# 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.