GATE2012ECE(5)

GATE 2012 ECE paper fully solved.Step by step, detailed solutions are given.Corrections and suggestions are appreciated. 

For reference material, quote the question and send a mail to arunsblog4ece@gmail.com.

25)  The electric field of a uniform plane electromagnetic wave in free space, along the positive
x direction, is given by

$$ \overrightarrow{E}=10(a_{y}+ja_{z})e^{-j25x}$$

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

Solution

E(r) = E₀ e^-jβx for a wave travelling in +x direction

β = 25

velocity, v = ω/ β

or           ω = v β

            2πf = v β

               f = v β / 2π

                 = (3 * 10⁸ * 25) / 2π

                 = 1.2 GHz (approx)

Electric field has two components E_y and E_z both of magnitude 10, but E_z  is out of phase from E_y by 90^o .

Therefore in trigonometric form they can be represented as

E_y = E₀ cos(ωt - βx)

E_z = - E₀ sin(ωt - βx)

E_r = √(E_y² + E_z²)

Substituting (ωt - βx) = 0, 45^o, 90^o ... we can see that resultant phasor is rotating in clockwise direction and therefore left circular polarization

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26)  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

Solution

At t = 0⁺ , ie when the switch is closed capacitors act as short circuits, with 12V across C1 as voltage source. Since impedance is zero, current = voltage / impedance  = 12V / 0 = infinity or an impulse.

Later on both capacitors will be at same voltage, since they are in parallel and there will not be any current flow.

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27)  The voltage gain v A of the circuit shown below is



     

Solution

The circuit shown is a Common Emitter amplifier with collector feedback.Since feedback is from collector it will be definitely be a negative feedback.

The voltage gain of negative feedback configuration can be written as

A_v = - R_f / R₁

Where R_f is the feedback resistance

                R₁ is the input resistance

|A_v|=  R_f / R₁

        = 100k / 10k

         = 10

  
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