Question 1 |
Consider a closed-loop control system with unity negative feedback and KG(s) in
the forward path, where the gain K=2. The complete Nyquist plot of the transfer
function G(s) is shown in the figure. Note that the Nyquist contour has been chosen
to have the clockwise sense. Assume G(s) has no poles on the closed right-half of
the complex plane. The number of poles of the closed-loop transfer function in the
closed right-half of the complex plane is _______


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Question 1 Explanation:

For K = 2, point A will be -0.8x2 = -1.6
Hence N = -2, P = 0
(By default Nyquist contoure is considered in clockwise direction)
P - N = 2
Number of closed loop pole in right side of the complex plane.
Question 2 |
The complete Nyquist plot of the open-loop transfer function G(s)H(s) of a feedback control system in the figure.

If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is

If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is
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4 | |
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Question 2 Explanation:
The given Nyquist plot is not matched according to the data given in the question.
Question 3 |
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 3 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 4 |
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 4 Explanation:

Number of poles (N_{P})= 6
Number of zeros (N_{Z}) = 3
Question 5 |
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 ______.

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Question 5 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}
There are 5 questions to complete.