Control Engineering
NPTEL Online Course
Indian Institute of Technology Madras
Solution 3
Note: Unless mentioned otherwise, all signals are causal, i.e., their value is zero for all time t < 0.
C(s)
1. Transfer function R(s) of given block diagram is
G3
R(s) + C(s)
G1 G2
_ _
H2
H1
G1 (G2 +G3 )
(a) 1+G2 H2 +G1 G2 H1
G1 (G2 +G3 )
(b) 1+G2 H2 +G1 G2 H1 +G1 G2 G3 H1 H2
G1 (G2 +G3 )
(c) 1+G2 H2 +G1 G2 H1 −G1 G2 G3 H1 H2 [Correct]
(d) none of these
Solution: Shift take-off point before the block G2
Shifting take-off point before the summing point
1
Separating the paths in the feedback path shown dotted
Shifting the summing point and then applying Associative law for interchanging the two summing
points
G3
G1 + C(s)
R(s)
1 + H1G1G 2 G2
_ _
H2
-H1H2G2
Solve parallel blocks
2
R(s) G1 G3 G2 C(s)
1 + H1G1G 2 1+
G2 1 + H 2G 2
_
-H1H2G2
Solve series blocks
R(s) G1 G G2 C(s)
1+ 3
_ 1 + H1G1G 2 G 2 1 + H2G 2
-H1H2G2
Now simplifying the feedback for block we get,
G1 (G2 +G3 )
C(s) (1+G2 H2 )(1+H1 G1 G2 )
= G1 (G2 +G3 )(−H1 H2 G2 )
R(s)
(1+G2 H2 )(1+H1 G1 G2 )
C(s) G1 (G2 + G3 )
=
R(s) 1 + G2 H2 + H1 G1 G2 + G1 G2 G3 H1 H2
Alternate Solution: Try to solve using SFG.
G3
G2
R(s) C(s)
1 G1 1 1 1
– H1 – H2
Number of forward paths=2
Mason gain formula is T F = T1 ∆1∆
+T2 ∆2
Gain of forward path T1 =G1 G2
Gain of second forward path T2 =G1 G3
Gain of feedback loop:
L1 =−G1 G2 H1
L1 =−G2 H2
L1 =G1 G3 H1 G2 H2
3
Now, ∆=1-(Sum of all individual loops)
Therefore, ∆=1-(−G1 G2 H1 −G2 H2 +G1 G3 H1 G2 H2 )
∆=1+G1 G2 H1 + G2 H2 − G1 G3 H1 G2 H2
∆1 =1 − 0=1
∆2 =1 − 0=1
C(s) G1 (G2 +G3 )
So the transfer function of given block diagram is R(s) = 1+G2 H2 +G1 G2 H1 −G1 G2 G3 H1 H2
C
2. The transfer function R2 from the block diagram shown in figure is given by
G2
R2
+ + C
+
G1 G3
R1 + _
H1
G2 +G1 G3
(a) 1+G3 H1 +G2
G3
(b) 1+G3 H1 [Correct]
(G1 +G2 )G3
(c) 1+(G1 +G2 )G3 H1
(d) none of these
C
Solution: For finding R2 , signal R1 must be zero, block diagram becomes
R2
+ C
+
G3
_
H1
Redraw above block diagram
R2 C
G3
+
- H1
Now,simplifying the feedback for block we get,
C G3
=
R2 1 + G3 H1
4
3. The signal flow graph for the following set of system equations given
(a) Y2 = G1 Y1 + G3 Y3
(a) Y3 = G4 Y1 + G2 Y2 + G5 Y3
(a) Y4 = G6 Y2 + G7 Y3
where Y1 is input and Y4 is output.
G4
G5
(a) G1 G2 G7 [Correct]
Y1 Y2 Y3 Y4
G3
G6
G4
G6
(b) G1 G2 G7
Y1 Y2 Y3 Y4
G3
G5
G4
(c) G1 G2 G7
Y1 Y2 Y3 Y4
G3 G6
G5
5
G5
(d) G1 G2 G7
Y1 Y2 Y3 Y4
G3 G6
G4
Solution: System node variables are Y1 , Y2 , Y3 , Y4 .
Consider equation 1 as given: Y2 = G1 Y1 + G3 Y3 This consists of three nodes Y1 , Y2 , Y3 . Signal
flow graph of equation 1 is
Y1 G1 Y2 Y3
G3
Consider equation 2 as given: Y3 = G4 Y1 + G2 Y2 + G5 Y3 This consists of three nodes Y1 , Y2 , Y3 .
Signal flow graph of equation 2 is
G5
G4
G2
Y1 Y2 Y3
Consider equation 3 as given:Y4 = G6 Y2 + G7 Y3 . This consists of three nodes Y2 , Y3 , Y4 . Signal
flow graph of equation 3 is
G7
Y2 Y3 Y4
G6
Combining all the three signal flow graphs, we get complete signal flow graph
6
G4
G5
G1 G2 G7
Y1 Y2 Y3 Y4
G3
G6
4. Finite steady-state error ( K is error coefficient )
(a) varies inversely with K [Correct]
(b) is independent of K
(c) varies directly with K
(d) None of these.
Solution:
This question is interesting in the sense that neither the type of system nor the nature of input
has been mentioned. Still the question talks about the steady-state error. But what is mentioned
is the word ’Finite’. This reveals both the type of system as well as the nature of input signal.
R
The steady state error is finite (i.e. neither nor 0); in type 0 system with step-input (1+K) , in
R R
type I system with ramp-input K and in type 2 system with parabolic-input K .In all the three
cases the steady-state error varies inversely with K.
5. Assume R(t) = 0.1t and it is desired that ess ≤ 0.005. The complete range of value of k for error
to be within specified limit for given system for unity feedback.
k
G(s) =
s(s + 1)
(a) k ≥ 20 [Correct]
(b) k ≤ 20
(c) k ≥ 200
(d) k ≤ 200
Solution: We know that
sR(s)
ess = lim
s→0 1 + G(s)H(s)
The input R(t) to the system in s domain will be
0.1
R(s) =
s2
Therefore,
( 0.1
s2 )s
ess = lim k
s→0 1+ s(s+1)
7
( 1s )0.1
ess = lim
s→0 1 (s + k
s (s+1) )
0.1
ess = k
1
But ess ≤ 0.005 Therefore,
0.1
ess ≤ 0.005 =⇒ ≤ 0.005
k
0.1
k≥ = 20
0.005
k ≥ 20
6. The impulse response of an RL circuit, with output measured across resistor is a
(a) Rising exponential function
(b) Decaying exponential function [Correct]
(c) Step function
(d) Parabolic function
Solution: Current,
V (s)
I(s) =
Z(s)
Where,
Z(s) = R + sL
Therefore,
V (s)
I(s) =
R + sL
And, V(s)=1. Therefore,
1
I(s) =
R + sL
Taking Inverse Laplace Transform,
1 −( R )t
i(t) = e L
L
which shows decaying exponential function.
7. When the input to a system was withdraw at t = 0 its output was found to decrease exponentially
from 1000 units to 500 units in 1.386 seconds. The time constant of the system is
(a) 0.5
(b) 0.693
(c) 1.386
(d) 2.00 [Correct]
8
Solution: Time response of the system is,
c(t) = C[e−t/T ]
At t=0,
c(0) = 1000 = C
From the above expression
500 = 1000(e−1.386/T )
e−1.386/T = 0.5
−1.386
= ln 0.5 = −0.6932
T
T =2
8. A unity feedback control system has a forward path transfer function equal to
42.25
s2 (s + 6.5)
The unit step response of this system starting from rest, will have its maximum value at a time
equal to
(a) 0 sec
(b) 0.56 sec
(c) 5.6 sec
(d) Infinite [Correct]
Solution:
Transfer function,
42.25
C(s) s2 (s+6.5)
= 42.25
R(s) 1 + s2 (s+6.5)
42.25
C(s) = R(s)
s2 (s + 6.5) + 42.25
For unit step response
1
R(s) =
" # " s#
1 42.25 1 42.25
C(s) = =
s s2 (s + 6.5) + 42.25 s s3 + 6.5s2 + 42.25
Using partial fraction expansion
0.071 − 0.91s 0.09 1
C(s) = − +
s2 − 0.8s + 5.8 (s + 7.3) s
Taking inverse Laplace transform
C(t) = 42.25 (−0.0107763 + 0.00144859i)exp((−2.37375i)t)((0.964502 + 0.264076i)exp(0.397056t)
+(1 + 0i)exp((0.397056 + 4.7475i)t)) − 0.00211606exp(−7.29411t) + 0.0236686
9
At t → ∞, C(t) = ∞.
Alternate solution:
Step Response
10 4
2
1.5
1
Amplitude
0.5
-0.5
-1
-1.5
0 5 10 15 20 25
Time (seconds)
The step response oscillates so maximum will be achieved as t → ∞.
9. The block diagrams shown in Fig. a and b are equivalent if ’X’ (in Fig. b) is equal to
R(s) 1 C(s)
R(s) C(s) s +1
X
s+2 + _
s +1
Fig (a) Fig (b)
(a) 1
(b) 2
(c) s + 1
(d) s + 2
(e) 2s + 3 [Correct]
Solution: From figure (b), transfer function of the system is
C(s) X −s − 1 + X
= −1 + =
R(s) (s + 1) (s + 1)
10
Comparing above with the transfer function of figure (a)
−s − 1 + X s+2
=
(s + 1) s+1
−s − 1 + X = s + 2
X = 2s + 3
10. For a unity feedback system, if the input is unit step and the output response is observed as, C(t) =
1 + 0.504e−3.07t − 1.504e−2.18t then the system is
(a) Over damped [Correct]
(b) Underdamped
(c) Undamped
(d) None of these
C(s)
Solution: To determine the and n we must have C.L.T.F. R(s) which can be obtained by taking
Laplace transform of C(t) and R(t), respectively.Therefore,
1 0.504 1.504
C(s) = + −
s s + 3.07 s + 2.18
The system has unit step input,i.e., R(t) = 1. Therefore,R(s) = 1/s and
1
C(s) + 0.504 − 1.504
= s s+3.07 s+2.18
R(s) 1/s
C(s) 0.504s 1.504s
=1+ −
R(s) s + 3.07 s + 2.18
C(s) (s + 3.07)(s + 2.18) + (0.504s)(s + 2.18) − (1.504s)(s + 3.07)
=
R(s) (s + 3.07)(s + 2.18)
C(s) s + 5.25s + 6.6926 + 0.504s2 + 1.0987s − 1.504s2 − 4.6172s
2
=
R(s) s2 + 5.25s + 6.6926
C(s) 6.6926 + 1.7315s
= 2
R(s) s + 5.25s + 6.6926
Comparing denominator with standard form, we get
ωn2 = 6.6926 =⇒ ωn = 2.587rad/sec
2ζωn = 5.25 =⇒ ζ = 1.0146
As ζ ≥ 1, hence the system is over damped.
11. Consider the system shown below, determine the value of k such that the damping ratio ζ is 0.5 .
Then obtain the settling time ts (2% criteria), in the unit-step response.
The value of k is and ts is secs. [Answer: 0.2 , Range: 0.175 to 0.225,
Answer: 2, Range: 1.9 to 2.1 ]
Solution:
Inner loop transfer function (G1 (s)) is
16
s+0.8 16
G1 (s) = 16 =
1 + k. s+0.8 s + 0.8 + 16k
11
C(s) G1 (s). 1s 16
= = 2
R(s) 1 + G1 (s). 1s s + (0.8 + 16k)s + 16
Comparing with a standard second order system, we have
2 × 0.5 × 4 − 0.8 3.2
ωn2 = 16 =⇒ ωn = 4 0.8 + 16k = 2ζωn =⇒ k = = = 0.2
16 16
4 4
ts = = =2
ζωn 2
12
