Feedback Characteristics of Control Systems

Question 1
The block diagram of a system is shown in the figure

If the desired transfer function of the system is
\frac{C(s)}{R(s)}=\frac{s}{s^{2}+s+1}
then G(s) is
A
1
B
s
C
1/s
D
\frac{-s}{s^{3}+s^{2}-s-2}
GATE EE 2014-SET-3   Control Systems
Question 1 Explanation: 
For all given values of G(s) in options, G(s) does not satisfy the desired condition. But option (B) is official answer given by IIT.
Question 2
The closed-loop transfer function of a system is T(s)=\frac{4}{(s^{2}+0.4s+4)}. The steady state error due to unit step input is _____.
A
0
B
0.15
C
0.75
D
1
GATE EE 2014-SET-2   Control Systems
Question 2 Explanation: 
Closed loop transfer function,
T(s)=\frac{4}{(s^2+0.4s+4)}
or, \frac{G(s)}{1+G(s)}
=\left ( \frac{4}{s^2+0.4s+4} \right )\;\;\;\;[H(s)=1]
or, (s^2+0.4s+4)=4+4G(s)
or, G(s)=\frac{4}{s(s+0.4)}
=Open llop transfer function ( for unity feedback system)
Given, input =u(t)=r(t)
\therefore \;\;\; R(s)=1/s
Steady state error
e_{ss}=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)H(s)} \right ]
\;\;=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)} \right ]\;\;\; [For H(s)=1]
\;\;=\lim_{s \to 0}\left [ \frac{s \times \frac{1}{s}}{1+\frac{4}{s(s+0.4)}} \right ]
\;\;=\lim_{s \to 0}\left [ \frac{s(s+0.4)}{s^2+0.4s+4} \right ]=0
\therefore Steady state error for step input, e_{ss}=0
Question 3
The open-loop transfer function of a dc motor is given as \frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}. When connected in feedback as shown below, the approximate value of K_a that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is
A
1
B
5
C
10
D
100
GATE EE 2013   Control Systems
Question 3 Explanation: 
\frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}=\frac{10}{1+\tau s}
\tau =10

Closed loop systems,
\frac{\omega (s)}{R(s)}=\frac{\frac{10K_a}{1+10s}}{1+\frac{10K_a}{1+10s}}
=\frac{10K_a}{1+10K_a+10s}
=\frac{\frac{10K_a}{1+10K_a}}{1+\frac{10}{1+10K_a}s}
According to quetion,
\frac{10}{1+10K_a}=\frac{1}{100}\times 10
1+10K_a=100
K_a=9.9
Question 4
As shown in the figure, a negative feedback system has an amplifier of gain 100 with \pm10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :
A
10\pm 1%
B
10\pm 2%
C
10\pm 5%
D
10\pm 10%
GATE EE 2010   Control Systems
Question 4 Explanation: 
G=100\pm 10%
\frac{\Delta G}{G}=10% or 0.1
H=\frac{9}{100}
Overall gain,
T=\frac{G}{1+GH}
T=\frac{100}{1+100 \times \frac{9}{100}}=10
Sensitivity w.r.t. 'G'=\frac{1}{1+GH} \times 10%
\;\;\;=\frac{1}{1+10 \times 1}\times 10%
\;\;\;=\frac{10}{11}\cong 1%
There are 4 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.