Time Response Analysis

Question 1
The damping ratio and undamped natural frequency of a closed loop system as shown in the figure, are denoted as \zeta and \omega _n, respectively. The values of \zeta and \omega _n are

A
\zeta =0.5 \text{ and }\omega _n=10 rad/s
B
\zeta =0.1 \text{ and }\omega _n=10 rad/s
C
\zeta =0.707 \text{ and }\omega _n=10 rad/s
D
\zeta =0.707 \text{ and }\omega _n=100 rad/s
GATE EE 2022   Control Systems
Question 1 Explanation: 
Reduced the block diagram:

Transfer function,
\frac{C(s)}{R(s)}=\frac{100/(s(s+10))}{1+(100/s(s+10))}=\frac{100}{s^2+10s+100}
Standard form,
T.F.=\frac{\omega _n^2}{s^2+2\xi \omega _ns+\omega _n^2}
On comparison : \omega =\sqrt{100}=10rad/sec and 2\xi \omega _n=10
\Rightarrow \xi=\frac{10}{2 \times 10}=0.5
Question 2
In the given figure, plant G_{P}\left ( s \right )=\dfrac{2.2}{\left ( 1+0.1s \right )\left ( 1+0.4s \right )\left ( 1+1.2s \right )} and compensator G_{C}\left ( s \right )=K\left [ \dfrac{1+T_{1}s}{1+T_{2}s} \right ]. The external disturbance input is D(s). It is desired that when the disturbance is a unit step, the steady-state error should not exceed 0.1 unit. The minimum value of K is _____________.
(Round off to 2 decimal places.)

A
12.25
B
14.12
C
9.54
D
6.22
GATE EE 2021   Control Systems
Question 2 Explanation: 
\begin{aligned} e_{s s} &=\lim _{s \rightarrow 0}\left[\frac{s R}{1+G_{C} G_{p}}-\frac{s D G_{p}}{1+G_{C} G_{P}}\right] \\ R(s) &=0 ; D(s)=\frac{1}{s} \\ \therefore \qquad\qquad e_{s s}&=\frac{2.2}{1+2.2 K}=0.1 \\ \therefore \qquad\qquad K_{\min }&=9.54 \end{aligned}
Question 3
Consider a closed-loop system as shown. G_{P}\left ( s \right )=\dfrac{14.4}{s\left ( 1+0.1s \right )} is the plant transfer function and G_{c}(S)=1 is the compensator. For a unit-step input, the output response has damped oscillations. The damped natural frequency is ___________________ \text{rad/s}. (Round off to 2 decimal places.)

A
10.9
B
4.62
C
12.02
D
8.05
GATE EE 2021   Control Systems
Question 3 Explanation: 

\begin{aligned} q(s)&=s^{2}+10 s+144=0 \\ \omega_{n}&=12 ; \xi=\frac{5}{12} \\ \omega_{d}&=\omega_{n} \sqrt{1-\xi^{2}} \\ \quad&=12 \sqrt{\frac{119}{144}}=10.90 \end{aligned}
Question 4
Consider a negative unity feedback system with the forward path transfer function \frac{s^2+s+1}{s^3+2s^2+2s+K}, where K is a positive real number. The value of K for which the system will have some of its poles on the imaginary axis is ________ .
A
9
B
8
C
7
D
6
GATE EE 2020   Control Systems
Question 4 Explanation: 
CE is
1+G(s)H(s)=0
\Rightarrow \, 1+\frac{s^{2}+s+1}{s^{3}+2s^{2}+2s+k}=0
\Rightarrow \, s^{3}+3s^{2}+3s+(1+K)=0
R.H. criteria:

9 - (1 + K) = 0
\Rightarrow \, \, K=8
Question 5
Which of the following option is correct for the system shown below?
A
4^{th} order and stable
B
3^{rd} order and stable
C
4^{th} order and unstable
D
3^{rd} order and unstable
GATE EE 2020   Control Systems
Question 5 Explanation: 
\begin{aligned} 1+\frac{20}{s^{2}(s+1)(s+20)}&=0\\ (s^{3}+s^{2})(s+20)+20&=0\\ s^{4}+20s^{3}+s^{3}+20s^{2}+20&=0\\ s^{4}+21s^{3}+20s^{2}+20&=0 \end{aligned}
Given system is fourth order system and unstable.

stablity status: since it has one missing term of 's' thus undoubtedly given transfer function is unstable.
Question 6
Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation.
\frac{d^2y(t)}{dt^2}+4y(t)=6r(t)
The poles of this system are at
A
+2j, -2j
B
+2, -2
C
+4, -4
D
+4j, -4j
GATE EE 2020   Control Systems
Question 6 Explanation: 
\begin{aligned}\frac{d^2 y(t)}{dt^{2}}+4y(t) &=6 r(t)\\ [s^{2}+4]Y(s)&=6 R(s) \\ \frac{Y(s)}{R(s)}&=\frac{6}{s^{2}+4} \\ \text{Poles: } s^{2}+4&=0 \\ s&=\pm j2 \end{aligned}
Question 7
The unit step response y(t) of a unity feedback system with open loop transfer function
G(s)H(s)=\frac{K}{(s+1)^{2}(s+2)}
is shown in the figure. The value of K is _______ (up to 2 decimal places).
A
4
B
8
C
10
D
12
GATE EE 2018   Control Systems
Question 7 Explanation: 
Closed loop transfer function,
\frac{C(s)}{R(s)}=\frac{\frac{K}{(s+1)^2(s+2)}}{1+\frac{K}{(s+1)^2(s+2)}}
\frac{C(s)}{R(s)}=\frac{K}{(s+1)^2(s+2)+K}
Given R(s)=\frac{1}{s}
C(s)=\frac{K}{s((s+1)^2(s+2)+K)}
\lim_{s \to 0}sC(s)=0.8
\frac{K}{2+K}=0.8
\Rightarrow \; K=8
Question 8
C onsider a unity feedback system with forward transfer function given by
G(s)=\frac{1}{(s+1)(s+2)}
The steady-state error in the output of the system for a unit-step input is _________(up to 2 decimal places).
A
0.25
B
0.45
C
0.66
D
0.85
GATE EE 2018   Control Systems
Question 8 Explanation: 
Steady state error for type-0 and step input,
e_{ss}=\frac{1}{1+k+p}
k_p=\lim_{s \to 0}\frac{1}{(s+1)(s+2)}=\frac{1}{2}
e_{ss}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}
\;\;=0.66 uinits
Question 9
Match the transfer functions of the second-order systems with the nature of the systems given below.
A
P-I, Q-II, R-III
B
P-II, Q-I, R-III
C
P-III, Q-II, R-I
D
P-III, Q-I, R-II
GATE EE 2018   Control Systems
Question 9 Explanation: 
P=\frac{15}{s^2+5s+15}
\omega _n=\sqrt{15}=3.872 rad/sec
2 \xi \times 3.872=5
\xi=\frac{5}{2 \times 3.872}=0.64 \;\;\;\;(Underdamped)
Q=\frac{25}{s^2+10s+25}
\omega _n=\sqrt{25}=5 rad/sec
2 \xi \times 5=10
\xi=1 \;\;\;\;\;(Critically \; damped)
Observing all the options, option (C) is correct.
Question 10
Which of the following systems has maximum peak overshoot due to a unit step input?
A
\frac{100}{s^{2}+10s+100}
B
\frac{100}{s^{2}+15s+100}
C
\frac{100}{s^{2}+5s+100}
D
\frac{100}{s^{2}+20s+100}
GATE EE 2017-SET-2   Control Systems
Question 10 Explanation: 
For maximum peak over shoot M_P\propto \frac{1}{\xi}
\xi=0.25 for option (C) which is least among all options. Therefore correct option is C.
There are 10 questions to complete.