# Control Systems

 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      Time Response Analysis
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      Time Response Analysis
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 characteristic equation of a system is

$s^3 + 3s^2 + (K + 2)s + 3K = 0$

In the root locus plot for the given system, as K varies from 0 to $\infty$, the break-away or break-in point(s) lie within
 A (-1,0) B (-2,-1) C (-3,-2) D (-$\infty$ ,-3)
GATE EC 2020      Root Locus
Question 3 Explanation:
$Q(s)=1+G(s)H(s)=0$
$s^{3}+3s^{2}+2s+ks+3k=0$
$-k=\frac{s^{3}+3s^{2}+2s}{s+3}$
$-\frac{\mathrm{d} k}{\mathrm{d} s}=\frac{(s+3)(3s^{2}+6s+2)-(s^{3}+3s^{2}+2s)}{(s+3)^{2}}=0$
$3s^{3}+6s^{2}+2s+9s^{2}+18s+6-s^{3}-3s^{2}-2s=0$
$2s^{3}+12s^{2}+18s+6=0$
$s=-0.46,-3.87,-1.65$

$\therefore$ Break-away point lies between (0, -1), i.e. (-1, 0)
 Question 4
For the given circuit, which one of the following is the correct state equation?
 A A B B C C D D
GATE EC 2020      State Space Analysis
Question 4 Explanation:
From source transformation,

KVL in loop 1,
$2i_{1}=2i+0.5\frac{\mathrm{d} i}{\mathrm{d} t}+V$
$\frac{\mathrm{d} i}{\mathrm{d} t}=-2V-4i+4i_{1}$
KCL at node (a),
$i=0.25\frac{\mathrm{d} V}{\mathrm{d} t}+\frac{V-i_{2}}{1}$
$\frac{\mathrm{d} v}{\mathrm{d} t}=-4V+4i+4i_{2}$
$\begin{bmatrix} v\\ i \end{bmatrix}=\begin{bmatrix} -4 & 4\\ -2 & -4 \end{bmatrix}\begin{bmatrix} v\\ i \end{bmatrix}+\begin{bmatrix} 0 &4 \\ 4 &0 \end{bmatrix}\begin{bmatrix} i_{1}\\ i_{2} \end{bmatrix}$
 Question 5
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      Time Response Analysis
Question 5 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 6
The pole-zero map of a rational function G(s) is shown below. When the closed counter $\Gamma$ is mapped into the G(s)-plane, then the mapping encircles.
 A the origin of the G(s)-plane once in the counter-clockwise direction. B the origin of the G(s)-plane once in the clockwise direction. C the point -1+j0 of the G(s)-plane once in the counter-clockwise direction. D the point -1+j0 of the G(s)-plane once in the clockwise direction.
GATE EC 2020      Frequency Response Analysis
Question 6 Explanation:
s-plane contour is encircling 2-poles and 3-zeros in clockwise direction hence the corresponding G(s) plane contour encircles origin 2-times in anti-clockwise direction and 3-times in clockwise direction.
Therefore, Effectively once in clockwise direction.
 Question 7
Consider a unity feedback system, as in the figure shown, with an integral compensator $\frac{K}{s}$ and open-loop transfer function

$G(s)=\frac{1}{s^2+3s+2}$

where $k \gt 0$. The positive value of K for which there are exactly two poles of the unity feedback system on the $j\omega$ axis is equal to ______ (rounded off to two decimal places).

 A 2.45 B 4.28 C 6 D 6.25
GATE EC 2019      Stability Analysis
Question 7 Explanation:
$\frac{Y(s)}{X(s)}=\frac{K}{s^{3}+3 s^{2}+2 s+K}$
Two poles of this system lie the system is moro:
System is marginally stable.
$\qquad k_{\text{mar}}=3 \times 2=6$
 Question 8
Let the state-space representation of an LTI system be $\dot{x}(t)=Ax(t)+Bu(t)$, $\dot{y}(t)=Cx(t)+du(t)$ where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let $B=[0\;0\;1]^T$ and d=0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is
$H(s)=\frac{1}{s^3+3s^2+2s+1}$?

 A A B B C C D D
GATE EC 2019      State Space Analysis
Question 8 Explanation:
\begin{aligned} X(t) &=A x(t)+B u(t) \\ y(t) &=C x(t) \\ B &=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \\ \frac{\gamma(s)}{U(s)} &=\\ \frac{\gamma(s)}{X_{1}(s)} \times \frac{X_{1}(s)}{U(s)}=& 1 \times \frac{1}{s^{3}+3 s^{2}+2 s+1} \\ x_{1}(s)\left(s^{3}+3 s^{2}+2 s+1\right]&=U(s) \\ x_{2} &=\dot{x}_{1}(t) \\ x_{2}(s) &=s x_{1}(s) \\ x_{3} &=\dot{x}_{2}(t) \\ \Rightarrow \quad x_{3}(s) &=s x_{2}(s)=s^{2} X_{1}(s) \\ \text { So, } \quad s x_{3}(s) &=-x_{1}(s)-2 x_{2}(s)-3 x_{3}(s)+u(s) \\ \dot{x}_{3}(t) &=-x_{1}(t)-2 r_{2}(t)-3 x_{3}(t)+u(t) \\ y(t) &=x_{1}(t) \\ \dot{x}(t) &=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3\end{array}\right] x(t)+\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] u(t) \\ y(t) &=[1 \quad 0 \quad 0] x(t) \end{aligned}
 Question 9
The block diagram of a system is illustrated in the figure shown, where X(s) is the input and Y(s) is the output. The transfer function $H(s)=\frac{Y(s)}{X(s)}$ is
 A $H(s)=\frac{s^2+1}{s^3+s^2+s+1}$ B $H(s)=\frac{s^2+1}{s^3+2s^2+s+1}$ C $H(s)=\frac{s+1}{s^2+s+1}$ D $H(s)=\frac{s^2+1}{2s^2+s+1}$
GATE EC 2019      Basics of Control Systems, Block Diagram and SFGs
Question 9 Explanation:
Using block diagram reduction, we get,

$\frac{Y(s)}{X(s)}=H(s)=\frac{s^{2}+1}{s^{3}+2 s^{2}+s+1}$
 Question 10
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      Time Response Analysis
Question 10 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,
$t \approx 4.50 \mathrm{sec}$

There are 10 questions to complete.