Unit 4 | unit 4 network functions

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Unit - 4

Definition of Network Function

Q1) For network below 1, pole - zero pattern is represented in fig 1, find numerical value of R, L and c for z (0) =1?

Fig 1 Pole zero plot

A1)

Calculating z(s) by taking L.T of fig 1)

z(s) = (R+SL)*1/CS

(R+SL)+1/CS = R+SL / SRC+S2LC+1

Z(S)= S+R/L / C [S2+R/L S+1/LC]

Given z(0)=1

z(0)=R/L / C/LC =1

Z(0)=R=1

.: R=1

Polls are given as

S2+RS/L+1/LC=0

S=-R/2L +J

Zeros are given as

S+R / L=0

From pole zero plot value of pole location is at -1 so

-R / L = -1

-1/ L=-1

L=1H

Also imaginary part of pole from plot is

1/LC-(R/2L)2 =11/4

1/C-(1/2)2=11/4

1/C-1/4=11/4

1/C=11/4+1/4

1/C=12/4 =3

C=1/3 F

Hence, value of R=1L=14 and c=1/3 f

Q2) FOR the following circuit, FIND G12(S) =v2/v1

A2) Taking Laplace transform of above we have

Fig 2 Laplace Transform of above circuit

Applying KCL at node v0

V0-V1(S) / S*1/2S/(S+1/2S) +V0+V2(S)/ (S*1/2S)/(S+1/2S)+V0/1/S=0

For Porr 2, We have,

v2-v0/(s*1/2s)/(s+1/2s)+v2/1/s =0 ---------(ii)

(v2-v0)(2s2+1) / s +sv2=0

v2 [2s2+1/s+s] = v0( 2s2+1)/ s

v0=v2(3s2+1) / (2s2+1)-----------(iii)

From (i) we have,

(v0-v1) /s (2s2+1) *sv0+(v0-v2)/5 (s2+1) =0

(2s2+1/5)v0-(2s2+1)/sv1+sv0+(2s2+1)/sv0-(s2+1)/s v2=0

(2s2+1)v1=(5s2+2)v0-(2s2+1)v2

Substitute value v0 from (iii) above

(2s2+1)v1 =[( 5s2+2)(3s2+1) /(2s2+1)-(2s2+1)]v2

(2s2+1)v1=(15s4+5s2+6s2+2-4s4+4s2) / (2s2+1 *v2

G12(s) = v2/v1 = (2s2+1) 2/11s4+15s2+1

Q3) Find the transfer function admittance ratio Y12(s) = I2(S) / V1(S)?

A3) Taking Laplace transform of above figure

Fig 3 Laplace Transform of above circuit

Applying KCL we get,

(V2-V1) / (3*1/2S) / (3+1/2S) * V2/ 1/2S+ V2/1/6 =0

(Y2-V1)(6S+1) / (6S+4) * (2S+6)V2V=0

BUT I2=-6V2

V2=-I2/6

We have ,

(6s+1) / (6s+4)*v2-(6s+1) / (6s+4)*v1 +(2s+6)v2=0

-[6s+1 / 6s+s+2s+6]I2 / 6 =V1(6S+1) / 6S+4

-[6S+1(2S+6)(6S+4)]I2= 6(6S+1)V1

-[(6S+1)*12S2+36S+8S+24] I2=6(6S+1)

I2 / V1 =-6(6S+1) / 12S2+50S+25

Y12(S) = I2(S) / V1(S) = -6(6S+1) / 12S2+50S+25

Q4) Find whether the following network function represent the driving point function.

a) f(s)= (s+1)/(s2+1) b) f(s) = 3s2+2s+1/ss3+9s2+3s+2 c) f(s)=(s2+1)2/s2(s+3)

A4)

a) f(s)=s+1/s2+1

It represents the driving point function:

b) F(S)= 3S2+2S+1 / 5S3+9S2+3S+2

It represents the driving point function.

b)F(s)=3s2+2s+1/5s3+9s2+3s+2

It represents the driving point function

c) f(s) = (s2+1)2/s2(5+3)

It has representative zeros, hence not valid

( s2+1)2=0

s2=+-1

s2=+-j,+-j

Q5) For the network shown, find driving point input impedance .plot the pole zero pattern for each as well.

Fig 4 (1) Circuit Diagram Fig 4(2) Circuit Diagram

A5) For Fig 4(1) taking L.T we have

Fig 5 Laplace for Fig 4(1) with roots

Z11= 1+ 1 /S/3+1/25+1/1/4S

=1+ 1/ 5/3+1/6S

=1+ 18S/6S2+3

Z11=6S2+3+18S/ 6S2+3 =2S2+6S+2/2S2+1

Zeros of equations are taking lt.of circuit b)

Fig 6 Laplace for Fig 4(2)

Z11= 1*4/5

1+4/5+2.8/3/2+8/5+1

=4/5+4 + 16/ 25+8+1

= 4/5+4+8/5+4+1

=s+12+4/(5+4)

Z11= S+16/(S+4)

For zeros of system

s+16=0

s=-16

For poles of system

s+4=0

s=-4

Q6) The Z21 – parameter for network shown is

A6)

The z- parameter is given as

v1 = Z11I1 + Z12I2

v2 = Z11I1 + Z22I2

So, applying KVL we get

v1=Zp(I1+I2) = ZpI1+ZpI2

v2= Zp(I2+I1) = ZpI1+ZpI2

Comparing with above equation we get

Z11= Z12= Z21= Z22=Zp

Q7) The h22 – parameter for network shown is

A7)

The h-parameter is given as

V1 = h11I1 + h12V2 (1)

I2 = h21I1 + h22V2 (2)

V1=R(I1+I2) (3)

V2=R(I1+I2) (4)

From (4) finding I2

=I2

V1=RI1+R[-I1]

V1=RI1+V2-RI1

V1=V2 (5)

Comparing (1) and (5)

h11=0

h12=1

From (2)

I2=

I2=-I1+ (6)

Comparing (6) and (2)

h21=-1

h22=

Q8) For the network below calculate y11&y22

A8)

Taking Laplace Transform of above circuit we get

For port 1

I1=s(V1-V3) (1)

For node V3

S(V3-V1)+V3+(V3-V2)2s=0

V3= + (2)

Substituting V3 from (2) in (1)

SV1-s[ + ]=I1

V1-V2=I1 (3)

From (3) and (1)

y11=

For port 2

+(V2-V3)2s=I2 (4)

Substituting V3 From (2) in (4) and solving we get

y22=

Q9) The y22,y11- parameter of the network are

A9)

Converting the upper part T-network to network we get

y11a== mho

y12a=- mho

y21a=- mho

y22a== mho

For the lower circuit from the question we can directly write

y11b=1 mho

y12b= mho

y21b=- mho

y22b=2+ = mho

The overall y-parameter of the circuit is given as

y11=y11a+y11b=+= mho

y12=y12a+y12b= =-= mho

y21=y21a+y21b=--= mho

y22=y22a+y22b= = mho

Q10) With the usual notation, a two-port series resistive network satisfies

A = D= . The Z11 at the network is?

A10)

The ABCD parameters are given as

V1 = AV2 - BI2 (1)

I1 = CV2 - DI2 (2)

The z- parameter is given as

v1 = Z11I1 + Z12I2 (3)

v2 = Z11I1 + Z22I2 (4)

Converting equation (1) in the form of (3). We substitute value of V2 from (2) in (1)

V2=+ (5)

Substituting V2 from (5) in (1) we get

V1=A[+ – BI2

V1=+- BI2 (6)

But we know A=

From (1) and (6)

Z11==4/3

Q11) For given circuit calculate h21

A11)

+ = I1 (1)

V1 - = I1

V1 = + I1

- + 3I1 = I2

3I1 + V2 = I2

From (1)

I2 = 3I1 + V2 – [ I1 + V2]

I2 = I1 + V2

h11 =

h12 =

h21 =

h22 =

Q12) The value of forward current gain Y21 from below is

A12)

Converting the T network to network we get

The y-parameter is given as

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

From port 2

+(V2-V1)=I2

Comparing with standard equation we get

y21=-3/5 mho

Q13) For the following network find which represents driving point function

A13)

It has no repeated zeros. Hence, not valid

No repeated zeros. Hence valid.

Q14) In a passive two port network, the open line impedance matrix is If the input port is interchanged with O/P port, then open circuit Impedance matrix will be?

A14)

If the ports are inter changed than we will get the inverse matrix for the above given matrix

Q15) Find Z parameter, is the network reciprocal?

A15)

The z- parameter is given as

v1 = Z11I1 + Z12I2

v2 = Z11I1 + Z22I2

V1=ZaI1+Zc(I1+I2)=(Za+Zc)I1+ZcI2

V2=ZcI1+(Zb+Zc)I2

Comparing above equations we get

Z11=Za+Zc

Z12=Zc

Z21=Zc

Z22=Zb+Zc

As Z12=Zc=Z21

Hence, its reciprocal network

Q16) Find Y21

A16)

The y-parameters are given as

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

Writing node equation for V2

I2=y2V2+y3(V2-V1)-gmV1

I2=(-y3-gm)V1+(y2+y3)V2

Comparing the above equations

Y21= -y3-gm

Q17) Find y parameters

A17)

The y-parameter is given as

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

Here Y parameter does not exist as V1=V2

Q18) With help of waveform explain all the standard test signals?

A18) The Impulse signal, Ramp signal, unit step and parabolic signals are used as the standard test signals. All these signals are explained below.

Impulse Signal: -

This signal has zero amplitude everywhere except at the origin. Fig below shown the representation of Impulse signal.

Fig: Unit Impulse Signal

The mathematical representations

A(t) = 0 for t ≠0

dt = A e

Where A represent energy or area of the Laplace Transform of Impulse signal is

L [A (t)] = A

UNIT IMPULSE SIGNAL:

If A = 1

(t) = 0 for t ≠0

L [(t)] = 1

The transfer function of a linear time invariant

System is the Laplace transform of the impulse response of the system. If a unit impulse signal is applied to system then Laplace transform of the output c(s) is the transfer function G(s)

As we know G(s) = c(s)/R(S)

r(t) = (t)

R(s) = L [(t)] = 1

:. G(S) = C(s)

Step signal: -

Step signal of size A is a signal that change from zero level to A in zero time and stays there forever.

Fig: Unit Step Signal

r(t)= A t >=0

=0 t<0

L[r(t)] = R(s) = A/s

UNIT STEP SIGNAL: - If the magnitude of the slip signal is I then it is called unit step signal.

u(t) = 1

t>=0

t<0

L[u(t)] = 1/s

Ramp Signal:

The vamp signal increase linearly with time from initial value of zero at t= 0 as shown in fig is below

Fig: Ramp Signal

r(t) = At t>=0

=0 t<0

A is the slope of the line The Laplace transform of ramp signal is

L[r(t)] = R(s) = A/s2

Parabolic Signal:

The instantaneous value of a parabolic signal varies as square of the time from an initial value of zero t=0. The signal representation in fig 14 below.