# GATE EC 2010

 Question 1
The eigen values of a skew-symmetric matrix are
 A always zero B always pure imaginary C either zero or pure imaginary D always real
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
A skew-synnmetric matrix is a square matrix whose transpose is equal to its negative value, i.e. $A^{\top}=-A$
The eigen values of a skew-symmetric matrix alwavs come in pairs $\pm \lambda$ (except in odd-dimensional case where there is an additional unpaired 0 eigen value). For a real skew symmetric matrix the non-zero eigen values are all pure imaginary and thus are of the form $i \lambda_{1},-i \lambda_{1}, i \lambda_{2}$
$-i \lambda_{2}, \ldots$ where each of the $\lambda_{K}$ is real.
 Question 2
The trigonometric Fourier series for the waveform $f(t)$ shown below contains
 A only cosine terms and zero values for the dc components B only cosine terms and a positive value for the dc components C only cosine terms and a negative value for the dc components D only sine terms and a negative value for the dc components
Signals and Systems   Fourier Series
Question 2 Explanation:
since f(t) is an even function, its trigonometric Fourier series contains only cosine terms.
D.C. component,
\begin{aligned} A_{0} &=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) d t=\frac{2}{T} \int_{0}^{T / 2} f(t) d t \\ &=\frac{2}{T}\left[\int_{0}^{T / 4} A d t+\int_{T / 4}^{T / 2}(-2 A) d t\right] \\ &=\frac{2}{T}\left[\frac{A T}{4}-2 A\left(\frac{T}{2}-\frac{T}{4}\right)\right] \\ &=\frac{2}{T}\left[-\frac{A T}{4}\right]=-\frac{A}{2} \end{aligned}
Therefore, the trigonometric Fourier series for the waveform f(t) contains only cosine terms and a negative value for the dc component.
 Question 3
A function $n(x)$ satisfied the differential equation $\frac{ d^{2}n(x)}{dx^{2}}-\frac{n(x)}{L^{2}}=0$ where L is a constant. The boundary conditions are : $n(0) = K$ and $n(\infty )$ = 0. The solution to this equation is
 A $n(x) = Kexp(x/L)$ B $n(x)=Kexp(-x/\sqrt{L})$ C $n(x)=K^{2}exp(-x/L)$ D $n(x) = Kexp(- x/L)$
Engineering Mathematics   Differential Equations
Question 3 Explanation:
\begin{aligned} \left(D^{2}-\frac{1}{L^{2}}\right) n(x) &=0 \\ D^{2}-\frac{1}{L^{2}} &=0 \\ D &=\pm \frac{1}{L} \end{aligned}
So, the solution is
\begin{aligned} n(x) &=C_{1} e^{x / L}+C_{2} e^{-x / L} \\ n(0) &=K \Rightarrow C_{1}+C_{2}=K \\ n(\infty) &=0 \Rightarrow C_{1}=0 \\ \Rightarrow \quad C_{2} &=K \\ \text { Therefore, } n(x) &=K \exp (-x / L) \end{aligned}
 Question 4
For the two-port network shown below, the short-circuit admittance parameter matrix is
 A $\begin{vmatrix} 4 & -2\\ -2&4 \end{vmatrix}S$ B $\begin{vmatrix} 1 & -0.5\\ -0.5&1 \end{vmatrix}S$ C $\begin{vmatrix} 1 & 0.5\\ 0.5&1 \end{vmatrix}S$ D $\begin{vmatrix} 4 & 2\\ 2&4 \end{vmatrix}S$
Network Theory   Two Port Networks
Question 4 Explanation:
Short-circuit admittance parameters for a two-port $\pi$-network are,
$\begin{array}{l} Y_{11}=Y_{a}+Y_{b} \\ Y_{12}=Y_{21}=-Y_{b} \\ Y_{22}=Y_{b}+Y_{c} \end{array}$

For the given network,
\begin{aligned} Y_{a}&=Y_{b}=Y_{c}=\frac{1}{0.5}=2 \mho\\ \text{So,}Y_{11}&=2+2=4 \mho\\ Y_{12}&=Y_{21}=-2 \mho\\ Y_{22}&=2+2=4 \mho \end{aligned}
 Question 5
For parallel RLC circuit, which one of the following statements is NOT correct?
 A The bandwidth of the circuit decreases if R is increased B The bandwidth of the circuit remains same if L is increased C At resonance, input impedance is a real quantity D At resonance, the magnitude of input impedance attains its minimum values
Network Theory   Sinusoidal Steady State Analysis
Question 5 Explanation:
Characteristic equation for a parallel RLC circuit is
$s^{2}+\frac{1}{R C} s+\frac{1}{L C}=0$
where, Bandwidth= $\frac{1}{RC}$
(i) It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased.
(ii) At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity.
(iii) In parallel RLC circuit, the admittance is minimum at resonance. Hence magnitude of input impedance attains its maximum value at resonance.
 Question 6
At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n-channel MOSFET is
 A $150 cm^{2}/V-s$ B $1350 cm^{2}/V-s$ C $1800 cm^{2}/V-s$ D $3600 cm^{2}/V-s$
Electronic Devices   BJT and FET Basics
 Question 7
Thin gate oxide in a CMOS process in preferably grown using
 A wet oxidation B dry oxidation C epitaxial oxidation D ion implantation
Electronic Devices   IC Fabrication
Question 7 Explanation:
Dry oxidation is prefferred for gate oxides.
 Question 8
In the silicon BJT circuit shown below, assume that the emitter area of transistor $Q_{1}$ is half that of transistor $Q_{2}$. The value of current $I_{0}$ is approximately
 A 0.5 mA B 2 mA C 9.3 mA D 15 mA
Analog Circuits   BJT Analysis
Question 8 Explanation:
The given circuit is a current mirror circuit in which the output curren Is a mirror image of the input current 1f both the transistors are identical.

To calculate $I_{i},$
$9.3 I_{i}+0.7=0-(-10)=10$
$\Rightarrow \quad I_{i}=1 \mathrm{mA}$
since the emitter area of transistor $Q_{1}$ is half that of transistor $Q_{2}$
$\mathrm{So} \quad I_{i}=I_{0} / 2$
Therefore, $I_{0}=2 \mathrm{mA}$
 Question 9
The amplifier circuit shown below uses a silicon transistor. The capacitors $C_{C}$ and $C_{E}$ can be assumed to be short at signal frequency and effect of output resistance $r_{0}$ can be ignored. If $C_{E}$ is disconnected from the circuit, which one of the following statements is true
 A The input resistance $R_{i}$ increases and magnitude of voltage gain $A_{v}$ decreases B The input resistance $R_{i}$ decreases and magnitude of voltage gain $A_{v}$ increases C Both input resistance $R_{i}$ and magnitude of voltage gain $A_{v}$ decreases D Both input resistance $R_{i}$ and the magnitude of voltage gain $A_{v}$ increases
Analog Circuits   BJT Analysis
Question 9 Explanation:
By disconnecting emitter bypass capacitor $C_{E}$
(i) Input impedance increases by a factor of $\left(1+\beta R_{E}\right)$
(ii) Magnitude of voltage gain $A_{V}$ decreases by the same factor.
 Question 10
Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is
 A $-\frac{R_{2}}{R_{1}}$ B $-\frac{R_{3}}{R_{1}}$ C $-\frac{R_{2}||R_{3}}{R_{1}}$ D $-(\frac{R_{2}+R_{3}}{R_{1}})$
Analog Circuits   Operational Amplifiers
Question 10 Explanation:

Assuming ideal OP-AMP, voltage at point A is zero. So, the given circuit can be considered as shown below:

$A_V=-\frac{R_2}{R_1}$
There are 10 questions to complete.