Question 1 |
Let p and q be real numbers such that p^{2}+q^{2}=1. The eigenvalues of the matrix \begin{bmatrix} p & q\\ q& -p \end{bmatrix}
are
1 and 1 | |
1 and -1 | |
j and -j | |
pq and -pq |
Question 1 Explanation:
Characteristic equation of A
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
Question 2 |
Let p\left ( z\right )=z^{3}+\left ( 1+j \right )z^{2}+\left ( 2+j \right )z+3, where z is a complex number.
Which one of the following is true?
Which one of the following is true?
\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right ) for all z | |
The sum of the roots of p\left ( z \right )=0 is a real number | |
The complex roots of the equation p\left ( z \right )=0
come in conjugate pairs | |
All the roots cannot be real |
Question 2 Explanation:
Since sum of the roots is a complex number
\Rightarrow absent one root is complex
So all the roots cannot be real.
\Rightarrow absent one root is complex
So all the roots cannot be real.
Question 3 |
Let f\left ( x \right )
be a real-valued function such that {f}'\left ( x_{0} \right )=0
for some x _{0} \in\left ( 0,1 \right ), and {f}''\left ( x \right )> 0
for all x \in \left ( 0,1 \right ). Then f\left ( x \right )
has
no local minimum in (0,1) | |
one local maximum in (0,1) | |
exactly one local minimum in (0,1) | |
two distinct local minima in (0,1) |
Question 3 Explanation:
x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
Question 4 |
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is
\text{10 V} in series with 12\:\Omega
| |
\text{65 V} in series with 15\:\Omega
| |
\text{50 V} in series with 2\:\Omega
| |
\text{35 V} in series with 2\:\Omega
|
Question 4 Explanation:
Given circuit can be resolved as shown below,
V_{T H}=15+50=65 \mathrm{~V}
\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
V_{T H}=15+50=65 \mathrm{~V}
\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
Question 5 |
Which one of the following vector functions represents a magnetic field \overrightarrow{B}?
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
10x\hat{X}+20y\hat{Y}-30z\hat{Z} | |
10y\hat{X}+20x\hat{Y}-10z\hat{Z} | |
10z\hat{X}+20y\hat{Y}-30x\hat{Z} | |
10x\hat{X}-30z\hat{Y}+20y\hat{Z} |
Question 5 Explanation:
If \vec{B} is magnetic flux density then \vec{\nabla} \cdot \vec{B}=0
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
Question 6 |
If the input x(t) and output y(t) of a system are related as y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right ), then the system is
linear and time-variant | |
linear and time-invariant | |
non-linear and time-variant | |
non-linear and time-invariant |
Question 6 Explanation:
\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}
Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1
\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.
Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1
\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.
Question 7 |
Two discrete-time linear time-invariant systems with impulse responses h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ] and h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ] are connected in cascade, where \delta \left [ n\right ] is the Kronecker delta. The impulse response of the cascaded system is
\delta \left [ n-2\right ]+\delta \left [ n+1 \right ] | |
\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ] | |
\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ] | |
\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ] |
Question 7 Explanation:
\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)
Question 8 |
Consider the table given:
\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}
The correct combination that relates the constructional feature, machine type and mitigation is
\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}
The correct combination that relates the constructional feature, machine type and mitigation is
P-V-X, Q-U-Z, R-T-Y | |
P-U-X, Q-S-Y, R-V-Z | |
P-T-Y, Q-V-Z, R-S-X | |
P-U-X, Q-V-Y, R-T-Z |
Question 8 Explanation:
P: Damper bars used in synchronous machine (U) to prevent hunting (X)
Q: Skewed rotor slots used in induction motor (S) t avoid magnetic locking (Y)
R: Compensating winding used in DC machine (V) to neutralize cross magnetizing effects of armature reaction (Z) under main poles (polar zone).
Q: Skewed rotor slots used in induction motor (S) t avoid magnetic locking (Y)
R: Compensating winding used in DC machine (V) to neutralize cross magnetizing effects of armature reaction (Z) under main poles (polar zone).
Question 9 |
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is N_{L}. The balanced Newton-Raphson method is used to carry out load flow study in polar form. \text{H, S, M, and R} are sub-matrices of the Jacobian matrix J as shown below:
\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}
The dimension of the sub-matrix M is
\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}
The dimension of the sub-matrix M is
N_{L}\times \left ( N-1 \right ) | |
\left ( N-1 \right )\times \left ( N-1-N_{L} \right ) | |
N_{L}\times \left ( N-1+N_{L} \right ) | |
\left ( N-1 \right )\times \left ( N-1+N_{L} \right ) |
Question 9 Explanation:
\left[\begin{array}{l} \Delta P \\ \Delta Q \end{array}\right]=J\left[\begin{array}{l} \Delta \delta \\ \Delta V \end{array}\right] \text { where } J=\left[\begin{array}{ll} H & S \\ M & R \end{array}\right]
For size of M
Row = No. of unknown variables of Q=N_{L}
Column = No. of variable which has \delta=N_{L}+\left(N-1-N_{L}\right)
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
For size of M
Row = No. of unknown variables of Q=N_{L}
Column = No. of variable which has \delta=N_{L}+\left(N-1-N_{L}\right)
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
Question 10 |
Two generators have cost functions F_{1} and F_{2}. Their incremental-cost characteristics are
\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}
\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}
They need to deliver a combined load of 260 \text{~MW}. Ignoring the network losses, for economic operation, the generations P_{1} and P_{2} (in \text{MW}) are
\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}
\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}
They need to deliver a combined load of 260 \text{~MW}. Ignoring the network losses, for economic operation, the generations P_{1} and P_{2} (in \text{MW}) are
P_{1}=P_{2}=130 | |
P_{1}=160, P_{2}=100 | |
P_{1}=140, P_{2}=120 | |
P_{1}=120, P_{2}=140 |
Question 10 Explanation:
\begin{aligned} I C_{1} &=I C_{2} \\ 40+0.2 P_{1} &=32+0.4 P_{2} \\ 0.4 P_{2}-0.2 P_{1} &=8 \\ P_{2}+P_{1} &=260 \end{aligned}
Solving equation (i) and (ii),
P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}
Solving equation (i) and (ii),
P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}
There are 10 questions to complete.