GATE Electronics and Communication Engineering - Control System
Exam Duration: 45 Mins Total Questions : 30
The block diagram shown in Fig. (a) is to be represented as in Fig. (b) for what value of H(s)?
-Q.png)
-Q.png)
- (a)
4s+1
- (b)
\(\frac { s+4 }{ s } \)
- (c)
4(s+1)
- (d)
5(s+4)
Here centre block of gain 4 is to shift to right side before \(\frac { 1 }{ s } \) block. The block will be \(\frac { 4 }{ { 1 }/{ s } } \)=4s
Now, this is parallel with unit gain i.e., =4s+1
-S.png)
-S.png)
The block diagram shown in figure is equivalent to
- (a)
-Q.png)
- (b)
-Q.png)
- (c)
-Q.png)
- (d)
-Q.png)
-S.png)
If the take-off point is taken before a block then gain is multiplied hence
-S.png)
Consider the system shown in figure. If the forward path gain is reduced by 10% in each system, then the variation in C1 and C2 will be respectively
- (a)
10% and 1%
- (b)
2% and 10%
- (c)
10% and zero
- (d)
5% and 1%
In open-loop system, gain is changed by 10% then C1 is also changed by 10%
But in closed-loop system, it is changed by
\(\frac { 10 }{ 10+1 } \)(feedback)=\(\frac { 10 }{ 11 } \)\(\approx \)1%
The position and acceleration error coefficients for the open-loop transfer function G(s)=\(\frac { K }{ { s }^{ 2 }(s+10)(s+100) } \) respectively are
- (a)
zero,infinity
- (b)
infinity and zero
- (c)
\(\frac { K }{ 100 } \) and zero
- (d)
infinity and \(\frac { K }{ 1000 } \)
Kp=\(\underset{{ s\rightarrow 0 }}{\lim} { G(s) } =\underset{{ s\rightarrow 0 }}{\lim}{ \frac { K }{ { s }^{ 2 }(s+10)(s+100) } =\alpha } \)
Ka=\(\lim _{ s\rightarrow 0 }{ { s }^{ 2 }G(s) } \)
=\(\underset{{ s\rightarrow 0 }}{\lim}{ \frac { { s }^{ 2 }K }{ { s }^{ 2 }(s+10)(s+100) } } \)
=0.001K=\(\frac { K }{ 1000 } \)
The transfer function G(s) is \(\frac { K }{ { s }^{ 2 }(1+sT) } \) . This open-loop system will be
- (a)
stable
- (b)
unstable
- (c)
marginally stable
- (d)
conditionally stable
This system has two poles at origin. If any system has double root at origin then system becomes unstable.
Consider the open-loop unity gain trans-function
G(s)=\(\frac { { e }^{ -sT } }{ s(s+1) } \)
For the system to be stable
- (a)
T<1
- (b)
T>1
- (c)
0<T<1
- (d)
None of these
e-st\(\approx \) 1-sT
s2+s(1-T)+1=0
T<1
For quadratic factor \(\frac { 1 }{ 1+2\varepsilon \left[ \frac { j\omega }{ { \omega }_{ n } } \right] +{ \left[ \frac { j\omega }{ { \omega }_{ n } } \right] }^{ 2 } } \) in Bode plot magnitude plot the high frequency asymtote is a straight line having the slope
- (a)
-20 dB/decade
- (b)
+20 dB/decade
- (c)
-40 dB/decade
- (d)
+40 dB/decade
It contains two poles, thus, slope is
2\(\times \)(-20 dB/decade)=-40 dB/decade
The transfer function of the system is given by
G(j\(\omega \))=\(\frac { K }{ (j\omega )(j\omega T+1) } ,\quad K<\frac { 1 }{ T } \)
Which one of the following is the Bode plot of this function?
- (a)
-Q.png)
- (b)
-Q.png)
- (c)
-Q.png)
- (d)
-Q.png)
Initial slope is -20 dB/decade and after \(\frac { 1 }{ T } \) slope, becomes -40 dB/decade. So, options (a) and (c) are satisfied. Now, if -20 dB line then it crosses 0 dB line at \(\omega =K<\frac { 1 }{ T } \).
This only satisfied by (c).
The value of H(s) equal to
- (a)
\(\frac { s+0.1 }{ { (s+0.1) }^{ 2 } } \)
- (b)
\(\frac { s+0.1 }{ { s(s+1) } } \)
- (c)
\(\frac { 0.2(s+1) }{ { { (s+0.1) }^{ 2 } } } \)
- (d)
None of these
H(j\(\omega \))=\(\frac { { K }^{ ' }(s+0.2) }{ { (s+1) }^{ 2 } } =\frac { K(10s+1) }{ { (s+1) }^{ 2 } } \)
The decibel shift of 6 dB corresponds to a linear gain of K=0.2
\(\Rightarrow \quad H(j\omega )=\frac { 0.2(s+0.1) }{ { (s+1) }^{ 2 } } \)
For open-loop pole zero plot sown in figure
The break in point is
- (a)
-1.29
- (b)
-2.43
- (c)
-3.43
- (d)
-1.43
Break-in point lies on root locus thus only -2.43 satisfies this condition.
The forward path transfer function of a unity feedback system is G(s)=\(\frac { K(s+5)(s+3) }{ s({ s }^{ 2 }-1) } \) Which one is the correct root loci?
- (a)
-Q.png)
- (b)
-Q.png)
- (c)
-Q.png)
- (d)
-Q.png)
In this system there are 3 finite pole and 2 finite zero. If finite zero is available loci will end at finite zero. One loci end at infinite. Since, number of zero is less than pole.
Transfer function of a phase lead network can be
- (a)
\(\frac { \alpha (1+sT) }{ (1+s\alpha T) } ;\quad \alpha <1\)
- (b)
\(\frac { \alpha (1+sT) }{ (1+s\alpha T) } ;\quad \alpha >1\)
- (c)
\(\frac { 1+sT }{ 1+s\beta T } ;\quad \beta >1\)
- (d)
\(\frac { 1+sT }{ 1+s\beta T } ;\quad \beta <1\)
\(\frac { \alpha (1+sT) }{ (1+s\alpha T) } ;\quad \alpha <1\)
\(=\frac { 1+sT }{ \left( \frac { 1 }{ \alpha } +sT \right) } ;\quad \alpha <1\quad \)
\(=\frac { \left( s+\frac { 1 }{ T } \right) }{ \left( s+\frac { 1 }{ \alpha T } \right) } ;\quad \alpha <1\)
Zero dominant so phase lead.
The pole-zero configuration given in figure below is
- (a)
PID controller
- (b)
Integrator
- (c)
Lag-lead compensator
- (d)
PD controller
Exhibits property of both lead-lag compensators.
A control system with PD controller is given. If velocity error constant Kv =1000 and \(\xi =0.5\), then value of KP and KD are
- (a)
KP=10, KD=0.9
- (b)
KP=1000, KD=9
- (c)
KP=10, KD=9
- (d)
KP=100, KD=0.9
Given G(s)=\(\frac { 100 }{ s(s+10) } \)
Gc(s)=KP+KDs
For unity feedback
\(\frac { E(s) }{ R(s) } =\frac { 1 }{ 1+G(s).{ G }_{ c }(s) } \)
\(=\frac { 1 }{ 1+\frac { ({ K }_{ P }+{ K }_{ D }s)100 }{ s(s+10) } } \)
\({ K }_{ v }=\lim _{ s\rightarrow 0 }{ sG(s){ G }_{ c }(s) } \)
\(=\lim _{ s\rightarrow 0 }{ \frac { s({ K }_{ P }+{ K }_{ D }s)100 }{ s(s+10) } } \)
1000=\(\frac { 100{ K }_{ P } }{ 10 } \)
i.e., KP=100
Now, characteristic equation
=s2+s(10+KD100)+100KP
Comparing with s2+2\(\xi { \omega }_{ n }s\)+\({ { { \omega }_{ n }^{ 2 } } }\)
We get \({ { { \omega }_{ n }^{ 2 } } }\)=100KP
i.e., \({ \omega }_{ n }\)=100 rad/s
Now, 10+KD100=\({ { 2\xi { \omega }_{ n } } }\)
\({ K }_{ D }=\frac { 2\times \frac { 1 }{ 2 } \times 100-10 }{ 100 } \) (\(\xi \)=0.5)
=0.9
A system is described by the equation
\(\overset { . }{ x } =\left[ \begin{matrix} 0 & 1 \\ -3 & -2 \end{matrix} \right] x+\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] u\\ Y=\left[ \begin{matrix} 3 & 2 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \end{matrix} \right] \)
The poles of the system is
- (a)
-1. -3
- (b)
+1, -3
- (c)
-1\(\pm \)j\(\sqrt { 2 } \)
- (d)
1\(\pm \)j\(\sqrt { 2 } \)
\(\left| (sI-A) \right| =0\)
i.e., \(\left| \begin{matrix} s & -1 \\ 3 & s+2 \end{matrix} \right| =0\)
s2+2s+3=0
s=-1\(\pm \)j\(\sqrt { 2 } \)
For the signal flow graph shown below.
The system dynamic equation will be
- (a)
\(\left[ \begin{matrix} \overset { . }{ { x }_{ 1 } } \\ \overset { . }{ { x }_{ 2 } } \\ \overset { . }{ { x }_{ 3 } } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ -5 & -3 & -2 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] x(t)\)
- (b)
\(\left[ \begin{matrix} \overset { . }{ { x }_{ 1 } } \\ \overset { . }{ { x }_{ 2 } } \\ \overset { . }{ { x }_{ 3 } } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -3 & -2 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] x(t)\)
- (c)
\(\left[ \begin{matrix} \overset { . }{ { x }_{ 1 } } \\ \overset { . }{ { x }_{ 2 } } \\ \overset { . }{ { x }_{ 3 } } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -3 & -5 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] x(t)\)
- (d)
None of the above
\(\overset { . }{ { x }_{ 1 } } \)=x2
\(\overset { . }{ { x }_{ 2 } } \)=x3
\(\overset { . }{ { x }_{ 3 } } \)=-5x2-3x2-2x3+x(t)
So,
\(\left[ \begin{matrix} \overset { . }{ { x }_{ 1 } } \\ \overset { . }{ { x }_{ 2 } } \\ \overset { . }{ { x }_{ 3 } } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ -5 & -3 & -2 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] x(t)\)
A linear time invariant system is described by
\(\frac { dx(t) }{ dt } =Ax(t)+Bu(t)\)
y(t)=Cx(t)
where \(A=\left[ \begin{matrix} 0 & 1 \\ -2 & -3 \end{matrix} \right] ,\quad B=\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] ,\quad C=\left[ \begin{matrix} 1 & 0 \end{matrix} \right] \)
The system is
- (a)
both controllable and observable
- (b)
controllable but unobservable
- (c)
not controllable but observable
- (d)
both uncontrollable and unobservable
\(A=\left[ \begin{matrix} 0 & 1 \\ -2 & -3 \end{matrix} \right] ,\quad B=\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] ,\quad C=\left[ \begin{matrix} 1 & 0 \end{matrix} \right] \)
QC=[B AB]
\(AB=\left[ \begin{matrix} 1 \\ -3 \end{matrix} \right] \)
QC=\(\left[ \begin{matrix} 0 & 1 \\ 1 & -3 \end{matrix} \right] \neq 0\)
Rank of QC=2=Order of QC
Hence, controllable.
Q0=[CT ATCT]
\({ [C }^{ T }]=\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] \), \({ A }^{ T }{ C }^{ T }=\left[ \begin{matrix} 0 & -2 \\ 1 & -3 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] \)
Q0=\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \neq 0\)
Rank of Q0=2=Order of QC
Hence, observable
For a first order instrument, a 5% settling time is equal to
- (a)
three times the time constant
- (b)
two times the time constant
- (c)
the time constant
- (d)
four times the time constant
In one time constant system reaches 63% of final value. In two times constant system reaches 63% of remaining value
=(63+23.31)%=86.31%
In three times constant system reaches 63% of remaining value
=(86.31+8.6)%=95%
The gain margin for the system with open-loop transfer function G(s)H(s)=\(\frac { 2(s+1) }{ { s }^{ 2 } } ,\) is
- (a)
\(\infty \)
- (b)
zero
- (c)
1
- (d)
\(-\infty \)
The gain margin is the reciprocal of magnitude |G(s)H(s)| at the frequency at which the phase angle is -1800
Given \(G(s)H(s)=\frac { 2(1+s) }{ { s }^{ 2 } } \)
\(G(j\omega )H(j\omega )=\frac { 2(1+j\omega ) }{ -{ \omega }^{ 2 } } \)
\(\angle G(j\omega )H(j\omega )=-{ 180 }^{ 0 }+\tan ^{ -1 }{ \omega } \)
\({ -180 }^{ 0 }=-{ 180 }^{ 0 }+\tan ^{ -1 }{ \omega } \)
\(\tan ^{ -1 }{ \omega } =0\)
\(\omega =0\)
\({ \left| G(j\omega )H(j\omega ) \right| }_{ \omega =0 }=\infty \)
\(GM=\frac { 1 }{ { \left| G(j\omega )H(j\omega ) \right| }_{ \omega =0 } } \)
\(=\frac { 1 }{ \infty } =0\)
\({ \ GM | }_{ dB }=10\log _{ 10 }{ 0 } =-\infty \)
An amplifier with resistive negative feedback has two left half-plane poles in its open-loop transfer function. The amplifier
- (a)
will always be unstable at high frequencies
- (b)
will be stable for all frequencies
- (c)
may be unstable, depending on the feedback factor
- (d)
will oscillate at low frequencies
An amplifier with resistive negative feedback has two left half-plane poles stabilises the gain system as gain reduces.
The open-loop transfer function of a ufb control system is
\(G(s)=\frac { K(1+2s)(1+4s) }{ { s }^{ 2 }({ s }^{ 2 }+2s+8) } \)
The position, velocity and acceleration error constants are respectively
- (a)
0, 0, 4K
- (b)
\(\infty ,\quad \frac { K }{ 8 } ,\quad 0\)
- (c)
0, 4K, \(\alpha\)
- (d)
\(\infty ,\quad \infty ,\quad \frac { K }{ 8 } \)
\(G(s)=\frac { K(1+2s)(1+4s) }{ { s }^{ 2 }({ s }^{ 2 }+2s+8) } \)
We know that
\({ K }_{ p }=\lim _{ s\rightarrow 0 }{ G(s)H(s)=\infty } \)
\({ K }_{ v }=\lim _{ s\rightarrow 0 }{ sG(s)H(s)=\infty } \)
\({ K }_{ a }=\lim _{ s\rightarrow 0 }{ { s }^{ 2 }G(s)H(s) } =\frac { K(1)(1) }{ 8 } =\frac { K }{ 8 } \)
A minimum phase system has fourteen poles and two zeros. The slope of its highest frequency asymptotes in its Bode plot is
- (a)
-240 dB/decade
- (b)
-280 dB/decade
- (c)
-320 dB/decade
- (d)
-40 dB/decade
There are 14 poles and two zeros.
Hence, highest frequency asymptote is
=-20(14)+20(2)
=-280+40
=-240 dB/decade
The zero-input response of a system given by the state space equation
\(\left[ \begin{matrix} \overset { . }{ { x }_{ 1 } } \\ \overset { . }{ { x }_{ 2 } } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \end{matrix} \right] \)and \(\left[ \begin{matrix} { x }_{ 1 }(0) \\ { x }_{ 2 }(0) \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] \)is
- (a)
\(\left[ \begin{matrix} { te }^{ t } \\ t \end{matrix} \right] \)
- (b)
\(\left[ \begin{matrix} { e }^{ t } \\ t \end{matrix} \right] \)
- (c)
\(\left[ \begin{matrix} { e }^{ t } \\ { te }^{ t } \end{matrix} \right] \)
- (d)
\(\left[ \begin{matrix} t \\ { te }^{ t } \end{matrix} \right] \)
\(\left[ \begin{matrix} \overset { . }{ { x }_{ 1 } } \\ \overset { . }{ { x }_{ 2 } } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \end{matrix} \right] \)
\(A=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(eAt={ L }^{ -1 }{ [sI-A] }^{ -1 }\)
\([sI-A]=\left[ \begin{matrix} s & 0 \\ 0 & s \end{matrix} \right] -\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} s-1 & 0 \\ -1 & s-1 \end{matrix} \right] \)
\({ [sI-A] }^{ -1 }=\frac { 1 }{ { (s-1) }^{ 2 } } \left[ \begin{matrix} s-1 & 0 \\ +1 & s-1 \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ s-1 } & 0 \\ \frac { 1 }{ { (s-1) }^{ 2 } } & \frac { 1 }{ s-1 } \end{matrix} \right] \)
\(\phi (t)={ \alpha }^{ -1 }{ [sI-A] }^{ -1 }=\left[ \begin{matrix} { e }^{ t } & 0 \\ { te }^{ t } & { e }^{ t } \end{matrix} \right] \)
Zero input response
\(=\phi (t)\left[ \begin{matrix} { x }_{ 1 }(0) \\ { x }_{ 2 }(0) \end{matrix} \right] \left[ \begin{matrix} { e }^{ t } & 0 \\ { te }^{ t } & { e }^{ t } \end{matrix} \right] \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} { e }^{ t } \\ { te }^{ t } \end{matrix} \right] \)
The asymptotic Bode plot of a transfer function is as shown in the figure. The transfer function G(s) corresponding to this Bode plot is
- (a)
\(\frac { 1 }{ (s+1)(s+20) } \)
- (b)
\(\frac { 1 }{ s(s+1)(s+20) } \)
- (c)
\(\frac { 100 }{ s(s+1)(s+20) } \)
- (d)
\(\frac { 100 }{ s(s+1)(1+0.05s) } \)
The corner frequencies are \({ \omega }_{ { C }_{ 1 } }=0,\)\({ \omega }_{ { C }_{ 2 } }=1,\)\({ \omega }_{ { C }_{ 3 } }=20\)
\(G(s)=\frac { K }{ s(1+s{ C }_{ 1 })(1+s{ C }_{ 2 }) } \)
\({ C }_{ 2 }=\frac { 1 }{ { \omega }_{ { C }_{ 2 } } } =\frac { 1 }{ 1 } =1\)
\({ C }_{ 2 }=\frac { 1 }{ { \omega }_{ { C }_{ 3 } } } =\frac { 1 }{ 20 } =0.05\)
\({ C }_{ 3 }=\frac { K }{ s(1+s)(1+0.05s) } \)
From graph,
G(s) at s=0.1=60 dB
\(20\log _{ 10 }{ \left| G(j\omega ) \right| } =60\)
\(20\log _{ 10 }{ \left| \frac { K }{ j\omega (1+j\omega )(1+0.05j\omega ) } \right| } =60\)
K=100
\(G(s)=\frac { 100 }{ s(1+s)(1+0.05s) } \)
The state space representation of a separately excited DC servo motor dynamics is given as
\(\left[ \begin{matrix} \frac { d\omega }{ dt } \\ \frac { d{ I }_{ a } }{ dt } \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 \\ -1 & -10 \end{matrix} \right] \left[ \begin{matrix} \omega \\ { I }_{ a } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 10 \end{matrix} \right] u\)
- (a)
\(\frac { 10 }{ { s }^{ 2 }+11s+11 } \)
- (b)
\(\frac { 1 }{ { s }^{ 2 }+11s+11 } \)
- (c)
\(\frac { 10s+10 }{ { s }^{ 2 }+11s+11 } \)
- (d)
\(\frac { 1 }{ { s }^{ 2 }+s+1 } \)
In DC servomotor \(\omega\) and u are output and input variables respectively.
Given, \(\left[ \begin{matrix} \frac { d\omega }{ dt } \\ \frac { d{ I }_{ a } }{ dt } \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 \\ -1 & -10 \end{matrix} \right] \left[ \begin{matrix} \omega \\ { I }_{ a } \end{matrix} \right] +\left[ \begin{matrix} 0 \\ 10 \end{matrix} \right] u\)
\(\frac { d\omega }{ dt } =-\omega +{ I }_{ a }\)
\(\frac { d{ I }_{ a } }{ dt } =-\omega -10{ I }_{ a }+10u\)
Taking Laplace transform,
sW(s)=-W(s)+Ia (s)
Ia(s)=(s+1)-10Ia(s)+10U(s)
W(s)=(-10-s)Ia(s)+10U(s)
=(-10-s)(s+1)W(s)+10U(s)
W(s)=(s2+11s+11)=10U(s)
Transfer function \(\frac { \omega (s) }{ U(s) } =\frac { 10 }{ { s }^{ 2 }+11s+11 } \)
The open-loop transfer function of a feedback system is
\(G(s)H(s)=\frac { K(1+s) }{ (1-s) } \)
The Nyquist plot of this system is
- (a)
-Q.png)
- (b)
-Q.png)
- (c)
-Q.png)
- (d)
-Q.png)
\(G(s)H(s)=\frac { K(1+s) }{ (1-s) } \)
\(\left| G(j\omega )H(j\omega ) \right| =K\)
\(\angle G(j\omega )=\tan ^{ -1 }{ \omega } -\tan ^{ -1 }{ -\frac { \omega }{ 1 } } \)
At \(\omega \rightarrow 0\)
\(G(j\omega )H(j\omega )=K\angle { 0 }^{ 0 }\)
At \(\omega \rightarrow \infty \quad G(j\omega )H(j\omega )=K\angle { 180 }^{ 0 }\)
At \(\omega \rightarrow 2\quad G(j\omega )H(j\omega )=K\angle { 127 }^{ 0 }\)
Consider the system shown in figure
The input waveform to yield a constant error is
- (a)
impulse
- (b)
step
- (c)
ramp
- (d)
parabolic
Here again not unity feedback.
Hence, \(Ge(s)=\frac { 10(s+10) }{ s(11s+{ 10 }^{ 2 }) } \)
Type 1 system hence ramp input will produce constant error.
The value of constant a which will increase the damping factor of the system to 0.7 is
- (a)
0.245
- (b)
0.1
- (c)
0.45
- (d)
None of these
Transfer function of internal loop,
\(=\frac { \frac { 8 }{ s(s+2) } }{ 1+\frac { 8as }{ s(s+2) } } =\frac { 8 }{ s(s+2)+8as } \)
Transfer function of total loop\(=\frac { 8 }{ { s }^{ 2 }+2s+8as+8 } \)
\(=\frac { 8 }{ { s }^{ 2 }+{ (2+8a) }^{ 3 }+8 } \)
\(\Rightarrow { { \quad \omega }_{ n } }=\sqrt { 8 } \quad \Rightarrow \quad 2\xi { { \omega }_{ n } }=2+8a\)
\(\Rightarrow \quad 2\times 0.7\times \sqrt { 8 } =2+8a\)
\(\Rightarrow\) 3.96=2a+8
\(\Rightarrow\) a=0.245
A system is described by the dynamic equations \(\overset { . }{ x } (t)=A.x(t)+B.u(t),\quad y(t)=C.x(t)\) where
\(A=\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3 \end{matrix} \right] ,\quad B=\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] ,\quad C=[1\quad 0\quad 0]\)
The eigen values of A are
- (a)
0.325, -1.662\(\pm\)j0.562
- (b)
2.325, 0.338\(\pm\)j0.562
- (c)
-2.325, -0.338\(\pm\)j0.562
- (d)
-0.325, 1.662\(\pm\)j0.562
\(sI-A=\left[ \begin{matrix} s & -1 & 0 \\ 0 & s & -1 \\ 1 & 2 & s+3 \end{matrix} \right] \)
\(\left| sI-A \right| ={ s }^{ 3 }+3{ s }^{ 2 }+2s+1,\)
\(\Rightarrow \quad s=-2.325,-0.338\pm j0.562\)
A system is described by the dynamic equations \(\overset { . }{ x } (t)=A.x(t)+B.u(t),\quad y(t)=C.x(t)\) where
\(A=\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3 \end{matrix} \right] ,\quad B=\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right] ,\quad C=[1\quad 0\quad 0]\)
The transfer-function relation between X(s) and U(s) is
- (a)
\(\frac { 1 }{ { s }^{ 3 }+3{ s }^{ 2 }+2s-1 } \left[ \begin{matrix} 1 \\ -s \\ { s }^{ 2 } \end{matrix} \right] \)
- (b)
\(\frac { 1 }{ { s }^{ 3 }+3{ s }^{ 2 }+2s+1 } \left[ \begin{matrix} 1 \\ s \\ { s }^{ 2 } \end{matrix} \right] \)
- (c)
\(\frac { 1 }{ { s }^{ 3 }+3{ s }^{ 2 }+2s+1 } \left[ \begin{matrix} 1 \\ -s \\ { s }^{ 2 } \end{matrix} \right] \)
- (d)
None of these
\(\frac { X(s) }{ U(s) } ={ (sI-A) }^{ -1 }B\)
\(=\frac { 1 }{ { s }^{ 3 }+3{ s }^{ 2 }+2s+1 } \left[ \begin{matrix} { s }^{ 3 }+3{ s }^{ 2 }+2 & s+3 & 1 \\ -1 & s(s+3) & s \\ -s & -2s-1 & { s }^{ 2 } \end{matrix} \right] \)
\(=\frac { 1 }{ { s }^{ 3 }+3{ s }^{ 2 }+2s+1 } \left[ \begin{matrix} 1 \\ s \\ { s }^{ 2 } \end{matrix} \right] \)
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