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
\left[\begin{array}{lll}-2 & -2 & 2\end{array}\right]^{T} | |
\left[\begin{array}{lll}3 & 0 & -1\end{array}\right]^{T} | |
\left[\begin{array}{lll}3 & 2 & 1\end{array}\right]^{T} | |
\left[\begin{array}{llll}1 & 2 & 3\end{array}\right]^{T} |
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}
\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
\frac{2 s+3}{s+1} | |
\frac{3 s+2}{s-1} | |
\frac{s+1}{3 \mathrm{~s}+2} | |
\frac{3 s+2}{s+1} |
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}
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
\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
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=\infty | |
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=0 | |
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=3 | |
\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s})=-5 |
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}
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}
G(s)=\frac{1}{s-1}
\infty, \infty | |
1,0 | |
1,-1 | |
-1,1 |
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
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
\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-r, b-p, and c-q | |
a-r, b-q, and c-p | |
a-p, b-r, and c-q | |
a-q, b-p, and c-r |
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
\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.
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