# GATE Electrical Engineering 2021

 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
 A 1 and 1 B 1 and -1 C j and -j D 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}
 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?
 A $\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right )$ for all z B The sum of the roots of $p\left ( z \right )=0$ is a real number C The complex roots of the equation $p\left ( z \right )=0$ come in conjugate pairs D 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.
 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
 A no local minimum in (0,1) B one local maximum in (0,1) C exactly one local minimum in (0,1) D 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 $(0,1)$
 Question 4
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

 A $\text{10 V}$ in series with $12\:\Omega$ B $\text{65 V}$ in series with $15\:\Omega$ C $\text{50 V}$ in series with $2\:\Omega$ D $\text{35 V}$ 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}
 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)
 A $10x\hat{X}+20y\hat{Y}-30z\hat{Z}$ B $10y\hat{X}+20x\hat{Y}-10z\hat{Z}$ C $10z\hat{X}+20y\hat{Y}-30x\hat{Z}$ D $10x\hat{X}-30z\hat{Y}+20y\hat{Z}$

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}
 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
 A linear and time-variant B linear and time-invariant C non-linear and time-variant D 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.
 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
 A $\delta \left [ n-2\right ]+\delta \left [ n+1 \right ]$ B $\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ]$ C $\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ]$ D $\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ]$

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,
$h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)$
 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
 A P-V-X, Q-U-Z, R-T-Y B P-U-X, Q-S-Y, R-V-Z C P-T-Y, Q-V-Z, R-S-X D 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).
 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
 A $N_{L}\times \left ( N-1 \right )$ B $\left ( N-1 \right )\times \left ( N-1-N_{L} \right )$ C $N_{L}\times \left ( N-1+N_{L} \right )$ D $\left ( N-1 \right )\times \left ( N-1+N_{L} \right )$

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}
 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
 A $P_{1}=P_{2}=130$ B $P_{1}=160, P_{2}=100$ C $P_{1}=140, P_{2}=120$ D $P_{1}=120, P_{2}=140$

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),
$P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}$
There are 10 questions to complete.