# Time Response Analysis

 Question 1
Consider the following closed loop control system

where $G(s)=\frac{1}{s(s+1)}$ and $C(s)=K\frac{s+1}{s+3}$. If the steady state error for a unit ramp input is 0.1, then the value of K is _______.
 A 10 B 20 C 30 D 40
GATE EC 2020   Control Systems
Question 1 Explanation:
Open loop transfer function for the system $=C(s)\times G(s)=\frac{K(s+1)}{ (s+3)}\times \frac{1}{s(s+1)}$
Since the system is type-1, so far a given unit ramp input steady state
$e_{ss}=\frac{1}{K_{V}}$
where, $K_{V}=\lim_{s\rightarrow 0}S\times \frac{K}{S(S+3)}=\frac{K}{3}$
so, $\; e_{ss}=\frac{1}{K/3}=\frac{3}{K}$
Given that, $\; e_{ss}=0.1$
So, $0.1=\frac{3}{K}\Rightarrow K=30$
 Question 2
A system with transfer function $G(s)=\frac{1}{(s+1)(s+a)}, a \gt 0$ is subjected an input $5\cos 3t$. The steady state output of the system is $\frac{1}{\sqrt{10}} \cos (3t-1.892)$. The value of $a$ is ____.
 A 3 B 4 C 5 D 6
GATE EC 2020   Control Systems
Question 2 Explanation:
Given that,
\begin{aligned}G(j\omega )&=\frac{1}{(1+j\omega )(\alpha +j\omega )} \\ \left |G(j\omega ) \right |&=\frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}} \\ \text{According to }& \text{question,}\\ \left | G(j\omega ) \right |_{\omega =3}&=\frac{1}{5\sqrt{10}}\\ \Rightarrow \frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}}&=\frac{1}{5\sqrt{10}} \\ \Rightarrow \frac{1}{\sqrt{10(a^{2}+9)}}&=\frac{1}{5\sqrt{10}} \alpha ^{2}+9=25 \\ \alpha ^{2}&=16 \\ \alpha &=4\end{aligned}
 Question 3
The loop transfer function of a negative feedback system is

$G(s)H(s)=\frac{K(s+11)}{s(s+2)(s+8)}$

The value of K, for which the system is marginally stable, is ________.
 A 160 B 120 C 180 D 200
GATE EC 2020   Control Systems
Question 3 Explanation:
Characteristic equation q(s) for the given open loop system will be
$q(s)=s^{3}+10s^{2}+16s+Ks+11K=0$
Using R-H criteria,

For System to be Marginally Stable
\begin{aligned}\frac{10(16+K)-11K}{10}&=0 \\ 160+10K-11K&=0 \\ K&=160\end{aligned}
 Question 4
Consider a causal second-order system with the transfer function

$G(s)=\frac{1}{1+2s+s^2}$

with a unit-step $R(s)=\frac{1}{s}$ as an input. Let C(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value $\lim_{t \to \infty }c(t)$, rounded off to two decimal places, is
 A 5.25 B 4.5 C 3.89 D 2.81
GATE EC 2019   Control Systems
Question 4 Explanation:
\begin{aligned} G(s)&=\frac{1}{s^{2}+2 s+1}=\frac{1}{(s+1)^{2}} \\ R(s)&=\frac{1}{s} \\ \alpha (s)&=G(s) R(s)=\frac{1}{s(s+1)^{2}} \end{aligned}
Using partial fraction expansion, we get,
\begin{aligned} C(s) & =\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+1)^{2}} \\ A\left(s^{2}+2 s+1\right)&+B\left(s^{2}+s\right)+C s=1 \\ A&=1 \\ A+B & =0 \Rightarrow B=-1 \\ 2 A+B+C & =0 \Rightarrow C=-1 \\ \therefore \quad C(s) & =\frac{1}{s}-\frac{1}{(s+1)}-\frac{1}{(s+1)^{2}} \\ \text{and}\quad c(t) & =\left(1-e^{-t}-t e^{-t}\right) u(t) \\ \lim _{t \rightarrow \infty} c(t) & =1 \end{aligned}
In order to reach 94 % of its steady-state value,
$\left(1-e^{-t}-t e^{-t}\right)=0.94$
By trial and error, we get,
$t \approx 4.50 \mathrm{sec}$
 Question 5
The open loop transfer function G(s)=$\frac{(s+1)}{s^{P}(s+2)(s+3)}$
Where p is an integer, is connected in unity feedback configuration as shown in figure.

Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is _________.
 A 1 B 2 C 3 D 4
GATE EC 2017-SET-1   Control Systems
Question 5 Explanation:
Steady state error of type 1 system, for step input is zero, for ramp input is $\frac{1}{K_{v}}$ and for parabolic input is infinity.
So, the given system must be of type- 1 and p=1
 Question 6
For the unity feedback control system shown in the figure, the open-loop transfer function G(s) is given as

$G(s)=\frac{2}{s(s+1)}$

. The steady state error $e_{ss}$ due to a unit step input is
 A 0 B 0.5 C 1 D $\infty$
GATE EC 2016-SET-3   Control Systems
Question 6 Explanation:
\begin{aligned} E(s) &=\frac{R(s)}{1+G(s) H(s)} \\ R(s) &=\frac{1}{s} ; G(s)=\frac{2}{s(s+1)}; \\ H(s) &=1 \\ E(s) &=\frac{\frac{1}{s}}{1+\frac{s}{s(s+1)}}=\frac{s+1}{s^{2}+s+2} \\ e_{s s} &=\lim _{s \rightarrow 0} s E(s) \\ &=\lim _{s \rightarrow 0} \frac{s(s+1)}{s^{2}+s+2} \\ e_{s s} &=0 \end{aligned}
 Question 7
In the feedback system shown below $G(s)=\frac{1}{(s^{2}+2s)}$. The step response of the closed-loop system should have minimum settling time and have no overshoot.

The required value of gain k to achieve this is ________
 A 0.5 B 1 C 2 D 3
GATE EC 2016-SET-2   Control Systems
Question 7 Explanation:
\begin{aligned} G(s)&=\frac{1}{s^{2}+2 s} \\ \frac{Y(s)}{R(s)}&=\frac{\frac{K}{s^{2}+2 s}}{1+\frac{K}{s^{2}+2 s}}=\frac{K}{s^{2}+2 s+K} \end{aligned}
Minimum settling time and no overshoot implies critical damping i.e.
\begin{aligned} \xi &=1 \\ \omega_{n} &=\sqrt{K} \\ \therefore \quad 2 \times \xi \cdot \omega_{n} &=2 \\ \omega_{n} &=1 \\ \Rightarrow \quad \sqrt{K} &=1 \text { or } K=1 \end{aligned}
 Question 8
The response of the system $G(s)=\frac{s-2}{(s+1)(s+3)}$ to the unit step input u(t) is y(t). The value of $\frac{dy}{dt}$ at t=$0^{+}$ is _________
 A 0 B 1 C 2 D 3
GATE EC 2016-SET-2   Control Systems
Question 8 Explanation:
\begin{aligned} L\left(\frac{d y}{d t}\right) &=(s Y(s)-y(0)) \\ Y(s) &=G(s) \times \frac{1}{s}=\frac{s-2}{s(s+1)(s+3)} \\ y(0)&=\lim _{s \rightarrow \infty} s Y(s) \\ \text{(Applying } & \text{initial value theorem) } \\ &=\lim _{s \rightarrow \infty} \frac{s-2}{(s+1)(s+3)} \\ &=\frac{\left(1-\frac{2}{s}\right)}{s\left(1+\frac{1}{s}\right)\left(1+\frac{3}{s}\right)} \\ y(0)&=0 \\ L\left(\frac{d y}{d t}\right) &=s Y(s)=\frac{s \times(s-2)}{s(s+1)(s+3)} \\ &=\frac{s-2}{(s+1)(s+3)} \\ \left.\frac{d y}{d t}\right|_{t=0} &=\lim _{s \rightarrow \infty} s L\left(\frac{d y}{d t}\right) \\ &=\lim _{s \rightarrow \infty} \frac{s \times(s-2)}{(s+1)(s+3)}\\ &=\frac{\left(1-\frac{2}{s}\right)}{\left(1+\frac{1}{s}\right)\left(1+\frac{3}{s}\right)}=1 \end{aligned}
 Question 9
The open-loop transfer function of a unity-feedback control system is given by

$G(s)=\frac{K}{s(s+2)}$

For the peak overshoot of the closed-loop system to a unit step input to be 10%, the value of K is____________
 A 1.6 B 2.2 C 2.8 D 4.2
GATE EC 2016-SET-1   Control Systems
Question 9 Explanation:
$G(s)=\frac{K}{s(s+2)} ; H(s)=1$
Characteristic equation $=1+G(s) H(s)=0$
\begin{aligned} 1+\frac{K}{s(s+2)} &=0 \\ s^{2}+2 s+K &=0 \Rightarrow \omega_{n}=\sqrt{K} \\ 2 \xi \omega_{n} &=2 ; \xi=\frac{1}{\sqrt{K}} \\ M_{p}&=e^{-\pi\xi/\sqrt{1-\xi^{2}}}-0.1\\ -\frac{\pi \xi}{\sqrt{1-\xi^{2}}} &=\ln (0.1) \\ \Rightarrow \quad\frac{\pi \xi}{\sqrt{1-\xi^{2}}} &=2.3 \\ \pi^{2} \xi^{2} &=(2.3)^{2}\left(1-\xi^{2}\right) \\ 15.16 \xi^{2} &=(2.3)^{2} \\ \Rightarrow \qquad \xi&=0.59 \\ \Rightarrow \quad \text{Also}, \quad K&=\frac{1}{\xi^{2}}=2.8 \end{aligned}
 Question 10
The output of a standard second-order system for a unit step input is given as $y(t)=1-\frac{2}{\sqrt{3}}e^{-t}cos(\sqrt{3}t-\frac{\pi }{6})$. The transfer function of the system is
 A $\frac{2}{(s+2)(s+\sqrt{3})}$ B $\frac{1}{s^2+2s+1}$ C $\frac{3}{s^2+2s+3}$ D $\frac{4}{s^2+2s+4}$
GATE EC 2015-SET-2   Control Systems
Question 10 Explanation:
The output of a second order system for unit step
input
$y(t)=1-\frac{e^{-\xi \omega_{n} t}}{\sqrt{1-\xi^{2}}} \sin \left(\omega_{d} t-\theta\right) \qquad\ldots(i)$
From given data
\begin{aligned} y(t)&=1-\frac{2}{\sqrt{3}} e^{-t} \cos \left(\sqrt{3} t-\frac{\pi}{6}\right)\\ \therefore \quad \alpha&=\xi \omega_{n}=1 \\ \text{and }\quad \omega_{d}&=\sqrt{3} \\ \omega_{n}&=\sqrt{\alpha^{2}+\omega_{d}^{2}}=2 \\ \xi&=\frac{1}{2} \end{aligned}
Therefore, the transfer function
$T(s)=\frac{\omega_{n}^{2}}{s^{2}+2 \xi \omega_{n} s+\omega_{n}^{2}}=\frac{4}{s^{2}+2 s+4}$
There are 10 questions to complete.