# Frequency Response Analysis

 Question 1
Consider a lead compensator of the form

$K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}, \quad \beta \gt 1, a \gt 0$

The frequency at which this compensator produces maximum phase lead is $4 \mathrm{rad} / \mathrm{s}$. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bodemagnitude plot of $K(\mathrm{~s})$, is $6 \mathrm{~dB}$. The values of $\alpha, \beta$, respectively, are
 A 1, 16 B 2, 4 C 3, 5 D 2.66, 2.25
GATE EE 2023   Control Systems
Question 1 Explanation:
$K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta}}$
Max. phase lead occur at, $\omega_{m}=\sqrt{a(a \beta)}$ $=\sqrt{\beta}$

Given : $\quad \omega_{\mathrm{m}}=4 \mathrm{rad} / \mathrm{sec}$
$a \sqrt{\beta}=4 \quad ...(1)$

Now, $\quad K(s)=\frac{1+\frac{s}{a}}{1+\frac{s}{\beta a}}$

Put, $s=j \omega$
$\mathrm{K}\left(\mathrm{j} \omega_{\mathrm{n}}\right)=\frac{1+\frac{\mathrm{j} \omega_{m}}{a}}{1+\frac{j \omega_{m}}{\beta a}}=\frac{j w_{m}}{a}$

Given : $M=6$
\begin{aligned} 20 \log \left(\frac{4}{a}\right) & =6 \\ \Rightarrow \quad a & =2 \end{aligned}

From eqn. (1), we get
\begin{aligned} 2 \sqrt{\beta} & =4 \\ \beta & =4 \end{aligned}
 Question 2
In the Nyquist plot of the open-loop transfer function

$\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1}$

corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in s-plane is mapped into a point at

 A $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\infty$ B $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=0$ C $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=3$ D $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=-5$
GATE EE 2023   Control Systems
Question 2 Explanation:
Nyquist Contour :

Given:
\begin{aligned} \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1} \\ \text { Put } \quad \mathrm{s} & =R e^{j \theta} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\operatorname{Lim}_{R \rightarrow \infty} \frac{3 R e^{j \theta}+5}{\operatorname{Re}^{j \theta}-1} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =3 \end{aligned}

 Question 3
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 3 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 4
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 4 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 5
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 5 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.

There are 5 questions to complete.