# GATE Electrical Engineering 2023

 Question 1
For a given vector $w=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{\top}$, the vector normal to the plane defined by $\mathbf{w}^{\top} x=1$ is
 A $\left[\begin{array}{lll}-2 & -2 & 2\end{array}\right]^{T}$ B $\left[\begin{array}{lll}3 & 0 & -1\end{array}\right]^{T}$ C $\left[\begin{array}{lll}3 & 2 & 1\end{array}\right]^{T}$ D $\left[\begin{array}{llll}1 & 2 & 3\end{array}\right]^{T}$
Engineering Mathematics   Calculus
Question 1 Explanation:
Given, $W^{T}=1$

$\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=1$

We have, vector normal to the plane $=\nabla F$
\begin{aligned} & =i \frac{\partial F}{\partial x}+\hat{j} \frac{\partial F}{\partial y}+\hat{k} \frac{\partial F}{\partial z} \\ & =\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}

$\therefore$ Normal vector $=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{T}$
 Question 2
For the block diagrm shown in the figure, the transfer function $\frac{Y(s)}{R(s)}$ is

 A $\frac{2 s+3}{s+1}$ B $\frac{3 s+2}{s-1}$ C $\frac{s+1}{3 \mathrm{~s}+2}$ D $\frac{3 s+2}{s+1}$
Control Systems   Mathematical Models of Physical Systems
Question 2 Explanation:
Signal flow graph:

Forward paths,

\begin{aligned} & \mathrm{P}_{1}=3, \quad \Delta_{1}=1 \\ & \mathrm{P}_{2}=\frac{2}{\mathrm{~S}}, \Delta_{2}=1 \end{aligned}

Loops: $L_1=\frac{1}{S}$

Using Masson's graph formula,

\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{P_{1} \Delta_{1}+P_{1} \Delta_{2}}{1-L_{1}} \\ & =\frac{3+\frac{2}{S}}{1-\frac{1}{S}} \\ & =\frac{3 S+2}{S-1} \end{aligned}

 Question 3
In the Nyquist plot of the open-loop transfer function

$\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1}$

corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in s-plane is mapped into a point at

 A $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\infty$ B $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=0$ C $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=3$ D $\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=-5$
Control Systems   Frequency Response Analysis
Question 3 Explanation:
Nyquist Contour :

Given:
\begin{aligned} \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1} \\ \text { Put } \quad \mathrm{s} & =R e^{j \theta} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =\operatorname{Lim}_{R \rightarrow \infty} \frac{3 R e^{j \theta}+5}{\operatorname{Re}^{j \theta}-1} \\ \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) & =3 \end{aligned}
 Question 4
Consider a unity-gain negative feedback system consisting of the plant $G(s)$ (given below) and a proportional-integral controller. Let the proportional gain and integral gain be 3 and 1, respectively. For a unit step reference input, the final values of the controller output and the plant output, respectively, are

$G(s)=\frac{1}{s-1}$
 A $\infty, \infty$ B $1,0$ C $1,-1$ D $-1,1$
Control Systems   Feedback Characteristics of Control Systems
Question 4 Explanation:
Given plant:

So, $\quad$ OLTF $=\frac{(3 s+1)}{s(s-1)}$

Closed loop transfer function,
\begin{aligned} \frac{Y(s)}{R(s)} & =\frac{3 s+1}{s^{2}+2 s+1} \\ Y(s) & =\frac{3 s+1}{s\left(s^{2}+2 s+1\right)} \quad\left[\because R(s)=\frac{1}{s}\right] \end{aligned}

Final value of plant,
$Y(\infty)=\operatorname{Lims}_{s \rightarrow 0} Y(s)=1$

From plant,
\begin{aligned} X(\mathrm{~s}) & =\left[\mathrm{R}(\mathrm{s})-\frac{\mathrm{X}(\mathrm{s})}{\mathrm{s}-1}\right]\left(3+\frac{1}{\mathrm{~s}}\right) \\ X(\mathrm{~s})\left[1+\frac{3 \mathrm{~s}+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right)\left[\because \mathrm{R}(\mathrm{s})=\frac{1}{\mathrm{~s}}\right] \\ X(\mathrm{~s})\left[\frac{\mathrm{s}^{2}+2 s+1}{\mathrm{~s}(\mathrm{~s}-1)}\right] & =\left(\frac{3 \mathrm{~s}+1}{\mathrm{~s}^{2}}\right) \\ \Rightarrow \quad X(\mathrm{~s}) & =\frac{(3 \mathrm{~s}+1)(\mathrm{s}-1)}{\mathrm{s}\left(\mathrm{s}^{2}+2 \mathrm{~s}+1\right)} \end{aligned}

$\therefore$ Final value of controller,
$x(\infty)=\operatorname{LimsX}_{s \rightarrow 0} \mathrm{~s}(\mathrm{~s})=-1$
 Question 5
The following columns present various modes of induction machine operation and the ranges of slip

$\begin{array}{ll} \textbf{A (Mode of operation)}& \textbf{B (Range of slip)}\\\\ \text{a. Running in generator mode}&\text{p) From 0.0 to 1.0}\\\\ \text{b. Running in motor mode} & \text{q) From 1.0 to 2.0}\\\\ \text{c. Plugging in motor mode} & \text{r) From - 1.0 to 0.0} \end{array}$
The correct matching between the elements in column A with those of column B is
 A a-r, b-p, and c-q B a-r, b-q, and c-p C a-p, b-r, and c-q D a-q, b-p, and c-r
Electrical Machines   Single Phase Induction Motors, Special Purpose Machines and Electromechanical Energy Conversion System
Question 5 Explanation:
Torque speed characteristic of $3-\phi$ IM :

$\mathrm{S} \gt 1 \Rightarrow$ Plugging mode
$0 \lt \mathrm{S} \lt 1 \Rightarrow$ Motoring mode
$\mathrm{S} \lt 0 \Rightarrow$ Generating Mode

There are 5 questions to complete.