# Compensators and Controllers

 Question 1
Which of the following statements is incorrect?
 A Lead compensator is used to reduce the settling time. B Lag compensator is used to reduce the steady state error. C Lead compensator may increase the order of a system. D Lag compensator always stabilizes an unstable system.
GATE EC 2017-SET-2   Control Systems
Question 1 Explanation:
In_ case of high type systems Lag compensator fails to give stability.
 Question 2
Which of the following can be pole-zero configuration of a phase-lag controller (lag compensator)?
 A A B B C C D D
GATE EC 2017-SET-1   Control Systems
Question 2 Explanation:
Phase lag controller transfer function is
$G_{C}(s)=\frac{s+Z}{s+P} ;|Z|\gt|P|$

 Question 3
The transfer function of a first-order controller is given as

$G_{c}(s)=\frac{K(s+a)}{s+b}$

where K, a and b are positive real numbers. The condition for this controller to act as a phase lead compensator is
 A $a \lt b$ B $a \gt b$ C $K \lt ab$ D $K \gt ab$
GATE EC 2015-SET-3   Control Systems
Question 3 Explanation:
\begin{aligned} G_{c}(s)&=\frac{(1+s \tau)}{(1+\alpha s \tau)} \quad ; \quad \alpha \lt 1\\ \text{Here,}\quad \tau&=\frac{1}{a}\\ \text{and}\quad \alpha \tau&=\frac{1}{b}\\ \text{or,}\quad \alpha&=\frac{a}{b} \lt 1\\ \text{or,}\quad a& \lt b \end{aligned}
 Question 4
A lead compensator network includes a parallel combination of R and C in the feed-forward path. If the transfer function of the compensator is $G(s)=\frac{s+2}{s+4}$ , the value of RC is ________.
 A 1 B 1.5 C 0.5 D 2
GATE EC 2015-SET-1   Control Systems
Question 4 Explanation:
$G_{C}(s)=\frac{s+2}{s+4} \quad\dots(i)$

Transfer function $=\frac{1+s \tau}{1+\alpha s \tau} \quad\dots(ii)$
Where,
$\tau=$ Lead time constant $=R_{1} C$
and $\alpha=\frac{R_{2}}{R_{1}+R_{2}}$
Comparing equation (i) and (ii), we get
$\tau=\frac{1}{2} \quad$ and $\quad \alpha \tau=\frac{1}{4}$
or, $\alpha=\frac{1}{2}$
$\therefore$ RC time constant =0.5
 Question 5
For the following feedback system $G(s)=\frac{1}{(s+1)(s+2)}$. The 2% settling time of the step response is required to be less than 2 seconds. Which one of the following compensators C(s) achieves this ?

 A $3 (\frac{1}{s+5})$ B $5 (\frac{0.03}{s}+1)$ C $2 (s+4)$ D $4 (\frac{s+8}{s+3})$
GATE EC 2014-SET-1   Control Systems
Question 5 Explanation:
Given open loop transfer function is
\begin{aligned} G(s)&=\frac{1}{(s+1)(s+2)} \\ \therefore T(s)&=\frac{G(s)}{1+G(s)}=\frac{1}{(s+1)(s+2)+1} \\ T(s)&=\frac{1}{s^{2}+3 s+3} &\ldots(i) \end{aligned}
Comparing equation (i) with standard transfer function
$\xi \omega_{n}=\frac{3}{2}=1.5$
$\therefore$ 2% settling time
$\tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{1.5}=2.67>2 \mathrm{sec}$
Thus, in order to make settling time $\left(\tau_{s}\right)$ less than 2 sec, PD controller should be used. Hence, option (C) is the correct choice.
Where, $C(s)=2(s+4)$
$\therefore$ New transfer function $=T^{\prime}(G)$
$=\frac{C(s) G(s)}{1+C(s) G(s)}=\frac{2(s+4)}{s^{2}+3 s+2+2 s+8}$
or $T^{\prime}(s)=\frac{2(s+4)}{s^{2}+5 s+10}$
$\therefore \quad \tau_{s}=\frac{4}{\xi \omega_{n}}=\frac{4}{5 / 2}=\frac{4}{2.5}=1.6$
 Question 6
The transfer function of a compensator is given as
$G_{c}(s)=\frac{s+a}{s+b}$

The phase of the above lead compensator is maximum at
 A $\sqrt{2}$ rad/s B $\sqrt{3}$rad/s C $\sqrt{6}$ rad/s D 1/$\sqrt{3}$ rad/s
GATE EC 2012   Control Systems
Question 6 Explanation:
For phase to be maximum,
\begin{aligned} \frac{\partial P}{\partial \omega}&=0 \\ \frac{1}{1+\frac{\omega^{2}}{a^{2}}}- \frac{1/b}{1+\frac{\omega^{2}}{b^{2}}}&=0 \\ \frac{1}{1+\omega^{2}}- \frac{1/b}{1+\frac{\omega^{2}}{4}}&=0\\ \frac{1}{1+\omega^{2}}-\frac{2}{\omega^{2}+4} &=0 \\ \omega^{2}+4-2-2 \omega^{2} &=0 \\ \omega^{2} &=2 \\ \omega &=\sqrt{2} \mathrm{rad} / \mathrm{sec} \end{aligned}
 Question 7
The transfer function of a compensator is given as
$G_{c}(s)=\frac{s+a}{s+b}$

$G_{c}(S)$ is a lead compensator if
 A a =1, b = 2 B a = 3, b = 2 C a = -3, b = -1 D a = 3, b = 1
GATE EC 2012   Control Systems
Question 7 Explanation:
$G_{c}(s)=\left(\frac{s+a}{s+b}\right)$
Phase $P=\tan ^{-1}(\omega / a)-\tan ^{-1} \omega / b$
for lead comparators phase must be + ve.
for this $\frac{\omega}{a} \gt \frac{\omega}{b} \Rightarrow a \lt b$
So $a=1, b=2$
 Question 8
A unity negative feedback closed loop system has a plant with the transfer function $G(s)=\frac{1}{s^{2}+2s+2}$ and a controller $G_{c}(s)$ in the feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is
 A $G_{c}(s)=\frac{s+1}{s+2}$ B $G_{c}(s)=\frac{(s+1)(s+4)}{(s+2)(s+3)}$ C $G_{c}(s)=\frac{s+2}{s+1}$ D $G_{c}(s)=1+\frac{2}{s}+3s$
GATE EC 2010   Control Systems
Question 8 Explanation:
\begin{aligned} e_{s s} &=\lim _{s \rightarrow 0} \frac{s R(s)}{1+G(s) G_{c}(s)} \\ r(t) &=u(t) \\ R(s) &=\frac{1}{s} \\ e_{ss} &=\lim _{s \rightarrow 0} \frac{s \cdot \frac{1}{s}}{1+G(s) G_{c}(s)} \\ e_{ss}&=\lim _{s \rightarrow 0} \frac{1}{1+G(s) G_{c}(s)} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+1}{s+2}, e_{s s}=\frac{2}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{s+2}{s+1}, e_{s s}=\frac{1}{3} \\ \text { Taking, } \quad G_{c}(s) &=\frac{(s+1)(s+4)}{(s+2)(s+3)}, e_{s s}=\frac{3}{5} \\ \text { Taking, } \quad & G_{c}(s)=1+\frac{2}{s}+3 s, e_{s s}=0 \end{aligned}
 Question 9
The magnitude plot of a rational transfer function G(s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?
 A Lead compensator B Lag compensator C PID compensator D Lead-lag compensator
GATE EC 2009   Control Systems
 Question 10
Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions $R_{2}C_{2} \gt R_{1}C_{1}$. The transfer functions $\frac{V_{o}}{V_{i}}$ represents a kind of controller. Match the impedances in Group I with the type of controllers in Group II

 A Q - 1,R - 2 B Q - 1,R - 3 C Q - 2,R - 3 D Q - 3,R - 2
GATE EC 2008   Control Systems
Question 10 Explanation:
$\begin{array}{c} \frac{V_{i}\left(R_{1} C_{1} S+1\right) 1}{R_{1}}=-\frac{V_{0}}{Z} \\ \Rightarrow \frac{V_{0}}{V_{i}}=-\frac{Z\left(R_{1} C_{1} S+1\right)}{R_{1}} \end{array}$
In case of Q, $\quad Z=\frac{R_{2} C_{2} s+1}{C_{2} S+1}$
In case of R, $\quad Z=\frac{R_{2}}{R_{2} C_{2} s+1}$
Considering Q
$\frac{V_{0}}{V_{i}}=-\frac{\left(R_{1} C_{1} S+1\right)}{R_{1}} \cdot \frac{\left(R_{2} C_{2} S+1\right)}{C_{2} S}$
Considering R ,
$\frac{V_{0}}{V_{i}}=-\frac{\left(R_{1} C_{1} S+1\right)}{R_{1}} \cdot \frac{R_{2}}{\left(R_{2} C_{2} s+1\right)}$
$\because$ Given that $R_{2} C_{2}>R_{1} C_{1}$
$\therefore \quad$ Considering R, controller is lag compensator.
and Considering Q , Controller is PID controller.
There are 10 questions to complete.