Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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.
Question 8 |
When a unit ramp input is applied to the unity feedback system having closed loop transfer
function
\frac{C(s)}{R(s)}=\frac{Ks+b}{s^{2}+as+b}(a \gt 0,b \gt 0,K\gt 0),
the steady state error will be
\frac{C(s)}{R(s)}=\frac{Ks+b}{s^{2}+as+b}(a \gt 0,b \gt 0,K\gt 0),
the steady state error will be
0 | |
\frac{a}{b} | |
\frac{a+K}{b} | |
\frac{a-K}{b} |
Question 8 Explanation:
Closed loop transfer function =\frac{Ks+b}{s^2+as+b}
Open loop transfer function = G(s)=\frac{Ks+b}{s^2+as+b-Ks-b}
G(s)=\frac{Ks+b}{s^2+as-Ks} =\frac{Ks+b}{s(s+a-K)}
Steady state error for ramp input given to type-1 system =1/K_V
where, velocity error coefficient,
K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}
Steady state error,
e_{ss}=\frac{a-K}{b}
Open loop transfer function = G(s)=\frac{Ks+b}{s^2+as+b-Ks-b}
G(s)=\frac{Ks+b}{s^2+as-Ks} =\frac{Ks+b}{s(s+a-K)}
Steady state error for ramp input given to type-1 system =1/K_V
where, velocity error coefficient,
K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}
Steady state error,
e_{ss}=\frac{a-K}{b}
Question 9 |
A second-order real system has the following properties:
a) the damping ratio \xi = 0.5 and undamped natural frequency \omega _{n}=10 rad/s,
b) the steady state value of the output, to a unit step input, is 1.02.
The transfer function of the system is
a) the damping ratio \xi = 0.5 and undamped natural frequency \omega _{n}=10 rad/s,
b) the steady state value of the output, to a unit step input, is 1.02.
The transfer function of the system is
\frac{1.02}{s^{2}+5s+100} | |
\frac{102}{s^{2}+10s+100} | |
\frac{100}{s^{2}+10s+100} | |
\frac{102}{s^{2}+5s+100} |
Question 9 Explanation:
Damping ratio \xi=0.5
Undamped natural frequency \omega _n=10 rad/sec
Steady state output toa unit step input C_{ss}=1.02
Hence steady state error e_{ss}=1.02-1.00=0.02
\because Characteristic equation is,
s^2+2 \xi \omega _ns+\omega _n^2=0
s^2+2 \times 0.5\times 10 s+100=0
s^2+10s+100=0
From options, if we take option (B) then
C_{ss}=\lim_{s \to 0}s\cdot C(s)
\;\; \; \; \;=\lim_{s \to 0 }s \times \frac{1}{s} \times \frac{102}{s^2+10s+100}
C_{ss}=1.02
Hence, Option (B) is correct answer.
Undamped natural frequency \omega _n=10 rad/sec
Steady state output toa unit step input C_{ss}=1.02
Hence steady state error e_{ss}=1.02-1.00=0.02
\because Characteristic equation is,
s^2+2 \xi \omega _ns+\omega _n^2=0
s^2+2 \times 0.5\times 10 s+100=0
s^2+10s+100=0
From options, if we take option (B) then
C_{ss}=\lim_{s \to 0}s\cdot C(s)
\;\; \; \; \;=\lim_{s \to 0 }s \times \frac{1}{s} \times \frac{102}{s^2+10s+100}
C_{ss}=1.02
Hence, Option (B) is correct answer.
Question 10 |
The unit step response of a system with the transfer function G(s)=\frac{1-2s}{1+s} is given by which one of the following waveforms?
A | |
B | |
C | |
D |
Question 10 Explanation:
T.F.=\frac{1-2s}{1+s}
C(s)=\frac{1-2s}{1+s}\cdot \frac{1}{s}
\;\;=\frac{A}{1+s}+\frac{B}{s}
For A,
A=\lim_{s \to -1}(s+1)\cdot \frac{1-2s}{s(1+s)}
\;\;=\frac{(1-2(-1))}{(-1)}=-3
For B,
B=\lim_{s \to 0}s\cdot \frac{1-2s}{s(1+s)}=1
C(s)=\frac{1}{s}-\frac{3}{1+s}
C(t)=(1-3e^{-t})u(t)
C(s)=\frac{1-2s}{1+s}\cdot \frac{1}{s}
\;\;=\frac{A}{1+s}+\frac{B}{s}
For A,
A=\lim_{s \to -1}(s+1)\cdot \frac{1-2s}{s(1+s)}
\;\;=\frac{(1-2(-1))}{(-1)}=-3
For B,
B=\lim_{s \to 0}s\cdot \frac{1-2s}{s(1+s)}=1
C(s)=\frac{1}{s}-\frac{3}{1+s}
C(t)=(1-3e^{-t})u(t)
There are 10 questions to complete.