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25 10
100
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)= 2
Indicating I, III & V harmonicsexhibiting half – wave symmetry(ii) For x2(t)
0 = G.C.D (2, 4, 6)= 2
Indicating 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(t–1)
L.T.IOutput is y(t) = s(t) – s(t–1)
111 1 tuetue tt
Alternative:Using laplace transform solve for transferfunction
1s
1
sX
sYsH
. Then find the output.
01. (c)Sol: Given circuit is
The above diagram can be redrawn as
By applying KCL at Node (1) we get
–i1 – 0.2v1 + i2 +10
v1 =0
–i1 + i2 = 0.1V1 ----- (1) andV1 = –25i1
Substitute in equation (1), we get–i1 + i2 = 0.1(–25i1)
i2 = –1.5i1 ---- (2)By applying KCL at same node,
25 + 21 i
100
V = 0
25 + 21 i
100V 11 i25V
25 +100
i25 1= –1.5i1
25 = 1i4
1 –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+
10
5000 = 0
i4 = – 5A
i1
A
C
E
i2
100
10
0.2v1
v1 +–25
B
i3
D
i4
F
v1
(0v)(Node-1)
25A
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01. (d)Sol: Given,
2t1;4t2
1t0;t2ti
= 0 ; other wise
F21
C
We know that
dttiC
1V
t
c
dtti2t
dtti2tVt
c
2dtt22t
0
; 0 t 1
dtti2tVt
c
2dtti22
1
24221
tdtt
2421
2
ttt
= 4t8t2 2 ; 1 t 2 The voltage across the capacitor in the interval
1 t 2 is,Vc(t) = –2t2 + 8t –4
0 1 2
2i
t
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01. (e)Sol: From the given circuit
VTH = 0 and RTH:
By applying KCL at node VT is,
10115
V
15
V4V TXT
15
5VV4V TXX
3VX = 4VX +VT
VX = –VT ---------(2)
Sub (2) in (1)
115
V
15
V3 TT
–2VT = 15
VT = –7.5
5.71
VR T
TH
Thevenin’s 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 M1
are marked as fm1, fb1, fb, fk1 and fk.
fm1= M1 21
2
dtxd
; fb1 = B1dt
dx1 ; fk1 = K1x1;
fb = Bdtd
(x1 – x) ; fk = K(x1 – x)
By Newton’s second law. fm1 + fb1+ fb + fk1 + fk = 0
M1 2dt
1x2d+B1
dt1dx
+Bdt
d(x1–x)+K1x1+K(x1–x) = 0
On taking Laplace transform of aboveequation with zero initial conditions we get,M1s
2X1(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(X
11
2
1
………. (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
22mdt
xdMf ;
dt
dxBf 22b
1b xxdt
dBf ; 1k xxKf
By Newton’s second law,
a
b
–7.5
M1
x1
fm1
fb1
fb
fk1
fk
M2
x
fm2
fb2
fb
fk
f(t)
+ –4Vx
a
b
155
10
Vx
–
+
VT
1AVT
+
–
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Mpc(tp)
1
2%5%
c(t)
0 td tr tp ts ts
0.5
t
fm2 + fb2 + fb +fk = f(t)
tfxx(K)xxdt
dB
dt
dxB
dt
xdM 1122
2
2
On taking Laplace transform of aboveequation with zero initial conditions we get,M2s
2X(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) [M2s
2 + (B2 +B)s +K]
–X(s) )s(F)KK(s)BB(sM
)KBs(
11
2
1
2
2K)(BsKB)s
2(B2s
2M)K
1K(s)B
1B(2s
1M
)K1
K(s)B1
B(2
s1
M
)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
nd
7.01t
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
e11
tt)t(c rd2
t
r
rn
drt
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Peak time (tp): It is the time taken by theresponse to reach the maximum value.
dp
p
t,0ttdt
)t(dc
Time period of damped oscillations isequal to the twice of the peak time = 2tp
Maximum (or) Peak overshoot (Mp): It isthe maximum error at the output
21
peM
1)t(cM pp ,
%100)(c
)(c)t(cM% p
p
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.., sn te = 5% (or) 2%
ns
3t for 5% tolerance band.
ns
4t for 2% tolerance band
If increases, rise time, peak time increasesand peak overshoot decreases.
02. (a).Sol: Given impulse response is, h(t) = e–t u(t).
Inverse of the system is hI(t) = k1(t) + k2 tand
1s1
sH
By applying Laplace transform,Hinv(s) = K1+sK2
By given H(s) . Hinv(s) = 1
11s
sKK 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) and
integrate from2
Tto
2
T 00
2/T
2/T
0
0
0
dt)tnf2cos()t(f
2/T
2/T 0n0
2n
0
0
)tnf2(cosa
dt)tnf2sin()tnf2cos(b 00n
dt)nf2sin()tnf2cos(b
2)tnf22(cos1
a
00n
0
0nn
20T
20T
2/T
2/T
0n
2/T
2/T
n0
0
0
0
dt)tnf22cos(2
adt
2
a
+ dt)tnf2sin()tnf2cos(b 0
2/T
2/T
0n
0
0
00T2a
0n
2/T
2/T
00
n
0
0
dttnf2costfT
2a
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
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Sampling theorem for band–limited signalsof finite energy.
A band–limited 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.
OR
A 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(t–2)]+(2t–2) [u(t–2)–u(t– 3)]+(–2t+10) [u(t–3)–u(t–4)]+2[u(t–4)–u(t–6)].
= 2 u(t) + u (t–2) [2t–2–2]+u(t–3)[–2t+10–2t+2] + u (t–4) [2–10+2t]
–2u(t–6)
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) = e–t u(t)Output of system is, y(t) = e–2t u(t) + e–3tu(t) Frequency response of the system is,
j1
1j3
1
j2
1
X
Y)(H
g(t)
t0
TsFig. 2
T
s(t)
0 2 4 6t
2
4
x(t)
t2 3 4 5 6
2
3
1
1
g(t)
0
Ts
t
Fig.1
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6j5j
j25j12
j3
2
j2
12
By applying Inverse Laplace Transform
I.R tue2tuet2th t3t2
03. (c)Sol: The differential equation is given by
)t(x)t(ydt
)t(dy 4
Apply Fourier transform to the abovedifferential equation
)(X)(Y)(Yj 4
j)(X
)(Y)(H
41
Given x(t) =sin4 t + cos (6 t+ /4)
H,)(H216
1= –tan-1( )4/
At 0 = 4 ; 2
0
416
1H
075.018.13
1
913.173
1
441tanH o
4034672 01 ..tan
At 0=6 2
0
616
1H
051.025.19
1
9456.370
1
5.1tan
4
6tanH 11
0
= –78.010= – 0.43
y(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
zXz8.0zXzzXz6.0zX4.0
zYz4.0zYz6.0zYz8.0zY321
321
321
321
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
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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)
nx
1
0
0
nq
nq
nq
8.06.04.0
100
010
1nq
1nq
1nq
3
2
1
3
2
1
nx4.0
nq
nq
nq
28.076.064.0ny
3
2
1
8.06.04.0
100
010
A
1
0
0
B ; 28.076.064.0C
4.0D
04.(a)Sol:
V3 = 2VApply KCl at Node V1
02
2V1
VV5.0
2V 1211
02
2VVV4V2 1
211
4V1 – 8 + 2V1 – 2V2 +V1 –2 = 07V1 – 2V2 = 10 ---- (1)Apply KCl at Node V2.
015.0
V
1
2V
1
VV 2212
01V22VVV 2212 3V4V 21 ----- (2)
Solve for (1) & (2)
V1 =2631
V,1323
2
i1 = A577.0A26
15
1
VV 21
04.(b)Sol: Use thevenin’s Theorem
by using Nodal analysis at Node A
5
12V = 12
V 12 = 60 V = 72
I1 =5
60
5
7212
= 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
i1
V3V2V1
I1
12V 12A
5
1
20V–
–
+
+
10 I1–+
+ –VTh
V
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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 = 16
RTh =2
V = 16
I = 8.6A
04. (c)Sol:
sV
sI
3s2ss
1s12sY
V(t) = (t); V(s) = 1
3s2ss
1s12sI
3s
C
2s
B
s
AsI
3s2s
1s12LtssILtA
0s0s
2
32
112
6
232
112sI2sLtB
2s
8C
3s
8
2s
6
s
2sI
i(t) = 2u (t) + 6e–2t u(t) –8e–3tu(t)if v(t) = u(t)V(s) = 1/s
3s2ss
1s12sI
2
3s
D
2s
C
s
B
s
A2
2
32
112sIsLtA 2
0s
312
112sI3sLtC 20s
3
8
19
212sI3sLtD
3s
3s2s
1s12dsd
LtB0s
23s2s
5s21s3s2s12
3
1
32
5612 2
I(s) = 2t u(t) + 1/3 u(t) –3e–2tu(t) + 8/3e–3t u(t)
04. (d)Sol: KCL at node 1 gives
i(t) = iR + iC
et U(t) = v(t) 2+j---(1)The Fourier transform is given by
Let U(t) = j1
1
Lv(t) = V(j)Taking the Fourier transform on both sides ofequation (1).
j2)j(Vj1
1
V(j) = j2j1
1
j2
1
j1
1
V(j) =2/j1
5.0
j1
1
Talking the inverse Fourier transformV(t) = L-1 -2tt eejV
volts.05. (a)
Sol: In many practical situations, a circuit isdesigned 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+–
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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. Byvarying 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 RR
RRR2RRV
dR
dP
0
RR
)R2RR(V
3LTh
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/dR2
L 0The maximum power transferred is obtained bysubstituting for
pmax =Th
2Th
R4
V--- (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
RV
IL
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–)
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05.(c)Sol: figures shows the transformed network of
Figure is further simplified to as shown inabove figure
Thevenin’s impedance’ ZTH’
1s2
1s1s1s1s
1s1sZZ xyTH
= 2
1s
Open-Circuit Voltage’VOC’
Voc = 1s21
s
1s
2
1s1s2
1
ZZ
VI
LTH
OC3R
s2
2s2ss
1s2
1
2
2s3s1s
s2
2s1s1s
s
2s
C
1s
B
1s
A2
1s
2
2s1s1s
s1sB
1s2s
s
121
1
2s2 2s1s
s2sC
2
12
22
1s
22
2s1s
s1s
ds
dA
1s2s
s
ds
d
1s
22s
1s12s
= 1s
22s
s2s
1s22s
2
2
21
22
Therefore,
2s
2
1s
1
1s
2sI
23R
or tue2tee2ti t2tt
3R
05. (d)Sol:
6s
K
2s
K
s
K
6s2s
3s1s2
s
sF 210
2
1
6s2s
3s1s2K
0s
0
4
1
6ss
3s1s2K
2s
1
4
5
2ss
3s1s2K
6s
2
6s
4
5
2s4
1
s2
1
s
sF
6s
s4
5
2s
s4
1
2
1sF
+
– 1S
1
S+1
S
1S
X
Y
S+1
+
– 1S
1
S R3
1
1/S
S
1
1
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+++–
++ G3
G4
G1
1H2G1
2G
1/G3H1
C(s)R(s)
H2
++
+ – ++
H1
H2
G1 G3
G4
R(s) C(s)
1H2G1
2G
if F(s) is an impedance function, i.e.,
,6s
s4
5
2s
s4
1
2
1sZ
:H81
241
KL;K
41
R;21
KR1
111100
H24
5
64
5K
L;4
5KR
2
2222
The foster form I synthesized network isshown in Fig. (a).
if F(s) is an admittance function, i.e.
6s
s4
5
2s
s4
1
2
1)s(Y
;F8
1
24
1K
C;4K
1R;2
K
1R
1
11
11
00
F24
5
64
5K
C;5
4
K
1R
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 G2G3
G4
H1H1
++++
G4
G1
H1/G3
C(s)R(s)
23212
32
HGGHG1
GG
1/8H 5/24H
5/41/41/2
ZRL (s)
YRC (s)2
41/8F 5/24F
4/5
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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 5th
order 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, and–j1, 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 =0s
Lt
sG(s)
=)3s()1s(s
)2s(20sLt
0s
=
31
220
=3
40
Steady state error, ess =vK
1=
40
3 = 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 =
32
10
=
35
Steady state error, ess =pK1
1
=
3
51
1
=53
3
=
8
3 = 0.375
1H2G1G2H3G2G1H2G13G2G1G
++
G4
C(s)R(s)
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(iii))2s()1s(s
10)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 unity
acceleration input,a
ss K
1e .
Acceleration error constant,Ka =
0sLt
s2 G(s) H(s)
=0s
Lt
s2 G(s)
=)2s()1s(s
10sLt
22
0s =
21
10
= 5
Steady state error, ess =aK
1=
5
1 = 0.2
06.(d)Sol: Given that, c(t) = 1+0.2e60t – 1.2 e–10t
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(s
s722s2.1s22s2.0600s702s
=)10s()60s(s
600
=)10s()60s(
600
s
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 the
system,600s70s
600
)s(R
)s(C2
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
=
600s70s
6002
On comparing we get, 2n = 70 and2n = 600 43.1
49.242
70
2
70
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(G1
1SM
G
25.6s2s
s2s
25.0)2s(s
251
12
2
Put js
25.62j
2jS
2
2MG
Put 1
.25.612j1
12j1S
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(GSM
H
25.6s2s
25.6
25.0)2s(s
251
25.0)2s(s
25
2
Put s = j
25.62j
25.6S
2MH
put 1
11.12j25.5
25.6
25.612j1
25.6S
2MH
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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 = G1G2
Gain of forward path-2, P2= –G3
Individual Loop GainThere is only one individual loop. Let the
individual loop gain be P11
Loop 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 = 1
Transfer Function, TBy Mason’s gain formula the transfer
function, T is given by,
T =1 PK K =
1
[P11 + P22]
=HG1
GGG
1
321
07.(c)Sol: The overall transfer function for the system is
)2s3s(
2
)s(R
)s(Cor
1.)3s(s
21
)3s(s
2
)s(R
)s(C2
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 be
determined as1Kand2K,1K 321
2s
1
)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
T
Ks
T
1s
T/K
1.)1sT(s
K1
)1sT(s
K
)s(R
)s(C
2
+–G1 G2
G3
H
C(s)R(s)+
–1
23 4 5 6
432
1 G1
–H
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The characteristic equation is
0T
Ks
T
1s2
T/12andT/K nn
KT2
1
T/K2
1.
T
1
2
1.
T
1
n
Let 1K be the forward path gain when
%60M 1p and the corresponding damping
ratio be .1Since
100e60%100eM21
1
21
1
111p
or )e(log1
)6.0(loge2
1
1
e
1.1
51.0or21
1
)1(51.0 2
12
221
or025.0or)1(026.0 2
121
21
158.01
Let 2K be the forward path gain when
%20M 2p and the corresponding damping
ratio be .2Since
100e20%100eM22
2
22
2
112p
From the above relation the value of 2 can becalculated as
447.02 Assuming time constant T to be constant
TK
1.
2
1and
TK
1.
2
1
2
2
1
1
1
TK2
TK
1.
2
1 2
12
1
Hence, .8
1
447.0
158.0
K
K22
2
1
1
2