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.