# Control Systems

 Question 1
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      Time Response Analysis
Question 1 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 2
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      Time Response Analysis
Question 2 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 3
Consider an even polynomial $p(s)$ given by
$p(s)=s^4+5s^2+4+K$,
where $K$ is an unknown real parameter. The complete range of$K$ for which $p(s)$ has all its roots on the imaginary axis is ________.
 A $-4\leq K\leq \frac{9}{4}$ B $-3\leq K\leq \frac{9}{2}$ C $-6\leq K\leq \frac{5}{4}$ D $-5\leq K\leq 0$
GATE EC 2022      Stability Analysis
Question 3 Explanation:
$p(s)=s^4+5s^2+4+k$
$\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 0 & 0 & \\ s^2& & & \\ s& & & \\ s^0& & & \end{matrix}$
$s^4+5s^2+(4+k)=0$
$4s^3+10s=0$
$(4s^2+10)=0$
$s=\pm j\sqrt{5/2}$
$\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 4 & 10 & \\ s^2& 5/2 & (4+k) & \\ s^1&10-\frac{4(4+k)}{5/2} & & \\ s^0& (4+k) & & \end{matrix}$
\begin{aligned} (4+k) & \gt 0\\ k & \gt -4\\ 10 \times \frac{5}{2}& \gt 4(4+K)\\ (4+K)& \lt \frac{25}{4}\\ K& \lt 9/4 \end{aligned}
$\Rightarrow$ All roots be on imaginary axis $-4\leq K\leqslant 9/4$
 Question 4
The root-locus plot of a closed-loop system with unity negative feedback and transfer function $KG(s)$ in the forward path is shown in the figure. Note that $K$ is varied from 0 to $\infty$.
Select the transfer function $G(s)$ that results in the root-locus plot of the closed-loop system as shown in the figure.

 A $G(s)=\frac{1}{(s+1)^5}$ B $G(s)=\frac{1}{(s^5+1)}$ C $G(s)=\frac{s-1}{(s+1)^6}$ D $G(s)=\frac{s+1}{(s^6+1)}$
GATE EC 2022      Root Locus
Question 4 Explanation:
Here 5 Root Locus branches are diverging from same point, this can possible only when if we have 5 poles in the system at the same point because Root Locus branch departs from open loop pole and
Number of Root Locus branches = Number of open loop poles or Number of zero (Whichever is greater).
Here, there are 5 multiple Real poles, and matching with option (A).
 Question 5
Consider a closed-loop control system with unity negative feedback and $KG(s)$ in the forward path, where the gain $K=2$. The complete Nyquist plot of the transfer function $G(s)$ is shown in the figure. Note that the Nyquist contour has been chosen to have the clockwise sense. Assume $G(s)$ has no poles on the closed right-half of the complex plane. The number of poles of the closed-loop transfer function in the closed right-half of the complex plane is _______

 A 0 B 1 C 2 D 3
GATE EC 2022      Frequency Response Analysis
Question 5 Explanation:

For K = 2, point A will be -0.8x2 = -1.6
Hence N = -2, P = 0
(By default Nyquist contoure is considered in clockwise direction)
P - N = 2
Number of closed loop pole in right side of the complex plane.
 Question 6
A unity feedback system that uses proportional-integral ($\text{PI}$ ) control is shown in the figure.

The stability of the overall system is controlled by tuning the $\text{PI}$ control parameters $K_{P}$ and $K_{I}$. The maximum value of $K_{I}$ that can be chosen so as to keep the overall system stable or, in the worst case, marginally stable (rounded off to three decimal places) is ______
 A 1.452 B 6.325 C 3.125 D 7.655
GATE EC 2021      Compensators and Controllers
Question 6 Explanation:
\begin{aligned} G H &=\left(\frac{s K_{p}+K_{I}}{s}\right)\left(\frac{2}{s^{3}+4 s^{2}+5 s+2}\right) \\ q(s) &=s^{4}+4 s^{3}+5 s^{2}+s\left(2+2 k_{p}\right)+2 k_{I} \\ \text{Necessary}:\qquad K_{p} & \gt -1 ; K_{I} \gt 0\\ &\begin{array}{l|ccc} s^{4} & 1 & 5 & 2 K_{I} \\ s^{3} & 4 & 2+2 K_{p} & \\ s^{2} & \frac{18-2 K_{p}}{4} & 2 K_{I} & \\ s^{1} & \left(9-K_{p}\right)\left(1+K_{p}\right)-8 K_{I} & & \\ s^{0} & 2 K_{I} & & \end{array}\\ \end{aligned}
Sufficient:
\begin{aligned} \frac{18-2 K_{p}}{4}& \gt 0\\ \Rightarrow \quad K_{p}& \lt 9\\ \therefore \quad-1& \lt K_{p} \lt 9\\ \left(18-2 K_{p}\right)\left(2+2 K_{p}\right)-32 K_{I}& \gt 0\\ 32 K_{I}& \lt 36+32 K_{p}-4 K_{p}^{2}\\ \therefore \qquad \qquad 0& \lt K_{I} \lt \frac{36+32 K_{p}-4 K_{p}^{2}}{32} \end{aligned}
\begin{aligned} \text{If }K_{p}&=-1 \Rightarrow k_{I}=0\\ \text{If }K_{p}&=9 \Rightarrow k_{I}=0\\ \end{aligned},
\begin{aligned} \therefore \qquad \qquad \frac{d K_{I}}{d K_{p}}&=0\\ \Rightarrow \qquad \qquad 32-8 K_{p}&=0=0 \Rightarrow K_{p}=4 \end{aligned}
$\therefore$ For $K_{p}=4, K_{I}$ is maximum, which is
$K_{I}=\frac{36+32 \times 4-64}{32}=3.125$
For $K_{p}=4, K_{I} \lt 3.125$ for stability
$\therefore K_{I \max }=3.125$

 Question 7
The electrical system shown in the figure converts input source current $i_{s}(t)$ to output voltage $v_{o}(t)$

Current i $i_{L}(t)$ in the inductor and voltage $v_{c}(t)$ across the capacitor are taken as the state variables, both assumed to be initially equal to zero, i.e., $i_{L}(0) =0$ and $v_{c}(0)=0$. The system is
 A completely state controllable as well as completely observable B completely state controllable but not observable C completely observable but not state controllable D neither state controllable nor observable
GATE EC 2021      State Space Analysis
Question 7 Explanation:
\begin{aligned} i_{s} &=v_{i}+v_{c} \\ \therefore \qquad v_{i} &=-v_{c}+i_{s} \\ v_{L} &=L i_{L} \\ v_{L} &=\left(i_{s}-i_{L}\right) \\ \therefore \qquad L i_{L} &=i_{s}-i_{L} \\ \therefore \qquad i_{L} &=-i_{L}+i_{s} \\ v_{o} &=v_{c} \\ X &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] X+\left[\begin{array}{ll} 1 \\ 2 \end{array}\right] U \\ Y &=[0 \quad 1] X+[0] \cup \\ A &=-I \\ Q_{C} &=\left[\begin{array}{ll} B & A B \end{array}\right]=\left[\begin{array}{ll} 1 & -1 \\ 1 & -1 \end{array}\right] \Rightarrow\left|Q_{c}\right|=0 \\ Q_{o} &=\left[\begin{array}{ll} C^{T} & A^{\top} C^{\top} \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array}\right] \Rightarrow\left|Q_{o}\right|=0 \end{aligned}
 Question 8
The complete Nyquist plot of the open-loop transfer function G(s)H(s) of a feedback control system in the figure.

If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is
 A 0 B 1 C 4 D 3
GATE EC 2021      Frequency Response Analysis
Question 8 Explanation:
The given Nyquist plot is not matched according to the data given in the question.
 Question 9
The block diagram of a feedback control system is shown in the figure

The transfer function $\dfrac{Y{\left ( s \right )}}{X {\left ( s \right )}}$ of the system is
 A $\frac{G_{1}+G_{2}+G_{1}G_{2}H}{1+G_{1}H}$ B $\frac{G_{1}+G_{2}}{1+G_{1}H+G_{2}H}$ C $\frac{G_{1}+G_{2}}{1+G_{1}H}$ D $\frac{G_{1}+G_{2}+G_{1}G_{2}H}{1+G_{1}H+G_{2}H}$
GATE EC 2021      Basics of Control Systems, Block Diagram and SFGs
Question 9 Explanation:

$\frac{Y}{R}=\frac{G_{1}(1-0)+G_{2}(1-0)}{1-\left[-G_{1} H\right]}=\frac{G_{1}+G_{2}}{1+G_{1} H}$
 Question 10
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 10 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, $0.1=\frac{3}{K}\Rightarrow K=30$

There are 10 questions to complete.