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}
Option (D) is correct option.
Let 1.001 = x
So in given data :
x^{1259} = 3.52
x^{2062} = 7.85
Again x^{3321} = x^{1259+2062}
= x^{1259x2062}
= 3.52 x 7.85
= 27.64
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.
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.
Which one of the following options is the closest in meaning to the word given
below ?
Latitude
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.
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 ?
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
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.
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
We will categorize the 8 bags in three groups as :
(i) A_{1}A_{2}A_{3} (ii) B_{1}B_{2}B_{3} (iii) C_{1}C<2
Weighting will be done as bellow :
1st weighting → A_{1}A_{2}A_{3} will be on one side of balance and B_{1}B_{2}B_{3} on the other. It
may have three results as described in the following cases.
Case 1 : A_{1}A_{2}A_{3} = B_{1}B_{2}B_{3}
This results out that either C_{1} or C_{2} will heavier for which we will have to perform
weighting again.
2^{nd} weighting " C_{1} is kept on the one side and C_{2} on the other.
if C_{1} > C_{2} then C_{1} is heavier.
C_{1} < C_{2} then C_{2} is heavier.
Case 2 : A_{1}A_{2}A_{3}> B_{1}B_{2}B_{3}it means one of the A_{1}A_{2}A_{3} will be heavier So we will perform next weighting as:
2nd weighting → A_{1} is kept on one side of the balance and A_{2} on the other.
if A_{1} = A_{2} it means A_{3} will be heavier
A_{1} > A_{2} then A_{1} will be heavier
A_{1} < A_{2} then A_{2} will be heavier
Case 3 : A_{1}A_{2}A_{3} < B_{1}B_{2}B_{3}
This time one of the B_{1}B_{2}B_{3} 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, = B_{2} B_{3} will be heavier
B_{1} > B2 B_{1} will be heavier
B_{1} < B_{2} B_{2} 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.
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
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%
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
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
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 r_{π} in the small signal model of the transistor is
Option (C) is correct.
Given i_{b} = 1 + 0.1 cos (1000πt)mA
So, I_{B} = DC component of ib
= 1mA
In small signal model of the transistor
VT → Thermal voltage
V_{T}= 25mV, I_{B, =} 1mA
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
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
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 intersymbol
interference, the maximum possible signaling rate in symbols per second is
For raised cosine spectrum transmission bandwidth is given as
BT = W(1 + α) α → Roll of factor
BT = R_{b/2} (1+α ) Rb → Maximum signaling rate
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
Electric field of the propagating wave in free space is given as
So, it is clear that wave is propagating in the direction (− 3a_{x} + 4a_{y}).
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.
The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10(a^{y}+ja^{z})e^{j25x. }The frequency and polarization of the wave, respectively, are
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.
Consider the given circuit
In this circuit, the race around
The given circuit is
Condition for the racearound
It occurs when the output of the circuit (Y_{1},Y_{2}) 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
A_{1} = B_{1} = 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, A_{1} = B_{1} = 1
And again as described for Option (B) race around doesn’t occur for the condition.
So, Option (A) will be correct.
The output Y of a 2bit comparator is logic 1 whenever the 2bit input A is greater than the 2bit input B. The number of combinations for which the output is logic 1, is
The i v characteristics of the diode in the circuit given below are
The current in the circuit is
Let v > 0.7 V and diode is forward biased. By applying Kirchoff’s voltage law
In the following figure, C_{1} and C_{2} are ideal capacitors. C_{1} has been charged to 12 V
before the ideal switch S is closed at t = 0. The current i (t) for all t is
The sdomain equivalent circuit is shown as below.
v_{C} (0) = 12 V
Taking inverse Laplace transform for the current in time domain,
i (t) = 12C_{eq} δ(t) (Impulse)
The average power delivered to an impedance (4 − j3)Ω by a current 5 cos (100 π t + 100)A is
In phasor form
Alternate method:
The unilateral Laplace transform of f (t) is^{ } The unilateral Laplace transform of tf (t) is
Using s domain differentiation property of Laplace transform.
With initial condition x(1) = 0.5, the solution of the differential equation
The diodes and capacitors in the circuit shown are ideal. The voltage v(t) across the diode D_{1} is
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
In the circuit shown
Parallel connection of MOS ⇒ OR operation
Series connection of MOS ⇒ AND operation
The pullup network acts as an inverter. From pull down network we write
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
Entropy function of a discrete memory less system is given as
where P_{k} is probability of symbol S_{k} .
For first two symbols probability is same, so
A coaxialcable 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
Option (B) is correct.
Characteristic impedance.
The radiation pattern of an antenna in spherical coordinates is given by
F(θ) = cos^{4}θ ; 0 θ π/2
The directivity of the antenna is
Option (B) is correct.
Characteristic impedance.
If then the region of convergence (ROC) of its z
transform in the z plane will be
.
In the sum of products function the prime implicants are
Prime implicants are the terms that we get by solving Kmap
A system with transfer function
is excited by sin(ωt). The steadystate output of the system is zero at
The impedance looking into nodes 1 and 2 in the given circuit is
We put a test source between terminal 1, 2 to obtain equivalent impedance
In the circuit shown below, the current through the inductor is
Given
If C is a counter clockwise path in the z plane such that z + 1 = 1, the value of
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
Probability density function of uniformly distributed variables X and Y is shown
as
If then the value of x^{x} is
Q. 36 to Q. 65 carry two marks each.
Q.
The source of a silicon (n_{i }= 10^{10} per cm^{3} ) nchannel MOS transistor has an area of 1 sq μm and a depth of 1 μm. If the dopant density in the source is 10^{19}/cm^{3}, the number of holes in the source region with the above volume is approximately
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
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 s_{11} and s_{21}.
The error at the receiver will be considered.
When : (i) s_{11} is transmitted and s_{21} received
(ii) s_{21} is transmitted and s_{11} received
So, probability for the 1st case will be as :
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
Option (C) is correct.
The input x(t) and output y(t) of a system are related as . The system is
Time invariance :
The feedback system shown below oscillates at 2 rad/s when
The Fourier transform of a signal h(t) is H(jω) = (2 cosω) (sin2ω)/ω. The value of
h(0) is
We know that inverse Fourier transform of sin c function is a rectangular function.
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
For controllability it is necessary that following matrix has a tank of n = 3.
a3 may be zero or not.
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
Consider the differential equation
The numerical value of is
The direction of vector A is radially outward from the origin, with where r^{2} = x^{2}+ y^{2}+ z^{2} and k is a constant. The value of n for which is
Option (A) is correct.
Divergence of A in spherical coordinates is given as
A fair coin is tossed till a head appears for the first time. The probability that the
number of required tosses is odd, is
Probability of appearing a head is 1/2. If the number of required tosses is odd, we
have following sequence of events.
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
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.
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 k_{p}/k_{f} (in rad/Hz) for the same maximum phase deviation is
The magnetic field among the propagation direction inside a rectangular waveguide
with the crosssection shown in the figure is
H_{z}= 3 cos (2.094 x 10^{2 } x)cos (2.618 x10^{2}y)cos (6.283 x10^{10}t βz)
The phase velocity v_{p} of the wave inside the waveguide satisfies
The circuit shown is a
Let y[n] denote the convolution of h[n] and g[n], where h[n] = (1/2)^{n}u[n] and g[n]
is a causal sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals
The state transition diagram for the logic circuit shown is
Let Q_{n+1} is next state and Q_{n} is the present state. From the given below figure.
The voltage gain Av of the circuit shown below is
DC Analysis :
In the given circuit
V_{A}− V_{B} = 6 V
So current in the branch will be
I_{AB =} 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
The maximum value of f (x) = x^{3}− 9x^{2}+ 24x + 5 in the interval [1,6] is
Given that
With 10 V dc connected at port A in the linear nonreciprocal twoport 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
With 10 V dc connected at port A in the linear nonreciprocal twoport 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
In the three dimensional view of a silicon nchannel MOS transistor shown below, δ = 20 nm. The transistor is of width 1 μm. The depletion width formed at every pn junction is 10 nm. The relative permittivity of Si and SiO_{2}, respectively, are 11.7 and 3.9, and ε_{0 = }8.9 x 10^{12 }F/m
Q.
The gate source overlap capacitance is approximately
In the three dimensional view of a silicon nchannel MOS transistor shown below, δ = 20 nm. The transistor is of width 1 μm. The depletion width formed at every pn junction is 10 nm. The relative permittivity of Si and SiO_{2}, respectively, are 11.7 and 3.9, and ε_{0 = }8.9 x 10^{12 }F/m
Q.
The sourcebody junction capacitance is approximately
Source body junction capacitance.
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
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
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
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
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
Note: For phase lead compensator zero is nearer to the origin as compared to pole,
so option (C) can not be true.
The transfer function of a compensator is given as
Q.
The phase of the above lead compensator is maximum at
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