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

Option (D) is correct option.
Let 1.001 = x
So in given data :
x¹²⁵⁹ = 3.52
x²⁰⁶² = 7.85
Again x³³²¹ = x^1259+2062
= x^1259x2062
= 3.52 x 7.85
= 27.64

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.

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.
2^nd 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.

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%

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

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 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

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

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 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.

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.

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

The s-domain 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)

In phasor form

Alternate method:

Using s -domain differentiation property of Laplace transform.

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

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

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

Option (B) is correct.
Characteristic impedance.

Option (B) is correct.
Characteristic impedance.

.

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