PART – II
NUMERICALS
Q.1 Write the
dynamic equation in respect of the mechanical system given in Fig.4. Then
using force-voltage analogy obtain the equivalent electrical network.
Legend
spring constants
viscous friction damping
coefficient
inertial constants of masses
displacements
F (t) .. Force. (14)
Ans : 
From figure, a
force F(t) is applied to mass M2.
Free body diagrams
for these two masses are
From these, the
following differential equations describing the dynamics of the system.
F(t) - K2( 
- 
) =M2
K2 (-) –K1 – B1
=M1 
From above we can
write down
M2 + K2( - ) = F(t)
M1 + K1 + B1 - K2 (-) = 0;
These two are
simultaneous second order linear differential equations. Manipulation of these
equations results in a single differential equations relating the response X2
(or X1) to input F(t).
Electrical
Analogous for the circuit:
The dynamic
equations of the system could also be obtained by writing nodal equations for
the electrical network.
f 1 
+ K1
dt
+ K2
dt = 0;
M2 
+ K2dt = F(t)
Force
Voltage Analogy:
The dynamical
equations of the system could also obtained by writing nodal equations for
electrical network.
K1
dt + R1 i1 + M1 
+ K2
dt = 0;
M2 i2
+ K2 dt = F(t)
Q.2 Determine the
transfer function 
for
the block diagram shown in Fig.5 by first drawing its signal flow graph and
then using the Mason’s gain formula. (14)
Ans :
SIGNAL FLOW
GRAPH OF THE BLOCK DIAGRAM:
Mason’s gain
formula
Overall system gain is given by
Pk –gain
of kth forward path
Δ = det of
the graph
= 1 – sum of
loop gains of all individual loops + (sum of gain products of all possible
combinations of two non-touching loops) –(sum of gain products of all possible
combinations of three non touching loops)+ ……
= the value of Δ
for that part of the graph not touching the kth forward path.
There are four
path gains
Individual loop
gains are
There are no non
touching loops
Δ = 1-(- G1G2H1-G5H2-G2G3H2-G4G5-G1G2G3-G4G2G3-G1G5)
P1
Δ1 + P2 Δ2 + P3 Δ3
+P4 Δ4
T =
------------------------------------------
Δ
G1G2G3 + G4G2G3 +G1G5
+ G4G5
=
-------------------------------------------------------------------------
1+
G1G2H1+G5H2+G2G3H2+G4G5+G1G2G3+G4G2G3+G1G5
Q.3
The open loop transfer functions of three systems are given as
(i) 
(ii) 
(iii) 
Determine
respectively the positional, velocity and acceleration error constants for
these systems. Also for the system given in (ii) determine the steady state
errors with step input 
, ramp input r(t) = t and acceleration
input 
. (10)
Ans :
4
(i)
----------------
(s+1)(s+2)
Positional error for unit step input
1 1 1
= lim
----------- = ----------- = ------------
s→0 1 + G(s) 1 + G(0) 1 + 
= G(0) is defined as
positional error constant
= 4/2 = 2 = 1/3
= lim s G(s) is
defined as velocity error constant 
s→0
velocity error for ramp input
1 1 1
= lim
---------------- = lim ------ = ------- = 
s→0 s( 1 + G(0)) s→0 sG(s)
Acceleration
error constant
= lim s2
G(s) is defined as velocity error constant 
s→0
1 1 1
= lim
------------- = lim --------- = ------- =
s→0
s2(1+G(s)) s→0 s2G(s)
(ii)
= lim G(s)H(s)
s→0
2
= lim
------------------ = ∞
s→0 s(s+4)(s+6)
2s 2 1
= lim sG(s)H(s) =
lim ---------------- = ------ = ------
s→0 s→0 s(s+4)(s+6) 24 12
2s2
= lim s2G(s)H(s)
= lim --------------- = 0
s→0 s→0 s(s+4)(s+6)
5
(iii) -------------------
s2(s+3)(s+10)
= lim G(s)H(s)
s→0
5
= lim
------------------ = ∞
s→0 s2(s+3)(s+10)
5 s
= lim sG(s)H(s) =
lim ---------------- = ∞
s→0 s→0 s2(s+3)(s+10)
5 s2 1
= lim s2G(s)H(s)
= lim --------------- = ------
s→0 s→0 s2(s+3)(s+10) 6
taking the second
system,
2
-------------------
s(s+4)(s+6)
step input u(t)
1 1
= ---------- =
----------- = 0
1 + 1 +
Ramp input t
1
= ------ = 12
parabola input t2/2
1
=
----- = ∞
Q.4 Describe a two phase a.c. servomotor and derive its transfer
function. (4)
Ans :
Working of AC Servomotor:
The symbolic representation of an AC servomotor as a control system
is shown in figure. The reference winding is excited by a constant voltage
source with a frequency n the range 50 to 1000Hz. By using frequency of 400Hz
or higher, the system can be made less susceptible to low frequency noise. Due
to this feature, ac drives are extensively used in aircraft and missile control
system in which the noise and disturbance often create problems.
The control winding is excited by the modulated control signal and
this voltage is of variable magnitude and polarity. The control signal of the
servo loop (or the system) dictates the magnitude and polarity of this voltage.
The control phase voltage is supplied from a servo amplifier and it
has a variable magnitude and polarity (+ or – 90o phase angle w.r.to
the reference phase). The direction of rotation of the motor reverses as the
polarity of the control phase signal changes sign.
It can be proved that using symmetrical components that the starting
torque of a servo-motor under unbalanced operation is proportional to E, the
rms value of the sinusoidal control voltage e(t) . A family of torque-speed
characteristics curves with variable rms control voltage is shown in figure.
All these curves have negative slope.
Note that the curve for zero control voltage goes through the origin
and the motor develops a decelerating torque.
From the torque speed characteristic shown above we can write
(1)
Where
=
torque
= a
positive constant = -ve of the slope of the torque-speed curve
= a
positive constant = torque per unit control voltage at zero speed
=
angular displacement
Further, for motor we have
(2)
Where J = moment of inertia of motor and load reffered to motor
shaft
f = viscous friction coefficient of the motor and
load referred to the motor shaft
form eqs.(1) and (2) we have
=
(3)
Taking the laplace transform on both sides, putting initial
conditions zero and simplifying we get
(4)
Where
=
motor gain constant
If the moment of inertia J is small. Then 
is small and for the frequency
range of relevance to ac servometer 
,then from eq (4) we can write the transfer
function as
(5)
It means that ac servometer works as an integrator. Following figure
gives the simplified block diagram of an ac servometer.
Q.5 For the system shown in the block diagram of Fig.7 determine the
values of gain 
and
velocity feedback constant 
so that the maximum overshoot with a unit
step input is 0.25 and the time to reach the first peak is 0.8 sec. Thus obtain
the rise time and settling time for 5% tolerance band. (10)
Ans :
k1/(s(s+2)
M(s) =
-------------------------------
1 + k1(1 + k2s)/(s(s+2))
k1 k1
=
-------------------------------- = ------------------------------
s2 + 2s + k1 + k1k2s s2
+ (2 + k1k2)s + k1
peak over shoot =
0.25
=
e –ξ 
/(
1- ξ 2)1/2
ξ = 0.403
= 0.8 sec =
----------------------
(
1- ξ2 )1/2
= 4.29 rad/sec
=2 = 18.4
(2 + 
) =2 ξ = 3.457
= 1.457 = 0.079
– 
((1- ξ2)1/2/
ξ)
=
------------------------------------ = 0.505 sec
((1- ξ2)1/2)
= 3/ ξ = 1.735 sec
Q.6 For the standard second order
system shown in Fig.8, with r (t) = u (t) explain how the time domain
specifications corresponding to resonant peak and bandwidth can be inferred. (4)
Ans :
C(s)
2
----- =
---------------------------
R(s) s2
+ 2 ξs+2
1
M(jw) =
-------------------------------
(1-w2/2)
+ j2 ξ(w/)
1
=
-----------------------------
(1-w2/2)2
+ 4 ξ2(
/2)
d -4(1-w2/2)(w/2) + 8 ξ2(w/2)
------ =
-----------------------------------------------------
dw
[(1-w2/2)2
+ 4 ξ2(w2/2)]2
= (1-2 ξ2)1/2
for this.,
1
= -----------------
2
ξ( 1- ξ2)1/2
Band width
For M = 1/1.414
is BW
1
-----------------------------------------
= 
(1-
2/2)2
+ 4 ξ2(2/2)
(4/4) -2(1-2
ξ2)(2/2) – 1 = 0
2/2 = (1-2
ξ2) 
(4
ξ4 - 4 ξ2 +2)1/2
= ( (1-2 ξ2) + (4
ξ4 -4 ξ2 + 2)1/2)1/2)
Q.7 The
characteristic equation of a closed loop control system is given as 
. For this system
determine the number of roots to the right of the vertical axis located at s =
- 2. (10)
Ans : s4 +10s3
+35s2 +50s +24 = 0
shift origin s =
-2
so, s = z-2
(z-2)4
+ 10(z-2)3 + 35(z-2)2 + 50(z-2) + 24
= (z2 +
4 -4z)2 + 10(z3 -6z2+12z -8) + 35(z2
+ 4 – 4z) + 50z -100 +24
= z4 +
16 + 16z2 +8z2 – 32z – 8z3 + 10z3
-80 – 60z2 +120z +35z2 -140z +140-50z -
100+24
=z4 +2z3
–z2 – 2z = 0
|
z4
|
1 -1 0
|
|
z3
|
2 -2 0
|
|
z2
|
0 0
|
A(z) = 2z3
– 2z
dA(z)
-------- = 6z2
– 2
dz
|
z3
|
2 -2 0
|
|
z2
|
3 -1 0
|
|
z1
|
-4/3
|
|
z0
|
-1
|
There is one sign
change so one root to right of s = -2
Q.8 Draw
the complete Nyquist plot for a unity feed back system having the open loop
function 
.
From this plot obtain all the information regarding absolute as well as relative
stability. (14)
Ans :
6
G(s)H(s) =
----------------------
s(1+0.5s)(6+s)
G(jw) at w = 0 = 
G(jw)
at w→∞ = 
6
=
---------------------------- = 6(-0.5w2 – jw)
(-0.5w2 + jw)(jw + 6)
Nyquist contour is
given by Semicircle around the origin represented by
varying from -900
through 00 to 900
Maps into
6
Lim
--------------------------------------- = ∞
→0 6 
(1+0.5)(1+1/6)
-900
through 00 to 900
mapping of
positive imaginary axis (w =0+ to ∞+)
calculate magnitude
and phase values of T.F
6
----------------------------
at various values of w
6jw(1+0.5jw)(1+1/6jw)
|
|
1
|
10
|
50
|
100
|
500
|
|
magnitude
|
-1.085
|
-39.89
|
-80.389
|
-98.39
|
-140.32
|
|
phase
|
234
|
132.37
|
99.16
|
94.59
|
90.91
|
Mapping of
infinite semicircular arc of the nyquist contour represented by
( varying from +90
through 0 to -90 as R→ ∞)
=
0 e-j3 Ǿ
-2700 through 00 to -2700
NYQUIST PLOT
The system is
stable and the relative stability is represented by phase margin and gain
margin.
Gain margin = 18.1
,Phase cross frequency = 3.46 rad/sec.
Phase margin =
57.2 , Gain cross frequency = 0.902 rad/sec
Q.9 Sketch
the root locus diagram for a unity feedback system with its open loop function
as 
. Thus
find the value of K at a point where the complex poles provide a damping factor
of 0.5. (14)
Ans :
K(s + 3)
G(s) =
-------------------------------------
s(s2
+2s +2)(s+5)(s+9)
Location of poles
and zeros
s2 +2s
+2 = 0 
= -1± j1
poles s=0, -1±j1,
-5,-9
zeros s = -3
±180(2q + 1)
angle of
asympototes = -----------------
(n
– m)
n =5, m = 1
± 45 , ± 135
-5-9-1-1-(-3)
centroid =
------------------ = -13/4
4
Break away point
C(S)
k(s+3)
----- =
----------------------------------------
R(s) s(s2
+ 2s+2)(s+5)(s+9) +k(s+3)
Characterstic
equation;
(s3 +
2s2 + 2s)(s2 + 14s +45) + k(s+3) =0
s5 +14s4
+45s3 +2s4 +28s3 +90s2 +2s3
+28s2 +90s)+k(s+3) =0
-(s5
+16s4 +75s3 +118s2+90s)
k =
----------------------------------------
(s+3)
dk -(s+3)(5s4
+64s3+22s2+236s+90)+(s5+16s4+75s3+118s2+90s)
---- =
----------------------------------------------------------------------------- =
0
ds
(s+3)2
(4s5+63s4+342s3+793s2+708s+270)=0
s = -7.407,
-3.522±j1.22,-0.64±j0.48
-7.4 is the
breaking point
Angle of
departure at A
=900
= 1800-tan-1(1/1)
= 1350
= tan-1(1/4)
= 14.030
= tan-1(1/8)
= 7.1250
A = tan-1
(1/2)= 26.560
1800 –
(900+1350+14.030+7.1250) +26.560=
-39.60
A* =39.60
To find k at
ξ =0.5
= cos-10.5
=60
product of length of vector from all poles to point
value of k
=--------------------------------------------------------------
product of length from all zeros to point
=
14.7 
Figure:
Root Locus Diagram
Q.10 Obtain
the transfer function of the two-phase servo–motor whose torque–speed curve is
shown in Fig.2. The maximum rated fixed-phase and control–phase voltages are
115 volts. The moment of inertia of the motor (including the effect of load)
is 
Kgm-
. Motor friction
(including the effect of load) is negligible. (7+7)
Ans : The
transfer function of a two phase servo motor is given by
Ө(s)
kc
----------- =
----------------------- ---- (1)
Ec(s)
Js2+ (f + kn) s
Where kc
= Torque per unit control voltage at zero speed = 5/115 = 1/23.
kn = -ve of the slope of the
torque speed curve
=5/4000 = 1/800.
J = Moment of Inertia of motor and load
= 7.77 x 10-4 kg-n /m2
f = viscous friction coefficient of
motor and load = 0
Putting these values in eqn (1) we have the
Ө(s)
1/23
T = ------------
= ----------------------------
Ec(s)
7.77 x 10-4 s2 + 1/800 s
Q.11 Obtain the unit – step response
of a unity feedback control system whose open – loop transfer function is 
. Obtain also the rise
time, peak time, maximum overshoot and settling time. (6+8)
Ans :
1 1
=
----------------- = -------------------------- ---(1)
s2+s+1 s2 + 2 ζ wn s +
wn2
1 1 A Bs+C
= ---- [ -------------------] = ------ + -----------------
s s2 + s +1 s s2
+ s + 1
A=1
1
1 1 Bs+C
---- [
-------------------] = ------ + -----------------
s
s2 + s +1 s s2 + s + 1
1
= s2 +s + 1 + (Bs+C)s
= (1+B) s2 +
(1+C) s + 1
B +
1 = 0 B= -1
C +
1 = 0 C= -1
1 s+1
C (s) = ---- -
[------------------]
s s2 + s +1
1 s+1
= ----- -
[---------------------]
s (s+1/2)2 + ¾
1 s+1/2 1/2
= ----- -
[---------------------] - [-------------------]
s (s+1/2)2 + ¾ (s+1/2)2 +
¾
1 s+1/2 ½ (3/4)
= ----- -
[---------------------] - [-------------------------]
s (s+1/2)2 + ¾ 3/4 [ (s+1/2)2 + ¾
]
c(t) = 1 – e-1/2t
cos 
– 
e-1/2t sin
wn
= 1
rise time = 
= 2.41 sec
peak time =
= 3.62 sec
% Mp = e (-1/2
∏ /(1- ¼)1/2 ) * 100 = 16.3%
= 4/(1/2) = 8 sec
Q.12 Consider the closed-loop system
given by 
.
Determine the values of 
and 
so that the system responds to a step
input with approximately 5% overshoot and with a settling time of 2 seconds
(use the 2% criterion). (7)
Ans :
C(s)
wn2
-------- =
------------------------
R(s) s2
+ 2
wns
+ wn2
%Mp=5% = e(-
/
) Given;
= 0.698
for 2% = 
= 2 sec
wn
= 2.89 rad/sec
Q.13 Obtain the unit-impulse response
of a unity feedback control system whose open loop transfer function is 
. (7)
Ans : 2s+1
G(s) = -----------
s2
G(s) 2s + 1
M(s) = ------------------
= ---------------------
1 +
G(s) s2 + 2s + 1
2s + 1
C(s) = --------------
s2 +
2s + 1
2s + 1 2s
1
= -------------- =
------------------- + -------------
(s + 1)2
(s + 1)2 (s + 1)2
= 2 [ (s+1)/(s+1)2
– 1/(s+1)2] + (1/(s+1)2
C(t) = 2 [ e-t –
t e-t] + t e-t
= (2-t)e-t
Q.14 Determine the range of K for
stability of a unity-feedback control system whose open-loop transfer function
is 
. (7)
Ans :
Characteristic equation:
s(s+1)(s+2) + k = 0
= (s2
+ s) (s+2) + k
= s3
+2s2 + s2 + 2s + k = s3 +3s2
+ 2s + k =0
Routh array:
|
s3
s2
s1
s0
|
1
2
3
k
(6 – k ) /3 0
k
|
For stability .,
6 – k
> 0 k < 6
range of k for
stability
0
< k < 6
Q.15 Comment on the stability of a
unity-feedback control system having the open-loop transfer function as 
. (7)
Ans :
10
G(s) =
-------------------
s(s-1)(2s
+ 3)
Poles s =
0, 1,-1.5
± 180 (2q +1)
Angle of
asymptotes = ----------------------
(n-m)
± 60 , ± 180
1 – 1.5
centroid =
------------- = -0.25
2
C(s)
k
------ =
--------------------------
R(s)
s(s-1)(2s + 3 ) + k
(s2-s)(2s
+ 3) + k =0
k = -(2s3
+ 3s2 – 2s2 -3s) = -(2s3 +s2 -3s)
dk/ds = -(6s2
+2s-3) = 0 
s
=0.5598
ROOT LOCUS FIGURE
From the root
locus, we find that for any gain the system has right half poles. So the
closed loop system is always unstable.
Q.16 Sketch the root loci for the system
with 
. (14)
Ans : K
G(s) =
------------------------------- H(s)=1
s(s+0.5)(s2 + 0.6s + 10)
Poles
s=0,-0.5, -0.3j
3.14
Asymptotes 45 deg, 135 deg
-0.5-0.3-0.3
Centroid =
------------------------ = -0.275
4
Break away points
s(s+0.5)(
+0.6s+10) + k = 0
K = -(s2
+0.5) (s2 +0.6s+10) = -(s4 + 0.6s3 + 10.5s2
+ 0.3s + 5)
dk/ds
= 0 s = -0.25
angle of departure
A = 90 + tan-1(3.14/0.2)
+ 180 – tan-1(3.14/0.3) = 268
A* = -268
FIGURE ROOT LOCUS
Q.17 A unity feedback control system
has the open-loop function as 
. Obtain the response of the system with a
controller transfer function as 
and with a bit step input. (8)
Ans :
5
. G(s) = ------------------
(2s+3)
s(s + 2)
G(s)
M(s) =
------------------
1+
G(s) H(s)
5
------------------ (2s+3)
s(s
+ 2)
=
-------------------------
5
1 +
------------------ (2s+3)
s(s + 2)
10s+15
=
------------------
s2+12s+15
C(s)
10s+15
----------- =
------------------
R(s) s2+12s+15
10s+15 1
C(s) =
------------------ . ----
s2+12s+15 s
10s+15
C(s) =
------------------
s(s2+12s+15)
10s+15
C(s) =
------------------
s(s+10.58)(s+1.4174)
applying
inverse laplace transform
c(t) = -0.9367*exp(-10.58*t)-
0.0636*exp(-1.4174*t)+ 1.0003
Q.18 Consider a unity feedback
control system with the following open-loop transfer function 
. Determine the value
of the gain K such that the phase margin is 
. What is the gain margin for this case? (8+6)
k
Ans : G(s)
= ------------------
s(s2 +
s + 4)
|
|
0.2
|
0.4
|
1
|
5
|
10
|
|
l G(jw) l
|
2
|
-3.73
|
-10
|
-40
|
59.7
|
|
< G(jw)
|
-92.9
|
-96
|
-108
|
-256
|
-264
|
For k= 1 GM =
12dB
Ø gc = 50 (deg)
-180 (deg)= -130(deg)
At -130 deg gain
is = -10.8
20 log k = 10.8
k = 1.79
For the
corresponding k we have GM = 6.98 dB
FIGURE
BODE PLOT
Q.18 Consider the system shown in
Fig.4. Draw the Bode-diagram of the open-loop transfer function 
with K = 1. Determine
the phase margin and gain margin. Find the value of K to reduce the phase
margin by 
. (14)
Ans :
k
G(s) =
----------------------
s(s+1) (s+10)
|
|
0.01
|
0.1
|
1
|
10
|
|
l
G(jw) l
|
20
|
0
|
-23.2
|
-63.2
|
|
<G(jw)
|
-91.1
|
-96.7
|
-141
|
-220
|
GM = 40.8
dB PM = 83.7
Ø gc = 73.7 – 180 =
-106.3
Gain at -106.3 is
-8.64
20 log k =
8.64 k = 1.54
FIGURE BODE PLOT
Q.19 For the
system whose signal flow graph is shown by Fig.1, find 
(8)
Ans:
There are no non
touching loops
Δ = 
P1
Δ1 + P2 Δ2
T =
----------------------
Δ
Transfer
function=
Q. 20 Determine the values of K and p
of the closed-loop system shown in Fig.2 so that the maximum overshoot in Unit
Step response is 25% and the peak time is 2 seconds. Assume that J=1 Kg-
. (14)
Ans :
k
-----------------
C(s)
(kp+Js)s
k k/J
-------- =
---------------------------- = ------------------------ =
--------------------
k
Js2 + kps + k s2
+ kp/Js +k/J
Y(s)
1 + --------------
(kp+Js)s
k
G(s) =
------------------ J = 1 kg-m2
s2
+kps + k
25% = peak over
shoot
which gives ξ
= 0.403
peak time =
---- = 2 
= 3.14/2
= 1.71 2 = k
k=2.94 2 ξ =kp
2 ξ
2 ξ
p =
--------- = ----------- = 0.47
k 2
Q.21 Obtain the Unit-step response of a unity-feedback
whose open-loop transfer function is
Find also the steady-state value of the Unit-Step
response. (10+4)
Ans :
5(s+20)
G(s) =
-----------------------------------------
s(s+4.59)(s2 +3.41s+16.35)
G(s) 5(s+20)
M(s) =
-------------- = ------------------------------------------------------
1
+ G(s) (s2+4.59s)(s2+3.41s+16.35)+5s+100
5(s+20)
=
-----------------------------------
s4+8s3+32s2+80.04s+100
C(s)
----- =
M(s) C(s) = 1/sM(s)
R(s)
5s+100
C(s) =
-------------------------------------------
s(s4+8s3+32s2+80.04s+100)
Applying inverse laplace transform
C(t) = 1+3/8e-tcos(3t)
– 17/24 e-tsin(3t) - 11/8e(-3t) cos(t) -13/8 e-3tsin(t)
s(5s+100)
C(s)= lim
c(t) = lim sC(s) = lim
-------------------------------------------
t→
∞ s→0 s→0 s(s4+8s3+32s2+80.04s+100)
100
= ------ =
1
100
Q.22 Consider
the characteristic equation
Using
the Routh’s stability criterion, determine the range of K for stability. (8)
Ans :
s4 +2s3+(4+k)s2+9s+25
= 0
Routh array is:
|
s4
|
1
(4+k) 25
|
|
s3
|
2
9 0
|
|
s2
|
(-1+2k)/2
25 0
|
|
s1
|
(18k-109)/(2k-1)
0
|
|
s0
|
25
|
For stability:
-1+2k>0
2k>1
k>1/2
18k -109
>0 k>109/18
so, the range of k
for stability is k>6.05
Q.23 Find the number of roots of
characteristic equation which lie in the right half of
s-plane for K =100. (6)
Ans :
For k= 100
The
Characteristic equation is given as
s4
+2s3+104s2+9s+25 =0
|
s4
|
1
104 25
|
|
s3
|
2
9 0
|
|
s2
|
49.5
25 0
|
|
s1
|
8.49
0
|
|
s0
|
25
|
As the first
columns are all positive
So., no roots are
in right half.
Q.24 Sketch
the root loci for the system shown in Fig.3 (14)
Ans :
k
G(s) =
----------------------
s(s+1)(s2+4s+13)
Poles
s=0, -1,-2±j3
n=4,m=0
± 180(2q+1)
angle of
asymptotes = ----------------- = ±45, ±135
(n-m)
-2-2-1
centroid =
-------------- = -1.25
4
4
M(s) =
------------------------
(s2+s)(s2+4s+13)+k
The Characteristic
equation is given as s4+4s3+13s2+s3+4s2+13s+k
= 0
k =
-(s4+5s3+17s2+13s)
dk/ds = -(4s3+15s2+34s+13)
= 0
so, s= -0.4664,
-1.6418±j2.067
so, the break away
point is -0.4664
Angle of
departures
For complex poles
Ø1= 90, Ø2 = 180
–(3/1), Ø3 =
180 – (3/2)
So., at A
=
– (Ø1+ Ø2+
Ø3)
=
-
and A* =
Figure
Root Locus
Q.25 The
forward path transfer function of a Unity-feedback control system is given as
Draw the Bode plot of G(s)
and find the value of K so that the gain margin of the
system
is 20 dB. (14)
k
Ans : G(s)
= -----------------------
s(1+0.1s)(1+0.5s)
k
G(jw) = -----------------------
jw(1+0.1jw)(1+0.5jw)
corner
frequencies:2,10.
|
|
1
|
2
|
10
|
20
|
100
|
|
Magnitude
db/dec
|
-20
|
-40
|
-60
|
-60
|
-60
|
|
phase
|
-122.27
|
-146.3
|
-216
|
-238
|
-263
|
Let k =1 then GM
at this k = 20 dB
Q.26 The loop transfer function of a
single feedback-loop control system is given as
Apply the Nyquist criterion and determine the range of
values of K for the system to be stable. (14)
Ans :
k 0.05k
G(s)H(s)= -------------------- = ---------------------
s(s+2)(s+10) s(1+0.5s)(1+0.1s)
mapping 
w → 0 to
+∞
0.05k
G(jw)H(jw)
= ------------------------------
-0.6w2
+ jw(1-0.05w2)
at w=
imaginary is zero
1-0.05w2
=0 =4.472
rad/sec.
at
GH = -.00417k
Mapping 
K
GH(s) = ------ =
0
at s= lim Rej Ө
s3 R→ ∞
When
Ө = /2
G(s)H(s) = 0
Ө =- /2
G(s)H(s) = 0
Mapping 
W → -
∞ to 0
It is mirror image
of mapping
Mapping 
For s= lim
Rej Ө
R→0
Ө = - /2
G(s)H(s) = ∞
Ө = /2
G(s)H(s) = ∞
FIGURE: NYQUIST PLOT
Range of values of
k for the system to be stable. Limiting value of k
-0.00417k = -1 k = 240
So when k<240 ,
the closed loop system is stable.
Q.27
(a) The transfer functions for a
single-loop non-unity-feedback control system are given as
Find
the steady-state errors due to a unit-step input, a unit-ramp input and a
parabolic input.
(9)
(b) Find also the impulse response
of the system described in part (a). (5)
Ans :
(a)
1 1
G(s) =
---------------- H(s) = --------
s2 +
s+2 (s+1)
R(s)
E(s) =
------------------ = lim sE(s)
1 +
G(s)H(s) s→0
for unit step
input
s*1/s
= lim
-----------------------------
s→0 1 + (1/s2+s+2)(1/s+1)
1
=
----------------------- = 2/3
1 +( ½)
for unit ramp
input
s*1/
= lim
------------------------ = ∞
s→0 1 + G(s)H(s)
for unit parabolic
input
s*1/s3
= lim
------------------- = ∞
s→0 1 + G(s)H(s)
C(s)
G(S) 1/(s2 + s+2)
----- =
--------------------- = ---------------------------------
R(s) 1+
G(S)H(s) 1 + (1/s2+s+2)(1/s+1)
(s+1)
=
--------------------------------
(s2+s+2)(s+1)+1
(s+1)
=
-----------------------------
s3+2s2+3s+3
(b). Impulse response of the system:
Q.28
Derive the transfer function of the op amp circuit
shown in Fig.3. Also, prove that the circuit processes the input signal by
‘proportional + derivative + integral’ action. (9)
Ans :
Q.29 The electro hydraulic position
control system shown in Fig.4 positions a mass M with negligible friction.
Assume that the rate of oil flow in the power cylinder is 
where x is the displacement of
the spool and 
is
the differential pressure across the power piston. Draw a block diagram of the
system and obtain therefrom the transfer function 
.
The system constants are given below.
Mass M = 1000 kg
Constants of the hydraulic actuator:
= 200 cm2/sec per cm of
spool displacement
= 0.5 cm2/sec per gm- 
t/ cm2
Potentiometer sensitivity KP = 1
volt/cm
Power amplifier gain KA =
500 mA/volt
Linear transducer constant K = 0.1 cm/mA
Piston area A = 100 cm2 (14)
Ans :
Q.30 A servo system is represented
by the signal flow graph shown in Fig.5. The nominal values of the parameters
are 
.
Determine the overall
transfer function
and its sensitivity to changes in under steady dc conditions, i.e., s = 0. (14)
Ans :
Fig: Signal flow graph
One forward path with gain 
Three feedback loops with gains
So,
since all loops are
touching.
Q.31 Determine
the values of K>0 and a>0 so that the system shown in Fig.6 oscillates at
a frequency of 2 rad/sec. (10)
Ans :
|
|
|
|
s3
|
1
(2+k) 0
|
|
s2
|
a
1+k 0
|
|
s1
|
(2+k)-(1+k)/a
0
|
|
s0
|
1+k
|
From the
Routh’s array ;we find that for the system to oscillate (2+k)a = 1+k
Oscillation frequency = 
These equations give a = 0.75 , k=2
Q.32 Consider the control system
shown in Fig 7 in which a proportional compensator is employed. A specification
on the control system is that the steady-state error must be less than two per
cent for constant inputs.
(i)
Find Kc that satisfies this specification. (5)
(ii) If the steady-state criterion cannot
be met with a proportional compensator, use a dynamic compensator 
. Find the range of 
that satisfies the
requirement on steady-state error. (9)
Ans : (i).
|
s3
|
1
5 0
|
|
s2
|
4
2+2Kc 0
|
|
s1
|
(18-2Kc)/4
0 0
|
|
s0
|
2+2Kc 0
|
|
|
|
The system is stable for Kc
< 9.
Kp
= 
D(s)G(s) =
Kc; =
1/(1+Kc); =
0.1 (10 %) is the minimum
possible value for
steady state error. Therefore less than 2 % is not
possible
with proportional
compensator.
(ii) Replace Kc by D(s) = 3 + Ki/s . The
closed-loop system is stable for 0 < Ki <3.
Any value in this
range satisfies the static accuracy requirements.
Q.33 Use the Nyquist
criterion to determine the range of values of K>0 for the stability of the
system in Fig. 8. (9)
Ans :
= 
The imaginary
part is equal to zero if 
= 0 ; this gives
w value as -tan
(0.8w); solving this equation for smallest positive value of w,
we get ,w =
2.4482
= 1+j0 ; 
= 0 and 
= -0.378+j0
The polar plot
spiral into the 
point
at the origin
The critical value
of K is obtained by letting G(j2.4482) equal to -1.This
gives
K=2.65
The closed loop
system is stable for K < 2.65.
Q.34 A unity-feedback system has
open-loop transfer function 
.
(i)
Using Bode plots of 
, determine the phase margin of
the system.
(ii) How should the gain be adjusted so that phase margin is 50°?
(iii) Determine the bandwidth of gain-compensated system.
The –3dB contour of the Nichols chart may be constructed using the
following
table. (10)
|
Phase, degrees
|
0
|
-30
|
-60
|
-90
|
-120
|
-150
|
-180
|
-210
|
|
Magnitude, dB
|
7.66
|
6.8
|
4.18
|
0
|
-4.18
|
-6.8
|
-7.66
|
-6.8
|
Ans :
(i). From the bode plot we get PM
= 11.4 deg
(ii). For the phase margin of 50
deg , we require that mag(G(jw)H(jw)) =1 and
ang(G(jw)H(jw)) = -130 deg
, for some value of w ,from the phase curve of
G(jw)H(jw) ,we find that
ang(G(jw)H(jw)) = -130 deg at w = 0.5. The magnitude
of G(jw)H(jw) at this
frequency is approximately 3.5 .The gain must be reduced by a
factor of 3.5 to achieve a
phase margin of 50 deg .
(iii).On a linear scale graph sheet , dB Vs phase curve of open loop frequency
response is
plotted with data coming from bode plot .On the same graph sheet ,the -3 dB
contour
using the given data is plotted . The intersection of the curves occurs at w =
0.911 rad /
sec . Therefore band width Wb = 0.911 (Nichols chart is not required).
Q.35 Discretize the PID controller
to obtain PID algorithm in
(i)
position form
(ii) velocity form.
What are the advantages of velocity PID algorithm over
the position algorithm? (14)
Ans :
The PID controller
can be written as
It can be
realized as
Position PID
algorithm:
If the S(k-1)
approximates the area under the e(t) curve up to t = (k-1)T,
Then the
approximation to the area under the e(t) curve up to t = kT is given by
S(k)=S(k-1)+Te(k)
Velocity PID
Algorithm :
The PID algorithm
can be realized by following equation.
subtracting above
two equations
Now only the
current change in the control variable
is calculated.
Compared to ‘position form’, the ‘velocity form’ provides a more
efficient way to program the PID algorithm from the standpoints of handling
the initialization when controller switched from ‘manual’ to ‘automatic’, and
of limiting the controller output to prevent reset wind-up. Practical
implementation of this algorithm includes the features of avoiding derivative
kicks/filtering measurement noise.
Q.36 The open-loop transfer function
of a control system is
(i) Draw the Bode plot
and determine the gain crossover frequency, and phase and gain margins.
(ii) A lead compensator
with transfer function
is now inserted in the
forward path. Determine the new gain crossover frequency,
phase margin and gain margin. (6+8)
Ans : (i)
From the bode plot
of G(jw)H(jw) ; we find that wg= 4.08 rad/sec:
PM
= 3.9 deg ; GM = 1.6 dB
(ii)
From the new bode plot ,we find that
Wg = 5 rad/sec ; PM = 37.6 deg ;
GM = 18 dB, The increase in
phase margin and gain margin implies that lead
compensation
increases margin of stability .The increase in Wg implies that lead
compensation increases the
speed of response.
PART – III
DESCRIPTIVES
Q.1 A
typical temperature control system for the continuously stirred tank is given
in Fig.6. The notations are 
for temperature, q for liquid flow and 
for the steam supplied
to the steam coil. Draw the block diagram of the system. (10)
Ans :
Q.2 Considering a typical feedback
control system, give the advantages of a P+I controller as compared to a purely
proportional controller. (4)
Ans :
P + I controller as compared to a purely proportional controller :
If the controller
is based on proportional logic, then in the new steady state, a non zero errormust exist to get a non
zero value of control signal 
. The operator must then reset the set
point to bring the output to the desired value. We need a controller that
automatically brings the output to the set point which is done by PI, which
gives a steady control signal with system error = 0.
Q.3 Explain the procedure to be followed
when in the Routh’s array all the elements of a row corresponding to 
are zeros. (4)
Ans : When all the elements of a row
corresponding to s4 are zeros, then write the auxiliary equation.
This is an equation formed by the coefficients of the row just above the row
having all zeros. Here it means make an equation from the row corresponding to
s5. Differentiate this equation and replace the entries in Routh’s
array by the coefficients of this differential equation. Then follow the usual
procedure.
Q.4 Justify the following statements
:
(i)
The impulse response of the standard second order
system can be obtained from its unit step response.
(ii)
The Bode plot of the standard second order
function 
has
a high frequency slope of 40 db / decade.
(iii)
The two phase a.c. servomotor has an inherent
braking effect under zero-control-voltage condition.
(iv)
An L.V.D.T. can be used for measuring the
density of milk. (4x3.5)
Ans :
(i) The impulse response of the standard second order system can
be obtained from its unit step response by integrating it.
(ii) 2
----------------------
s2
+ 2ξs
+ 2
put s = jw
1
---------------------------
1 +
j(2ξ)
–2/2
let w/= u (normalized
frequency)
1
=
-----------------
1 + j2ξu
–u2
dB = 20 log (
1/(1-u2 + j2ξu))
At low frequency
such that u<<1 magnitude may be approximately dB = -20 log 1 = 0
For high frequency
such that u>>1 the mag may be approximately dB = -20 log(u2)=
-40log u Therefore an approximately mag plot for T.F consists of two straight
line asymptotes, one horizontal line at 0 dB for u<=1and other line with a
slope -40dB/dec for u>=1.
(iii)
In two phase AC
servo motor there are two supplies. One with constant voltage supply to the
reference phase and second one is control phase voltage which is given through
servo amplifier.
Torque speed
characteristics
All these curves
have –ve slope and these always depends on control voltage.
Note that the
curve for zero control voltage goes through the origin and the motor develops a
decelerating torque which means braking.
Q.5 Write
notes on any TWO of the following:
(i) Disturbance rejection.
(ii) Turning method based on the process
reaction curve.
(iii) Phase lag compensation. (2
x 7)
Ans :
(i) Disturbance rejection :
The
degree of disturbance rejection may be expressed by the ratio of 
Gyd (the closed loop
transfer function between the disturbance d and the output y ) and 
(the feed forward
transfer function between the disturbance d and the output y) or
To improve the
disturbance rejection, make 
small over a wide frequency range
(ii) Phase lag compensator:
It has a pole at
-1/ b
and a zero and -1/
with zero located to the left of pole on the negative real axis.
(s + 1/ )
(s) = -------------
(s + 1/b)
(s) R2
+ 1/sC
------- =
---------------------
(s) R1
+ R2 + 1/sC
= R2C b = (R1
+ R2)/R
Q.6 Compare the response of a P+D
controller with that of a purely proportional controller with unit step input,
the system being a type–1 one. (6)
Ans :
k
G(s) =
------------------ Let us consider a type 1 system
s(s+a)
If considering
only proportional controller we can be able to improve the gain of the system.
The steady state error due to step commands can theoretically be eliminated
from proportional control systems by intentionally misadjusting the input
value.
By increasing the
loop gain the following behavior exists.
1. Steady state
tracking accuracy.
2 Disturbance
rejection
3. Relative
stability. i.e. rate of decay of the transients.
All aspects of
system behavior are improved by high gain proportional.
By P.D.
controller:
We can
able to improve the steady state accuracy using a P.D. controller, improvement
of relative stability.
The
speed of response is also very fast using this type of controller.
Q.7 Write
short notes on the following:
(7+7)
(i) Controller tuning
(ii) Phase-lead compensation
Ans :
(i) Controller tuning
Considering the basic control
configuration, wherein the controller input is the error between the desired
output and the actual output.
METHOD 1.
This method is applicable only if dynamic model of plant is not
available and step response of the plant is S-shaped curve .Then tuning done by
Zeigler –Nichols method.
Procedure: Obtaining the step response, from that find the dead
time (L) and time constant (T)
Then the values of controller is given by
Proportional gain (
) =1.2(T/L)
Integral time constant =2L
Derivative time constant=0.5L
Note: There are no specific tuning method available if plant model
is not known and step response is not S-shaped curve.
METHOD 2.
This method is applicable only if dynamic model of plant is known
and no integral term in the transfer function.
Procedure: Find the critical gain (K) and critical time period (T)
at which the system is oscillating using routh array or root locus.
Then the controller parameters are given by
Proportional gain = 0.6 K
Integral time constant=0.5 T
Derivative time constant= 0.125 T
METHOD 3.
This method is applicable only dynamic model of the plant, having an
integral term in the transfer function. If the critical period and the gain
cannot be calculated, then tuning is done through root locus method.
(ii) Phase lag compensator
Consider
the following circuit. This circuit
Has the following
transfer function
Eo
T2 ( 1 + T 1 s)
----------- = -------------------------- (1)
E1 T1 ( 1 + T 2 s)
Where
T1 = R1 C1 and T2 = R2/
(R1 + R2) T1.
Obviously T1
> T2. For getting the frequency response of the network, but s =
jw i.e.,
(2)
And phase ф
= tan-1 wT1 - tan-1 wT2
Let us consider the polar plot for this transfer function as shown
in figure below. We can observe that at low frequencies, the magnitude is
reduced being T2 / T1 at w=0.
Next, let us
consider the bode plot the transfer function as shown in figures below.
We observe here
that phase ф is always positive. From magnitude plot we observe that
transfer function has zero db magnitude at w= 1 / T2 .
We can
put dф / dw =0 to get maximum value of ф which occurs at some
frequency wm
i.e., wm =
1 / (
)
фmax
= tan-1 [ T1 / T2 ]1/2 - tan-1
[ T2 / T1 ]1/2
In this network we
have an attenuation of T2 / T1 therefore, we can use an
amplification of T1 / T2 to nullify the effect of
attenuation in the phase network.
Q.8 Write
short notes on the following :
(i) Constant M and N circles.
(ii) Stepper Motor.
(2 x 7 = 14)
Ans.
(i) Constant M and N circles:
The magnitude of closed loop transfer function with unity feedback
can be shown to be in the form of circle for every value of M. These circles
are called M-circles.
If the phase of closed loop transfer function with unity
feedback is α, then it can be shown that tan α will
be on the form of circle for every value of α. These
circles are called N- circles.
The M and N circles are used to find the closed loop
frequency response graphically form the open loop frequency response G(jw) without
calculating the magnitude and phase of the closed loop transfer for at each
frequency .
For M circles:
Consider the closed loop transfer function of unity
feedback system.
C(s) / R(s) = G(s) / 1+G(s)
Put s = jw;
C(jw) /R(jw) =G(jw) / 1 + G(jw)
( X + M2 / M2-1
)2 + γ2 = ( M2 / M2-1 )2 ---- (1)
The equation of circle with centre at (X1,Y1)
and radius r is given by
( X - X1)2 + (Y – Y1
)2 = r2 -----(2)
On comparing eqn (1) and (2)
When M = 0;
X1 = - M2
/ M2-1 =0
Y1 = 0
R = M / M2-1 = 0;
When M = ∞
X1 = - M2 / M2-1 = -1
Y1= 0;
R = M / M2-1 =
1/M = 0
When M = 0 the magnitude circle becomes a point at (0, 0)
When M=∞, the magnitude circle becomes a point at (-1, 0)
From above analysis it is clear that magnitude of closed loop
transfer function will be in the form of circles when M ¹ 1 and when M=1
, the magnitude is a straight line passing through (-1/2 , 0).
Family of N- circles:
For constant N circles
Tan =
N = 
Constant N-circles have centre as
Family of N- circles
(ii) Stepper
Motor:
A stepper motor transforms electrical pulses into equal increments
of shaft motion called steps. It has a wound stator and a non-excited rotor.
They are classified as variable reluctance, permanent magnet or hybrid,
depending on the type of rotor.
The no of teeth or poles on the rotor and the no of poles on the
stator determine the size of the step (called step angle). The step angle is
equal to 360 divided by no of step per revolution.
Operating
Principle:
Consider a stepper motor having 4-pole stator with 2-phase windings.
Let the rotor be made of permanent motor with 2 poles. The stator poles are
marked A, B, C and D and they excited with pulses supplied by power
transistors. The power transistors are switched by digital controllers a
computer. Each control pulse applied by the switching device causes a stepped
variation of the magnitude and polarity of voltage fed to the control windings.
Stepper
motors are used in computer peripherals, X-Y plotters, scientific instruments,
robots, in machine tools and in quartz-crystal watches.
Q.9 Define the transfer function of a linear time-invariant system
in terms of its differential equation model. What is the characteristic
equation of the system? (5)
Ans :
The input-
relation of a linear time invariant system is described by the following
nth-order differential output equation with constant real coefficients:
“Transfer
function is defined as the ratio of Laplace transform of output variable to
the input variable assuming all initial condition to be zero.”i.e.
T(s)=Y(s)/U(s)=sm+bn-1sm-1+ …………………..+b0/sn+an-1sn-1+…….+a0
The characteristic
equation is given by
Q.10 Define the terms:
(i)
bounded-input, bounded-output (BIBO) stability,
(ii)
asymptotic stability. (2+2)
Ans :
(i) Bounded Input Bounded Stability :With
zero initial conditions,
the system is said to be bounded input bonded output (BIBO) stable, if its
output y(t) is bounded to a bounded input u(t).
(ii) Asymptotic Stability : A linear
time-invariant system is asymptotically stable if for any set of finite 
,there exists a positive
number M,which depends on , such that
Q.11 Discuss
the compensation characteristics of cascade PI and PD compensators using root
locus plots. Show that
(i) PD compensation is suitable for systems having
unsatisfactory transient response, and it provides a limited improvement in
steady-state performance.
(ii) PI compensation is suitable for systems with
satisfactory transient response but unsatisfactory steady-state response.
(7+7)
Ans. PI
Compensation : When a PI compensator is used in
cascade, it increases the type of the system. With the root loci, we can see
that steady state error is reduced once the type of the original system is
increased. No satisfactory improvement in the transient performance is noted.
kp ,kv
and ka values increase with the type of the system.
PD Compensation: When a PD compensator is used in cascade (forward path),it results
in the addition of an open loop zero in the original system.. Since addition of
open loop zeros results in the leftward movement of the root loci, this
effectively means the increase in the value of the damping ratio which improves
the transient response but no satisfactory improvement in the steady state
response is seen
Q.12 State and explain the Nyquist stability criterion. (5)
Ans :
If the contour of
the open-loop transfer function G(s)H(s) corresponding to the Nyquist contour
in the s-plane encircles the point (-1 +j0) in the counterclockwise
direction as many times as the number of right half s-plane poles
of G(s)H(s),the
closed-loop system
is stable.
In the commonly
occurring case of the open-loop stable system ,the
closed-loop system
is stable if the counter of G(s)H(s) does not encircle (-1 +j0)
point, i.e., the
net encirclement is zero.
Q.13 When
is a control system said to be robust? (4)
Ans :
A control system
is said to be robust when
(i) It has low
sensitivities
(ii) It is stable
over a wide range of parameter variations; and
(iii) The
performance stays within prescribed (but practical) limit bounds in presence
of
changes
in the parameters of the controlled system and disturbance input.
is _______.
(B)
(D)
, the meeting point of the asymptotes on
the real axis occurs at _________.
(B) 
(D)
then the steady state
value of the error works out as ______.
(B) 
(D)
. (B)
.
. (D)
s.
has
of a closed-loop system
encloses the 
point
in the 
, where 
is the Laplace transform 
is equal to 
.
(B) 
(D) 
,
its characteristic equation is given by 
(B)
(D)
the value of K for a damping
ratio of 0.5 is 
is
equal to 
. (B)
.
. (D)
.
is a two-dimensional state vector and A is
a matrix given by 
for a unit-step input is 
is
, then the time constant
of the system is
second. 
second.
(B)
(D)
. Steady-state error to unit-ramp input is 
with respect to parameter 
is 
is critically damped. The value
of the parameter p is
for input 
rad/sec, is
