EC GATE 2015 PAPER 01 - Online Test
Q1. 
For the NMOSFET in the circuit shown, the threshold voltage is Vth, where Vth > 0. The source voltage VSS is varied from 0 to VDD. Neglecting the channel length modulation, the drain current ID as a function VSS is represented by
Answer : Option A
Explaination / Solution:
We have the following conditions
Since it is in saturation, current ID is given by
Thus, ID - VSS graph shows Parabolic relation for Vss < VDD
- Vth and zero for Vss > VDD
- Vth. Only graph shown in option (A) satisfies this result.
Q2. If 
then the value of x is
then the value of x is
Answer : Option A
Explaination / Solution:
Q3. Operators 
and ⟶ are defined by: 
Find the value 
and ⟶ are defined by: Find the value
Answer : Option C
Explaination / Solution:
Q4. A cube of side 3 units is formed using a set of smaller cubes of side 1 unit. Find the proportion of the number of faces of the smaller cubes visible to those which are NOT visible.
Answer : Option C
Explaination / Solution:
Consider the cube of side 1 unit shown below.
Now, a set of similar cubes as shown above is joint to form a cube of side 3 units, we have to determine the ratio of faces of smaller cubes visible to those which are not visible . Complete surface area of the cube of side 3 units determined as
A = 6 × (3)2
=54
Again, the area of one face of cube of side 1 unit is
A1 = (1)2 = 1
So, the total number of visible faces is
Nvisible = (A/A1)= 54
Again, total number of smaller cubes required to form the big cube is
Number of smaller cubes = volume of cube of side unit/volume of cube of side unit
= (3)3/(1)3
= 27
So, the total number of faces of smaller cubes is given as
Ntotal = 6 × (number of smaller cubes)
= 6 × 27 = 162
Therefore number of invisible faces is
Ninvisible = Ntotal - Nvisible
= 162 - 54 = 108
Hence, the desired ratio is
(Nvisible /Ninvisible) = 54/108 = 1/2
Q5. Consider the signal 
where 
denotes the Hilbert transform of m(t) and the bandwidth of m(t) is very small compared to fc. The signal s(t) is a
where denotes the Hilbert transform of m(t) and the bandwidth of m(t) is very small compared to fc. The signal s(t) is a
Answer : Option C
Explaination / Solution:
We have the signal,
Here, s(t) represents SSB - Lower side band, and thus a band pass signal.
Q6. A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size Δ of the delta modulator are 20,000 samples per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is
Answer : Option A
Explaination / Solution:
For preventing slope overload, we should have
(slope of m(t) ≤ (slope of sampling)
For sinusoidal signal, if
Q7.
The longitudinal component of the magnetic field inside an air-filled rectangular waveguide made of a perfect electric conductor is given by the following expression 
The cross-sectional dimensions of the waveguide are given as a = 0.08 m and b = 0.033 m. The mode of propagation inside the waveguide is
The cross-sectional dimensions of the waveguide are given as a = 0.08 m and b = 0.033 m. The mode of propagation inside the waveguide is
Answer : Option C
Explaination / Solution:
We have the expression,
and the cross sectional dimension is
a = 0.08 m, b = 0.033m
Now, we compare the equation for TEm,n mode whose Hz is given as
Hence, given mode is TE21 mode
Q8. In an 8085 microprocessor, the shift registers which store the result of an addition and the overflow bit are, respectively
Answer : Option B
Explaination / Solution:
The shift registers A and F store the result of an addition and the overflow bit.
Q9. A 3-input majority gate is defined by the logic function M(a,b,c) = ab + bc + ca. Which one of the following gates is represented by the function 
Answer : Option B
Explaination / Solution:
3 input majority gate is given as
M(a,b,c) = ab + bc + ca
We have to obtain 
We obtain truth table for the function as
So, the function is odd number of 1’s detector. This function represent the 3-input XOR gate.
Q10. The Boolean expression 
converted into the canonical product of sum (POS) form is
converted into the canonical product of sum (POS) form is
Answer : Option A
Explaination / Solution:
We have the SOP Boolean form,
Hence, in POS form, we have
