# 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.