Control System
Exam Duration: 45 Mins Total Questions : 30
The equivalent transfer function of three parallel blocks \({ G }_{ 1 }(s)=\frac { 1 }{ s+2 } ,{ G }_{ 2 }(s)=\frac { 1 }{ s+3 } ,{ G }_{ 3 }(s)=\frac { s+2 }{ s+5 } is\)
- (a)
\(-\frac { ({ s }^{ 3 }+{ 9s }^{ 2 }+31s+37) }{ (s+2)(s+3)(s+5) } \)
- (b)
\(\frac { 1 }{ (s+3)(s+5) } \)
- (c)
\(\frac { { s }^{ 3 }+{ 9s }^{ 2 }+31s+37 }{ (s+2)(s+3)(s+5) } \)
- (d)
\(\frac { { 3s }^{ 2 }+20s+31 }{ (s+2)(s+3)(s+5) } \)
The transfer of a system is given as
\(G(s)=\frac { k }{ s(sT+1) } \)
The Bode plot of this function is
- (a)
- (b)
- (c)
- (d)
None of the above
The transfer function \(P(s)=\frac { s+4 }{ (s+1)(s+2)(s-1) } \) represents
- (a)
a unstable system
- (b)
a stable system
- (c)
critically stable system
- (d)
None of the above
The close-loop transfer function of a control system is given by \(\frac { C(s) }{ R(s) } =\frac { 1 }{ 1+s } \).For the input r(t)=sint, the steady state value of c(t) is equal to
- (a)
\(\frac { 1 }{ \sqrt { 2 } } cost\)
- (b)
1
- (c)
\(\frac { 1 }{ \sqrt { 2 } } sint+\frac { 1 }{ 2 } { e }^{ -t }\)
- (d)
\(\frac { 1 }{ \sqrt { 2 } } sin(t-\frac { \pi }{ 4 } )+\frac { 1 }{ 2 } { e }^{ -t }\)
The damping frequency of a system is 4.7 rad/s.The second overshoot will occur at
- (a)
2.0 s
- (b)
1.34 s
- (c)
0.66 s
- (d)
1.49 s
If the closed-loop transfer function of unity negative feedback system is given by
T(s)=\(\frac { a_{ n-1 }s+{ a }_{ n } }{ { s }^{ n }+{ a }_{ 1 }{ s }^{ n-1 }+...+a_{ n-1 }s+{ a }_{ n } } \)
Then the steady state error for a unit ramp is
- (a)
\(\frac { { a }_{ n } }{ a_{ n-1 } } \)
- (b)
\(\frac { { a }_{ n } }{ a_{ n-2 } } \)
- (c)
\(\frac { a_{ n-2 } }{ { a }_{ n } } \)
- (d)
zero
A phase lag lead network introduces in the output
- (a)
lag at high frequencies and lead at low frequencies
- (b)
lag att low frequencies and lead at high frequencies
- (c)
lag at all frequencies
- (d)
None of the above
PD controller improves
- (a)
transient response
- (b)
steady stte response
- (c)
Both (a) and (b)
- (d)
Neither (a) nor (b)
Consider the system with
\(A=\left[ \begin{matrix} 0 & -2 \\ 0 & -3 \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ 1 \end{matrix} \right] ,C=\left[ \begin{matrix} 0 & 1 \end{matrix} \right] \)
Then, the system is
- (a)
controllable and observable
- (b)
uncontrollable and unobservable
- (c)
controllable and unobservable
- (d)
observable and uncontrollable
For the signal flow graph shown below.
the system dynamic equation will be
- (a)
\(\left[ \begin{matrix} { \dot { X } }_{ 1 } \\ { \dot { X } }_{ 2 } \\ { X }_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ -5 & -3 & -2 \end{matrix} \right] \left[ \begin{matrix} { X }_{ 1 } \\ { X }_{ 2 } \\ { X }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] X\left( t \right) \)
- (b)
\(\left[ \begin{matrix} { \dot { X } }_{ 1 } \\ { \dot { X } }_{ 2 } \\ { X }_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -3 & 2 \end{matrix} \right] \left[ \begin{matrix} { X }_{ 1 } \\ { X }_{ 2 } \\ { X }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] X\left( t \right) \)
- (c)
\(\left[ \begin{matrix} { \dot { X } }_{ 1 } \\ { \dot { X } }_{ 2 } \\ { X }_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -3 & -5 \end{matrix} \right] \left[ \begin{matrix} { X }_{ 1 } \\ { X }_{ 2 } \\ { X }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] X\left( t \right) \)
- (d)
None of the above
In the root locus for open-loop transfer function G(s) H(s)=\(\frac { k(s+7) }{ (s+4)(s+6) } \), the breakaway and break-in points are located respectively at
- (a)
-5.27 and -8.73
- (b)
-2 and -1
- (c)
-7.73 and -4.27
- (d)
-4.27 and -7.73
A closed-loop control system has an open-loop gain of 100. Its feedback loop has a negative feedback gain of 0.005. Its sensitivity will be
- (a)
0.67
- (b)
0.866
- (c)
1.414
- (d)
2.0
The approximate bode magnitude plot of a minimum phase system is shown in the figure. The transfer function of the system is
- (a)
\({ 10 }^{ 8 }\frac { { \left( s+0.1 \right) }^{ 3 } }{ { \left( s+10 \right) }^{ 2 }(s+100) } \)
- (b)
\({ 10 }^{ 7 }\frac { { \left( s+0.1 \right) }^{ 3 } }{ { (s+10)\left( s+100 \right) } } \)
- (c)
\({ 10 }^{ 7 }\frac { { \left( s+0.1 \right) }^{ 3 } }{ { \left( s+10 \right) }^{ 2 }(s+100) } \)
- (d)
\({ 10 }^{ 9 }\frac { { \left( s+0.1 \right) }^{ 3 } }{ { (s+10)\left( s+100 \right) }^{ 2 } } \)
The system block diagram is given below. Find \(\frac { C(s) }{ N(s) } \), if R(s)=0
- (a)
\(\frac { 5(s+8) }{ { s }^{ 2 }+6s+40 } \)
- (b)
\(\frac { s(s+1) }{ { s }^{ 2 }+16s+40 } \)
- (c)
\(\frac { 5s(s+8) }{ { s }^{ 2 }-4s-8 } \)
- (d)
\(\frac { s(s+1) }{ { s }^{ 2 }-4s-8 } \)
The matrix of any state-space equations for the transfer function C(s)/R(s) of the system, shown below in figure is
- (a)
\(\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right] \)
- (b)
\(\left[ \begin{matrix} 0 & 1 \\ 0 & -1 \end{matrix} \right] \)
- (c)
\(\left[ -1 \right] \)
- (d)
\(\left[ 3 \right] \)
The impulse response of an initially relaxed linear system is e-2t u(t). To produce a response of te-2t u(t), the input must be equal to
- (a)
2e-t u(t)
- (b)
\(\frac { 1 }{ 2 } { e }^{ -2t }u\left( t \right) \)
- (c)
e-2t u(t)
- (d)
e-t u(t)
A system is described by the state equation
\(\dot { X } =AX+BU\)
The output is given by Y = CX
where, \(A=\left[ \begin{matrix} -4 & -1 \\ 3 & -1 \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ 1 \end{matrix} \right] ,C=\left[ \begin{matrix} 1 & 0 \end{matrix} \right] \)
Transfer function G(s) of the system is
- (a)
\(\frac { s }{ { s }^{ 2 }+5s+7 } \)
- (b)
\(\frac { 1 }{ { s }^{ 2 }+5a+7 } \)
- (c)
\(\frac { s }{ { s }^{ 2 }+3s+2 } \)
- (d)
\(\frac { 1 }{ { s }^{ 2 }+3s+2 } \)
The closed-loop transfer function of a control system is given by \(\frac { C(s) }{ R(s) } =\frac { 1 }{ 1+s } \) For the input r(t) = sint, the steady state value of c(t) is equal to
- (a)
\(\frac { 1 }{ \sqrt { 2 } } \cos { t } \)
- (b)
1
- (c)
\(\frac { 1 }{ \sqrt { 2 } } \sin { t } \)
- (d)
\(\frac { 1 }{ \sqrt { 2 } } \sin { \left( t-\frac { \pi }{ 4 } \right) } \)
The damping frequency of a system is 8 rad/s. The rise time for a \(\theta \) of \({ 60 }^{ ° }\) is
- (a)
0.4 s
- (b)
0.261 s
- (c)
7.5 s
- (d)
15 s
The maximum phase shift that can be provided by a lead compensator with transfer function
\({ G }_{ e }(s)=\frac { 1+6s }{ 1+2s } \)
- (a)
\(30°\)
- (b)
\(15°\)
- (c)
\(60°\)
- (d)
\(45°\)
Determine error ratio of the given diagram.
\({ G }_{ 1 }(s)=\frac { 4 }{ (s+2) } \\ { G }_{ 2 }(s)=\frac { (s+8) }{ (s+10) } \\ H(s)=\frac { s }{ (s+4) } \)
- (a)
\(\frac { (s+2)(s+10)(s+4) }{ { s }^{ 3 }+12{ s }^{ 2 }+36s+80 } \)
- (b)
\(\frac { (s+4)(s+10)(s+4) }{ { s }^{ 3 }+12{ s }^{ 2 }+36s+80 } \)
- (c)
\(\frac { (s+8)(s+10)(s+2) }{ { s }^{ 3 }+12{ s }^{ 2 }+36s+80 } \)
- (d)
\(\frac { (s+8)(s+6)(s+2) }{ { s }^{ 3 }+12{ s }^{ 2 }+36s+80 } \)
For a unity feedback system whose OLTF is
\(G(s)=\frac { 50 }{ (1+0.1s)(1+2s) } \)
Find the velocity error constant.
- (a)
0
- (b)
-1
- (c)
\(\infty \)
- (d)
1
A linear time invariant system initially at rest, when subjected to a unit step input, gives a response y(t) = te-t t>0. The transfer function of the system is
- (a)
\(\frac { 1 }{ { \left( s+1 \right) }^{ 2 } } \)
- (b)
\(\frac { 1 }{ { s\left( s+1 \right) }^{ 2 } } \)
- (c)
\(\frac { s }{ { \left( s+1 \right) }^{ 2 } } \)
- (d)
\(\frac { 1 }{ { s\left( s+1 \right) } } \)
Let s(t) be the step response of a linear system with zero initial conditions then the response of this system to an input u(t) is
- (a)
\(\int _{ 0 }^{ t }{ s\left( t-\tau \right) } u\left( \tau \right) d\tau \)
- (b)
\(\frac { d }{ dt } \left[ \int _{ 0 }^{ t }{ s\left( t-\tau \right) } u\left( \tau \right) d\tau \right] \)
- (c)
\(\int _{ 0 }^{ t }{ s\left( t-\tau \right) } \left[ \int _{ 0 }^{ t }{ u\left( { \tau }_{ 1 } \right) d{ \tau }_{ 1 } } \right] d\tau \)
- (d)
\(\int _{ 0 }^{ t }{ s{ \left( t-\tau \right) }^{ 2 } } u\left( \tau \right) d\tau \)
The polar plot of an open-loop stable system is shown below. The closed-loop system is
- (a)
always stable
- (b)
marginally stable
- (c)
unstable with one pole on the RH s-plane
- (d)
unstable with two poles on the RH s-plane
A second order system has overshoot of 50% and period of oscillaton 0.2s in step response. The resonant peak is
- (a)
2.377
- (b)
30.63
- (c)
5
- (d)
3
The gain and phase crossover frequencies in rad/s are respectively
- (a)
0.632 and 1.26
- (b)
0.632 and 0.485
- (c)
0.485 and 0.632
- (d)
1.26 and 0.632
Based on the above results the gain and phase margins of the system will be
- (a)
-7.09 dB and \(87.5°\)
- (b)
7.09 dB and \(87.5°\)
- (c)
7.09 dB and -\(87.5°\)
- (d)
-7.09 and -\(87.5°\)
A lead compensator is a
- (a)
low-pass filter
- (b)
high-pass filter
- (c)
band-pass filter
- (d)
band-stop filter
A linear system is equivalently represented by two sets of state equations
\(\dot { X } =AX+BU\) and \(\dot { W } =CW+DU\)
The eigen values of the representations are also computed as \(\left[ \lambda \right] \) and \(\left[ \mu \right] \). Which one of the following statements is true?
- (a)
\(\left[ \lambda \right] =\left[ \mu \right] \) and X = W
- (b)
\(\left[ \lambda \right] =\left[ \mu \right] \) and \(X\neq W\)
- (c)
\(\left[ \lambda \right] \neq \left[ \mu \right] \) and X = W
- (d)
\(\left[ \lambda \right] \neq \left[ \mu \right] \) and \(X\neq W\)