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 ofK 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.