# GATE EC 2012

 Question 1
The current $i_{b}$ through the base of a silicon npn transistor is $1 + 0.1 cos(10000 \pi t)$ mA. At 300 K, the $r_{\pi}$ in the small signal model of the transistor is
 A 250 $\Omega$ B 27.5 $\Omega$ C 25 $\Omega$ D 22.5 $\Omega$
Analog Circuits   BJT Analysis
Question 1 Explanation:
We know that
$\begin{array}{l} r_{\pi}=(\beta+1) r_{e} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{I_{e}} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{(\beta+1) I_{b}} \\ r_{\pi}=\frac{V_{T}}{I_{b}} \end{array}$
Where $I_{b}$ is d.c. current through base so
$I_{b}=1 \mathrm{mA}$
$V_{T}=25 \mathrm{mV}$ at room temperature
So,$r_{\pi}=\frac{25 \times 10^{-3}}{1 \times 10^{-3}}=25 \Omega$
 Question 2
The power spectral density of a real process X(t) for positive frequencies is shown below. The values of $E[X^{2}(t)]$ and |E[X(t)]|, respectively, are
 A 6000/$\pi$,0 B 6400/$\pi$,0 C $6400/\pi,20(\pi /\sqrt{2})$ D $6000/\pi,20(\pi /\sqrt{2})$
Communication Systems   Random Processes
Question 2 Explanation:

We know that $E\left[X^{2}(t)\right]$ represent the total power in random signal. So
\begin{aligned} P=E\left[X^{2}(t)\right]&=2 \times \frac{1}{2 \pi} \int_{9 \times 10^{3}}^{11 \times 10^{3}} S_{x}(\omega) d \omega \\ E\left[X^{2}(t)\right]&=\frac{1}{\pi}\left[400+\frac{1}{2} \times 6 \times 2 \times 10^{3}\right] \\ E\left[X^{2}(t)\right]&=\frac{6400}{\pi} \end{aligned}
At $\omega=0,$ there is no any frequency component present, hence dc value of the process is zero.
 Question 3
In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is
 A 1750 B 2625 C 4000 D 5250
Communication Systems   Digital Communications
Question 3 Explanation:
\begin{aligned} \mathrm{BW} &=\frac{R_{b}}{2}(1+\alpha) \\ 3500 &=\frac{R_{b}}{2}(1+0.75) \\ R_{b} &=4000 \mathrm{bits} / \mathrm{sec} \end{aligned}
In base band transmission $\rightarrow$
Symbol rate = Bit rate =4000 symbols/sec
 Question 4
A plane wave propagating in air with $\vec{E}=(8\hat{a}_{x}+6\hat{a}_{y}+5\hat{a}_{z})e^{j(\omega t+3x-4y)}$ V/m is incident on a perfectly conducting slab positioned at 0 $\leq$ x. The E field of the reflected wave is
 A $(-8\hat{a}_{x}-6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t+3x+4y)} V/m$ B $(-8\hat{a}_{x}+6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t+3x+4y)} V/m$ C $(-8\hat{a}_{x}-6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t-3x-4y)} V/m$ D $(-8\hat{a}_{x}+6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t-3x-4y)} V/m$
Electromagnetics   Plane waves and Properties
Question 4 Explanation:

Incident wave propagation $\rightarrow-3 \hat{a}_{x}+4 \hat{a}_{y}$
Reflected wave propagation $\rightarrow 3 \hat{a}_{x}+4 \hat{a}_{y}$
$E_{i}$ incident direction $\rightarrow 8 \hat{a}_{x}+6 \hat{a}_{y}+5 \hat{a}_{z}$
$\left.\begin{array}{l}E_{i} \text { tangential } \rightarrow 6 \hat{a}_{y}+5 \hat{a}_{z} \\ E_{r} \text { tangential } \rightarrow-6 \hat{a}_{y}-5 \hat{a}_{z}\end{array}\right\} E_{\text {tang }}=0$
$\left.\begin{array}{rl}E_{i} \text { normal } & =8 \hat{a}_{x} \\ E_{r} \text { normal } & =8 \hat{a}_{x}\end{array}\right\} E_{\text {normal }}=E_{\max }$
$E_{r}$ reflected direction $=8 \hat{a}_{x}-6 \hat{a}_{y}-5 \hat{a}_{z}$
 Question 5
The electric field of a uniform plane electromagnetic wave in free space, along the positive $x$ direction, is given by $\vec{E}=10(\hat{a}_{y}+j\hat{a}_{z})e^{-j25x}$. The frequency and polarization of the wave, respectively, are
 A 1.2 GHz and left circular B 4 Hz and left circular C 1.2 GHz and right circular D 4 Hz and right circular
Electromagnetics   Plane waves and Properties
Question 5 Explanation:
Out of phase by $90^{\circ}$and equal amplitude the wave is circularly polarized.
Check the trace of E with the time which gives left circularly polarized.
\begin{aligned} a_{y}&=\sin \omega t \\ a_{z}&=\cos \omega t \end{aligned}
 Question 6
Consider the given circuit.

In this circuit, the race around
 A does not occur B occurs when CLK = 0 C occurs when CLK = 1 and A = B = 1 D occurs when CLK = 1 and A = B = 0
Digital Circuits   Sequential Circuits
Question 6 Explanation:
Given flipflop is S-R flipflop with A= Sand B = R.
In S-R flipflop race around condition does not occur.
 Question 7
The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is
 A 4 B 6 C 8 D 10
Digital Circuits   Combinational Circuits
Question 7 Explanation:
Output will be 1 if A>B.
If B=00 then there will be three combinations for which output will be 1 i.e. when A=01,10 , or 11.
If B=01 there will be two conditions i.e. A=10 and 11.
If B=10 there will be one condition i.e. A=11.
So total 6 combinations are there for which output will be 1 .
 Question 8
The i-v characteristics of the diode in the circuit given below are $i=\left\{\begin{matrix} \frac{v-0.7}{500}A &v\geq 0.7V \\ 0 A & v \lt 0.7 V \end{matrix}\right.$

The current in the circuit is
 A 10 mA B 9.3 mA C 6.67 mA D 6.2 mA
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 8 Explanation:
Applying KVL in given circuit we have
$10-1000 i-v=0 \qquad\ldots(1)$
since 10 V is greater than 0.7 V so current will flow through diode and is
$i=\frac{v-0.7}{500}\qquad\ldots(2)$
from (1) and (2) we have,
\begin{aligned} 10-2(v-0.7)-v &=0 \\ 10+1.4-3 v &=0 \\ 3 v &=11.4 \\ v &=\frac{11.4}{3}=3.8 v\\ \text{so}\quad i &=\frac{v-0.7}{500}=\frac{3.8-0.7}{500} \\ &=6.2 \mathrm{mA} \end{aligned}
 Question 9
In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current $i(t)$ for all $t$ is
 A zero B a step function C an exponentially decaying function D an impulse function
Network Theory   Transient Analysis
Question 9 Explanation:
Since there is no resistance so time constant will be zero. That means as soon as the switch will be closed voltages at $C_{1}$ and $C_{2}$ will become equal and capacitor allows sudden change of voltage only if impulse of current will pass through it.
 Question 10
The average power delivered to an impedance (4 - j3) $\Omega$by a current 5cos(100$\pi$ t +100) A is
 A 44.2 W B 50W C 62.5W D 125W
Network Theory   Basics of Network Analysis
Question 10 Explanation:
Average power is same as RMS power.
$\begin{array}{l} P=I_{\mathrm{rms}}^{2} R=\left(\frac{5}{\sqrt{2}}\right)^{2} \times 4 \\ =\frac{25}{2} \times 4=50 \mathrm{W} \end{array}$
Note: Power is consumed only by resistance i.e. by real part of impedance.
There are 10 questions to complete.