# 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.

There are 5 questions to complete.