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}




There are 5 questions to complete.