GATE PAPERS: 2013

1) Consider the following statements regarding the complex Poynting vector Pv for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(Pv) denotes the real part of Pv.S denotes a spherical surface whose centre is at the point source, and nt denotes the unit surface normal on S. Which of the following statements is TRUE?

(A) Re(Pv) remains constant at any radial distance from the source

(B) Re(Pv) increases with increasing radial distance from the source

(D) Re(P) .n dS S ## v t decreases with increasing radial distance form the source

Ans) Power radiated from any source is constant.

Hence (C) is correct option.

2) A transmission line of characteristic impedance 50 Ω is terminated by a 50 Ω load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be 4 π radians. The phase velocity of the wave along the line is

(A) 0.8 108 m/s # (B) 1.2 108 m/s #

Ans) We have d = 2mm and f = 10 GHz

Phase difference (2π /λ)d= π/4;

or = λ = 8d = 8*2 mm = 16 mm

v =f λ= 10 *109 *16 *10−3

=1.6 108 m/ sec

Hence (C) is correct option.

3) An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is ________ bits / second.

(A) 1 (B) 2

Ans) Quantized 4 level require 2 bit representation i.e. for one sample 2 bit are required.

Since 2 sample per second are transmitted we require 4 bit to be transmitted per second.

Hence (D) is correct option.

4) The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

(A) G(s) H(s) = k (s (s + 1)/ (s + 2) (s + 3))
(B) G(s) H(s) = k ( (s + 1)/ s (s + 2) (s + 3) 2)
(C) G(s) H(s) k(1/ s (s − 1) (s + 2) (s + 3) )
(D) G(s) H(s) = k ((s + 1)/s (s + 2) (s + 3))

Ans) For given plot root locus exists from −3 to 3, So there must be odd number of
poles and zeros. There is a double pole at s =− 3
Now poles = 0, − 2, − 3, − 3
zeros =− 1
Thus transfer function G(s)H(s)
G (s) H (s) = k (s + 1)/ s (s + 2) (s + 3) 2

Hence (B) is correct option.

5) A system is defined by its impulse response h(n) = 2n(n − 2). The system is

(A) stable and causal (B) causal but not stable
(C) stable but not causal (D) unstable and non-causal

Ans) Function h(n) =an u (n) stable if a < 1 and Unstable if a H 1 We have Here a h (n) = 2n u (n − 2);Here a = 2 therefore h(n) is unstable and since h(n) = 0 for n < 0 Therefore h(n) will be causal. So h(n) is causal and not stable.

Hence (B) is correct option

6) If the unit step response of a network is (1 − e− αt),then its unit impulse response is

(A) αe− αt
(B) α−1 e− αt

(C) (1 − α−1) eαt

(D) (1 − α) e− αt

Ans) Impulse response =d/ dt (step response)
= d/ dt (1-αe− αt )

= 0 + αe− αt = αe− αt

Hence (A) is correct option.

7) The output Y in the circuit below is always ‘1’ when

(A) two or more of the inputs P, Q, R are ‘0’
(B) two or more of the inputs P, Q, R are ‘1’
(C) any odd number of the inputs P, Q, R is ‘0’
(D) any odd number of the inputs P, Q, R is ‘1’

Ans)

If any two or more inputs are ‘1’ then output y will be 1.

Hence (B) is correct option.

8) In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude | vo/vi | at 10 M rad/s is

In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude | vo/vi | at 10 M rad/s is

(A) maximum (B) minimum

Ans) For the parallel RLC circuit resonance frequency is,

Hence (A) is correct option.

9) Drift current in the semiconductors depends upon

(A) only the electric field

(B) only the carrier concentration gradient

(C) both the electric field and the carrier concentration

(D) both the electric field and the carrier concentration gradient

Ans) Drift current Id = qnμn E

It depends upon Electric field E and carrier concentration n

Hence (C) is correct option.

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

(A) reverse bias region below the breakdown voltage

(B) reverse breakdown region

(D) forward bias constant current mode

Ans) Zener diode operates in reverse breakdown region

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

Hence (B) is correct option.

11) The circuit shown below is driven by a sinusoidal input vi = vp cos (t/RC ) The steady state output vo

(A) (vp/3)cos (t/RC) (B) (vp/3) sin(t/RC)

Ans) The given circuit is shown below

Hence (A) is correct option.

12) Consider a closed surface S surrounding volume V.If rv is the position vector of a point inside S, with nt the unit normal on S, the value of the integral ## 5rt.nv dS is

(A) 3V (B) 5V

Ans)

Consider a closed surface S surrounding volume V.If rv is the position vector of a point inside S, with nt the unit normal on S, the value of the integral ## 5rt.nv dS is

Hence (D) is correct option

13) The modes in a rectangular waveguide are denoted by TEmn/ TMmn 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

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

Ans) TM11 is the lowest order mode of all the TMmn modes.

Hence (A) is correct option.

14) The solution of the differential equation

(A) x = ce−ky

(B) x = kecy

(D) y = ce−kx

Ans)

The solution of the differential equation dy/dx = ky , y (0) = c is

Hence (C) is correct answer

15) 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

P. Power efficient transmission of signals

Q. Most bandwidth efficient transmission of voice signals

R. Simplest receiver structure

S. Bandwidth efficient transmission of signals with Significant dc component

Column-II

1. Conventional AM

2. FM

3. VSB

4. SSB-SC

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

Ans)

In FM the amplitude is constant and power is efficient transmitted. No variation

in power.

There is most bandwidth efficient transmission in SSB- SC. because we transmit

only one side band.

Simple Diode in Non linear region ( Square law ) is used in conventional AM that

is simplest receiver structure.

In VSB dc. component exists.

Hence (B) is correct option.

16) The differential equation 100 d 2 y/ dt 2 - 20(dy/dt) + y = x (t) describes a system with an in put x(t) and an output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform

The differential equation 100 d 2 y/ dt 2 - 20(dy/dt) + y = x (t) describes a system with an in put x(t) and an output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform

Ans)

Hence (A) is correct option.

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

Ans) We have G(jω) = 5 + jω
Here σ = 5. Thus G(jω) is a straight line parallel to jω axis.
Hence (A) is correct option.

18)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

Ans) For an even function Fourier series contains dc term and cosine term (even and odd
harmonics).
Hence (C) is correct option.

19)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

Ans) For the output to be high, both inputs to AND gate should be high.
The D-Flip Flop output is the same, after a delay.
Let initial input be 0; (Consider Option A)
then Q = 1 (For 1st D-Flip Flop). This is given as input to 2nd FF.
Let the second input be 1. Now, considering after 1 time interval; The output of 1st
Flip Flop is 1 and 2nd FF is also 1. Thus Output = 1.
Hence (A) is correct option.

20)The logic function implemented by the circuit below is (ground implies logic 0)

(A) F = AND^P,Qh (B) F = OR ^P,Qh

Ans) Hence (D) is correct option.

F = S1S0I0+ S1S0I1+ S1S0I2+ S1S0I3

I0 = I3 = 0

F = PQ + PQ = XOR(P, Q) ( S1 = P, S0 = Q )

21)The circuit below implements a filter between the input current ii and the output
voltage vo. Assume that the opamp is ideal. The filter implemented is a

(A) low pass filter (B) band pass filter

Ans)The given circuit is shown below :

Hence (D) is correct option.

22)A silicon PN junction is forward biased with a constant current at room temperature.
When the temperature is increased by 10ºC, the forward bias voltage across the
PN junction

(A) increases by 60 mV (B) decreases by 60 mV
(C) increases by 25 mV (D) decreases by 25 mV

Ans) For every 1cC increase in temperature, forward bias voltage across diode decreases
by 2.5 mV. Thus for 10cC increase, there us 25 mV decreases.
Hence (D) is correct option.

23)In the circuit shown below, the Norton equivalent current in amperes with respect
to the terminals P and Q is

(A) 6.4 − j 4.8 (B) 6.56 − j 7.87

Ans) Replacing P − Q by short circuit as shown below we have

Hence (A) is correct option.

24)In the circuit shown below, the value of RL such that the power transferred to RL
is maximum is

(A) 5 Ω (B) 10 Ω

Ans) Power transferred to RL will be maximum when RL is equal to the thevenin
resistance. We determine thevenin resistance by killing all source as follows :

Hence (C) is correct option.

25)The value of the integral

where c is the circle z = 1 is given by

(A) 0 (B) 1/10
(C) 4/5 (D) 1

Ans)

So poles are outside the unit circle.
Hence (A) is correct option.

GATE PAPERS

Friday, 1 February 2013

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