# Feedback Characteristics of Control Systems

 Question 1
Consider a unity-gain negative feedback system consisting of the plant $G(s)$ (given below) and a proportional-integral controller. Let the proportional gain and integral gain be 3 and 1, respectively. For a unit step reference input, the final values of the controller output and the plant output, respectively, are

$G(s)=\frac{1}{s-1}$
 A $\infty, \infty$ B $1,0$ C $1,-1$ D $-1,1$
GATE EE 2023   Control Systems
Question 1 Explanation:
Given plant:

So, $\quad$ OLTF $=\frac{(3 s+1)}{s(s-1)}$

Closed loop transfer function,
\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{3 s+1}{s^{2}+2 s+1} \\ Y(s) & =\frac{3 s+1}{s\left(s^{2}+2 s+1\right)} \quad\left[\because R(s)=\frac{1}{s}\right] \end{aligned}

Final value of plant,
$Y(\infty)=\operatorname{Lims}_{s \rightarrow 0} Y(s)=1$

From plant,
\begin{aligned} X(\mathrm{~s}) & =\left[\mathrm{R}(\mathrm{s})-\frac{\mathrm{X}(\mathrm{s})}{\mathrm{s}-1}\right]\left(3+\frac{1}{\mathrm{~s}}\right) \\ X(\mathrm{~s})\left[1+\frac{3 \mathrm{~s}+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right)\left[\because \mathrm{R}(\mathrm{s})=\frac{1}{\mathrm{~s}}\right] \\ X(\mathrm{~s})\left[\frac{\mathrm{s}^{2}+2 s+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right) \\ \Rightarrow \quad X(\mathrm{~s}) & =\frac{(3 \mathrm{~s}+1)(\mathrm{s}-1)}{\mathrm{s}\left(\mathrm{s}^{2}+2 \mathrm{~s}+1\right)} \end{aligned}

$\therefore$ Final value of controller,
$x(\infty)=\operatorname{LimsX}_{s \rightarrow 0} \mathrm{~s}(\mathrm{~s})=-1$
 Question 2
The block diagram of a system is shown in the figure

If the desired transfer function of the system is
$\frac{C(s)}{R(s)}=\frac{s}{s^{2}+s+1}$
then G(s) is
 A 1 B s C 1/s D $\frac{-s}{s^{3}+s^{2}-s-2}$
GATE EE 2014-SET-3   Control Systems
Question 2 Explanation:
For all given values of G(s) in options, G(s) does not satisfy the desired condition. But option (B) is official answer given by IIT.

 Question 3
The closed-loop transfer function of a system is $T(s)=\frac{4}{(s^{2}+0.4s+4)}$. The steady state error due to unit step input is _____.
 A 0 B 0.15 C 0.75 D 1
GATE EE 2014-SET-2   Control Systems
Question 3 Explanation:
Closed loop transfer function,
$T(s)=\frac{4}{(s^2+0.4s+4)}$
or, $\frac{G(s)}{1+G(s)}$
$=\left ( \frac{4}{s^2+0.4s+4} \right )\;\;\;\;[H(s)=1]$
or, $(s^2+0.4s+4)=4+4G(s)$
or, $G(s)=\frac{4}{s(s+0.4)}$
=Open llop transfer function ( for unity feedback system)
Given, input =u(t)=r(t)
$\therefore \;\;\; R(s)=1/s$
$e_{ss}=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)H(s)} \right ]$
$\;\;=\lim_{s \to 0}\left [ \frac{sR(s)}{1+G(s)} \right ]\;\;\; [For H(s)=1]$
$\;\;=\lim_{s \to 0}\left [ \frac{s \times \frac{1}{s}}{1+\frac{4}{s(s+0.4)}} \right ]$
$\;\;=\lim_{s \to 0}\left [ \frac{s(s+0.4)}{s^2+0.4s+4} \right ]=0$
$\therefore$ Steady state error for step input, $e_{ss}=0$
 Question 4
The open-loop transfer function of a dc motor is given as $\frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}$. When connected in feedback as shown below, the approximate value of $K_a$ that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is
 A 1 B 5 C 10 D 100
GATE EE 2013   Control Systems
Question 4 Explanation:
$\frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}=\frac{10}{1+\tau s}$
$\tau =10$

Closed loop systems,
$\frac{\omega (s)}{R(s)}=\frac{\frac{10K_a}{1+10s}}{1+\frac{10K_a}{1+10s}}$
$=\frac{10K_a}{1+10K_a+10s}$
$=\frac{\frac{10K_a}{1+10K_a}}{1+\frac{10}{1+10K_a}s}$
According to quetion,
$\frac{10}{1+10K_a}=\frac{1}{100}\times 10$
$1+10K_a=100$
$K_a=9.9$
 Question 5
As shown in the figure, a negative feedback system has an amplifier of gain 100 with $\pm$10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :
 A $10\pm 1$% B $10\pm 2$% C $10\pm 5$% D $10\pm 10$%
GATE EE 2010   Control Systems
Question 5 Explanation:
$G=100\pm 10$%
$\frac{\Delta G}{G}=10$% or 0.1
$H=\frac{9}{100}$
Overall gain,
$T=\frac{G}{1+GH}$
$T=\frac{100}{1+100 \times \frac{9}{100}}=10$
Sensitivity w.r.t. $'G'=\frac{1}{1+GH} \times 10$%
$\;\;\;=\frac{1}{1+10 \times 1}\times 10$%
$\;\;\;=\frac{10}{11}\cong 1$%

There are 5 questions to complete.