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.
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