Question 1 |
The transfer function of a real system, H(s), is given as:
H(s)=\frac{As+B}{s^2+Cs+D}
where A, B, C and D are positive constants. This system cannot operate as
H(s)=\frac{As+B}{s^2+Cs+D}
where A, B, C and D are positive constants. This system cannot operate as
low pass filter. | |
high pass filter | |
band pass filter. | |
an integrator. |
Question 1 Explanation:
Put s=0, H(0)=\frac{A \times 0+B}{0+C \times 0+D}=\frac{B}{D}
So, the system pass low frequency component. Put s=\infty , H(\infty )=0
For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.
So, the system pass low frequency component. Put s=\infty , H(\infty )=0
For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.
Question 2 |
For an ideal MOSFET biased in saturation, the magnitude of the small signal
current gain for a common drain amplifier is
0 | |
1 | |
100 | |
infinite |
Question 2 Explanation:
For ideal MOSFET, i_G=0
Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty
Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty
Question 3 |
The most commonly used relay, for the protection of an alternator against loss of
excitation, is
offset Mho relay. | |
over current relay. | |
differential relay | |
Buchholz relay. |
Question 4 |
The geometric mean radius of a conductor, having four equal strands with each strand
of radius 'r', as shown in the figure below, is
4r | |
1.414r | |
2r | |
1.723r |
Question 4 Explanation:
Redraw the configuration:
\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}
Where, r'=0.7788r
Hence, GMR=1.723r
\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}
Where, r'=0.7788r
Hence, GMR=1.723r
Question 5 |
The valid positive, negative and zero sequence impedances (in p.u.), respectively, for
a 220 kV, fully transposed three-phase transmission line, from the given choices are
1.1, 0.15 and 0.08 | |
0.15, 0.15 and 0.35 | |
0.2, 0.2 and 0.2 | |
0.1, 0.3 and 0.1 |
Question 5 Explanation:
We have,
X_0 \gt X_1=X_2
(for 3-\phi transposed transmission line)
X_0 \gt X_1=X_2
(for 3-\phi transposed transmission line)
Question 6 |
The steady state output (V_{out}), of the circuit shown below, will
saturate to +V_{DD} | |
saturate to -V_{EE} | |
become equal to 0.1 V | |
become equal to -0.1 V |
Question 6 Explanation:
Redraw the circuit:
From circuit,
\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}
Hence, V_{out}=-V_{EE}
From circuit,
\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}
Hence, V_{out}=-V_{EE}
Question 7 |
The Bode magnitude plot of a first order stable system is constant with frequency.
The asymptotic value of the high frequency phase, for the system, is -180^{\circ}. This
system has
one LHP pole and one RHP zero at the same frequency | |
one LHP pole and one LHP zero at the same frequency | |
two LHP poles and one RHP zero | |
two RHP poles and one LHP zero. |
Question 7 Explanation:
The given system is non-minimum phase system
Therefore, transfer function, T.F=\frac{s-1}{s+1}
Hence, one LHP pole and one RHP zero at the same frequency.
Hence, one LHP pole and one RHP zero at the same frequency.
Question 8 |
A balanced Wheatstone bridge ABCD has the following arm resistances:
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
30\Omega \pm 3\Omega | |
30\Omega \pm 0.9\Omega | |
3000\Omega \pm 90\Omega | |
3000\Omega \pm 3\Omega |
Question 8 Explanation:
The condition for balanced bridge
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
Question 9 |
The open loop transfer function of a unity gain negative feedback system is given by G(s)=\frac{k}{s^2+4s-5}.
The range of k for which the system is stable, is
The range of k for which the system is stable, is
k \gt 3 | |
k \lt 3 | |
k \gt 5 | |
k \lt 5 |
Question 9 Explanation:
Characteristic equation:
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
Question 10 |
Consider a 3 x 3 matrix A whose (i,j)-th element, a_{i,j}=(i-j)^3. Then the matrix A will be
symmetric. | |
skew-symmetric. | |
unitary | |
null. |
Question 10 Explanation:
for \; i=j\Rightarrow a_{ij}=(i-i)^3=0\forall i
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
There are 10 questions to complete.