Question 1 |
The pole-zero map of a rational function G(s) is shown below. When the closed counter \Gamma is mapped into the G(s)-plane, then the mapping encircles.
the origin of the G(s)-plane once in the counter-clockwise direction. | |
the origin of the G(s)-plane once in the clockwise direction. | |
the point -1+j0 of the G(s)-plane once in the counter-clockwise direction. | |
the point -1+j0 of the G(s)-plane once in the clockwise direction. |
Question 1 Explanation:
s-plane contour is encircling 2-poles and 3-zeros in clockwise direction hence the corresponding G(s) plane contour encircles origin 2-times in anti-clockwise direction and 3-times in clockwise direction.
Therefore, Effectively once in clockwise direction.
Therefore, Effectively once in clockwise direction.
Question 2 |
For an LTI system, the Bode plot for its gain is as illustrated in the figure shown. The number of system poles N_p and the number of system zeros N_z in the frequency range 1Hz\leq f\leq 10^7Hz is
N_p=5,N_z=2 | |
N_p=6,N_z=3 | |
N_p=7,N_z=4 | |
N_p=4,N_z=2 |
Question 2 Explanation:
Number of poles (N_{P})= 6
Number of zeros (N_{Z}) = 3
Question 3 |
The figure below shows the Bode magnitude and phase plots of a stable transfer function G(s)=\frac{n_{o}}{s^{2}+d_{2}s^{2}+d_{1}+d_{0}}
Consider the negative unity feedback configuration with gain k in the feedforward path. The closed loop is stable for k \lt k_{o} . The maximum value of k_{o} is ______.
Consider the negative unity feedback configuration with gain k in the feedforward path. The closed loop is stable for k \lt k_{o} . The maximum value of k_{o} is ______.0.1 | |
0.2 | |
0.3 | |
0.4 |
Question 3 Explanation:
For G(s)
M_{\mathrm{dB}}\left(\omega_{p c}\right)=20 \mathrm{dB}
When cascaded with k ,
\begin{aligned} \mathrm{GM}_{\mathrm{dB}} &=-20 \mathrm{dB}-20 \log _{10}(k) \gt 0 \mathrm{dB} \\ 20+20 \log _{10}(k) & \lt 0 \\ 20 \log _{10}(k) & \lt -20 \\ k & \lt 10^{-1}=0.10\\ \text{So,}\quad k_{0}&=0.10 \end{aligned}
M_{\mathrm{dB}}\left(\omega_{p c}\right)=20 \mathrm{dB}
When cascaded with k ,
\begin{aligned} \mathrm{GM}_{\mathrm{dB}} &=-20 \mathrm{dB}-20 \log _{10}(k) \gt 0 \mathrm{dB} \\ 20+20 \log _{10}(k) & \lt 0 \\ 20 \log _{10}(k) & \lt -20 \\ k & \lt 10^{-1}=0.10\\ \text{So,}\quad k_{0}&=0.10 \end{aligned}
Question 4 |
For a unity feedback control system with the forward path transfer function
G(s)=\frac{K}{s(s+2)}
The peak resonant magnitude M_{r}, of the closed-loop frequency response is 2. The corresponding value of the gain K (correct to two decimal places) is _________.
G(s)=\frac{K}{s(s+2)}
The peak resonant magnitude M_{r}, of the closed-loop frequency response is 2. The corresponding value of the gain K (correct to two decimal places) is _________.
5.38 | |
14.92 | |
20.48 | |
25.84 |
Question 4 Explanation:
Maximum resonant peak,
\begin{aligned} M_{r} &=\frac{1}{2 \xi \sqrt{1-\xi^{2}}}=2 \\ 2 \xi \sqrt{1-\xi^{2}} &=\frac{1}{2} \\ \xi^{2}\left(1-\xi^{2}\right) &=\frac{1}{16} \\ \xi^{4}-\xi^{2}+\frac{1}{16} &=0 \\ \xi^{2}&=\frac{1}{2} \pm \sqrt{\frac{1-\frac{1}{4}}{4}}=\frac{1}{2} \pm \frac{\sqrt{3}}{4} \\ \text { As } M_{r}=2 \gt 1, \xi& \lt \frac{1}{\sqrt{2}} \text { and } \xi^{2} \lt \frac{1}{2} \\ \text { So. } \quad\xi^{2}&=\frac{1}{2}-\frac{\sqrt{3}}{4}\\ Given, \quad G(s)&=\frac{K}{s(s+2)}=\frac{\omega_{n}^{2}}{s\left(s+2 \xi \omega_{n}\right)} \\ \text{So} \quad\omega_{n} &=\sqrt{K} \\ 2 \xi \sqrt{K} &=2\\ \sqrt{K} &=\frac{1}{\xi} \\ K &=\frac{1}{\xi^{2}}=\frac{1}{\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\right)} \\ &=\frac{4}{2-\sqrt{3}}=14.928 \end{aligned}
\begin{aligned} M_{r} &=\frac{1}{2 \xi \sqrt{1-\xi^{2}}}=2 \\ 2 \xi \sqrt{1-\xi^{2}} &=\frac{1}{2} \\ \xi^{2}\left(1-\xi^{2}\right) &=\frac{1}{16} \\ \xi^{4}-\xi^{2}+\frac{1}{16} &=0 \\ \xi^{2}&=\frac{1}{2} \pm \sqrt{\frac{1-\frac{1}{4}}{4}}=\frac{1}{2} \pm \frac{\sqrt{3}}{4} \\ \text { As } M_{r}=2 \gt 1, \xi& \lt \frac{1}{\sqrt{2}} \text { and } \xi^{2} \lt \frac{1}{2} \\ \text { So. } \quad\xi^{2}&=\frac{1}{2}-\frac{\sqrt{3}}{4}\\ Given, \quad G(s)&=\frac{K}{s(s+2)}=\frac{\omega_{n}^{2}}{s\left(s+2 \xi \omega_{n}\right)} \\ \text{So} \quad\omega_{n} &=\sqrt{K} \\ 2 \xi \sqrt{K} &=2\\ \sqrt{K} &=\frac{1}{\xi} \\ K &=\frac{1}{\xi^{2}}=\frac{1}{\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\right)} \\ &=\frac{4}{2-\sqrt{3}}=14.928 \end{aligned}
Question 5 |
The Nyquist stability criterion and the Routh criterion both are powerful analysis tools for
determining the stability of feedback controllers. Identify which of the following statements
is FALSE:
Both the criteria provide information relative to the stable gain range of the system. | |
The general shape of the Nyquist plot is readily obtained from the Bode magnitude plot
for all minimum-phase systems. | |
The Routh criterion is not applicable in the condition of transport lag, which can be
readily handled by the Nyquist criterion. | |
The closed-loop frequency response for a unity feedback system cannot be obtained
from the Nyquist plot. |
Question 6 |
A unity feedback control system is characterized by the open-loop transfer function
G(s)=\frac{10K(s+2)}{s^{3}+3s^{2}+10}
The Nyquist path and the corresponding Nyquist plot of G(s) are shown in the figures below.
If 0 \lt K \lt 1, then the number of poles of the closed-loop transfer function that lie in the right half of the s-plane is
G(s)=\frac{10K(s+2)}{s^{3}+3s^{2}+10}
The Nyquist path and the corresponding Nyquist plot of G(s) are shown in the figures below.
If 0 \lt K \lt 1, then the number of poles of the closed-loop transfer function that lie in the right half of the s-plane is
0 | |
1 | |
2 | |
3 |
Question 6 Explanation:
Given that, the open loop transfer function of a unity feedback system is,
G(s) =\frac{10 k(s+2)}{s^{3}+3 s^{2}+10}
From the given Nyquist plot, for 0 \lt k \lt 1 the encirclements about the point (-1+j 0) is,
\begin{array}{l} N=0 \\ N=P-Z \end{array}
P= number of open loop poles in the right half of s-plane
Z= number of closed loop poles in the right half of s-plane
To determine the value of P :
Applying R-H criteria to G(s)
G(s)=\frac{10 k(s+2)}{s^{3}+3 s^{2}+10}
\begin{array}{c|cc} s^{3} & 1 & 0 \\ s^{2} & 3 & 10 \\ s^{1} & -10 / 3 & 0 \\ s^{0} & 10 & 0 \end{array}
Two sign changes are there in the first column of the RH table. So, two open loop poles are there in right half of s-plane.
So,
P=2
\begin{array}{l} N=P-Z \\ 0=2-Z \\ Z=2 \end{array}
Z=2 indicates, there are two closed loop poles in right half of s-plane.
G(s) =\frac{10 k(s+2)}{s^{3}+3 s^{2}+10}
From the given Nyquist plot, for 0 \lt k \lt 1 the encirclements about the point (-1+j 0) is,
\begin{array}{l} N=0 \\ N=P-Z \end{array}
P= number of open loop poles in the right half of s-plane
Z= number of closed loop poles in the right half of s-plane
To determine the value of P :
Applying R-H criteria to G(s)
G(s)=\frac{10 k(s+2)}{s^{3}+3 s^{2}+10}
\begin{array}{c|cc} s^{3} & 1 & 0 \\ s^{2} & 3 & 10 \\ s^{1} & -10 / 3 & 0 \\ s^{0} & 10 & 0 \end{array}
Two sign changes are there in the first column of the RH table. So, two open loop poles are there in right half of s-plane.
So,
P=2
\begin{array}{l} N=P-Z \\ 0=2-Z \\ Z=2 \end{array}
Z=2 indicates, there are two closed loop poles in right half of s-plane.
Question 7 |
The Nyquist plot of the transfer function
G(s)=\frac{K}{(s^{2}+2s+2)(s+2)}
does not encircle the point (1+j0) for K=10 but does encircle the point (-1+j0) for K=100. Then the closed loop system (having unity gain feedback) is
G(s)=\frac{K}{(s^{2}+2s+2)(s+2)}
does not encircle the point (1+j0) for K=10 but does encircle the point (-1+j0) for K=100. Then the closed loop system (having unity gain feedback) is
stable for K = 10 and stable for K = 100 | |
stable for K = 10 and unstable for K = 100 | |
unstable for K = 10 and stable for K = 100 | |
unstable for K = 10 and unstable for K = 100 |
Question 7 Explanation:
Given that,
G(s)=\frac{K}{\left(s^{2}+2 s+2\right)(s+2)}
P= open loop poles in RHS of s-plane
Z= closed loop poles in RHS of s-plane
N= number of encirclements about the
\begin{array}{l} \text { point }(-1+j 0) \\ \text { For } K=10 ; \\ N=P-Z=0 \Rightarrow Z=0: \text { stable } \\ \text { For } K=100 ; \\ N=P-Z=1 \Rightarrow Z \neq 0: \text { unstable } \end{array}
G(s)=\frac{K}{\left(s^{2}+2 s+2\right)(s+2)}
P= open loop poles in RHS of s-plane
Z= closed loop poles in RHS of s-plane
N= number of encirclements about the
\begin{array}{l} \text { point }(-1+j 0) \\ \text { For } K=10 ; \\ N=P-Z=0 \Rightarrow Z=0: \text { stable } \\ \text { For } K=100 ; \\ N=P-Z=1 \Rightarrow Z \neq 0: \text { unstable } \end{array}
Question 8 |
Consider a stable system with transfer function
G(s)=\frac{s^{p}+b_{1}S^{p-1}+....+b_p}{s^{q}+a_{1}S^{q-1}+....+a_{q}}
where b_{1},...b_{p} and a_{1},...a_{q} are real valued constants. The slope of the Bode log magnitude curve of G(s) converges to -60 dB/decade as \omega \rightarrow \infty . A possible pair of values for p and q is
G(s)=\frac{s^{p}+b_{1}S^{p-1}+....+b_p}{s^{q}+a_{1}S^{q-1}+....+a_{q}}
where b_{1},...b_{p} and a_{1},...a_{q} are real valued constants. The slope of the Bode log magnitude curve of G(s) converges to -60 dB/decade as \omega \rightarrow \infty . A possible pair of values for p and q is
p=0 and q=3 | |
p=1 and q=7 | |
p=2 and q=3 | |
p=3 and q=5 |
Question 8 Explanation:
Final slope = -60 dB/decade,
which indicates that, q-p=3
Among the given options, option (A) satisfies this condition
Among the given options, option (A) satisfies this condition
Question 9 |
The asymptotic Bode phase plot of G(s)=\frac{1}{(s+0.1)(s+10)(s+p_{1})}, with k and p_{1} both positive, is shown below. The value of p_{1} is ________
0.5 | |
1 | |
1.5 | |
2.5 |
Question 9 Explanation:
\begin{aligned} G(s)=\frac{K}{(s+0.1)(s+10)\left(s+p_{1}\right)} \\ \angle G(s)=-\tan ^{-1} \frac{\omega}{0.1}-\tan ^{-1} \frac{\omega}{10}-\tan ^{-1} \frac{\omega}{p_{1}} \\ \left.\angle G(s)\right|_{\omega=1}=-\tan ^{-1} \frac{1}{0.1}-\tan ^{-1} \frac{1}{10}-\tan ^{-1} \frac{1}{p_{1}} \\ =-135^{\circ} \\ -\tan ^{-1} 10-\tan ^{-1} 0.1-\tan ^{-1} \frac{1}{p_{1}}=-135^{\circ} \\ -84.28^{\circ}-5.71^{\circ}-\tan ^{-1} \frac{1}{p_{1}}=-135^{\circ} \\ -\tan ^{-1} \frac{1}{p_{1}}=-135^{\circ}+90^{\circ}\\ \tan ^{-1} \frac{1}{p_{1}} =45^{\circ} \\ \frac{1}{p_{1}} =1 \Rightarrow p_{1}=1 \end{aligned}
Question 10 |
In the feedback system shown below G(s)=\frac{1}{(s+1)(s+2)(s+3)}
The positive value of k for which the gain margin of the loop is exactly 0 dB and the phase margin of the loop is exactly zero degree is ________
The positive value of k for which the gain margin of the loop is exactly 0 dB and the phase margin of the loop is exactly zero degree is ________
40 | |
50 | |
60 | |
70 |
Question 10 Explanation:
1+G(s) H(s)=1+\frac{K}{(s+1)(s+2)(s+3)}=0
(s+1)\left(s^{2}+5 s+6\right)+K=0
s^{3}+5 s^{2}+6 s+s^{2}+5 s+6+k=0
\Rightarrow s^{3}+6 s^{2}+11 s+6+K=
Gain margin =0dB and phase margin =0^{\circ}
It implies marginal stable system
By Routh Array
\begin{array}{r|rr} s^{3} & 1 & 11 \\ s^{2} & 6 &(6+k) \\ s & \frac{66-6-K}{6} & 0 \\ s^{\circ} & 6+K & \end{array}
For marginal stable system,
60-K=0
K=60
(s+1)\left(s^{2}+5 s+6\right)+K=0
s^{3}+5 s^{2}+6 s+s^{2}+5 s+6+k=0
\Rightarrow s^{3}+6 s^{2}+11 s+6+K=
Gain margin =0dB and phase margin =0^{\circ}
It implies marginal stable system
By Routh Array
\begin{array}{r|rr} s^{3} & 1 & 11 \\ s^{2} & 6 &(6+k) \\ s & \frac{66-6-K}{6} & 0 \\ s^{\circ} & 6+K & \end{array}
For marginal stable system,
60-K=0
K=60
There are 10 questions to complete.