(iii)

K m

Example 01:Perform Impedance scaling to the following network.



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R=1 ohm+ +

V2 C =1F V1

-

-

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 thatC

new= C

old/K

m

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

Therefore, Rnew= Km.Rold

= 105

* 1 Rnew= 100K

R=100k+ +

V2 C =10 uf V1

-

-

Fig(ii) scaled ckt.

The transfer function for fig. (i) ,Told(s) = 1/(s+1)

1R

new

CnewAnd, Tnew= = 1/s+11S +

Rnew

Cnew

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 magnitudeof the impedance , i.e

ZL= ( = XL) = LS = jWL

ZL=WL is a constant.

Similarly,

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



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Z c =1

is constant.wc

To do so any change in w must be compensated by corresponding change in L and cIf, w = old corner frequency

= new corner frequency.

= Kfw

Where,

Kf= frequency scaling factor.

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

o

Expansion

o = 10 3

o = 10

Compression

3

o = 10 o = 1

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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3. Both magnitude and Frequency scaling:

It is not necessary that we scale magnitude and scale in frequency separately. We can doboth at once. Cobining all the above equations.

Rnew= KmRold.(Viii)

Lnew= Km/kf. Lold(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,1

R0ld

Cold 1

Told(S) = 1 =s +1s +Rold

Cold

1R

new

Cnew 10

And, Tnew(s) = 1 =s +10s +Rnew

Cnew

Example 02:R=1/10

1

1F



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Perform frequency scaling with o= 1

Example 03:R1

C1v1 R2 v2

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

1

s +R1C1T(s) = .(i)1 1

s + +R1C1 R2C1

But given ,

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

Comparing equation (i) and (ii)

1/R1C1= 0.5 R1C1=

2 ..(iii)

Again, ( 1/R1+ 1/R

2)1/C

1=

3..(iv) Let , C1= 1 F

For equation (iii) R1 1 = 2

R1= 2 Therefore fromequation (iv)

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

Therefore, R2= 2/5 In order to perform magnitude scalingR

1old= 2

R2old= 2/5 = 0.4Cold= 1 FSay, C1new= 10 F

Then, Km= Cold/Cnew= 1F/ 10F Km= 10

5

Therefore, Rnew= kmR2old= 105 0.4 =

40 k The selected ckt will be :



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200k

40.4Fv1 40k v2

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:

C2= 1/10 F R2= 1/100

_

R1= 1 C1= 1 F+

Perform magnitude scaling to the ckt given.

Note: Take C new as the new value of capacitor for C old where C old represents thelargest value in the circuit.

Solution:

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

C2old= 1/10 F.

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

new= C

old/k

m

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

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

C2new= 1 FSimilarly,

R1old= km. R1old= 105 1 = 100 kR2new= km. R2old= (1/100). 10

5= 1 k.

1k

_

100k 10 uF+



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Fig: Magnitude Scaling Ckt.



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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.

I(s) I1(s) I2(s)

V(s) 1- port V1(s) 2- portn/w n/w

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) = 1Then, z(s) = 3s+2

= Ls +R3H

V(s) 2

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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(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) Z 2(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: S5 +4S

4 +3S

3 +3S

2 +3S

1 +2

S6 +4S

4 +2S

3 +3S

2 +3K

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

S5 +4S

4+3S

3 +3S

2 +3S

Example:S

6 +4S

4+2S

3 +3S

2 +3K

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 onlycapacitive 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(s

2+1)

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

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

2+1)/s

2s2 +5A = .s

s(s +1) s =0

2s2 +5 (s

2 +1)B = .

s(s2 +1) s s

2

2(1) +5= = 3(1)

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

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

(s+1)(s+4) s(s +4)+2(s +4)(ii) z(s) =

=

(s+1)(s+3)s(s +3)+1(s +3)



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=s 2+6s +8s

2+4s +3

2s+5=1 +

(s+1)(s+3)

=z1(s)+z 2(s)

2s+5 AWhere z2(s) = =(s+1)(s+3) s +1

3 / 2 1 / 2Therefore, z(s) = 1 + +s +1 s +3

It is not PRF.

8s3+4s

2 +3s+1(iii) z(s) =

8s3 +3s

s2+2s+8(iv) Y(s) =s(s +4)

+ B = 3 / 2 + 1 / 2

S +3 s +1 s +3

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

a s n +a sn 1+a s n2 +.............+as +a n n1 n2 1 0F (s)=

b sm +b sm1 +b s

m2 +............+b s +b

m m 1 m 2 1 0

=P(s)Q(s)

=N (s)D(s)

=Z (s)P(s)

Design of LC Ckt . (Loss less ckt):Consider a impedance function as

Z(s) =E

n

(s)

+

On

(s)

Em(s)+Om(s)

Where En(s) and Om(s) denote the even parts of numerator and denominator respectively

and On(s) and On(s) denote odd part.

s5 +s4 +s

3+s +1 N (s)Z(s) = =

s6+s5

+s4 +s

3+s

2 +s +1 Q(s)

(s

4

+

s

2

+

1)

(s5 +s

3 +

5)+En(s) On(s)=

(s6 +s

4 +s

2+1) (s

5 +s

3 +1)

+

Em(s) Om(s)

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

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

En(s) +On(s) Em(s) Om(s)Now, z(s) =

Em(s)+Om(s) Em(s)Om(s)

=En(s)Em(s)+On(s)Em(s)En(s)Om(s)On(s)Om(s)Em

2(s)Om

2(s)



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=En(s)Em(s)+On(s).Om(s) +On(s)Em(s)En(s)Om(s)

Em2(s)+Om

2(s) Em

2(s)Om

2(s)

En(s)Em(s)On (s).Om(s)= Re[ z(s)] = (ii)Em

Therefore from equation (i) and (ii).

En(s)Em(s)On(s).Om(s)=0Em

2(s)Om

2(s)

En(s)Em(s)On(s).Om(s)=

0En(s)Em(s)=On(s).Om(s)E

m

(s)=

On

(s).(iii)

Om(s) Em(s)

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

Properties of LC Ckt:

ansn +an2s

n2 +an 2sn4 +.............+a 01. F (s)=

bms m +bm2s m2 +bm4s m4 + +b0

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

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 deno minator must differ by the order of

2 all the way through . Example:s4

+17s

2+

165s

0s

3 +4s

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)(s

2+5)/(s

2+2)(s

2+10) k>0 It is

not LC ckt function because,1. There is neither pole or zero at the origin though the pole zero are alternatively placed on

the imaginary axis.

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

(ii) Z(s) = z(s2+1)(s

2+9)/s(s

2+4)

(iii) Z(s) = k s(s2+4)/(s

2+1)(s

2+3) , k> 0

(iv) Z(s) = s5+4s

3+5/(4s

4+s

2)

Date: 2065/5/12

Design of LC ckt by Fosters Method:

In this case ,

k0 2kisF(s) = + +............ +ks.(i)S s

2 +wi 2

This equation may represent z(s) or Y(s)Case I : ( i.e when F(s) = z(s))

22


2(s) Om (s)


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Then,

k0 2kisZ(s) = + +............ +ksS s

2 +wi 2

Here,- ko/s will represent a capacitive reactance of 1/koF.

- 2ki(s)/(s2+w

2) will represent LC parallel combination.

Having capacitor of value 1/2kiF and inductor of value 2ki/wi2. Thus the final circuit will

be:2ki/wi

2

1/k0k

.......

1/2kiz(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

k0 2kisY(s) = + +............ +ksS s

2 +wi 2

Here,

- KO/s represents admittance of inductor having value of 1/koH.- K s represent admittance of capacitor having value K F.- 2ki(s)/s2+w2 represents admittance of series LC combination having inductor of value

1/2kiH and capacitor value wi2/2ki

The ckt can be realize as :

1/2ki

1/k k

wi2/2ki

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

method forLC ckt.

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

s(s2 +4)

F (s)=2(s

2+1)(s

2 +9)

Solution:

It is Fosters series n/w



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s(s2 +4)

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

2 +1)(s2 +9)

s(s2 +4) As BsNow, z(s)= =

+

2(s2 +1)(s

2 +9) s2+1 s2+9

s(s2

+4)

(s2 +9)Where, A = .

2(s2 +1)(s

2+9) s s

2= 1

1 +4= =3 / 162(1 +9)

Therefore, A = 3/16

s(s2 +4) (s

2+9)And B = .

2(s2 +1)(s

2 +9) s s

2 = 9

9 +4 5 5= =

=

2(9 +1) 2 8 16

Therefore, B = 5/16

(3 / 16)s (5 / 6)sz(s)= + =z1(s)+z 2(s)

s2 +1 s 2 +9

The ckt will be as follows.

L 1= 3/16 H L 2 = 5/144 H

z(s)C1 =16/3 F

C2 = 16/5 F

The first part of z(s) ( i.e z1(s) ) represents parallel LC combination having inductor L1ofvalue 3/16 H and capacitor of value 16/3 F.

The 2nd

part of z(s) (i.e z2(s) ) represents parallel LC combination having inductor L2of

value 5/144 H and capacitor C2of value 16/5 F.

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

s(s2 +4)

F (s)=2(s

2+1)(s

2 +9)

Solution:It is Fosters parallel n/w

s(s2 +4)

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

2 +1)(s

2 +9)

s(s2 +4) As BsNow, z(s)= =

+2(s

2+1)(s

2 +9) s

2+ 1 s

2+9

s(s2 +4) (s

2 +9)Where, A = .

2(s2 +1)(s

2 +9) s s2= 1

24



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1 +4= =3 / 16

2(1 +9)

Therefore, A = 3/16

s(s2 +4) (s

2 +9)And B = .

2(s2

+1)(s2+9) s s

2 = 9

9 +4 5 5=

= =

2(9 +1) 2 8 16

Therefore, B = 5/16(3 / 16)s (5 / 6)s Y (s)= +

=Y (s)+Y (s)1 2

s2 +1 s

2 +9

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.

=2(s

2 +1)(s

2 +9)

Example 03: Design Foster parallel n/w for the functionF(s)s(s

2 +4)

Solution:It is Foster Parallel ,

= =2(s2 +1)(s

2 +9)

F (s) Y (s)s(s

2+4)

=2s4+20s

2+18

s3 +4s

S3+4s )2s

4+20s

2+18( 2s

2s4+8s

2

12s2+18

12s2+18Therefore, Y(s) = 2s +

s3 +4s

12s2+18= 2s +s(s

2+4)

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

12s2 +18 A Bs 9 / 2 (15 / 2)sNow Y2(s) = 2s + = + = +

s(s2 +4) s s

2 +4 s s

2 +4

9 / 2 (15 / 2)sY(s) = 2s + +

=Y (s)+Y (s) +Y(s)1 2 3

s s2 +4

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



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Y2(s) =9 / 2

, So, L1= 2/9 Hs

(15 / 2).sAnd Y

3

(s) = s2 +4

L2= 2/15 H

C2= 8/15 FTherefore, The final ckt will be

L 2 =2/15 H

L 1= 2/9HC1 = 2 F

C2 = 8/15 F

Fig. Fosters parallel n/w of LC ckt.

Assignment:

2(s

2

+

1)(s

2

+

9)1. z(s) =s(s

2 +1)

2(s2 +2)(s

2+4)

2. Y(s) =(s

2 +3)(s

2+1)

Date: 2065/5/17

Continued Fraction method or cauer method for LCCkt 1. case- I

It is removed by successive removal of pole at . The ckt will be as follows:L1 L2....

C1 C2 Cn

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

L1 L2

....

C1 C2 C3 CnV(s)

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

Example 01: Synthesis the following function in cauer form.



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Z(s) =2s5 +12s

3 +16s

s4 +4s

2 +3

Solution:In cauer n/w we proceed as follows:

S4+4s

2+3) 2s

5+ 12s

3+ 16s (2s z1(s)

2s5+8s

3+6s

4s3+10s) s

4+ 4s

2+3 (s/4 Y2(s)

S4+ 10s2/4

3S2/2+3) 4S

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

4S3+8S

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

3s2/2

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

2H 8/3 2/3

1/4 3/4Z(s)

Fig. Cauer n/w for LC series ckt

Example: 02: Y(s) =2s

5+

12s

3

+16s

s4 +4s

2+3

Y(s) 2F 8/3 F 2/3 F

Fig: Cauer n/w for LC parallel ckt.

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

s(s2 +2)(s

2 +4) s(s

2+2)(s

2 +4)

(i) Y(s) = (ii) Z(s) =(s

2 +1)(s

2 +3) (s

4 +1)(s

2 +3)

Cauer II:This is the case of removal of pole at origin.

C1 C2....

L1 L2 Ln

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



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

4 +4s

2 +3

Z(s) =2s

5 +12s3 +16s

Solution:

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

3 +4s2 +s

4

Z(s) =16s+12s

3 +2s

5

16s+12s3

+2s5

) 3+4s2

+ s4

(3/16s z1(s)3+9s

2/4+ 3s

4/8

7s2/4+5s

4/8)16s+12s

3+2s

5(64/7s Y2(s)

16s+40s3/7

44s3/7+2s

5) 7s

2/4+5s

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

7s2/4+44s

4/88

3s4/44) 44s

3/7 +2s

5((44)

2/21s Y4(s)

44s3/7

2s5) 3s

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

3s2/44

16/3 176/40 88/3

Z(s)7/64

21/44.44

Fig. Cauer II n/w for LC ckt

s4 +4s

2+3

Example:02: Y(s) =2s

5 +12s

3+16s

7/64 21/1936

Y(s) 176/49 88/316/3

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

R-C one port n/w: (R-C impedance /R-L admittance)1. Foster 1

stmethod:

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



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1/k0 k

.......

1/k1z(s)

Foster method defines F(s) as

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

Here,

- ko/s represent capacitive reactance having capacitor of value 1/koF.- krepresent resistor of value k.- ki/(s+ i) represents RC parallel in which the resister has a value of ki/ iand 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 rea l 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, Ris present

= 0, Ris 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:01 State giving reasons which of the following if not RC impedance.(s+1)(s+4)(s+9)(a) Z(s) =s(s +2)(s +5)

(s+1)(s+8)(b) Z(s) =(s+2)(s+4)

(s+2)(s+4)(c) Z(s) =(s+1)

(s+1)(s+2)(d) Z(s) =s(s +3)

Example:02: Synthesis the following function in Foster series form: F(s) =6(s

+

2)(s

+

4)

s(s +3)Solution:

Since it is foster series function z(s) =6(s+2)(s+4)

s(s +3)

This is the RC impedance n/w.

Now,

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



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

6(s+2)(s+4)Ko= .s = (6.2.4)/3 = 16s(s +3) s =0

K2= 2

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

The component values are as follows:

16/s 1/cos c0= 1/16 F

R R = 62/(s+3)

R1= 2/3 and C1= F

The ckt will be:2/31/16 6

.......

1/2z(s)

Date: 2065/5/19

F(s) =6(s+2)(s+4)s(s +3)

F(s) = z(s) =6(s+2)(s+4)s(s +3)= 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

Lo

L1 L2

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

method in this case

- ko/s represents inductor of value 1/ko

- krepresents inductor of value 1/ko

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

value i/k .



31/126


Properties:Same as RC- impedance.

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

F(s) =6(s+2)(s+4)s(s +3)

Solution:

Since it is Foster parallel,F(s) = Y(s) =6(s+2)(s+4)

s(s +3)

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

The ckt will be:

2/3

1/161/6

1/2

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) =6(s+2)(s+4)s(s +3)

Solution:

F(s) = z(s) =6(s+2)(s+4) =6s2 +36s+48

s(s +3) s2 +3s

Now,

S2+3s)6S

2+36s+48(6 Z1(s)

S2+18s

18s+48) s

2

+ 3s (s/18 Y2(s)S

2+8s/3

The ckt will be:

s/3) 18s+ 4s (54 Z3(s)18s

48) s/3 (s/3.48 Y4(s)

s/3

6 54

1/18 1/144

Fig. Caure 1 n/w31



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Cauer 2 n/w:

Example: 02: Realise the given function in cauer 2 n/w F(s) =6(s

+

2)(s

+

4)

s(s +3)

Solution:In this case,

F(s) = Y(s) = 6(s+2)(s+4)

s(s +3)In this case circuit will be :

1/18 1/144

1/541/6

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) + ..+ k s

k1 k2ko k

....k1/1 k2/2

z(s)

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 ki and inductor of value

ki/ i.

This method of synthesis is know as foster series (1st

) 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.



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2. Foster parallel method:In this case,

F(s) = Y(s) = ko+k1s/(s+ i) + k2s/(s+ 2) + .+ k

The ckt will be as follows:

1/k 1/k2 k

Y(s) 1/k 0 1 k2/2

k1/1

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) =4(s

+

1)(s

+

3)

. Realise the above function in (a) Foster series(s+2)(s+6)

(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) =4(s

+

1)(s

+

3)

(s+2)(s+6)

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

Z(0) = (413)/(26 ) = 1 = k o. i.e Rois present .Z() = 4 = Ri , Lis missing.

4(s+1)(s+3)z (s) /s =(s+2)(s+6)

4(s+1)(s+3)K

1 =

.(s+2)s(s +2)(s +6)

=4(2 +1)(2 +3)2(2 +6)

1 k1 k2=

+ +

s s +2 s +6

s = 2

=

4(s+1)(s+3)K2 = .(s+6)s(s +2)(s +6) s = 6

=4(6 +1)(6 +3)6(6 +2)

K2= 5/2

z(s)/s =1 +(1 / 2).s+(5 / 2).s

ss +2 s +6



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Downloaded from www.bhawesh.com.np1/2

5/21

1/4 5/12

z(s)

Fig. Foster series n/w

(b)Foster parallel:In this case,

F(s) = Y(s) =4(s+1)(s+3)(s+2)(s+6)

Which yields R-C admittance n/w.

(1 / 2).s

(5 /

2).sY(s) = 1 + +s +2 s +6

2 2 / 5Y(s) 1

1 1 2 / 5

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

4s2+16s+12

(a) caure 1 n/w. (b) cauer 2 n/w. s 2+8s+12Solution:

(a) cauer 1 n/w:

In this case

4(s+1)(s+3) 4s2+16s+12F(s) = z(s) = =

(s+2)(s+6) s 2+8s+12

S2+8s+12 ) 4s

2+ 16s+12 ( 4

4s2+ 32s+4s

-ve

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

Z(s) =12 +16s+4s2

12 +8s+s2

34



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12+8s +s2) 12+16s+4s

2( 1

Z1(s) 12+ 8s+s2

8s +3s2) 12+8s+s

2 ( 3/2s Y2(s)

12+9s/2

7s/2 +s

2

) 8s+3s

2

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

2/7

5s2/7 ) 7s/2+s

2( 49/10s Y4(s)

7s/2

s2) 5s

2/7 ( 5/7 Z5(s)

5s2/7

1 16/7

2/310/49 5/7

Fig. cauer 1 n/w

(b) Cauer 2 n/w:

In this case,

F(s) = Y(s) =4s2 +16s+12 =12 +16s+4s

2

s2 +8s +12 12 +8s+s

2

1/2ki1/k k

wi2/2ki

Fig. Cauer 2 n/w

Assignment: 03

(s+2)(s+4)1. F(s) = Find the n/w of the form (a) Foster series (b) Foster parallel.(s+1)(s+3)

(s+1)(s+3)2. Realize the n/w function F(s) = (a) 1st Foster method. (b) 2nd

foster

(s+2)(s+4)

method.

(s+2)(s+4)3. Realise the n/w function Y(s) = as a cauer n/w.(s+1)(s+3)

(s+1)(s+3)4. z(s) = Realise the function in foster and cauer n/w.(s+2)(s+2)



36/126


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

+

2)(s

+

4)

(s+1)(s+6)

Two port n/w:1. Z-Parameter

2. Y Parameter3. ABCD Parameter4. Transformation of one parameter to other

5. T and n/w6. Interconnection of two port n/wa. Cascade b. series c. parallel.

Date: 2065/5/24

Chapter: 4

Low pass Filter Approximations:T(jw) T(jw)

11

PB SB

w Wwo=1 Wp 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, s

3. Low transition band ratio, ws/wp4. Simple network.The approximation Method are:

1. Butterworth2. Chebyshev3. Inverse chebyshev4. Ellipse or Cauer5. Bessel Thomson

1. Butterworth low pass approximation: Generally signal become contaminated withhigh

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 T(jw) 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 sharproll off (n-pole roll off). Where n will be l arge 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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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 calle dconjugate of T(jw)

Thus (i) and (ii) givesT(jw) T*(jw) = T(jw)

2= Re[T(jw)]

2+ jIm[T(jw)]

2

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

The function T(s)2 (or T(jw)

2 ) 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

T (s)2 = T(s) . T(-s)

=(2+s)/(s3

+2s2

+2s+3) (2-s)/(-s3

+2s2

2s+3)=..

The magnitude square function is an even function which can be represented by using anumerator and denominator polynomial that are both even, i.e

A(w2)

T (jw)2

=B(w

2)

2 A +A w2 +A w

4 +........... +A w2n

0 2 4 2nT (jw) =

B0+B2w2+B4w

4+........... +B2nw

2n

2 A0

T (jw) =B0+B2w2+B4w

4 +........... +B2nw2n

Here A2= A4= A 2n = 0 (assumption).The choice has been made as per our inspection on the roll off that was directly dependent onthe 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 beknow as All pole function.

Special case:We assume ,

B2= B4 = 0

B2n= (1/w0)2n. B0 and A0= B0Now , putting these assumption in the equation (i) we get,

2 AoT (jw) =

B0 +B2nw2n

=B

o

1 2nB0 + B0

w0

37



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1=

1 2n 2n1 + w

w0

1T (

jw)2 = .(ii)

w 2n1 +w

0

In generalize condition,wo= 1

2 1T (

jw) = .(iii)1 +(w)2n

1T (

jw) 2 = (iv)1 +(w)2n

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:

1T (jw) 2 =

1 +(w)2n= (1+w

2n)-1/2

2n 2 2n 2

= 1+ . w + (1/2) . (w ) /2! - ..

1 . w2n

In Taylor series,1

T (jw) =1 w2n ..(v)2

Again we know ,

1 T (jw)2=

1 +(w)2n

Putting jw = s

1 1 1 1T (s)2= =

= =

2n s2n 1 +s

2

n

1 +(1)

n

s

2n

s 1 +1 + 2n n

j j (1)1

T (s) 2 = (vi)1 +(1)

ns

2n

Which gives the butterworth response in s-domain

Evaluation of T(s) for BU Response:(i) For n = 1 equation (vi) becomes



39/126

1T (s) 2 =1 s

2

s2=1

s = 1T (s)

2= 1/(1-s)(1+s)

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

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


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

Date: 2065/5/29

Butterworth transfer function (continued )

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

1T (s) 2 =

1 +(1)2s

4

jw

45135

315225

1

=1+s4To get the poles ,1+s

4= 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 = -0.0707 j0.707 = s1, s21

T(s) =(ss1)(ss2

)

1=

(s+0.707 j0.707)(s+0.707 +j0.707)

1=

s2 + 2s+1

(iii) For n = 3

39



40/126


1T (s) =

1 +(1)3s

6

1=

1 s6

To get the pole

1-s6= 0

S6= 1

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

S = 1 0, 60, 120, 180, 240, 300

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

= 1 120, 180, 240

Or, = 1 120, 1 180, 1 240S1= -0.5 + j0.866

S2= -1+j0

S3= -0.5 - 0.866j

1 T (s) =(ss1)(ss2)(ss3

)

1=

(s+1)(s0.5 0.866j)(s+0.50.866j)

1=

(s+1)(s2+s +1)

jw

120 60

180

240320

Order and cutoff frequency for Butterworth:

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

1T (s)

2=

w2n

1 +w

oAlso the attenuation formula is given by ;= -20log

40


T (s)


41/126



42/126

1= -20log10 2 n

w1+

wo

12 n

2w= -20log10 1 +w

o

w 2n= 10log10 1 + .(i)wo

w 2n /10 = log101 +w

o2n

10/10 =1+wwo2 n

w = 10/10-1

wo

w /10 1/2n= (10 -1)w

o

ww =

1

(10/ 10 1)2n

Now at w = wp, = maxwpwo= .(ii)1

(10max/ 10 1)2 n

and at w = ws , = minw

swo= ..(iii)1

(10min/ 10 1)2nequating (i) and (ii) can be equated as:

wp ws1 = 1

(10max/ 101) 2 n (10min/ 10 1)2n1

wp (10max/ 10 1)2n= 1

wo

(10min/ 101)

2n

w p 2n (10max/ 10 1)= min/ 10

wo (10 1)

Taking log on both sides,

wp 2n (10max/ 101)20 log =log min/ 10

wo (10 1)

(10max/ 10 1) w pn = log / 2 log

min/ 10

(10 1)w

o



43/126


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 followingspecifications of LPF.

max= 0.5 dB

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

Determine the order and cut off frequency for thefilter. Solution:

n = 4.83 5wo= 1234.12 rad/sec

Note: Always choose higher value of n ( i.e the o rder of filter )because it provides largerroll off which decreases attenuation.

Date: 2065/6/2

2. Chebyshev Approximation Method For LPF :

T(jw)

1C-R

WWo

T(jw)

1 BU -R

WWo

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

The generalize low pass filter can be represented by

1Tn(jw) 2 = .(i)1+[F(w)]

2

nFor 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 forwhich the concept of Lissagious figure is required.

Lissagious figure:



44/126


Horizontal plate

Line of axis

sinVertical plate

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

yn=2

n=3

n=4

x

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 p roves the multiple frequencies.From (ii),KT = cos

-1x

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-1

xx = cos

Then,y = cos4

x 4 y

0 1 0 122.5 0.924 90 045 0.707 180 -1

67.5 0.383 270 0

90 0 360 1



45/126


46/126


Properties of magnitude response for Chebyshev:We know that,

Tn(jw)2 = 1

1+ 2cn

2(w)

1Tn(jw) =

1+ 2 cn

2(w)

Where, cn(w) = cosn cos-1w w 1= coshn cosh

-1w w 1 and 1

1. At w = 0,

Cn(0) = cosn /2 ; 0,1,2.

Tn(jw) = 1 for n = odd

1= for n = even

1+ 2

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

1Tn ( jw) =

1+ 2

1 1

w ww=1 w=1

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 Tn(jw) dB

But,

1

1 1 2

T (jw) =

=n 2 2

1+ 2 cn

2(w) 1+ cn (w)

1

1 2

= - 20log2 2

1+ cn (w)

1

= -10log 1+ 2 cn2(w)

= 10 log 1+ 2 cn2(w) (vii)

= 10log 1+ 2(cos ncos 1w) 2 w 1for w > 1,

= 10 log

Now ,

1+ 2 (cosh ncosh

1w)

2.............(ix)



47/126


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 = 10max/10

1

=(10max/ 10

1)2 .(xi)

Date: 2065/6/7

Here

we know that

w = wnp, then, 2 cn

2(w) =1

1c (w ) = =cosh(ncosh1

w ) [since w >1]n np np np1Cosh

-1(ncosh

-1whp) =

1

Cosh-1

(ncosh-1

whp) =

1 Cosh-1

whp = 1/n. cosh-1

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

-1( ))(xii)

Wnp= cosh [1/n. cosh-1

({10max/10

-1}1/2

)]

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

2cn

2(ws)

2 cn

2(ws) = 10

min/10

1

2(cosh ncos

-1ws)

2 = 10

min/10

-1

Or, ( cosh ncosh-1

ws)2= (10

min/10

-1)/ (10max/10

-1) n

cosh-1

ws= cosh-1

[(10min/10

-1)/ (10max/10

-1)]1/2

n = {cosh-1

[(10 min/10

-1)/ (10 max/10

-1)]1/2

}/cosh-1

ws ..(xiii)

Example: Given wp= 1 , ws= 2.33 ,max= 0.5dB ,min= 22 dB. Calculate n for

Butterworth 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 5n for BU = 5

For Chebyshev the order is given by ,

n= cosh-1

[(10min/10

-1)/(10max/10

-1)]/cosh-1

(2.33)= 2.89 3

46



48/126


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 choosechebyshev filter.

Chebyshev poles location and network function:

We know 1T (jw)2=

..(i)1+

2 cn2 (w)

Substituting s = jw equation (i) becomes,

1T (s)2= ..(ii)

1+ 2 cn

2(s/j)

To determine the poles,

1+ 2 cn

2(s/j) =0

1c

n(s/j)= j (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 j

1

Thus, comparing , weget, Cosnu . cosh nv = 0 -

sinnu. Sinhnv = 0

[ from equ. (iii)]

[ cosjnv = coshnv[ sinjnv = jsinhv]

The minimum value ofCoshnv = 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

1

Or sinhnvk

=

Nvk= sinh-1

( 1

)

Vk =1/n. sinh-1

( 1

)47



49/126


Again, we know that

Cos-1

(s/j) = x = u +jvs/j = cosx = cos(u+jv)in general,

sk= jcos(uk+jv)

=j[cosuk.cosjv sinu k. sinjv]

=j[cosuk. coshv jsinu k. sinhv ]

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

Sk= sin[(2k +1) /2n] sinhv + jcos[(2k+1)/2n]coshv Or , 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 onthe ellipse.

Date: 2065/6/9

Example:01 Obtained the 4th

order network function of a low pass chebyshev filter withmax= 0.75 dBSolution: n = 4 max= 0.75 dBNow = ( 10

max/10-1)

1/2whp= cosh (1/n. cosh

-1

(1/)) = (100.75/10

-1)1/2

= 0.434

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

(1/)) =

Pole location is given by48



50/126


Sk= sinuksinhv + jcosuk coshv

Where, 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

S7 S0

S6

S1

S5S2

S4S3

Home Assignment:

Example:02: Determine the network function for 3rd

order chebyshev LPF withmax=0.75dB ( =p; pass band attenuation)

Date: 2065/6/14

Inverse chebyshev low pass approximation:

49



51/126


T (jw) T (jw)

Ideal LPF

T (jw)

W W

BU-ResponseT (jw)

WChebyshev-respone

Winverse-Chebyshev-response

21- T (jw)

W

Fig: intermediate stage to obtain inverse chebyshev response.

2Tic(jw)

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 by

1T (jw)2

=

1+ 2cn2(w)

2 11-T (jw) = 1-c

1+ 2 cn

2(w)

2 cn

2(w)

=1+

2 cn

2(w)

Now replace w by 1/w



52/126


2 2 cn

2(1/ w)T

IC(jw) = (i)1+

2 cn

2(1 / w)

Where,

TIC(jw) 2 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 nThus equation (i) becomes

2 .1

TIC(j.1)

2=

1+ 2

2 .1T

IC(j.1) = . (ii)

1+ 2

We know that ,

min= -20log TIC(j.1) dB (iii)

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

= min= -20logTIC(j.1) dB

2 1 / 2= - 20log 2

1+

21+ = 10log 2

1min = 10 log [ 1+ }

2

1Or , 10

min/10-1 =

21

=(10min/ 10 10)2 . (iv)Again in general, the attenuation formula can be written as:

2 cn

2(1 / w)

= -10log

1+ 2 c 2 (1 / w)n1

= 10 log 1 +

2 c 2 (1 / w)

n

Now at w = wp =

max

Then above equation becomes

1= max= 10 log 1+

2 c 2 (1 / w )

n p

1

(10max/10 1) = 2 cn2(1 / wp)

1 1c2(1 / w ) =

.

n p

2 (10max/ 10 1)



53/126


cn2(1 /w p) =

cn(1 /wp) =

cn(1 /wp) =

(10min/ 101)

(10max/ 101)

(10min/ 101).(v)

(10max/ 101)coshn cosh

-1(1/wp) . (vi)

[ wp< 1, 1/wp> 1]

Thus equating equation (v) and (vii) 1

(10min/ 10 1) 2Coshn cosh

-1(1/w ) =

p

(10max/ 10 1)1

1 (10min/ 10 1) 2cosh

(10max/ 10 1)n = (vii)cosh

1(1 / wp)

Which gives the required order for the inverse chebyshev filter.

Now , for half power frequency i.e at w = wp

TIC(j.1) = 1/2TIC(j.1)

2 =

Which means,

2 cn

2(1 /wP)= 1

1cn

2(1 / wnp) =

2

1cn(1 / w) =

Coshn.cosh-1

(1/wnp) =cosh-1

1

n coshn.cosh-1

( (1 /wnp)= cosh-1

( 1

)

cosh-1

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

( 1

)

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

( 1

)]

1wnp = < 1 .(viii)

1 1 1 cosh cosh (

)n

Which gives the desire half power frequency.

Example: 01

Given, max= 0.5 dBmin= 22 dB

wp= 0.9n = ?

wnp= ?



54/126


Assignment:

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

Pole zero location for inverse chebyshev:We know that ,

2 2

2 cn (1 / w)

TIC(jw) =1+

2

cn2

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

Where, z(s) z(-s) |s = jw=2 cn

2(1 / w)

P(s) P(-s)|s = jw= 1 + 2 cn

2(1 / w)

For zero location:

2 cn

2(1 / wk)

0

cn2

(1 /wk )

= 0 cn(1 /wk)=0Cosn cos

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

odd) ncos-1

(1/wk) = k/21/wk= cos(k/2n) which gives the zero for inverse chebyshev.Wk= sec(k/2n)

For poles:

1+ 2 cn

2(1 /wk)=0

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. Zero locationFig. Pole location

Example:01Given,

min= 18 dBmax= 0.25 dBws= 1.4 rad/sec

wp= 1 rad/secFind out the pole and zero for inverse chbyshev response.



55/126


Chapter 5

Frequency transformation:

Frequency transformation is important because the prototype LPF with any type ofapproximation can be converted into high pass band pass , band stops filters within the samecharacteristics easily.

T(jw)

10.707

Wc

The effect of frequency transformation are:

1. Magnitude response 2. Network function3. Location of poles and zeroes. 4. Network elements.

Types of transformation:

1. LP to LP transformation

Transformation

WW0 0

New LPFOld LPF

Replace s by wo/o.s

i.ew0= 1 ( in normalized case)

s s/0T

LP(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

Putting ss

0

54



56/126


sL

oldXL = L old = s =Lnew.S0 0L

new

= Lold

/0

3. For capacitor:

Xc= 1/cs

sPutting s0

1 1 1Xc = = =

sC

old C

new

.sC

old0 0 .s

Cnew = Cold/s

2 LP to HP Transformation:

Transformation

WW0 0

LPF with W0 HPF with 0

0In this case we replace sw0.s

0

Or , s [Since w0= 1]s

0THP(s) = TLP(s) 0=

TLP(

S )s = sExample if TLP(s) = 1/(s+1)

1 s

Then, THP(s) =0 = 0 +

s+1s

(1)For resistor:

No change

(2) For inductor:

XL= LS

Putting s

0

s



57/126


XL= L.

0

s

1

= 1.s

L01

Comparing with 1/CS1

.s

L01

C =

L0

(3) For capacitor:

Xc= 1/cs

Putting s

0

s

1 s1XL = = = .S =LS

0 c0 C0.c

s

1Comparing .S with LS

C 01

L =

C0

Date: 2056/6/15

(3) LP to BP Transformation:

T (jw)LP

Transmission

p WW

Ws

TBP(j)

L U

In this case,

sw.s2+

2

0u L

Here, u L=B And w0= 1

s.s2 +

2Bs

Where 02= L. u



58/126


(1) For resistor- no change

(2) For inductor:

XL= LSThe new value of inductive reactance is given by:

s2

+ 0 2XL = L. Bs

L L02 L 1XL = .s + = .s+

B Bs B B

L0 2 .s

The new component are inductor and capacitor in series.

L B

B Lo2

(3) For capacitor:The new capacitive reactance form LP to BP is given by :

1 1 1 1

= = = =c 1s 2+ 02 cs

2 +c0

2 c c 02

c. Bs Bs Bs + Bs Bs + B

c02 .s

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

B C

Co2 B

LP to BS Transformation:

T (jw)LP

Transmission

p WW Ws

TBS(j)

L U

57



59/126


BsIn this case s is replaced by .w0

s2 + 0 2

BsBut w0= 1, s

s2+ 0 2

(1) For resistor :

Resistor value remain same.

(2) For inductor:

XL= LSBs 1 1 1

XL=L. = . = = 1 1s2 + 0

2 s

2 + 0

2 s

2 0

2

+ s +LBS LBS LBs LB LB

02 .s

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

LB 1

o2 LB

(3) For capacitor:

Xc= 1/cs1 s

2 + 2 s2 2 1 1

0 0Xc= = = + = .s+Bs CBS

CBs CBs CB CB

c.s2 + 0

2 0

2.s

CB 1

o2 CB

1Example:01: If T(s) = , then change the above function from LP to BP. Given , L=s +1

10 and u = 20.Solution:

1Then, TLP(s) = , L= 10 , u = 20

s +1

We know ,

02= L. u= 10. 20 =

200 For Lpto BP we replace

s2 + 02 s

2 +2000 s2 +200s = =

B (20 10)s 10s

58



60/126


Thus,

1 10sTLP s

2 +200=TBP(s)= 2 = 2

s +10s+200s = s +200

10s +110s

10(s)TBP(s) =

s2 +10s+200

Example:02: Obtain the transfer function of the 4thorder Butter worth HPF with0= 210

4rad/sec.

1TLP(s) =

s4 +2.61313s

3 +3.41921s

2 +2.61313s+1

We know that ,

0s s

1

=4

3

2

0 0 0

0+ 2.61313 +3.41921 +2.61313 +1

s

s

s

s

Example:03:The filter shown in the figure below is a 4th

order 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.

+

A C+

V1 -}B }D }E V2

-

Solution:

For LP to BP conversion , we replaces

2 + 0 2

s Bs

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

Now for section A:L = 1.2817

Which changes to series LC component as shown below:L BB Lo2

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

and the new value of capacitor is = B/L 20 = 150/(1.2817400

2) = 731.45 F.

For section B:

59



61/126


C= 1. 9093

Which changes for LP to BP As:

B CCo

2 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 L

Lo2 B

For section D:

B C

Co2 B

For section E:R = 1 R = 1

+ 8. 54 mH 9. 41 mH731. 45uF 663. 66uF

+ 12.72 uF 1 V 491uH 893. 71mH

- 6. 99 mF

-

Date: 2065/6/16



62/126


Doubly Terminated LC-Ladder ckt:

I 1R 1

1 Loss less

I 2+

+V s zi n Ladder

- V 1R 2

V 2

-

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= zinWhich 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

(s).(s)=(R

1

zin

)2

(R1+zin)2

(R1zin)(s) = .....................(vii)(R1+zin)

From equation (vii) , we get1(s)z

in=R1. 1st z

in ..(viii)1+(s)1+(s) nd

zin=R1.1(s) -----------2

zin

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)61



63/126


For butterworth response:

2 2 1 N (s)N (s)T (jw =H (jw) = = [since w0= 1]

1 +w2n D(s)D(s)

1 w2n(s) 2 =1 H (s) 2 =1

H (jw)2 =1 =

1 +w2

n

1+w2

n

w2 n w2n sn.(s)

n(s).(s) = = = (ix)

1 +w2n

D(s).D(s) D(s).D(s)

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

s

1 (s) 1 s +1 s +1sz

in1

=R1

. =1. =. 1 +(s) s s +1+s1

+s +1

1Zin1= . .............(a)2s+1

Zin2= 2s+1 .(b)Zin2= 2s+ 1 = Ls + Ri.e L = 2, and R = 1

The ckt will be 2R 1

1

+ R 2 1 vs

-

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

1

+ 1v2vs-

2

-

For n = 2

D(s) = s2+2s + 1

sn sn(s) = =

D(s) s2 + 2s +1

1 (s) 1 s2 /(s

2 +2

s +1) (s2 + 2s +1s

2)z

in1 = = =

1 +(s) 1 +s2 /(s

2 + 2s+1) (s

2 + 2s+1 +s

2)



64/126

Downloaded from

www.bhawesh.com.np

( 2s +1)zin1 = ..(a)(2s

2+ 2s+1)

Similary,

2s2

+ 2s +1Z

in2

= .(b)

2s+1Taking equation (b)

2 s+1) 2s2+2 s +1 (2.s z1(s)

2s2+2 s

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

1) 1 (1 z3(s)1

The ckt will be as follows:+

1 1 . 4 1

+ 1v2vs-

1 . 4 1

-2

1 1 . 4 1

vs + 1 . 4 1 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 ,

H2(0)H (jw) 2 = = T (jw)2

1 +w2n

Generally we take,

R11 and R1R2

R1 I1

+ ZinVs-

I2

LC R2 V2

63



65/126


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

2

Vs

From which we get ,

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) theButterworth response is given by,

H

(

j

w

)2

=

H

2(0)1

+w2n

Here, n = 3, R1= 1 & R2= 22 4R2.R1 4.1.2 8

H (0) = (R2 +R1)2 =(1 +2)

2 =9

8 / 9H (jw)2 =

1+

w

2

n

The reflection coefficient function is

(jw)2 =1

H (jw) 2

8 / 9 1 +w2n

8 / 9 1 / 9 +w2n

=1 = =

1 +w2n 1 +w

2n 1 +w

2n

2

1 / 9 (s/j)23

(jw) =

1 +w23

2 1 / 9 (s)6

Or, (s) = =1 s6

(1 / 3 +s3)(s).(s) =

D(s)

=1 / 9 +(s/j)61 +w

6

(1 / 3)2 (s

3)

2

1 s6

.(1 / 3 s3)

D(s)

=(1 / 3 s)(1 / 3 +s) 1 s

6

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

64


R2T(0) =R2 +R1

Now we know

4R1 V2(s)H (s)

2 = .R2 Vs(s)

H (s) 2 = 4R1. T (s) 2R2 2

R1H (S ) =2 .T(s)

R2

R1 R1 R2H (0) =2 .T(0) =2 .

R2 R2 R1 +R2

R2.R1H (0) = 2

R1 +R2


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(s)=1 / 3 +s3

D(s)

For n = 3,

D(s) = s3+2s

2+ 2s+1 (from table)

The first impedance is ,

1 / 3 +s3

1 (s) 1 s 3+2s2 +2s+1Z

in1

=1 (s) = 1 / 3 +s3

1 +s3+2s2 +2s+1

2s2 +2s+2 / 3Z

in1

= .(a)2s

3+2s

2+2s+4 / 3

2s3+2s

2+2s+4 / 3Z

in2

= ..(b)2s

2 +2s+2 / 3

Now using continued fraction method for equation (b)

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

3+2s

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

2s3+2s

2+2/3.s

4/3.s+4/3 ) 2s2+2s+ 2/3 (3/2. s Y2(s)

2s2+2s

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

4/3.s

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

1 21+

-2/3 2

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.



67/126


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:-

Rf

R-+ Vo

Vo =

Rf

.ViR

(2) Non-investing:-

Rf

R-+ Vo

Vi

RFVo =1 + Vi

R

(3)Integration:-



68/126


c

R-

Vi + Vo

1 1 1Vo = .Vi=

ViRCS RC S

If R1=1 & C = 1, then

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

Vi S

(4) Differentiator:-

Rf

-c + Vo

Vi

Vi O Vo O=1 R

CS

Vo = (CRS )Vi

If Ro= 1& Co= 1, ThenVo

= SVi

(5) Summer:-Rf

R1V1 -

+V2 Vo

R1

Vo = RF

(V1 +V2)Ri

(6) Subtract or (Difference A/F)



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Rf

R1V1 -

+V2 VoR1

Rf

Vo =RF

(V2 V1)Ri

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

(1) Investing type:-

Z2

V1 Z1

- V2

+

From fig.

V1(S)R(S) = =

Z2

V1(S) Z1

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

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

1I.e. T(S)= =

KRCS S

1

K =RC

If R=1, then,

C=1/K

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

1/k

R =1

- V2+

Thus the design will be

(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) = Z

2

Z1



70/126


Z

2 = k( S +

a1)Z1

1

y2 y1Or =

=k (S +a )

11 y2

y1

If y 2= 1, then,

Y1= KS+Ka1

1

Y2

Y1

1/ka1

1

-+ V2

1/ka1

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

K(d) T(S) =

S +P1

Let we can write,Z 2 K=Z1 S +P1

y K1

y2 =(S +P1)y 1

1y2=(S +P1)

K

y1=1, then

S +P S P1 1

y2= = +K K K

ks

(e) T(S) =s +p

1

Z2 1= 1 P V1Z1 1+

KK

S

If Z2=1, then,

1/k

k/p1

- 2V

1 +

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

1

1/k k/p1-

+ V2



71/126


1 P1Z = +

1 K KS

(f)

Z2

Z1

y1

y2

T(S)= KS

+

q1

S +P1

s +q1

=K s +p1

s +q1

=ks +p1

Let y1= ks + kq1

Then, y2= s + p1

1

k1/p1

V1 -+ V2

1/ka1

Fig: Design for T(S) =

(k

(s

+

q

)1)

s +p1

# 2nd

approach of above problem (Do Yourself)

(2)Non-investing type:-

V1 -+

Z2

Z1

(a) T(S) =k((s

+

q1)

)

s +p1

Comparing,

z s +q2 1

1 + =k z s +p1 1

V2

1 1a -p /p1

- V2+

Where, q1>p1

1/a-p1

1



72/126


z s +q2 1

=k 1z s +p1 1

ks +kq1s p1=s +p1

z2 =s(k 1)+(kq1p1)

z1 (s +p1)

For, k = 1

z2 q1p1 k (s +q1)

z1 = s +p1 T(s) = (s +p1) for k = 1

y1 1=y2 s p1+

q1p1 q1p1

If y1= 1, then

s p1y2 =

+

q1p1 q1p1

For, k 1

z2 =s(k 1)+(kq1p1)

z1 (s +p1)1200

We assume,

Kq1= p1

K = p1/q1

s(p1q1)

z2

s (k

1) q1

z1 =(s +p1) = (s +p1)

z2

p1

q1= z1 q1

1 +p1/ s

p1q1Ifz2 =

q1

Then,z1 =1 +p1/ s

# 2nd

approachy1

=? (Do Yourself)y2

-V2V1 +

Design for T(s) =k((s

+

q1)

)

s +p1

for k1& p1>q1

Example:- 01

Realize 1st

order inverting which satisfy the following T(s)

1000T(s) =

s +1000

We know that,

z2 = 1000

z s +1001



73/126


74/126


75/126


R 1To get poles, S2+ S + =0L LC

&, for loss less ckt , ie, if R= 0,

Then, S2+ 1/Lc =0

1 1Or, s= j = jwo where,wo =

LC Lc

Poles are imaginary and conjugate,

Quality factor:-(Q)

=WoLQ

R

It is defined as the ratio of inductive reactance at frequency Woto the

resistance. Now,

1 L 1 LQ = .

=

LC RR C

Wo RAlso, = Q L

Wo 2T (s)= . (i)

WoS2 + S +Wo

2

Q

This is the standard from & the design parameter is Wo & Q.

V1 Wo1.Q V2

To get the actual poles:- S2

+

Wo.S+Wo

2

=0Q

Let, the poles be, jthen,

D(S) = (S++j)(s+j)

D(S) = S2 +2+(2+2)=0 (ii)

Comparing equation (i) and (ii)

Wo2= & 2 +

2 =Wo

2

Q

=Wo

2Q

&=Wo 1 1

4Q2

Typing Biquad current:-

74



76/126


Gw 20A typical Biquad ckt can be represented as, T(s) =

w0 2

s2 + s +w

0Q

Where, G = Gain & choice of inverting and non inverting.

In normalized case, i.e. for w0= 1

GT (s)= .. (i)

s

s

2

+ +1Q

Equation (i) can be implemented if G & Q ate given,

Let us go for inverting type of design

Gi. e. T(s)=

ss

2 + +1

Q

v2 G= v1 2 ss + +1

Q

sGv = s2+ +1 v

1 2Q

1Gv = s s 2 + + 1 v

1 2

Q

Gv1 v2v2 = 1 1

s s + s s +

Q Q

v2 v2v2 = 1 1s + s s +

Q Q

vs Gv 11

v2 = [1]. (ii)

1 1 ss + s +

QQ

The equation (ii) is cascade Realization costing of 3 steps:

Stage:-1

1

1s +

Q

Gv +

v ii (a)2 1

1s +

Q

Stage:-2

1

..ii (b) s



77/126


1

-+

Stage:-3

(-1) = ii(c)

For stage 1, we need more analysis:

Z3

V1 Z2

-

Z1 +

From figure

z3 z3v = .v .v3 1 2

z1 z2

v3 =z3 1 v2 + 1 v1 .. (iii)z2 z1

From equation ii (a)

-+

V3

1v3= [1v2 +(G)]v11

s +Q

1 v 12 v = + v . (iV)

3 1 1 1 1s +

Q G

Comparing eqn(iii) & (iV)

1

z3= 1s +Q

z2=1 (a resistor)

1z

2= (a resistor)G



78/126


1For,z3 =

s +1

Q

Or, y3 =s +1

Q

The ckt for Z3will be

1

Q

11/Q

QV1 -

+V2 V31

The overall ckt will be,

11 1

1/Q Q- 1 1

- -+ V2

1 + +

Fig: This is ring of 3 ckt and is popularly known as two Thomas Biquid.

Two Thomas Biquid:-

C1 C2R5

R3R1V1 - R4 R5

- -V 2 + 3 V2V +R2 V4 +

Fig:-General Two Thomas Biquid.

From figure,

1R1c1

sv2v3 =

1

R1+

R2

c1s

1v4= v3 (ii)

R4c2s

v2= v4 (iii)

From eqn(ii) & (iii)

1v2= v3 (iv)R4c2s

+v1

(i)

R3

Again, from eqn(i) & (iv)



79/126


R1R4c2s.v2 =

c1s v2 v1+

1R1+R

2R

3

c1s

1

v2 R3R4c1c2T (s)= = (V)v1 2 1 1s + s+

Rc

1 R2R4c1c2

But the standard form of Biquid is

v2 Gw02T (s)= = (Vi)v 2 w0 2

1s + .s+w0Q

Comparing eqn (V) & (Vi)

1w2 =

0

R2R4c1c2

1w2 = (Vii)0 R

2

R4

c1

c2

1Also, =Gw2

0R2R4c1c2

1 1or, =G.R

2

R4

c1

c2

R2

R4

c1

c2

R2or, G= .(Viii)R3

Finally,

w0 1=Q R1c1

1 1or, =

R Rc c R1c12 4 1 2

Q

R2 c1 1

Q = (ix)R2

R4

c1

c2

With,

c1 =c2 =1

& R2= R4=

1 We get,

W0= 1

G =1

R

3

R3 =1

G

Q = R1

78



80/126


R1= Q

The important property of the Biquid ckt is that it can be orthogonally turned. It means

(a) R2can be adjusted to a specified Value of w0.

(b)R1can then be adjusted to give specified of Q without changing w0, which has been already adjusted.

(3) Finally, R3can be adjusted to give the desired Value of G fir the ckt without changing w0 & Q which

has already been set.

These three steps are known as tuning algorithm.

Sallen key Biquad circuit:-

C1

R1R2Va

-+ V2

V1-

+

C2 RA RB

Fig: Sallen-key Biquad

From fig (i),

v2 RA=1 + =k . .(i)v1 RB

v2

Va = (ii)k

Applying Nodel Analysis at mode a,

Va Vb

V

a 0+ =0

(iii)R2 1 S

C2

Applying Nodal analysis at node b,

v2

vb v1 vb k vb v2+ +

=0R1 R2 1

c1s

1 1 1 v v2 1

Or,Vb + + v2c1s =0

R1 R2R

3 R2k R1

1 1 v1 1Or,Vb +

+c1

s

v2

+c1

s =0

R1 R2 R1 R2k

Similarly, rearranging eqn(iii)

v2 v2

k v3 k + =0R

2R

2 1

c2s

1 c2sv

b

+ v2 =0R

2

k k R2

79



81/126


1 c2sOr, vb +=R

2 v2 .(Vi)R

2

k k

Thus from eqn(V) & (Vi)

1 c2s 1 1 v1 1R

2 + + +c1 s v 2 v2 +c1 s =0

R2kk R1 R2

R

1 R2k

1 c s 1 1 1 v2

1

Or,R2 + + +c1s +c1 sv2 =

R2k k R1 R2 R2k R2

1k

v2 R1R2c1c2

T (s)= =

v1 2 1 1 (1 k) 1s + + + s+

R1c

1 R2c1R

2

c2 R1R2c1c2

Gw 20

T (s)= . (Viii)w 2

0 s2 + s +w 0

Q

Comparing eqn(Vii) and (Viii)

G = k

1w0 = R

1

R2

c1

c2

w0 1 1 1 k=

+ +

QR1c1 R2c1 R2c2

Design I (equal elements Values):-In this case,

R2= R1= R = 1 &

C1= C2= C = 1

For which, W0

= 1

K = 3 1

=1 +R

B

Q RANow, let us take,

RA= 1 then,

RB= 2- 1/Q

In this case the final ckt will be,

C1= 1

R1= 1 R2= 1-V2

+V1-+

C2= 1

RA=R1B= 2-1/Q

80



82/126


83/126


84/126


0.1F 0.1F

1.6k 1.6k - 1.6k -+ 1.6k V2V1

- ++

2360.1F 2k1.6k 1 1.6k

Fig: 4th

order butter worth active Salleney biquad with equal element

design forW=21000rad/ sec &C=0.1F.

Gain adjustment (EqualiZation in Sallen key:-

KWo2

T (S )= .. (i)Wo

S 2 + .S+Wo2

QIn Butterworth,

T (jo) = 1 or 0 dBBut in equation (i) T (jo) = k (k>/1) which needs to be equalized.

RaR1

- Vb- + V1 Rb+ V1

Fig:- i(a) fig:- i(b)

If H is considered to be the gain provided by fig i(b) which is such

that, H. k = 1Also it is to noted that, in Sallen key,

G = K

Also, H. G = 1

H =1/G

Vb RbNow, T (S) = =

V1 Ra +Rb

Also, we should remember that,R

a

Rb =R

Ra +Rb 1Now, solving the above equation by setting R=1, we get Ra=

1/H In term of G the Value of R a& Rbis

Ra =G

GRb =

G

1

83



85/126


In term Q R aand Rbcan be expressed as,

Ra =31

Q

G =K =3

1

Q

=3Q1Rb

2Q1

Gain Enhancement (Increment) in Sallen key:-

C1

R1 R2-

V2

+V1 -+

Ra

C2 RBRA

Rb

We have, gain,

K =3 1

=1 +R

A

Q RB

But, sometimes for given Q the Value of gain will be Very small and amplification to our need.

Although the separate ckt for gain enhancement can be used, the Sallen key ckt itself can be modified tocompensate the gain, using additional arrangement of two resistor as, shown in the fig (ii)

Let, C1= C2= C

& R1= R2= R then,

T(S) = of sallen key will be,

k

R2 C2T(S) =2 3 k 1

S + .S+RC R

2C

2

RbWhere,=Ra +Rb

Wo 3 k

=Q RC

1Q =

3 k

For a given Value of Q the gain k can be increa sed to our requirement by proportionally decreasing

the new factor.



86/126


High pass sallen key:-

In this case,

GS2

THP(S) =

WoS 2 +

.S+Wo2

Q

Applying RC-CR transformation in active low pass Sallen key biquid, in the non-inverting terminals weget the following final ckt THP(S) as,

1/

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