Question 1 |
The damping ratio and undamped natural frequency of a closed loop system as
shown in the figure, are denoted as \zeta and
\omega _n, respectively. The values of \zeta and \omega _n are
\zeta =0.5 \text{ and }\omega _n=10 rad/s | |
\zeta =0.1 \text{ and }\omega _n=10 rad/s | |
\zeta =0.707 \text{ and }\omega _n=10 rad/s | |
\zeta =0.707 \text{ and }\omega _n=100 rad/s |
Question 1 Explanation:
Reduced the block diagram:
Transfer function,
\frac{C(s)}{R(s)}=\frac{100/(s(s+10))}{1+(100/s(s+10))}=\frac{100}{s^2+10s+100}
Standard form,
T.F.=\frac{\omega _n^2}{s^2+2\xi \omega _ns+\omega _n^2}
On comparison : \omega =\sqrt{100}=10rad/sec and 2\xi \omega _n=10
\Rightarrow \xi=\frac{10}{2 \times 10}=0.5
Transfer function,
\frac{C(s)}{R(s)}=\frac{100/(s(s+10))}{1+(100/s(s+10))}=\frac{100}{s^2+10s+100}
Standard form,
T.F.=\frac{\omega _n^2}{s^2+2\xi \omega _ns+\omega _n^2}
On comparison : \omega =\sqrt{100}=10rad/sec and 2\xi \omega _n=10
\Rightarrow \xi=\frac{10}{2 \times 10}=0.5
Question 2 |
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 2 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 3 |
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 3 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 4 |
The open loop transfer function of a unity gain negative feedback system is given by G(s)=\frac{k}{s^2+4s-5}.
The range of k for which the system is stable, is
The range of k for which the system is stable, is
k \gt 3 | |
k \lt 3 | |
k \gt 5 | |
k \lt 5 |
Question 4 Explanation:
Characteristic equation:
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
Question 5 |
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 5 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 6 |
The state space representation of a first-order system is given as
\overset{\bullet }{x}=-x+u
y=x
where,x is the state variable, u is the control input and y is the controlled output. Let u=-Kx be the control law, where K is the controller gain. To place a closed-loop pole at -2, the value of K is ___________________.
\overset{\bullet }{x}=-x+u
y=x
where,x is the state variable, u is the control input and y is the controlled output. Let u=-Kx be the control law, where K is the controller gain. To place a closed-loop pole at -2, the value of K is ___________________.
1 | |
2 | |
4 | |
6 |
Question 6 Explanation:
\dot{x}=-x-K x=x(-k I-I)
Characteristic equation,
\begin{aligned} |S I+K I+I| &=0 \\ |(S+1+K) I| &=0 \\ \therefore \qquad\qquad S+1+K &=0 \\ S &=-1-K \\ -2 &=-1-K \\ K &=1 \end{aligned}
Characteristic equation,
\begin{aligned} |S I+K I+I| &=0 \\ |(S+1+K) I| &=0 \\ \therefore \qquad\qquad S+1+K &=0 \\ S &=-1-K \\ -2 &=-1-K \\ K &=1 \end{aligned}
Question 7 |
In the given figure, plant G_{P}\left ( s \right )=\dfrac{2.2}{\left ( 1+0.1s \right )\left ( 1+0.4s \right )\left ( 1+1.2s \right )} and compensator G_{C}\left ( s \right )=K\left [ \dfrac{1+T_{1}s}{1+T_{2}s} \right ]. The external disturbance input is D(s). It is desired that when the disturbance is a unit step, the steady-state error should not exceed 0.1 unit. The minimum value of K is _____________.
(Round off to 2 decimal places.)
(Round off to 2 decimal places.)
12.25 | |
14.12 | |
9.54 | |
6.22 |
Question 7 Explanation:
\begin{aligned} e_{s s} &=\lim _{s \rightarrow 0}\left[\frac{s R}{1+G_{C} G_{p}}-\frac{s D G_{p}}{1+G_{C} G_{P}}\right] \\ R(s) &=0 ; D(s)=\frac{1}{s} \\ \therefore \qquad\qquad e_{s s}&=\frac{2.2}{1+2.2 K}=0.1 \\ \therefore \qquad\qquad K_{\min }&=9.54 \end{aligned}
Question 8 |
Consider a closed-loop system as shown. G_{P}\left ( s \right )=\dfrac{14.4}{s\left ( 1+0.1s \right )} is the plant transfer function and G_{c}(S)=1 is the compensator. For a unit-step input, the output response has damped oscillations. The damped natural frequency is ___________________ \text{rad/s}. (Round off to 2 decimal places.)
10.9 | |
4.62 | |
12.02 | |
8.05 |
Question 8 Explanation:
\begin{aligned} q(s)&=s^{2}+10 s+144=0 \\ \omega_{n}&=12 ; \xi=\frac{5}{12} \\ \omega_{d}&=\omega_{n} \sqrt{1-\xi^{2}} \\ \quad&=12 \sqrt{\frac{119}{144}}=10.90 \end{aligned}
Question 9 |
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 9 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 10 |
For the closed-loop system shown, the transfer function \dfrac{E\left ( s \right )}{R\left ( s \right )} is
\frac{G}{1+GH} | |
\frac{GH}{1+GH} | |
\frac{1}{1+GH} | |
\frac{1}{1+G} |
Question 10 Explanation:
\begin{aligned} \frac{E(s)}{R(s)} &=\frac{R(s)-H \times C(s)}{R(s)}=1-H \times \frac{C(s)}{R(s)} \\ &=1-\frac{H G}{1+G H}=\frac{1+G H-G H}{1+G H} \\ &=\frac{1}{1+G H} \end{aligned}
There are 10 questions to complete.