# Time Response Analysis

 Question 1
A closed loop system is shown in the figure where $k \gt 0$ and $\alpha \gt 0$. The steady state error due to a ramp input $\left(R(s)=\alpha / s^{2}\right)$ is given by

 A $\frac{2 \alpha}{k}$ B $\frac{\alpha}{k}$ C $\frac{\alpha}{2 k}$ D $\frac{\alpha}{4 k}$
GATE EC 2023   Control Systems
Question 1 Explanation:
Given, $\quad$ input is $r(t)=\alpha t u(t)$
$R(s)=\frac{\alpha}{s^{2}}$

From the figure,
$G(s) H(s)=\frac{K}{s(s+2)}$
Now steady state error for Ramp input is
$e_{s s} =\frac{\alpha}{K_{v}}, \text { where } \alpha \text { is the magnitude of Ramp input }$
\begin{aligned} K_{v} & =\lim _{s \rightarrow 0}[s G(s) H(s)] \\ K_{v} & =\lim _{s \rightarrow 0}\left[\frac{s \times K}{s(s+2)}\right]=\frac{K}{2}\\ \therefore \quad e_{s s} & =\frac{\alpha \times 2}{K} \\ e_{s s} & =\frac{2 \alpha}{K} \end{aligned}
 Question 2
The block diagram of a closed-loop control system is shown in the figure. R(s),Y(s) and D(s) are the Laplace transforms of the time-domain signals $r(t),y(t),$ and $d(t)$, respectively. Let the error signal be defined as $e(t)=r(t)-y(t)$. Assuming the reference input $r(t)=0$ for all $t$, the steady-state error $e(\infty )$, due to a unit step disturbance $d(t)$, is _________ (rounded off to two decimal places).

 A -0.2 B -0.1 C 0 D -0.3
GATE EC 2022   Control Systems
Question 2 Explanation:
$Y(s)=\frac{R(s)\cdot \frac{10}{s(s+10)}}{1+\frac{10}{s(s+10)}} +\frac{D(s)\cdot \frac{10}{s(s+10)}}{1+\frac{10}{s(s+10)}}$
When, $r(t)=0\; and \; d(t)=U(t)\underleftrightarrow{L.T.}\frac{1}{s}$
$e(\infty )=-\lim_{s \to 0}\frac{s\cdot \frac{1}{s} \times 1}{s(s+10)+10} =-1/10=-0.1\;\;\;(\because e(t)=r(t)-y(t))$

 Question 3
Two linear time-invariant systems with transfer functions
$G_1(s)=\frac{10}{s^2+s+1}, G_2(s)=\frac{10}{s^2+s\sqrt{10}+10}$
have unit step responses $y_1(t)$ and $y_2(t)$, respectively. Which of the following statements is/are true?
 A $y_1(t)$ and $y_2(t)$ have the same percentage peak overshoot. B $y_1(t)$ and $y_2(t)$ have the same steady-state value. C $y_1(t)$ and $y_2(t)$ have the same damped frequency of oscillation. D $y_1(t)$ and $y_2(t)$ have the same 2% settling time.
GATE EC 2022   Control Systems
Question 3 Explanation:
\begin{aligned} G_1(s)&=\frac{10}{s^2+s+1}\\ \omega _{n1}^2&=1\\ 2\xi _1 \times 1&=1\\ \xi _1&=1/2\\ G_2(s)&=\frac{10}{s^2+s\sqrt{10}+10}\\ \omega _{n2}^2&=10\\ 2\xi _1 \times \sqrt{10}&=\sqrt{10}\\ \xi _2&=1/2\\ \end{aligned}
$\because M_P$ depends on $\xi$ only
$y_1(t) \; and \; y_2(t)$ have same percentage peak overshoot.
$\omega _{d_1}=\omega _{n_1}\sqrt{1-\xi ^2}$
$\Rightarrow \omega _{d_1}\neq \omega _{d_2} \;\; \because \omega _{n_1}\neq \omega _{n_2}$
damped frequency of oscillation is not same.
\begin{aligned} T_s &=\frac{4}{\xi \omega _n}(2%\; setting \; time) \\ T_{s_1} &\neq T_{s_2} \;\because \omega _{n_1}\neq \omega _{n_2}\\ C_1(s) &= \frac{10 \times 1/s}{s^2+s+1}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_1(s) \\ &= 10\\ C_2(s) &= \frac{10 \times 1/s}{s^2+\sqrt{10}s+10}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_2(s) \\ &= \frac{10}{10}1=1 \end{aligned}
Steady - state value is not same.
 Question 4
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 4 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 5
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 5 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}

There are 5 questions to complete.