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 |
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 2 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 3 |
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 3 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 4 |
Consider a negative unity feedback system with the forward path transfer function \frac{s^2+s+1}{s^3+2s^2+2s+K}, where K is a positive real number. The value of K for which the system
will have some of its poles on the imaginary axis is ________ .
9 | |
8 | |
7 | |
6 |
Question 4 Explanation:
CE is
1+G(s)H(s)=0
\Rightarrow \, 1+\frac{s^{2}+s+1}{s^{3}+2s^{2}+2s+k}=0
\Rightarrow \, s^{3}+3s^{2}+3s+(1+K)=0
R.H. criteria:
9 - (1 + K) = 0
\Rightarrow \, \, K=8
1+G(s)H(s)=0
\Rightarrow \, 1+\frac{s^{2}+s+1}{s^{3}+2s^{2}+2s+k}=0
\Rightarrow \, s^{3}+3s^{2}+3s+(1+K)=0
R.H. criteria:
9 - (1 + K) = 0
\Rightarrow \, \, K=8
Question 5 |
Which of the following option is correct for the system shown below?
4^{th} order and stable | |
3^{rd} order and stable | |
4^{th} order and unstable | |
3^{rd} order and unstable |
Question 5 Explanation:
\begin{aligned}
1+\frac{20}{s^{2}(s+1)(s+20)}&=0\\ (s^{3}+s^{2})(s+20)+20&=0\\ s^{4}+20s^{3}+s^{3}+20s^{2}+20&=0\\ s^{4}+21s^{3}+20s^{2}+20&=0
\end{aligned}
Given system is fourth order system and unstable.
stablity status: since it has one missing term of 's' thus undoubtedly given transfer function is unstable.
Given system is fourth order system and unstable.
stablity status: since it has one missing term of 's' thus undoubtedly given transfer function is unstable.
Question 6 |
Consider a linear time-invariant system whose input r(t) and output y(t) are related by
the following differential equation.
\frac{d^2y(t)}{dt^2}+4y(t)=6r(t)
The poles of this system are at
\frac{d^2y(t)}{dt^2}+4y(t)=6r(t)
The poles of this system are at
+2j, -2j | |
+2, -2 | |
+4, -4 | |
+4j, -4j |
Question 6 Explanation:
\begin{aligned}\frac{d^2 y(t)}{dt^{2}}+4y(t) &=6 r(t)\\ [s^{2}+4]Y(s)&=6 R(s) \\ \frac{Y(s)}{R(s)}&=\frac{6}{s^{2}+4} \\ \text{Poles: } s^{2}+4&=0 \\ s&=\pm j2 \end{aligned}
Question 7 |
The unit step response y(t) of a unity feedback system with open loop transfer function
G(s)H(s)=\frac{K}{(s+1)^{2}(s+2)}
is shown in the figure. The value of K is _______ (up to 2 decimal places).
G(s)H(s)=\frac{K}{(s+1)^{2}(s+2)}
is shown in the figure. The value of K is _______ (up to 2 decimal places).
4 | |
8 | |
10 | |
12 |
Question 7 Explanation:
Closed loop transfer function,
\frac{C(s)}{R(s)}=\frac{\frac{K}{(s+1)^2(s+2)}}{1+\frac{K}{(s+1)^2(s+2)}}
\frac{C(s)}{R(s)}=\frac{K}{(s+1)^2(s+2)+K}
Given R(s)=\frac{1}{s}
C(s)=\frac{K}{s((s+1)^2(s+2)+K)}
\lim_{s \to 0}sC(s)=0.8
\frac{K}{2+K}=0.8
\Rightarrow \; K=8
\frac{C(s)}{R(s)}=\frac{\frac{K}{(s+1)^2(s+2)}}{1+\frac{K}{(s+1)^2(s+2)}}
\frac{C(s)}{R(s)}=\frac{K}{(s+1)^2(s+2)+K}
Given R(s)=\frac{1}{s}
C(s)=\frac{K}{s((s+1)^2(s+2)+K)}
\lim_{s \to 0}sC(s)=0.8
\frac{K}{2+K}=0.8
\Rightarrow \; K=8
Question 8 |
C onsider a unity feedback system with forward transfer function given by
G(s)=\frac{1}{(s+1)(s+2)}
The steady-state error in the output of the system for a unit-step input is _________(up to 2 decimal places).
G(s)=\frac{1}{(s+1)(s+2)}
The steady-state error in the output of the system for a unit-step input is _________(up to 2 decimal places).
0.25 | |
0.45 | |
0.66 | |
0.85 |
Question 8 Explanation:
Steady state error for type-0 and step input,
e_{ss}=\frac{1}{1+k+p}
k_p=\lim_{s \to 0}\frac{1}{(s+1)(s+2)}=\frac{1}{2}
e_{ss}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}
\;\;=0.66 uinits
e_{ss}=\frac{1}{1+k+p}
k_p=\lim_{s \to 0}\frac{1}{(s+1)(s+2)}=\frac{1}{2}
e_{ss}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}
\;\;=0.66 uinits
Question 9 |
Match the transfer functions of the second-order systems with the nature of the systems
given below.
P-I, Q-II, R-III | |
P-II, Q-I, R-III | |
P-III, Q-II, R-I | |
P-III, Q-I, R-II |
Question 9 Explanation:
P=\frac{15}{s^2+5s+15}
\omega _n=\sqrt{15}=3.872 rad/sec
2 \xi \times 3.872=5
\xi=\frac{5}{2 \times 3.872}=0.64 \;\;\;\;(Underdamped)
Q=\frac{25}{s^2+10s+25}
\omega _n=\sqrt{25}=5 rad/sec
2 \xi \times 5=10
\xi=1 \;\;\;\;\;(Critically \; damped)
Observing all the options, option (C) is correct.
\omega _n=\sqrt{15}=3.872 rad/sec
2 \xi \times 3.872=5
\xi=\frac{5}{2 \times 3.872}=0.64 \;\;\;\;(Underdamped)
Q=\frac{25}{s^2+10s+25}
\omega _n=\sqrt{25}=5 rad/sec
2 \xi \times 5=10
\xi=1 \;\;\;\;\;(Critically \; damped)
Observing all the options, option (C) is correct.
Question 10 |
Which of the following systems has maximum peak overshoot due to a unit step input?
\frac{100}{s^{2}+10s+100} | |
\frac{100}{s^{2}+15s+100} | |
\frac{100}{s^{2}+5s+100} | |
\frac{100}{s^{2}+20s+100} |
Question 10 Explanation:
For maximum peak over shoot M_P\propto \frac{1}{\xi}
\xi=0.25 for option (C) which is least among all options. Therefore correct option is C.
\xi=0.25 for option (C) which is least among all options. Therefore correct option is C.
There are 10 questions to complete.