Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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
Question 7 |
Loop transfer function of a feedback system is G(s)H(s)=\frac{s+3}{s^{2}(s-3)}. Take the Nyquist contour in the clockwise direction. Then, the Nyquist plot of G(s)H(s) encircles -1+ j0
once in clockwise direction | |
twice in clockwise direction | |
once in anticlockwise direction | |
twice in anticlockwise direction |
Question 7 Explanation:
Nyquist plot of G(s)H(s)=\frac{s+3}{s^2(s-3)} is as shown below
From the Nyquist plot G(s)H(s) encircle -1+j0 once in clockwise direction.
Alternate Solution:
Characteristic equation,
1+G(s)H(s)=0
s^2(s-3)+(s+3)=0
s^3-3s^2+s+3=0
using Routh's array
\left.\begin{matrix} s^3\\ s^2\\ s^1\\ s \end{matrix}\right|\begin{matrix} 1 &1 \\ -3& 3\\ 2& 0\\ 3& \end{matrix}
There are two sign changes, hence two poles in right side of s-plane exist.
Z=2, \; P=1
N=P-Z=-1
One encirclement in clockwise direction.
From the Nyquist plot G(s)H(s) encircle -1+j0 once in clockwise direction.
Alternate Solution:
Characteristic equation,
1+G(s)H(s)=0
s^2(s-3)+(s+3)=0
s^3-3s^2+s+3=0
using Routh's array
\left.\begin{matrix} s^3\\ s^2\\ s^1\\ s \end{matrix}\right|\begin{matrix} 1 &1 \\ -3& 3\\ 2& 0\\ 3& \end{matrix}
There are two sign changes, hence two poles in right side of s-plane exist.
Z=2, \; P=1
N=P-Z=-1
One encirclement in clockwise direction.
Question 8 |
Consider the following asymptotic Bode magnitude plot (\omega is in rad/s).
Which one of the following transfer functions is best represented by the above Bode magnitude plot?
Which one of the following transfer functions is best represented by the above Bode magnitude plot?
\frac{2s}{(1+0.5s)(1+0.25s)^{2}} | |
\frac{4(1+0.5s)}{s(1+0.25s)} | |
\frac{2s}{(1+2s)(1+4s)} | |
\frac{4s}{(1+2s)(1+4s)^{2}} |
Question 8 Explanation:
From the given Bode plot, it is evident that there are 3(three) poles in the transfer function, out of which there are double poles at corner frequency near but less than \omega=8 rad/sec and one pole is near but greater than \omega=0.5 rad/sec. The initial slope is +20 dB/dec. Therefore one zero exist at s=0. So from all the given options, option (A) satisfies all the conditions. Therefore Option (A) is correct.
Question 9 |
The phase cross-over frequency of the transfer function G(s)=\frac{100}{(s+3)^{3}} in rad/s is
\sqrt{3} | |
1/\sqrt{3} | |
3 | |
3\sqrt{3} |
Question 9 Explanation:
G(s)=\frac{100}{(s+1)^3}
G(j\omega )=\frac{100}{(1+j\omega )^3}
\;\;=\frac{100}{1+(j\omega )^3+3(j\omega )^2+3j\omega }
\;\;=\frac{100}{(1-3\omega^2 )+j(3\omega-\omega^3)}
\;\;=\frac{100[(1-3\omega^2 )+j\omega(3-\omega^2)^2]}{[ (1-3\omega^2 )+\omega^2(3-\omega^2)^2]}
For phase corssover frequency \omega_{ph} Img[G(j \omega )]=0;
Hence, \omega (3-\omega ^2)=0
\omega =0; \pm \sqrt{3}
Therefore, \omega _{ph}=\sqrt{3} rad/sec
G(j\omega )=\frac{100}{(1+j\omega )^3}
\;\;=\frac{100}{1+(j\omega )^3+3(j\omega )^2+3j\omega }
\;\;=\frac{100}{(1-3\omega^2 )+j(3\omega-\omega^3)}
\;\;=\frac{100[(1-3\omega^2 )+j\omega(3-\omega^2)^2]}{[ (1-3\omega^2 )+\omega^2(3-\omega^2)^2]}
For phase corssover frequency \omega_{ph} Img[G(j \omega )]=0;
Hence, \omega (3-\omega ^2)=0
\omega =0; \pm \sqrt{3}
Therefore, \omega _{ph}=\sqrt{3} rad/sec
Question 10 |
Nyquist plots of two functions G_1(s) \;and \; G_2(s) are shown in figure.
Nyquist plot of the product of G_1(s) \;and \; G_2(s) is
Nyquist plot of the product of G_1(s) \;and \; G_2(s) is
A | |
B | |
C | |
D |
Question 10 Explanation:
G_1(s)=\frac{1}{s}
G_2(s)=s
G_1(s) \times G_2(s)=\frac{1}{s}\cdot s=1
G_2(s)=s
G_1(s) \times G_2(s)=\frac{1}{s}\cdot s=1
There are 10 questions to complete.
