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Electronics And Communication - ECE 2012 GATE Paper (Practice Test)


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65 Questions MCQ Test GATE Past Year Papers for Practice (All Branches) | Electronics And Communication - ECE 2012 GATE Paper (Practice Test)

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) for Electronics and Communication Engineering (ECE) 2023 is part of GATE Past Year Papers for Practice (All Branches) preparation. The Electronics And Communication - ECE 2012 GATE Paper (Practice Test) questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Electronics And Communication - ECE 2012 GATE Paper (Practice Test) MCQs are made for Electronics and Communication Engineering (ECE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) below.
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Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 1

Q. 1 - Q.  5 carry one mark each.

Q.

If (1.001)1259 = 3.52 and (1.001)2062 = 7.85, then (1.001)3321

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 1

Option (D) is correct option.
Let 1.001 = x
So in given data :
x1259 = 3.52
x2062 = 7.85
Again x3321 = x1259+2062
= x1259x2062
= 3.52 x 7.85
= 27.64

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 2

Choose the most appropriate alternate from the options given below to complete
the following sentence :
If the tired soldier wanted to lie down, he..................the mattress out on the
balcony.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 3

Choose the most appropriate word from the options given below to complete the
following sentence :
Give the seriousness of the situation that he had to face, his........was impressive.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 4

Which one of the following options is the closest in meaning to the word given
below ?
Latitude

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 5

One of the parts (A, B, C, D) in the sentence given below contains an ERROR.
Which one of the following is INCORRECT ?
I requested that he should be given the driving test today instead of tomorrow.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 6

Q. 6 - Q. 10 carry two marks each

Q.

One of the legacies of the Roman legions was discipline. In the legious, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them.


Which one of the following statements best sums up the meaning of the above
passage ?

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 7

Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 7

Let no. of notes of Rs.20 be x and no. of notes of Rs. 10 be y .
Then from the given data.
x + y = 14
20x + 10y = 230
Solving the above two equations we get
x = 9, y = 5
So, the no. of notes of Rs. 10 is 5.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 8

There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 8

We will categorize the 8 bags in three groups as :
(i) A1A2A3 (ii) B1B2B3 (iii) C1C<2
Weighting will be done as bellow :
1st weighting → A1A2A3 will be on one side of balance and B1B2B3 on the other. It
may have three results as described in the following cases.
Case 1 : A1A2A3 = B1B2B3

This results out that either C1 or C2 will heavier for which we will have to perform
weighting again.
2nd weighting " C1 is kept on the one side and C2 on the other.
if C1 > C2 then C1 is heavier.
C1 < C2 then C2 is heavier.

Case 2 : A1A2A3> B1B2B3it means one of the A1A2A3 will be heavier So we will perform next weighting as:
2nd weighting → A1 is kept on one side of the balance and A2 on the other.
if A1 = A2 it means A3 will be heavier
A1 > A2 then A1 will be heavier
A1 < A2 then A2 will be heavier

Case 3 : A1A2A3 < B1B2B3
This time one of the B1B2B3 will be heavier, So again as the above case weighting
will be done.
2nd weighting " B1 is kept one side and B2 on the other
if B, = B2 B3 will be heavier
B1 > B2        B1 will be heavier
B1 < B2        B2 will be heavier
So, as described above, in all the three cases weighting is done only two times to
give out the result so minimum no. of weighting required = 2.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 9

The data given in the following table summarizes the monthly budget of an average
household.

The approximate percentages of the monthly budget NOT spent on savings is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 9

Total budget = 4000 + 1200 + 2000 + 1500 + 1800
= 10,500
The amount spent on saving = 1500
So, the amount not spent on saving
= 10,500 − 1500 = 9000
So, percentage of the amount
= (9000/10500) x100%
= 86%

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 10

A and B are friends. They decide to meet between 1 PM and 2 PM on a given day.
There is a conditions that whoever arrives first will not wait for the other for more
than 15 minutes. The probability that they will meet on that days is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 10

The graphical representation of their arriving time so that they met is given as
below in the figure by shaded region.

 

So, the area of shaded region is given by

So, the required probability = 1575/3600 =7/16

 

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 11

Q. 11 - Q. 35 carry one mark each

Q.

The current ib through the base of a silicon npn transistor is 1 + 0.1 cos (10000 π t)mA
At 300 K, the rπ in the small signal model of the transistor is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 11

Option (C) is correct.
Given ib = 1 + 0.1 cos (1000πt)mA
So, IB = DC component of ib
= 1mA
In small signal model of the transistor

                              VT → Thermal voltage

                                   VT= 25mV, IB, = 1mA

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 12

The power spectral density of a real process X(t) for positive frequencies is shown
below. The values of E[X2(t)] and |E[X(t)]| , respectively, are

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 12

The mean square value of a stationary process equals the total area under the
graph of power spectral density, that is

= 1/π[area under the triangle]+integration of delta function ]

E[X(t)] is the absolute value of mean of signal X(t) which is also equal to value
of X(ω) at (ω = 0).
From given PSD

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 13

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

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 13

For raised cosine spectrum transmission bandwidth is given as

BT = W(1 + α)              α → Roll of factor
BT = Rb/2 (1+α )        Rb → Maximum signaling rate

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 14

A plane wave propagating in air with  is incident on a perfectly conducting slab positioned at x 0. The E field of the reflected wave is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 14

Electric field of the propagating wave in free space is given as

So, it is clear that wave is propagating in the direction (− 3ax + 4ay).
Since, the wave is incident on a perfectly conducting slab at x = 0. So, the reflection
coefficient will be equal to −1.

i.e

Again, the reflected wave will be as shown in figure.

i.e. the reflected wave will be in direction 3ax + 4ay . Thus, the electric field of the
reflected wave will be.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 15

The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10(ay+jaz)e-j25x. The frequency and polarization of the wave, respectively, are

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 15

The field in circular polarization is found to be

propagating in +ve x -direction.
where, plus sign is used for left circular polarization and minus sign for right
circular polarization. So, the given problem has left circular polarization.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 16

Consider the given circuit

In this circuit, the race around

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 16

The given circuit is

Condition for the race-around
It occurs when the output of the circuit (Y1,Y2) oscillates between ‘0’ and ‘1’
checking it from the options.
1. Option (A): When CLK = 0

it won’t oscillate for CLK = 0.
So, here race around doesn’t occur for the condition CLK = 0.
2. Option (C): When CLK = 1, A = B = 1
A1 = B1 = 0 and so Y1 = Y2 = 1
And it will remain same for the clock period. So race around doesn’t occur for the
condition.
3. Option (D): When CLK = 1, A = B = 0
So, A1 = B1 = 1
And again as described for Option (B) race around doesn’t occur for the condition.
So, Option (A) will be correct.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 17

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 17

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 18

The i -v characteristics of the diode in the circuit given below are

The current in the circuit is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 18

Let v > 0.7 V and diode is forward biased. By applying Kirchoff’s voltage law

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 19

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

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 19

The s-domain equivalent circuit is shown as below.

                                              vC (0) = 12 V

Taking inverse Laplace transform for the current in time domain,
i (t) = 12Ceq δ(t)                             (Impulse)

                                             

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 20

The average power delivered to an impedance (4 − j3)Ω by a current 5 cos (100 π t + 100)A is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 20

In phasor form

Alternate method:

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 21

The unilateral Laplace transform of f (t) is  The unilateral Laplace transform of tf (t) is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 21

Using s -domain differentiation property of Laplace transform.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 22

With initial condition x(1) = 0.5, the solution of the differential equation

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 22

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 23

The diodes and capacitors in the circuit shown are ideal. The voltage v(t) across the diode D1 is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 23

Option (A) is correct.
The circuit composed of a clamper and a peak rectifier as shown.

Clamper clamps the voltage to zero voltage, as shown

The peak rectifier adds +1 V to peak voltage, so overall peak voltage lowers down
by −1 volt.
So, vo = cosωt − 1

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 24

In the circuit shown

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 24

Parallel connection of MOS ⇒ OR operation
Series connection of MOS ⇒ AND operation
The pull-up network acts as an inverter. From pull down network we write

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 25

A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount ε and decreases that of the second by ε. After encoding, the entropy of the source

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 25

Entropy function of a discrete memory less system is given as

where Pk is probability of symbol Sk .
For first two symbols probability is same, so

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 26

A coaxial-cable with an inner diameter of 1mm and outer diameter of 2.4 mm is
filled with a dielectric of relative permittivity 10.89. Given μ0 = 4πx 10-7 H/m,

the characteristic impedance of the cable is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 26

Option (B) is correct.
Characteristic impedance.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 27

The radiation pattern of an antenna in spherical co-ordinates is given by

F(θ) = cos4θ ; 0 θ π/2

The directivity of the antenna is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 27

Option (B) is correct.
Characteristic impedance.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 28

If then the region of convergence (ROC) of its z
-transform in the z -plane will be

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 28

.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 29

In the sum of products function  the prime implicants are

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 29

Prime implicants are the terms that we get by solving K-map

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 30

A system with transfer function

is excited by sin(ωt). The steady-state output of the system is zero at

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 30

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 31

The impedance looking into nodes 1 and 2 in the given circuit is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 31

We put a test source between terminal 1, 2 to obtain equivalent impedance

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 32

In the circuit shown below, the current through the inductor is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 32

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 33

Given 

If C is a counter clockwise path in the z -plane such that |z + 1| = 1, the value of

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 33

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 34

Two independent random variables X and Y are uniformly distributed in the
interval [−1,1]. The probability that max[X,Y] is less than 1/2 is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 34

Probability density function of uniformly distributed variables X and Y is shown
as

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 35

If  then the value of xx is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 35

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 36

Q. 36 to Q. 65 carry two marks each.

Q.

The source of a silicon (ni = 1010 per cm3 )  n-channel MOS transistor has an area of 1 sq μm and a depth of 1 μm. If the dopant density in the source is 1019/cm3, the number of holes in the source region with the above volume is approximately

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 36

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 37

A BPSK scheme operating over an AWGN channel with noise power spectral density
of N0/2, uses equiprobable signals 

over the symbol interval (0,T). If the local oscillator in a coherent receiver is ahead in phase by 45c with respect to the received signal, the probability of error in the resulting system is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 37

In a coherent binary PSK system, the pair of signals s1(t) and s2(t) used to represent
binary system 1 and 0 respectively.

Now here the two message points are s11 and s21.
The error at the receiver will be considered.
When : (i) s11 is transmitted and s21 received
(ii) s21 is transmitted and s11 received

So, probability for the 1st case will be as :

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 38

A transmission line with a characteristic impedance of 100 Ω is used to match a
50 Ω section to a 200 Ω section. If the matching is to be done both at 429MHz and
1GHz, the length of the transmission line can be approximately

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 38

Option (C) is correct.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 39

The input x(t) and output y(t) of a system are related as . The system is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 39

Time invariance :

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 40

The feedback system shown below oscillates at 2 rad/s when

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 40

 

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 41

The Fourier transform of a signal h(t) is H(jω) = (2 cosω) (sin2ω)/ω. The value of
h(0) is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 41

 

We know that inverse Fourier transform of sin c function is a rectangular function.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 42

The state variable description of an LTI system is given by

where y is the output and u is the input. The system is controllable for

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 42

For controllability it is necessary that following matrix has a tank of n = 3.

a3 may be zero or not.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 43

Assuming both the voltage sources are in phase, the value of R for which maximum
power is transferred from circuit A to circuit B is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 43

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 44

Consider the differential equation

The numerical value of  is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 44

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 45

The direction of vector A is radially outward from the origin, with  where r2 = x2+ y2+ z2 and k is a constant. The value of n for which  is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 45

Option (A) is correct.
Divergence of A in spherical coordinates is given as

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 46

A fair coin is tossed till a head appears for the first time. The probability that the
number of required tosses is odd, is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 46

Probability of appearing a head is 1/2. If the number of required tosses is odd, we
have following sequence of events.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 47

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2
are equally sized. The device M1 is in the linear region if

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 47

Given the circuit as below :

Since all the parameters of PMOS and NMOS are equal.

Given that M1 is in linear region. So, we assume that M2 is either in cutoff or
saturation.

 

 

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 48

The signal m(t) as shown is applied to both a phase modulator (with kp as the
phase constant) and a frequency modulator (with kf as the frequency constant)
having the same carrier frequency

The ratio kp/kf (in rad/Hz) for the same maximum phase deviation is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 48

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 49

The magnetic field among the propagation direction inside a rectangular waveguide
with the cross-section shown in the figure is

Hz= 3 cos (2.094 x 10 x)cos (2.618 x102y)cos (6.283 x1010t -βz) 

The phase velocity vp of the wave inside the waveguide satisfies

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 49

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 50

The circuit shown is a

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 50

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 51

Let y[n] denote the convolution of h[n] and g[n], where h[n] = (1/2)nu[n] and g[n]
is a causal sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 51

 

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 52

The state transition diagram for the logic circuit shown is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 52

Let Qn+1 is next state and Qn is the present state. From the given below figure.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 53

The voltage gain Av of the circuit shown below is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 53

DC Analysis :

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 54

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 54

In the given circuit
VA− VB = 6 V
So current in the branch will be
IAB = 6/2 = 3 A

We can see, that the circuit is a one port circuit looking from terminal BD as
shown below

For a one port network current entering one terminal, equals the current leaving
the second terminal. Thus the outgoing current from A to B will be equal to the
incoming current from D to C as shown

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 55

The maximum value of f (x) = x3− 9x2+ 24x + 5 in the interval [1,6] is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 55

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 56

Given that

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 56

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 57

With 10 V dc connected at port A in the linear nonreciprocal two-port network
shown below, the following were observed :

(i) 1Ω connected at port B draws a current of 3A
(ii) 2.5 Ω connected at port B draws a current of 2A

Q.

With 10 V dc connected at port A, the current drawn by 7 Ω connected at port B
is

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 58

With 10 V dc connected at port A in the linear nonreciprocal two-port network
shown below, the following were observed :

Q.

For the same network, with 6 V dc connected at port A, 1Ω connected at port B
draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 59

In the three dimensional view of a silicon n-channel MOS transistor shown below, δ = 20 nm. The transistor is of width 1 μm. The depletion width formed at every p-n junction is 10 nm. The relative permittivity of Si and SiO2, respectively, are 11.7 and 3.9, and ε0 = 8.9 x 10-12 F/m

Q.

The gate source overlap capacitance is approximately

 

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 59

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 60

In the three dimensional view of a silicon n-channel MOS transistor shown below, δ = 20 nm. The transistor is of width 1 μm. The depletion width formed at every p-n junction is 10 nm. The relative permittivity of Si and SiO2, respectively, are 11.7 and 3.9, and ε0 = 8.9 x 10-12 F/m

Q.

The source-body junction capacitance is approximately

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 60

Source body junction capacitance.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 61

Statement for Linked Answer Question 61 and 62

 An infinitely long uniform solid wire of radius a carries a uniform dc current of
density J

Q.

The magnetic field at a distance r from the center of the wire is proportional to

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 61

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 62

An infinitely long uniform solid wire of radius a carries a uniform dc current of
density J

Q.

A hole of radius b(b < a) is now drilled along the length of the wire at a distance
d from the center of the wire as shown below.

 

The magnetic field inside the hole is

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 62

Assuming the cross section of the wire on x -y plane as shown in figure.

Since, the hole is drilled along the length of wire. So, it can be assumed that the
drilled portion carriers current density of −J .
Now, for the wire without hole, magnetic field intensity at point P will be given as

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 63

A binary symmetric channel (BSC) has a transition probability of 1/8. If the
binary symbol X is such that P(X = 0) = 9/10, then the probability of error for
an optimum receiver will be

Correct answer will be   uploaded soon  

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 64

Statement for Linked Answer Question 64 and 65

The transfer function of a compensator is given as

Q.

Gc (s) is a lead compensator if

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 64

Note: For phase lead compensator zero is nearer to the origin as compared to pole,
so option (C) can not be true.

Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 65

The transfer function of a compensator is given as

Q.

The phase of the above lead compensator is maximum at

Detailed Solution for Electronics And Communication - ECE 2012 GATE Paper (Practice Test) - Question 65

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