EC EQUATIONS

: 2 : ESE_Offline & Online Test


25 10100

i1 i2

i3 i4

0.2V1V1+

25AA

BC

DE F

01. (a)Sol:

For half wave symmetry periodic signalshould contain odd harmonic components(f0, 3f0, 5f0 ,)(i) For x1(t),0 = G.C.D (2, 6, 10)

= 2Indicating I, III & V harmonicsexhibiting half wave symmetry(ii) For x2(t)0 = G.C.D (2, 4, 6)

= 2Indicating I, II & III harmonicsnot satisfying half wave symmetry.

01. (b).Sol. Given step response of an LTI system is,

tue1ts tSince, x(t) = rect (t 0.5) = u(t) u(t1)

L.T.IOutput is y(t) = s(t) s(t1)

111 1 tuetue ttAlternative:Using laplace transform solve for transferfunction

1s1

sXsY

sH . Then find the output.

01. (c)Sol: Given circuit is

The above diagram can be redrawn as

By applying KCL at Node (1) we geti1 0.2v1 + i2 + 10

v1=0

i1 + i2 = 0.1V1 ----- (1) andV1 = 25i1Substitute in equation (1), we geti1 + i2 = 0.1(25i1)

i2 = 1.5i1 ---- (2)By applying KCL at same node,

25 + 21 i100V = 0

25 + 21 i100V 11 i25V

25 +100

i25 1= 1.5i1

25 = 1i41 1.5i1

i1 = 20Ai2 = 1.5(20) = 30Ai2 = 30Av1 = 500 VBy applying KCL at Node B,i1 + 0.2 v1 + i3 = 0i3 = i1 0.2v1 = (20) 0.2500i3 = 20 100 = 80 Ai3 = 80 A

i3 25+i4+

105000

= 0

i4 = 5A

i1

A

C

E

i2

100

10

0.2v1

v1 +25

B

i3

D

i4

F

v1

(0v) (Node-1)

25A

: 3 : COV 1 Solutions


01. (d)Sol: Given,

2t1;4t21t0;t2ti

= 0 ; other wise

F21C

We know that

dttiC1V

t

c

dtti2 t

dtti2tV tc

2dtt22t

0

; 0 t 1 dtti2tV tc

2dtti2 21

2422

1

t dtt

2421

2

ttt = 4t8t2 2 ; 1 t 2

The voltage across the capacitor in the interval1 t 2 is,Vc(t) = 2t2 + 8t 4

0 1 2

2i

t



01. (e)Sol: From the given circuit

VTH = 0 and RTH:

By applying KCL at node VT is, 101

15V

15V4V TXT

155VV4V TXX

3VX = 4VX +VT VX = VT ---------(2)Sub (2) in (1)

115V

15V3 TT

2VT = 15VT = 7.5

5.71

VR TTH

Thevenins equivalent is

01. (f)Sol: In the given system, applied force f(t) is the

input and displacement x is the output.Let, Laplace transform of f(t) =L{f(t)}= F(s)Laplace transform of x = L {x} = X(s)Laplace transform of x1= L {x1}= X1(s)The system has two nodes and they are massM1, and M2. The differential equationsgoverning the system are given by forcebalanced equations at these nodes.

Let the displacement of mass M1, be x1.

The free body diagram of mass M1 is shown infigure. The opposing forces acting on mass M1are marked as fm1, fb1, fb, fk1 and fk.

fm1= M1 21

2

dtxd

; fb1 = B1 dtdx1 ; fk1 = K1x1;

fb = B dtd (x1 x) ; fk = K(x1 x)

By Newtons second law. fm1 + fb1+ fb + fk1 + fk = 0

M1 2dt1x

2d+B1

dt1dx +B

dtd (x1x)+K1x1+K(x1x) = 0

On taking Laplace transform of aboveequation with zero initial conditions we get,M1s2X1(s)+B1sX1(s)+Bs[X1(s)X(s)]+ K1X1(s)+ K [X1(s) X(s)]= 0X1(s) [M1 s2 + (B1+B)s + (K1 +K)] X(s)[Bs+K] = 0X1(s) [M1 s2 + (B1+B)s + (K1 +K)] = X(s)[Bs+K]X1(s) = )KK(s)BB(sM

KBs)s(X11

21

. (1)The free body diagram of mass M2 is show

in figure.

The opposing forces acting on M2 are markedas fm2, fb2, fb, fk.

2

2

22m dtxdMf ;

dtdxBf 22b

1b xxdtdBf ; 1k xxKf

By Newtons second law,

a

b

7.5

M1

x1

fm1fb1fbfk1fk

M2

x

fm2fb2fbfk

f(t)

+ 4Vx

a

b

155

10

Vx

+

VT

1AVT

+



Mpc(tp)

12%5%

c(t)

0 td tr tp ts ts

0.5

t

fm2 + fb2 + fb +fk = f(t)

tfxx(K)xxdtdB

dtdxB

dtxdM 1122

2

2

On taking Laplace transform of aboveequation with zero initial conditions we get,M2s2X(s)+B2sX(s) + Bs[X(s) X1(s)]

+K [X(s) X1(s)] = F(s)X(s) [M2s2 + (B2 +B)s + K] X1(s) [Bs + K]

= F(s) (2)Substituting for X1(s) from equation (1) inequation (2) we get,X(s) [M2s2 + (B2 +B)s +K]

X(s) )s(F)KK(s)BB(sM)KBs(

112

1

2

2K)(BsKB)s2

(B2s2

M)K1

K(s)B1

B(2s1

M

)K1K(s)B1B(2

s1M

)s(F)s(X

01.(g)Sol: A control system is a combination of

elements arranged in a planned manner wherein each element causes an effect to produce adesired output. This cause and effectrelationship is governed by a mathematicalrelation. In control system the cause actsthrough a control process which in turn resultsinto an effect.

Control systems are used in manyapplications for example, systems for thecontrol of position, velocity, acceleration,temperature, pressure, voltage and current etc.

Control systems can be broadly divided intwo types.(i) open loop control system

The accuracy of an open loop systemdepends on the calibration of the input. i.e.,output is independent of input.Eg: Traffic control system

Traffic control by means of traffic signalsoperated on a time basis constitutes an openloop control system. The sequence of controlsignals are based on a time slot given for eachsignal. The time slots are decided based on atraffic study. The system will not measure thedensity of the traffic before giving the signals.Since the time slot does not changes according

to traffic density, the system is open loopsystem.(ii) Closed loop control system:

The output of a system depends on input.Eg: Bread Toaster.

Traffic control system can made as aclosed loop system if the time slots of thesignals are decided based on the density oftraffic. In closed loop traffic control system,the density of the traffic is measured on all thesides and the information is fed to a computer.The timings of the control signals are decidedby the computer based on the density oftraffic. Since the closed loop systemdynamically changes the timings, the flow ofvehicles will be better than open loop system.

01.(h)Sol: Time domain specifications (or) transient

response parameters:

Delay time (td): It is the time taken by theresponse to change from 0 to 50% of itsfinal/steady state value.

5.0tt

)t(cd

n

d7.01

t

Rise time (tr): It is the time taken by theresponse to reach from 0 to 100%. Generally10% to 90% for over damped and 5% to 95%for critically damped system is defined.

1)tsin(1

e11tt

)t(c rd2t

r

rn

drt



Peak time (tp): It is the time taken by theresponse to reach the maximum value.

dp

pt,0

ttdt)t(dc

Time period of damped oscillations isequal to the twice of the peak time = 2tpMaximum (or) Peak overshoot (Mp): It isthe maximum error at the output

21p eM

1)t(cM pp ,

%100)(c)(c)t(c

M% pp

If the input is doubled, then the steadystate and peak values double, thereforemagnitude of Mp doubles but % Mp remainsconstant.

Settling time (ts): It is the time taken by theresponse to reach 2% or 5% toleranceband as shown in the fig above.

i.e.., snte = 5% (or) 2%

n

s

3t for 5% tolerance band.

n

s

4t for 2% tolerance bandIf increases, rise time, peak time increasesand peak overshoot decreases.

02. (a).Sol: Given impulse response is, h(t) = et u(t).

Inverse of the system is hI(t) = k1(t) + k2 tand

1s1

sH By applying Laplace transform,

Hinv(s) = K1+sK2By given H(s) . Hinv(s) = 1

1

1ssKK 21

To make this K1= K2 = 1

02. (b)Sol: Fourier coefficient an:

0n

0n0n )tfn2(sinb)tfn2(cosa)t(f

Multiply both sides by cos (2nf0t) andintegrate from

2T

to2

T 00

2/T

2/T0

0

0

dt)tnf2cos()t(f

2/T

2/T 0n0

2n

0

0

)tnf2(cosa

dt)tnf2sin()tnf2cos(b 00n

dt)nf2sin()tnf2cos(b2

)tnf22(cos1a

00n

0

0nn

20T

20T

2/T

2/T0

n

2/T

2/T

n0

0

0

0

dt)tnf22cos(2

adt2

a

+ dt)tnf2sin()tnf2cos(b 02/T

2/T0n

0

0

00T2a

0n

2/T

2/T0

0n

0

0

dttnf2costfT2

a

02. (c)Sol: Sampling Theorem:

Sample the signal g(t) instantaneously andat a uniform rate, once every Ts seconds.

We refer to Ts as the sampling period, andto its reciprocal 1/Ts as the sampling rate.

This ideal form of sampling is calledinstantaneous sampling.

g(t)

t0



Sampling theorem for bandlimited signalsof finite energy.

A bandlimited signal of finite energy,which has no frequency components higherthan W hertz, is completely described byspecifying the values of the signal at instantsof time separated by 1/2W seconds.

ORA band-limited signal of finite energy,

which has no frequency components higherthan W hertz, may be completely recoveredfrom a knowledge of its samples taken at therate of 2W per second.

Flat-Top sampling:The analog signal g(t) is sampled

instantaneously at the rate 1/Ts ,and theduration of each sample is lengthened to T .

The flat-top sampled signal s(t) is shown inFig.2

By using flat-top samples, amplitudedistortion as well as a delay of T/2are introduced. This effect is similar to thevariation in transmission with frequency thatis caused by the finite size of the scanningaperture in television and facsimile.Accordingly, the distortion caused bylengthening the samples, is referred to as theaperture effect. This distortion may becorrected by connecting an equalizer in

cascade with the low-pass reconstructionfilter.

02. (d)Sol: Given signal x(t) is shown below

x(t) = 2 [u(t) u(t2)]+(2t2) [u(t2)u(t 3)]+(2t+10) [u(t3)u(t4)]+2[u(t4)u(t6)].

= 2 u(t) + u (t2) [2t22]+u(t3)[2t+102t+2] + u (t4) [210+2t]

2u(t6)

6tu24tr23tr42tr2tu2)t(x

03. (a)Sol:

As we know u(t)*u(t) = r(t) = tu(t) 4tu1tu*2tu21tututh*tx 4tu*2tu21tu*2tu24tu*

1tu1tu*1tu4tu*tu1tu*tu

6tr23tr25tr2tr4tr1tr 6tr25tr4tr3tr22tr1tr

03. (b).Sol: Given

Input of system is, x(t) = et u(t)Output of system is, y(t) = e2t u(t) + e3tu(t) Frequency response of the system is,

j11

j31

j21

XY)(H

g(t)

t0

TsFig. 2

T

s(t)

0 2 4 6 t24

x(t)

t2 3 4 5 6

2

31

1

g(t)

0

Ts

t

Fig.1



6j5j

j25j12

j32

j212

By applying Inverse Laplace Transform

I.R tue2tuet2th t3t2

03. (c)Sol: The differential equation is given by

)t(x)t(ydt

)t(dy 4Apply Fourier transform to the abovedifferential equation

)(X)(Y)(Yj 4

j)(X

)(Y)(H41

Given x(t) =sin4 t + cos (6 t+ /4) H,)(H 216

1= tan-1( )4/

At 0 = 4 ; 20 4161H

075.018.13

1913.173

1

441tanH o

4034672 01 ..tan

At 0=6 20 6161H

051.025.19

19456.370

1

5.1tan4

6tanH 110

= 78.010= 0.43y(t) = 0.075 sin(4 t 72.340)

+0.05cos(6 t+ /4 78.010)= 0.075sin (4 t 0.4 )

+0.05cos (6 t+ ). 4304

03. (d)Sol: Apply z transform to the given differential

equation

zXz8.0zXzzXz6.0zX4.0zYz4.0zYz6.0zYz8.0zY

321

321

321321

z4.0z6.0z8.01z8.0zz6.04.0

zXzY

zH

No. of delays = No. of state variables = 3q1(n), q2 (n), q3(n) are state variables.

H() ooo Htsin.H Sin 0t



z

-1

z-1

z-1

y[n]

0.6

1

0.8

0.6

0.8

x[n]

0.4

1 0.4

q3(n)=q2(n+1)

q2(n)=q1(n+1)

q1(n)

q3(n+1)

nxnq8.0nq6.0nq4.01nq 3213 nq1nq 32 nq1nq 21 nq8.0nqnq6.01nq4.0ny 1233 nq8.0nqnq6.0

nxnq8.0nq6.0nq4.04.0ny

123

321

nq8.0nqnq6.0nx4.0

nq32.0nq24.0nqn16.0ny

123

3211

nx4.0nq28.0nq76.0nq64.0ny 321 Q(n+1) = AQ(n) + Bx(n)

y(n) = CQ(n) + Dx(n)

nx100

nqnqnq

8.06.04.0100010

1nq1nq1nq

3

2

1

3

2

1

nx4.0nqnqnq

28.076.064.0ny

3

2

1

8.06.04.0100010

A

100

B ; 28.076.064.0C

4.0D

04.(a)Sol:

V3 = 2VApply KCl at Node V1

02

2V1

VV5.0

2V 1211

02

2VVV4V2 1211 4V1 8 + 2V1 2V2 +V1 2 = 07V1 2V2 = 10 ---- (1)Apply KCl at Node V2.

015.0

V1

2V1

VV 2212

01V22VVV 2212 3V4V 21 ----- (2)

Solve for (1) & (2)V1 = 26

31V,1323

2

i1 = A577.0A2615

1VV 21

04.(b)Sol: Use thevenins Theorem

by using Nodal analysis at Node A

512V

= 12

V 12 = 60 V = 72

I1 = 560

57212 = 12 A

10 I1 = 10 12 = 120 VFinding Vth by applying KVL in Loop

0.5

2 V+

+

2 1

1

1

0.5 1 A

2 V

i1V3V2V1

I1

12V 12A

5

1

20V

+

+

10 I1+

+ VTh

V



72 ( 120) Vth 20 = 072 + 120 20 = Vth

Vth = 172 VFinding Rth by disabling all sources.No Independent source, so by applying currentof 1 A across 1 & 2

By applying KVL in Loop. 5 I1 10 I1 V I1 = 05 + 10 V + 1 = 0V = 16RTh = 2

V= 16

I = 8.6A04. (c)Sol:

sVsI

3s2ss1s12

sY

V(t) = (t); V(s) = 1 3s2ss

1s12sI

3s

C2s

Bs

AsI

3s2s1s12LtssILtA

0s0s

232112

6232112

sI2sLtB2s

8C

3s8

2s6

s

2sI

i(t) = 2u (t) + 6e2t u(t) 8e3tu(t)if v(t) = u(t)V(s) = 1/s 3s2ss

1s12sI 2

3sD

2sC

s

Bs

A2

232112

sIsLtA 20s

312112

sI3sLtC 20s

38

19212

sI3sLtD3s

3s2s1s12

dsdLtB

0s

23s2s

5s21s3s2s12

31

325612 2

I(s) = 2t u(t) + 1/3 u(t) 3e2tu(t) + 8/3e3t u(t)04. (d)Sol: KCL at node 1 gives

i(t) = iR + iCet U(t) = v(t) 2+j---(1)

The Fourier transform is given by

Let U(t) = j11

Lv(t) = V(j)Taking the Fourier transform on both sides ofequation (1).

j2)j(Vj11

V(j) = j2j11

j21

j11

V(j) =2/j1

5.0j1

1

Talking the inverse Fourier transformV(t) = L-1 -2tt eejV volts.

05. (a)Sol: In many practical situations, a circuit is

designed to provide power to a load. There areapplications in areas such as communicationswhere it is desirable to maximize the powerdelivered to a load. We now address theproblem of delivering the maximum power to

I1 5

1A

+

10 I1+

1

V

16

172V 4+



a load when given a system with knowninternal losses. It should be noted that this willresult in significant internal losses greater thanor equal to the power delivered to the load.

The Thevenin equivalent is useful in findingthe maximum power a linear circuit can deliver to aload. We assume that we can adjust the loadresistance RL. If the entire circuit is replaced by itsThevenin equivalent except for the load, as shownin fig, the power delivered to the load is

L

2

LTh

ThL

2 RRR

VRiP

--- (1)For a given circuit, VTh and RTh are fixed. By

varying the load resistance RL, the power deliveredto the load varies as sketched in fig (b) . we noticefrom fig (b) that the power is small for small orlarge values of RL but maximum for some value ofRL between 0 and . We now want to show that thismaximum power occurs when RL is equal to RTh.This is known as the maximum power theorem.

Maximum power is transferred to the loadwhen the load resistance equals the Theveninresistance as seen from the load (RL= RTh).To prove the maximum power transfer theorm. Wedifferentiate P with respect to RL and set the resultequal to zero. We obtain.

4

LTh

LThL2

LTh2Th

L RRRRR2RRV

dRdP

0RR)R2RR(V 3

LTh

LLTh2Th

This implies that0 = (RTh + RL 2RL) = (RTh RL)

Which yields

RL = RTh ---(2)Showing that the maximum power transfer takesplace when the load resistance RL equals theThevenin resistance RTh. We can readily confirmthat equation (2) RL = RTh gives the maximumpower by showing that d2P/dR2L 0The maximum power transferred is obtained bysubstituting for

pmax =Th

2Th

R4V

--- (3)

Equation (3) applies only whenRL= RTh, where RL RTh we compute thepower delivered to the load using equation (1).

05. (b)Sol: When switch is in position 1 for long time. It

will go to steady state condition inductor willacts as a short circuit .

Current through inductor A22550

RVIL

A20i0i LL When switch moved to position (2)Inductor acts as a current source of 2A

By converting inductor current and inductorparallel network into series it will look like

+VTh

RTh a

b

RL

i

Fig. (a) The circuit used formaximum power transfer

RL

Pmax

RTh

P

Fig. (b) Powerdelivered to the loadas a function of RL

100/S25

iL(0) 2s

100/S

25

2S

+ 4

50V25

iL(0)



05.(c)Sol: figures shows the transformed network of

Figure is further simplified to as shown inabove figure

Thevenins impedance ZTH

1s2

1s1s1s1s

1s1sZZ xyTH

=

21s

Open-Circuit VoltageVOC

Voc = 1s21

s

1s2

1s1s2

1

ZZV

ILTH

OC3R

s2

2s2ss1s2

1

2

2s3s1s s2 2s1s1s

s

2sC

1sB

1sA

2

1s2

2s1s1ss1sB

1s2ss

121

1

2s2 2s1ss2sC

2122

2

1s

22

2s1ss1s

dsdA

1s2s

s

dsd

1s22s

1s12s

= 1s22ss2s

1s22s2

2212

2 Therefore,

2s2

1s1

1s2

sI 23R

or

tue2tee2ti t2tt3R 05. (d)Sol:

6sK

2sK

s

K6s2s3s1s2

s

sF 210

2

16s2s3s1s2K

0s0

41

6ss3s1s2K

2s1

45

2ss3s1s2K

6s2

6s

45

2s41

s21

s

sF

6s

s45

2s

s41

21

sF

+

1S

1

S+1

S

1S X

Y

S+1

+

1S

1

S R3

1

1/S

S

1

1



+++

+

+ G3

G4

G11H2G1

2G

1/G3H1

C(s)R(s)

H2

++

+ ++

H1

H2

G1 G3

G4

R(s) C(s)

1H2G12G

if F(s) is an impedance function, i.e.,

,6s

s45

2s

s41

21

sZ

:H81

241

KL;K41R;

21KR

1

111100

H245

645

KL;45KR

2

2222

The foster form I synthesized network isshown in Fig. (a).

if F(s) is an admittance function, i.e.

6s

s45

2s

s41

21)s(Y

;F81

241

KC;4K1R;2

K1R

1

11

11

00

F245

645

KC;54

K1R

2

22

22

06.(a)Sol:

Step1: Splitting the summing point andrearranging the branch points

Step 2: Eliminating the feedback path

Step 3: Shifting the branch point after theblock

Step 4: Combining the blocks in cascadeand eliminating feedback path

Step5: Combining the blocks in cascade andeliminating feedback path

++

++

+G1 G2 G3

H1

C(s)R(s)

H2

G4

++

+ +

++

C(s)R(s)

H2

G1 G2 G3

G4

H1H1

++++

G4

G1

H1/G3

C(s)R(s)23212

32HGGHG1

GG

1/8H 5/24H

5/41/41/2

ZRL (s)

YRC (s)2 4

1/8F 5/24F4/5



Step 6: Eliminating forward path

412123212

321 GHGGHGGHG1

GGG)s(R)s(C

06.(b)Sol: The characteristic equation of the system is

s5+4s4+8s3+8s2+7s+4=0.The given characteristic polynomial is 5thorder equation and so it has 5 roots.s5: 1 8 7s4: 1 2 1s3: 1 1s2: 1 1s1 : s0: 1

When 0, there is no sign change in thefirst column of routh array. But we have arows of all zeros (s1 row or row-5) and so thereis a possibility of roots on imaginary axis. Thiscan be found from the roots of auxiliarypolynomial. Here the auxiliary polynomial isgiven by s2 row.

The auxiliary polynomial is, s2 + 1 = 0; s2 = 1 or s = 1 = j1The roots of auxiliary polynomial are +j1, andj1, lying on imaginary axis. The roots ofauxiliary polynomial are also roots ofcharacteristic equation. Hence two roots ofcharacteristic equation are lying on imaginaryaxis and so the system is limitedly ormarginally stable. The remaining three roots ofcharacteristic equation are lying on the lefthalf of s-plane.

06.(c)Sol: (i) )3s()1s(s

)2s(20)s(G

Let us assume unity feedback system, H(s) = 1

The open loop system has a pole atorigin. Hence it is a type-1 system. Insystems with type number 1, thevelocity (ramp) input will give a constantsteady state error.

The steady state error with unit

velocity input, ess =vK

1

Velocity error constant, Kv=0s

Lt

sG(s) H(s) Kv =

0sLt

sG(s)

= )3s()1s(s)2s(20

sLt0s

=

31220

=

340

Steady state error, ess =vK

1=

403

= 0.075

(ii) )3s()2s(10)s(G

Let us assume unity feedback system, H(s) = 1

The open loop system has no pole atorigin. Hence it is a type-0 system. Insystems with type number 0, the stepinput will give a constant steady stateerror.

The steady state error with unit step input,

ess =pK1

1

Position error constant,Kp =

0sLt

G(s) H(s)

sGLt0s

= )3s()2s(10Lt

0s = 3210 = 3

5

Steady state error, ess =pK1

1 =

351

1

=

533 = 8

3 = 0.375

1H2G1G2H3G2G1H2G13G2G1G ++

G4

C(s)R(s)



(iii) )2s()1s(s10)s(G 2

Let us assume unity feedback system, H(s) = 1.The open loop system has two poles atorigin. Hence it is a type-2 system. Insystems with type number-2, theacceleration (parabolic) input will give aconstant steady state error.The steady state error with unityacceleration input,

ass K

1e .

Acceleration error constant,Ka =

0sLt

s2 G(s) H(s) =

0sLt

s2 G(s)

= )2s()1s(s10

sLt 22

0s = 2110 = 5

Steady state error, ess =aK

1=

51

= 0.2

06.(d)Sol: Given that, c(t) = 1+0.2e60t 1.2 e10t

On taking Laplace transform of c(t) we get,C(s) = )10s(

12.1)60s(12.0

s

1

= )10s()60s(s)60s(s2.1)10s(s2.0)10s()60s(

=

)10s()60s(ss722s2.1s22s2.0600s702s

= )10s()60s(s600

= )10s()60s(

600s

1

Since input is unit step, R(s) = 1/s C(s) = R(s) )10s()60s(

60

= R(s)600s70s

6002

The closed loop transfer function of thesystem,

600s70s600

)s(R)s(C

2

The damping ratio and natural frequency ofoscillation can be estimated by comparing thesystem transfer function with standard form ofsecond order transfer function.

2nn

2

2n

s2s)s(R)s(C

=

600s70s600

2 On comparing we get, 2n = 70 and

2n = 600 43.149.242

70270

n

n = 600 = 24.49 rad/sec.07. (a)Sol: (i) The sensitivity of overall transfer function

(M) w.r.t. forward path transfer function (G) isgiven by

)s(H)s(G11SMG

25.6s2ss2s

25.0)2s(s251

12

2

Put js

25.62j2jS 2

2MG

Put 1 .

25.612j112j1S 2

2MG

398.02j25.5

2j1

(ii) The sensitivity of overall transfer function(M) w.r.t. feedback path transfer function(H) is given by

)s(H)s(G1)s(H)s(GSMH

25.6s2s25.6

25.0)2s(s251

25.0)2s(s25

2

Put s = j

25.62j25.6S 2

MH

put 1 11.1

2j25.525.6

25.612j125.6S 2

MH



1 1

3 4 521

G1 G2 C(s)R(s)6

1

1 2 3 4 5 6

G3

C(s)R(s) 1 1 1

1 2 3 4 5 6R(s) C(s)1 1

1 G1 G2

G3

H

07.(b)Sol: The nodes are assigned at input, output, at

every summing point and branch point asshown in figure.

The signal flow graph of the above system isshown in figure.

Forward Paths:There are two forward paths K = 2Let the forward path gains be P1 and P2

Gain of forward path-1, P1 = G1G2Gain of forward path-2, P2= G3

Individual Loop GainThere is only one individual loop. Let the

individual loop gain be P11Loop gain of individual loop-1,P11 = G1H

Gain Products of Two Non-touching LoopsThere are no combinations of non-touching

loops.Calculations of and K : = 1 [P11] = 1 + G1H

Since there are no part of the graph which isnon-touching with forward path-1 and 2,1 = 2 = 1Transfer Function, T

By Masons gain formula the transferfunction, T is given by,

T = 1 PK K =

1 [P11 + P22]

=HG1GGG

1

321

07.(c)Sol: The overall transfer function for the system is

)2s3s(

2)s(R)s(C

or

1.)3s(s21

)3s(s2

)s(R)s(C

2

It is noted that the denominator of the aboveexpression can be factored as [(s+1)(s+2)]

)2s)(1s(2)s(R)s(C

As the input is a unit steps/1)s(R

)2s)(1s(2

.

s

1)s(C The R.H.S. of the above expression can beexpanded into partial fraction as follows:

)2s(3K

)1s(K

s

K)2s)(1s(

2.

s

1 21

The coefficients 321 KandK,K can bedetermined as

1Kand2K,1K 321

2s1

)1s(2

s

1)s(C Taking inverse Laplace transform on bothsides, .tuee21)t(c t2t

07.(d)Sol: The overall Transfer function is given by

TK

sT1

s

T/K

1.)1sT(sK1

)1sT(sK

)s(R)s(C

2

+G1 G2

G3

H

C(s)R(s)+

1

23 4 5 6

4321 G1

H



The characteristic equation is

0TK

sT1

s2

T/12andT/K nn

KT21

T/K21

.

T1

21

.

T1

n

Let 1K be the forward path gain when

%60M 1p and the corresponding dampingratio be .1Since

100e60%100eM21

121

1

111p

or )e(log1

)6.0(loge2

1

1e

1.1

51.0or21

1

)1(51.0 2122

21

or

025.0or)1(026.0 212121 158.01

Let 2K be the forward path gain when%20M 2p and the corresponding damping

ratio be .2Since

100e20%100eM22

222

2

112p

From the above relation the value of 2 can becalculated as

447.02 Assuming time constant T to be constant

TK1

.

21

andTK

1.

21

22

11

1TK2

TK1

.

21 2

12

1

Hence, .81

447.0158.0

KK 2

2

2

1

1

2

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