# Frequency Response Analysis

 Question 1
The open loop transfer function of a unity gain negative feedback system is given as
$G(s)=\frac{1}{s(s+1)}$
The Nyquist contour in the s-plane encloses the entire right half plane and a small neighbourhood around the origin in the left half plane, as shown in the figure below. The number of encirclements of the point $(-1+j0)$ by the Nyquist plot of $G(s)$, corresponding to the Nyquist contour, is denoted as $N$. Then $N$ equals to

 A 0 B 1 C 2 D 3
GATE EE 2022   Control Systems
Question 1 Explanation:
Given: P = 1 (Because, Nyquist contour encircle one pole i.e s = 0)
We have, N = P - Z
N = 1-Z
Characteristic equation
\begin{aligned} 1+G(s)H(s) &=0 \\ 1+\frac{k}{s(s+1)}&=0 \\ s^2+s+k&= 0 \end{aligned}
R-H criteria:
$\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k\\ 1 & 0\\ k & \end{matrix}$
Hence, z = 0 (because no sign change in first column of R-H criteria)
(where, z = closed loop pole on RHS side of s-plane)
Therefore, N = 1
 Question 2
An LTI system is shown in the figure where
$G(s)=\frac{100}{s^2+0.1s+10}$
The steady state output of the system, to the input $r(t)$, is given as $y(t)=a+b\sin (10t+\theta )$. The values of $a$ and $b$ will be

 A a=1, b=10 B a=10, b=1 C a=1, b=100 D a=100, b=1
GATE EE 2022   Control Systems
Question 2 Explanation:
We knaow, $y(t)=A|G(j\omega ) \sin (\omega t+\phi )$
Here, $G(j\omega )=\frac{100}{-\omega ^2+j0.1\omega +100}$
Put $\omega =0$
$|G(j\omega )|=1$
Now, put $\omega =10$ rad/sec
$|G(j\omega )|=\left | \frac{100}{-100+i1+100} \right |=100$
Therefore, $y(t)=1+0.1 \times 100 \sin(10t+\theta ) =1+10 \sin (10t+\theta )$
On comparision $: a=1 ,b=10$
 Question 3
The Bode magnitude plot of a first order stable system is constant with frequency. The asymptotic value of the high frequency phase, for the system, is $-180^{\circ}$. This system has

 A one LHP pole and one RHP zero at the same frequency B one LHP pole and one LHP zero at the same frequency C two LHP poles and one RHP zero D two RHP poles and one LHP zero.
GATE EE 2022   Control Systems
Question 3 Explanation:
The given system is non-minimum phase system Therefore, transfer function, $T.F=\frac{s-1}{s+1}$
Hence, one LHP pole and one RHP zero at the same frequency.
 Question 4
The Bode magnitude plot for the transfer function $\frac{V_{o}\left ( s \right )}{V_{i}\left ( s \right )}$ of the circuit is as shown. The value of R is _____________$\Omega$. (Round off to 2 decimal places.)

 A 0.1 B 0.2 C 0.25 D 0.05
GATE EE 2021   Control Systems
Question 4 Explanation:
From response plot
\begin{aligned} M_{r} &=26 \mathrm{~dB}=20 \\ \therefore \qquad \qquad \frac{1}{2 \xi \sqrt{1-\xi^{2}}} &=20 \\ \therefore \qquad \qquad \xi &=0.025 \end{aligned}
From electrical network
\begin{aligned} \chi&=\frac{R}{2} \sqrt{\frac{C}{L}}=0.025 \\ \therefore \qquad \qquad R&=0.10 \Omega \end{aligned}
 Question 5
A stable real linear time-invariant system with single pole at p, has a transfer function $H(s)=\frac{s^2+100}{s-p}$ with a dc gain of 5. The smallest positive frequency, in rad/s at unity gain is closed to:
 A 8.84 B 11.08 C 78.13 D 122.87
GATE EE 2020   Control Systems
Question 5 Explanation:
\begin{aligned} H(s)&=T.F.=[latex]\frac{s^{2}+100}{s-p} \\ \text{D.C. gain }&= 5 \\ \Rightarrow \; \; \frac{100}{-P}&=5=P=-20 \\ H(j\omega )&=\frac{-\omega ^{2}+100}{j\omega +20} \\ \left |H(j\omega ) \right |&=\frac{-\omega ^{2}+100}{\sqrt{\omega ^{2}+400}} \\ \frac{-\omega ^{2}+100}{\sqrt{\omega ^{2}+400}}&=1 \\ \Rightarrow \; \; \omega &=8.84 \text{rad/sec.} \end{aligned}
 Question 6
Consider a negative unity feedback system with forward path transfer function $G(s)=\frac{K}{(s+a)(s-b)(s+c)}$, where K, a, b, c are positive real numbers. For a Nyquist path enclosing the entire imaginary axis and right half of the s-plane in the clockwise direction, the Nyquist plot of $(1 + G(s))$, encircles the origin of $(1 + G(s))$-plane once in the clockwise direction and never passes through this origin for a certain value of K. Then, the number of poles of $\frac{G(s)}{1+G(s)}$ lying in the open right half of the s-plane is _________ .
 A 1 B 2 C 3 D 4
GATE EE 2020   Control Systems
Question 6 Explanation:
\begin{aligned}O.L.T.F = G(s) &=\frac{K}{(s+a)(s-b)(s+c)} \\ N &= P - Z ; \; \;P = 1 \\ -1 &= 1 - Z ; \; \; N = -1 \\ Z &= 2\end{aligned}
 Question 7
The asymptotic Bode magnitude plot of a minimum phase transfer function G(s) is shown below.

Consider the following two statements.

Statement I: Transfer function G(s) has three poles and one zero.
Statement II: At very high frequency ($\omega \rightarrow \infty$), the phase angle $\angle G(j\omega)=-\frac{3\pi}{2}$.

Which one of the following options is correct?
 A Statement I is true and statement II is false. B Statement I is false and statement II is true. C Both the statements are true D Both the statements are false
GATE EE 2019   Control Systems
Question 7 Explanation:
$G(s)=\frac{k}{s\left ( 1+\frac{s}{1} \right )\left ( 1+\frac{s}{20} \right )}$
Transfer function shows 2 poles and no zeros. So statement I is false.

$\angle G(j\omega )=-90-tan^{-1}\omega -tan^{-1}\frac{\omega }{20}$
$\angle G(j\omega )|_{\omega \rightarrow \infty }=-270^{\circ}=-\frac{3\pi}{2} \;rad$
So statement II is true.
 Question 8
The open loop transfer function of a unity feedback system is given by
$G(s)=\frac{\pi e^{-0.25s}}{s}$
In G(s) plane, the Nyquist plot of G(s) passes through the negative real axis at the point
 A (-0.5, j0) B (-0.75, j0) C (-1.25, j0) D (-1.5, j0)
GATE EE 2019   Control Systems
Question 8 Explanation:
$\angle G(j\omega )=-0.25 \times \frac{180\omega _{pc}}{\pi}-90^{\circ}=-180^{\circ}$
$-0.25 \times \frac{180\omega _{pc}}{\pi}=-90^{\circ}$

$\omega _{pc}=\frac{4\pi}{2}=2\pi$
$|G(j\omega )|_{\omega =\omega _{pc}}=\left | \frac{\pi e^{-0.25s}}{s} \right |_{\omega =\omega _{pc}}$
$=\frac{\pi}{2\pi}=0.5$
$\therefore$ Point is $(-0.5,j0)$
 Question 9
Consider the unity feedback control system shown. The value of K that results in a phase margin of the system to be $30^{\circ}$ is _______.
 A 0.5 B 1.04 C 2.09 D 4.029
GATE EE 2017-SET-1   Control Systems
Question 9 Explanation:
Forward path transfer function,$G(s)=\frac{Ke^{-s}}{s}$
Given,
Phase margin $=30^{\circ}$
Phase margin $=180^{\circ}+\phi$
$30^{\circ}=180^{\circ}+\phi$
$\phi =-150^{\circ}$
$\phi =\angle G(j\omega )_{\omega =\omega _{gc}}$
[where $\omega _{gc}$ is gain crossover frequency ]
$\angle G(j\omega )=-90^{\circ}-57.3\omega ^{\circ}$
At $\omega =\omega _{gc}$
$|G(j\omega )|=1$
$\frac{k \times 1}{\omega }=1$
$\Rightarrow \omega =\omega _{gc}=K \; rad/sec.$
$\therefore \;\; \angle G(j\omega )_{\omega =\omega _{gc}}=-90^{\circ}-57.3K^{\circ}$
$-90^{\circ}-57.3K^{\circ}=-150^{\circ}$
$-57.3K^{\circ}=60^{\circ}$
$K=\frac{60}{57.3}=1.047$
 Question 10
The transfer function of a system is given by $\frac{V_{o}(s)}{V_{i}(s)}=\frac{1-s}{1+s}$
Let the output of the system be $v_{o}(t)=V_{m} sin(\omega t + \varphi )$ for the input $v_{i}(t)=V_{m} sin(\omega t )$. Then the minimum and maximum values of $\varphi$ (in radians) are respectively
 A $\frac{-\pi }{2} \; and \; \frac{\pi }{2}$ B $\frac{-\pi }{2} \; and \; 0$ C $0 \; and \; \frac{\pi }{2}$ D $-\pi$ and 0
GATE EE 2017-SET-1   Control Systems
Question 10 Explanation:
$\frac{V_o(s)}{V_i(s)}=\frac{1-s}{1+s}=H(s)$
For the minimum and maximum values of $'\phi '$
$H(j\omega )=\frac{1-j\omega }{1+j\omega }$
$\angle H(j\omega )=-2 tan^{-1}\omega$
At $\omega =0; \;\; \; \angle H(j\omega )=0^{\circ}$
At $\omega =\infty ; \;\;\; \angle H(j\omega )=-\pi$
There are 10 questions to complete.