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 _______

0 | |

1 | |

2 | |

3 |

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

0 | |

1 | |

4 | |

3 |

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 ______.

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 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}

Question 6 |

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 6 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 7 |

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 8 |

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 8 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 9 |

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 9 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 10 |

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 10 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

There are 10 questions to complete.