Question 1 |
For A=\begin{bmatrix} 1 & tan x\\ -tan x &1 \end{bmatrix}, the Determinant of A^{T}A^{-1} is
sec^{2}x | |
cos4x | |
1 | |
0 |
Question 1 Explanation:
Long Method:
\begin{aligned} A &=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \\ A^{T} &=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ \operatorname{ad} A &=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|}[a d j(A)]^{T} \\ &=\frac{1}{1+\tan ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ &=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \end{aligned}
Here,
\begin{aligned} A^{\top} A^{-1} &=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ &=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right] \\ A^{\top} A^{-1} &=\left[\begin{array}{cc} \frac{1-\tan ^{2} x}{\sec ^{2} x} & \frac{-2 \tan x}{\sec ^{2} x} \\ \frac{2 \tan x}{\sec ^{2} x} & \frac{1-\tan ^{2} x}{\sec ^{2} x} \end{array}\right] \\ \left|A^{T} A^{-1}\right| &=\left(\frac{1-\tan ^{2} x}{\sec ^{2} x}\right)^{2}+\left(\frac{2 \tan x}{\sec ^{2} x}\right)^{2} \\ &=\frac{1+\tan ^{4} x-2 \tan ^{2} x+4 \tan ^{2} x}{\sec ^{4} x}\\ &=1 \end{aligned}
Or
Short Method:
\begin{aligned} \text{since, }|A B|&=|A||B| \\ \left|A^{T} A^{-1}\right| &=\left|A^{T}\right|\left|A^{-1}\right| \\ &=|A| \times \frac{1}{|A|}=1 \\ &\left(\text { Note }:\left|A^{T}\right|=|A| \text { and }\left|A^{-1}\right|=\frac{1}{|A|} \right) \end{aligned}
\begin{aligned} A &=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \\ A^{T} &=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ \operatorname{ad} A &=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|}[a d j(A)]^{T} \\ &=\frac{1}{1+\tan ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ &=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \end{aligned}
Here,
\begin{aligned} A^{\top} A^{-1} &=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ &=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right] \\ A^{\top} A^{-1} &=\left[\begin{array}{cc} \frac{1-\tan ^{2} x}{\sec ^{2} x} & \frac{-2 \tan x}{\sec ^{2} x} \\ \frac{2 \tan x}{\sec ^{2} x} & \frac{1-\tan ^{2} x}{\sec ^{2} x} \end{array}\right] \\ \left|A^{T} A^{-1}\right| &=\left(\frac{1-\tan ^{2} x}{\sec ^{2} x}\right)^{2}+\left(\frac{2 \tan x}{\sec ^{2} x}\right)^{2} \\ &=\frac{1+\tan ^{4} x-2 \tan ^{2} x+4 \tan ^{2} x}{\sec ^{4} x}\\ &=1 \end{aligned}
Or
Short Method:
\begin{aligned} \text{since, }|A B|&=|A||B| \\ \left|A^{T} A^{-1}\right| &=\left|A^{T}\right|\left|A^{-1}\right| \\ &=|A| \times \frac{1}{|A|}=1 \\ &\left(\text { Note }:\left|A^{T}\right|=|A| \text { and }\left|A^{-1}\right|=\frac{1}{|A|} \right) \end{aligned}
Question 2 |
The contour on the x-y plane, where the partial derivative of x^{2}+y^{2} with respect to y is equal to the partial derivative of 6y + 4x with respect to x, is
y=2 | |
x=2 | |
x+y=4 | |
x-y=0 |
Question 2 Explanation:
Partial derivative w.r.t.
\frac{\partial}{\partial y}\left(x^{2}+y^{2}\right)=2 y
Partial derivative w.r.t. x
\frac{\partial}{\partial x}(6 y+4 x)=4
From given condition
\begin{aligned} 2 y&=4 \\ \Rightarrow y\quad &=2 \end{aligned}
\frac{\partial}{\partial y}\left(x^{2}+y^{2}\right)=2 y
Partial derivative w.r.t. x
\frac{\partial}{\partial x}(6 y+4 x)=4
From given condition
\begin{aligned} 2 y&=4 \\ \Rightarrow y\quad &=2 \end{aligned}
Question 3 |
If C is a circle of radius r with centre z_{0}, in the complex z-plane and if n is a non-zero integer, then \oint_{C}\frac{dz}{(z-z_{0})^{n+1}} equals
2\pi nj | |
0 | |
\frac{nj}{2\pi } | |
2\pi n |
Question 3 Explanation:
By Cauchy integral formula
\begin{array}{l} \oint \frac{f(z)}{\left(z-z_{0}\right)^{n+1} d z}=\frac{2 \pi i f^{*}\left(z_{0}\right)}{n !} \\ \oint \frac{d z}{\left(z-z_{0}\right)^{n+1}}=\frac{2 \pi i}{n !} \cdot 0=0 \end{array}
\begin{array}{l} \oint \frac{f(z)}{\left(z-z_{0}\right)^{n+1} d z}=\frac{2 \pi i f^{*}\left(z_{0}\right)}{n !} \\ \oint \frac{d z}{\left(z-z_{0}\right)^{n+1}}=\frac{2 \pi i}{n !} \cdot 0=0 \end{array}
Question 4 |
Consider the function g(t)=e^{-t}sin(2\pi t)u(t) where u(t) is the unit step function. The area under g(t) is _______.
0.15 | |
0.34 | |
0.5 | |
1 |
Question 4 Explanation:
\begin{aligned} g(t) &=e^{-t} \sin (2 \pi t) u(t) \\ \text { Area } &=\int_{-\infty}^{\infty} g(t) d t \\ &=\int_{-\infty}^{\infty} \sin (2 \pi t) u(t) e^{-t} d t=\left.G(s)\right|_{s=0} \\ G(s) &=\frac{2 \pi}{(s+1)^{2}+4 \pi^{2}} \\ \text { Area } &=\left.\frac{2 \pi}{(s+1)^{2}+4 \pi^{2}}\right|_{S=0}=0.155 \end{aligned}
Question 5 |
The value \sum_{n=0}^{\infty }n(\frac{1}{2})^{n} is ______.
1 | |
2 | |
3 | |
4 |
Question 5 Explanation:
\begin{aligned} \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} &=\sum_{n=-\infty}^{\infty} n\left(\frac{1}{2}\right)^{n} u(n) \\ &=Z \cdot T\left[n\left(\frac{1}{2}\right)^{n} u(n)\right]_{z=1} \\ =& \frac{\frac{1}{2} Z^{-1}}{\left(1-\frac{1}{2} z^{-1}\right)^{2}}=\left.\frac{\frac{Z}{2}}{\left(z-\frac{1}{2}\right)^{2}}\right|_{z=1} \\ =& \frac{1 / 2}{1 / 4}=2 \end{aligned}
Question 6 |
For the circuit shown in the figure, the Thevenin equivalent voltage (in Volts) across terminals a-b is ________.
12 | |
10 | |
8 | |
16 |
Question 6 Explanation:
\begin{aligned} V_{Th}&=V_{6\;\omega}\\ \frac{V_{A}-12}{3}+\frac{V_{A}}{6} &=V_{6 \Omega} \\ V_{A}\left[\frac{1}{3}+\frac{1}{6}\right] &=1+4 \\ V_{A} \frac{3}{6} &=5 \\ V_{T h} &=V_{6 \Omega}=V_{A}=10 \mathrm{V} \end{aligned}
Question 7 |
In the circuit shown, the voltage V_{X} (in Volts) is _______.
7 | |
8 | |
9 | |
10 |
Question 7 Explanation:
\begin{aligned} \frac{V_{x}}{20}+\frac{V_{x}-V_{y}}{10}+0.5 V_{x}&=5 \mathrm{A} \\ V_{x}\left[\frac{1}{20}+\frac{1}{10}+0.5\right]&=5+\frac{V_{y}}{10} \\ 13 \mathrm{V}_{x}&=100+2 \mathrm{V}_{y} \quad\ldots(i)\\ \text { and also, } V_{y}&=0.25 \mathrm{V}_{x} \quad\dots(ii) \end{aligned}
Solving equations (i) and (ii), we have
\begin{aligned} 52 V_{y} &=100+2 V_{y} \\ 50 V_{y} &=100 \Rightarrow V_{y}=2 \mathrm{V} \\ V_{x} &=4 V_{y}=8 \mathrm{V} \end{aligned}
Question 8 |
At very high frequencies, the peak output voltage V_{0} (in Volts) is ________.
0 | |
0.5 | |
1 | |
1.5 |
Question 8 Explanation:
Circuit contains balanced Wheat-stone bridge. Also at high frequencies capacitor can be considered as short circuits.
Redrawing the circuit
\begin{aligned} V_{0} &=1 \sin \omega t \times 1 k \Omega=0.5 \sin \omega t \\ \text { Peak output } &=0.5 \mathrm{V} \end{aligned}
Redrawing the circuit
\begin{aligned} V_{0} &=1 \sin \omega t \times 1 k \Omega=0.5 \sin \omega t \\ \text { Peak output } &=0.5 \mathrm{V} \end{aligned}
Question 9 |
Which one of the following processes is preferred to form the gate dielectric (SiO_{2}) of MOSFETs ?
Sputtering | |
Molecular beam epitaxy | |
Wet oxidation | |
Dry oxidation |
Question 9 Explanation:
Dry oxidation is preferred for gate oxides.
Question 10 |
If the base width in a bipolar junction transistor is doubled, which one of the following statements will be TRUE?
Current gain will increase | |
Unity gain frequency will increase. | |
Emitter-base junction capacitance will increase | |
Early Voltage will increase |
Question 10 Explanation:
W_{B} doubled (increased) \rightarrow early effect is still present but its effect less severe relative to previous W_{B} . Slope I_{C} versus V_{C E} decreases.
There are 10 questions to complete.