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
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
0 | |
1 | |
2 | |
3 |
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
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
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=1, b=10 | |
a=10, b=1 | |
a=1, b=100 | |
a=100, b=1 |
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
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
one LHP pole and one RHP zero at the same frequency | |
one LHP pole and one LHP zero at the same frequency | |
two LHP poles and one RHP zero | |
two RHP poles and one LHP zero. |
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.
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.)
0.1 | |
0.2 | |
0.25 | |
0.05 |
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}
\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:
8.84 | |
11.08 | |
78.13 | |
122.87 |
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
_________ .
1 | |
2 | |
3 | |
4 |
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?
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?
Statement I is true and statement II is false. | |
Statement I is false and statement II is true. | |
Both the statements are true | |
Both the statements are false |
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.
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
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
(-0.5, j0) | |
(-0.75, j0) | |
(-1.25, j0) | |
(-1.5, j0) |
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)
-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 _______.
0.5 | |
1.04 | |
2.09 | |
4.029 |
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
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
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
\frac{-\pi }{2} \; and \; \frac{\pi }{2} | |
\frac{-\pi }{2} \; and \; 0 | |
0 \; and \; \frac{\pi }{2} | |
-\pi and 0 |
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
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.