GATE2011ECE(1)

GATE 2011

1) The modes in a rectangular waveguide are denoted by TE_mn ,TM_mn where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE?

(A) The TM10 mode of the wave does not exist

(B) The TE10 mode of the wave does not exist

(C) The TM10 and the TE10 modes both exist and have the same cut-off

frequencies

(D) The TM10 and TM01 modes both exist and have the same cut-off frequencies

Solution

For TM₁₀ mode , the electric and magnetic field components will vanish. So it will not exist

Answer is (A)

(B) – incorrect because TE₁₀ exists

(C) – incorrect because TM₁₀  does not exist

(D) - incorrect because TM₀₁ cannot exist.For TM wave in rectangular waveguide m and n cannot be zero.Cutoff frequencies are same for same modes of TE and TM.

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2)  The Column-I lists the attributes and the Column-II lists the modulation systems. Match the attribute to the modulation system that best meets it 

          Column-I                                                            Column-II

P Power efficient transmission of signals          1 Conventional AM

Q Most bandwidth efficient transmission of       2 FM

voice signals

R Simplest receiver structure                          3 VSB

S Bandwidth efficient transmission of

signals with Significant dc component              4 SSB-SC

(A) P-4;Q-2;R-1;S-3            (B) P-2;Q-4;R-1;S-3

(C) P-3;Q-2;R-1;S-4            (D) P-2;Q-4;R-3;S-1

 

Solution

Power efficient transmission of signals – FM 

Most bandwidth efficient transmission of voice signals – SSB with Suppressed Carrier because there is only one sideband(bandwidth efficiency ) and no carrier(power efficiency) .

Simplest receiver structure – AM

Bandwidth efficient transmission of signals with Significant dc component – Vestigial Side Band(VSB) is used when low frequencies(DC) contain significant information(as in case of TV signals).

Answer is (B) 

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3)   For the transfer function G(jw) = 5 + jw , the corresponding Nyquist plot for positive frequency has the form

 

Solution

Magnitude of G(jw) = √(5² + w²)

Angle of G(jw) = tan^-1(w/5)

when w = 0,

Magnitude of G(jw) = √(5² ) = 5

Angle of G(jw) = 0

when w = ∞,

Magnitude of G(jw) = ∞

Angle of G(jw) = +90

So answer is (A) where magnitude starts from 5 (angle = 0 ie X axis) and move towards ∞ as w approaches ∞ (direction is vertical because angle approaches 90 ie positive Y axis).

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4)  The trigonometric Fourier series of an even function does not have the

(A) dc term           (B) cosine terms

(C) sine terms       (D) odd harmonic terms

Solution

even symmetry - no sine terms will be there, cosine and DC components will be present.

Answer is (C) 

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5)  When the output Y in the circuit below is ‘1’, it implies that data has 

 

(A) changed from 0 to 1                 (B) changed from 1 to 0

(C) changed in either direction        (D) not changed

 

 Solution

The method we follow is to go back from output to input side

AND gate

For output to be 1 both inputs should be 1.

Let us assume Q₁ be output of first D flip flop and be Q₂ output of second flipflop

So Q₁ = Q₂ = 1

Similarly assume terms D₁,D₂,Q’₁,Q₂’ for inputs and complement outputs of the two flipflops respectively

Second D Flip flop

Q₂ = 1

Implies D₂ input of previous clock cycle is 1

First D Flip flop

D₂ input of previous clock cycle is 1 implies Q’₁ of previous cycle is 1 (since they are interconnected)

Q’₁ =1 means  Q₁ = 0 , that happen only when D₁=0(at cycle before above mentioned previous) .

At current clock cycle Q₁=1 , so previously D₁=1

So we can see that D₁ changed from 0 to 1

Answer is (A)

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6)   A Zener diode, when used in voltage stabilization circuits, is biased in
 

(A) reverse bias region below the breakdown voltage
 

(B) reverse breakdown region
 

(C) forward bias region
 

(D) forward bias constant current mode

 Solution

At breakdown region voltage across zener will be constant, and this property is used in voltage stabilization circuits

Answer is (B)

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