Question 1 |
A stable real linear time-invariant system with single pole at p, has a transfer function H(s)=\frac{s^2+100}{s-p}
with a dc gain of 5. The smallest positive frequency, in rad/s at unity
gain is closed to:
8.84 | |
11.08 | |
78.13 | |
122.87 |
Question 1 Explanation:
\begin{aligned}
H(s)&=T.F.=[latex]\frac{s^{2}+100}{s-p} \\ \text{D.C. gain }&= 5 \\ \Rightarrow \; \; \frac{100}{-P}&=5=P=-20 \\ H(j\omega )&=\frac{-\omega ^{2}+100}{j\omega +20} \\ \left |H(j\omega ) \right |&=\frac{-\omega ^{2}+100}{\sqrt{\omega ^{2}+400}} \\ \frac{-\omega ^{2}+100}{\sqrt{\omega ^{2}+400}}&=1 \\ \Rightarrow \; \; \omega &=8.84 \text{rad/sec.}
\end{aligned}
Question 2 |
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 2 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 3 |
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 3 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 4 |
Consider a negative unity feedback system with forward path transfer function
G(s)=\frac{K}{(s+a)(s-b)(s+c)}, where K, a, b, c are positive real numbers. For a Nyquist
path enclosing the entire imaginary axis and right half of the s-plane in the clockwise
direction, the Nyquist plot of (1 + G(s)), encircles the origin of (1 + G(s))-plane once
in the clockwise direction and never passes through this origin for a certain value of
K. Then, the number of poles of \frac{G(s)}{1+G(s)} lying in the open right half of the s-plane is
_________ .
1 | |
2 | |
3 | |
4 |
Question 4 Explanation:
\begin{aligned}O.L.T.F = G(s) &=\frac{K}{(s+a)(s-b)(s+c)} \\ N &= P - Z ; \; \;P = 1 \\
-1 &= 1 - Z ; \; \; N = -1 \\
Z &= 2\end{aligned}
Question 5 |
Which of the options is an equivalent representation of the signal flow graph shown here?
A | |
B | |
C | |
D |
Question 5 Explanation:
Simplifying given signal flow graph
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 |
Consider a state-variable model of a system
\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} 0 &1 \\ -\alpha &-2\beta \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}+\begin{bmatrix} 0\\ \alpha \end{bmatrix}r
y=[1\;\;0]\begin{bmatrix} x_1\\ x_2 \end{bmatrix}
where y is the output, and r is the input. The damping ratio \xi and the undamped natural frequency \omega _n (rad/sec) of the system are given by
\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} 0 &1 \\ -\alpha &-2\beta \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}+\begin{bmatrix} 0\\ \alpha \end{bmatrix}r
y=[1\;\;0]\begin{bmatrix} x_1\\ x_2 \end{bmatrix}
where y is the output, and r is the input. The damping ratio \xi and the undamped natural frequency \omega _n (rad/sec) of the system are given by
\xi =\frac{\beta }{\sqrt{\alpha }}; \; \omega _n=\sqrt{\alpha } | |
\xi =\sqrt{\alpha }; \; \omega _n=\frac{\beta }{\sqrt{\alpha }} | |
\xi =\frac{\sqrt{\alpha }}{\beta }; \; \omega _n=\sqrt{\beta } | |
\xi =\sqrt{\beta }; \; \omega _n=\sqrt{\alpha } |
Question 7 Explanation:
Characteristic equation is,
|sI-A|=0
|sI-A|=\begin{vmatrix} s &-1 \\ \alpha &s+2\beta \end{vmatrix}
\;\;=s^2+2s\beta +\alpha =0
\therefore \;\; \omega _n^2=\alpha
\;\;\; \omega _n=\sqrt{\alpha }
\;\;\;2\xi \omega _n=2\beta
\;\;\;\xi =\frac{\beta }{\sqrt{\alpha }}
|sI-A|=0
|sI-A|=\begin{vmatrix} s &-1 \\ \alpha &s+2\beta \end{vmatrix}
\;\;=s^2+2s\beta +\alpha =0
\therefore \;\; \omega _n^2=\alpha
\;\;\; \omega _n=\sqrt{\alpha }
\;\;\;2\xi \omega _n=2\beta
\;\;\;\xi =\frac{\beta }{\sqrt{\alpha }}
Question 8 |
The transfer function of a phase lead compensator is given by
D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}
The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}
The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
\sqrt{\frac{3}{T^2}} | |
\sqrt{\frac{1}{3T^2}} | |
\sqrt{3T} | |
\sqrt{3T^3} |
Question 8 Explanation:
T(s)=\frac{1+3Ts}{1+Ts}
Frequency at which \angle T(j\omega ) is maximum,
(\omega _m)=\frac{1}{T\sqrt{\alpha }}
\alpha =\frac{1}{1/3}=3
\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Frequency at which \angle T(j\omega ) is maximum,
(\omega _m)=\frac{1}{T\sqrt{\alpha }}
\alpha =\frac{1}{1/3}=3
\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Question 9 |
The asymptotic Bode magnitude plot of a minimum phase transfer function G(s) is shown below.
Consider the following two statements.
Statement I: Transfer function G(s) has three poles and one zero.
Statement II: At very high frequency (\omega \rightarrow \infty), the phase angle \angle G(j\omega)=-\frac{3\pi}{2}.
Which one of the following options is correct?
Consider the following two statements.
Statement I: Transfer function G(s) has three poles and one zero.
Statement II: At very high frequency (\omega \rightarrow \infty), the phase angle \angle G(j\omega)=-\frac{3\pi}{2}.
Which one of the following options is correct?
Statement I is true and statement II is false. | |
Statement I is false and statement II is true. | |
Both the statements are true | |
Both the statements are false |
Question 9 Explanation:
G(s)=\frac{k}{s\left ( 1+\frac{s}{1} \right )\left ( 1+\frac{s}{20} \right )}
Transfer function shows 2 poles and no zeros. So statement I is false.
\angle G(j\omega )=-90-tan^{-1}\omega -tan^{-1}\frac{\omega }{20}
\angle G(j\omega )|_{\omega \rightarrow \infty }=-270^{\circ}=-\frac{3\pi}{2} \;rad
So statement II is true.
Transfer function shows 2 poles and no zeros. So statement I is false.
\angle G(j\omega )=-90-tan^{-1}\omega -tan^{-1}\frac{\omega }{20}
\angle G(j\omega )|_{\omega \rightarrow \infty }=-270^{\circ}=-\frac{3\pi}{2} \;rad
So statement II is true.
Question 10 |
The characteristic equation of a linear time-invariant (LTI) system is given by
\Delta (s)=s^4+3s^3+3s^2+s+k=0
The system is BIBO stable if
\Delta (s)=s^4+3s^3+3s^2+s+k=0
The system is BIBO stable if
0 \lt k \lt \frac{12}{9} | |
k \gt 3 | |
0 \lt k \lt \frac{8}{9} | |
k \gt 6 |
Question 10 Explanation:
Routh array is
For BIBO stability,
\frac{\left ( \frac{8}{3}-3k \right )}{\left ( \frac{8}{3} \right )} \gt 0
\Rightarrow k \lt \frac{8}{9}
and k \gt 0
\therefore \;\; 0 \lt k \lt \frac{8}{9}
For BIBO stability,
\frac{\left ( \frac{8}{3}-3k \right )}{\left ( \frac{8}{3} \right )} \gt 0
\Rightarrow k \lt \frac{8}{9}
and k \gt 0
\therefore \;\; 0 \lt k \lt \frac{8}{9}
There are 10 questions to complete.