ISRO Scientist Electronics Mock Test - 2 - Electronics and Communication Engineering (ECE) MCQ

Concept:

Let us assume the Fourier transform of x(t) is X(ω)

then as per property of duality,

x(t) X(ω)

X(t) 2π x(-ω)

Calculation:

We know that,

for, a = 1

e^-|t| ....(1)

applying differentiation in frequency property,

x(t) X(ω)

Now,

t x(t) j

.....(2)

applying property of duality in eqn (2) we get,

t = -ω

Concept:
The current leaving a current/voltage source must be equal to current entering the source in order to justify the law of conservation of charge.

∴ For loop 2:
y = y + I

I = 0
Similarly, for Loop 1:
i = i - I
I = 0
∴ The current 'I' though resister 10 Ω will be:
I = 0 A

Concept:

• The given circuit contains both NMOS and PMOS. So it is a CMOS implementation.
• For CMOS implementation if NMOS transistors are in series then PMOS transistors corresponding to their NMOS counterparts will be in parallel.
• Similarly, if NMOS transistors are in parallel then PMOS transistors corresponding to their NMOS counterparts will be in series.
• The output will be the negate of the function implemented by the NMOS transistors.

According to De morgan's theorem:

Calculation:
In the given circuit, signals A, B, and C̅ are in series for NMOS implementation. ∴ The output will be:

Taking the partial derivative on both the sides, we get:

In partial derivative, we take the other variable as a constant.

Concept:
The boundary condition for the tangential magnetic field is given by:

Also, the tangential component of the magnetic field inside a perfect conductor is zero.
Calculation:
Given:

The boundary condition for tangential magnetic field is:

The tangential component of the magnetic field inside a perfect conductor is zero.

i.e.

Concept:
Recombination rate (R) is defined as the ratio of the excess minority carrier concentration to the minority carrier lifetime, i.e.
R = Excess minority carried concentration/Minorty Carrier lifetime\
R = Δ_P/τ_p or Δn/τ_n
Consider a p-type sample with hole concentration p₀ and electron concentration n₀, where
p₀ ≃ N_A
n_o = (ni²)/Na
n_i = Intrinsic carrier concentration
When Illuminated with light, excess electron-hole pairs are generated.

• When light is fallen on the semiconductor, because of the photon energy at the surface of the semiconductor, a large number of covalent bonds will be broken due to the photon energy, and electron-hole pairs will be generated.
• Since excess carriers are generated in pairs, we can write:

Excess electron generated = excess hole generated?
Δn = Δp

• But the percentage increase in minority carriers will be high than the percentage increase in the majority carrier concentration.
• In our case, the minority carrier concentration is electrons, that’s why we have shown it in the diagram. There will be holes generation also but the percentage increase will be less.
• Now, these minority carriers, in our case electrons generated, undergo diffusion from high concentration to lower concentration.
• So for a uniformly doped semiconductor, a diffusion current will flow.
• Also the recombination of these minority carriers will take place because majority charge carriers are too present in the sample.

Recombination Rate R = Δ_P/τ_n or Δn/τ_p
Δn or Δp = Excess minority carrier concentration
τ_n or τ_p = Minority carrier lifetime
In our case: R = Δ_n/τ_n 
R ∝ Δn
∴ The recombination rate is directly proportional to excess minority carrier concentration.

Concept:
An ideal op-amp follows the virtual ground concept, i.e. the positive and negative terminals of the Op-amp remains at the same potential:
V₊ = V_-
Also, the Inputs terminals current of an ideal opamp is zero.
Calculation:
Using KCL at the negative terminal of the Opamp, we can write:
(V₁ - V₂)/R = V₂/ 2R + (V₂ - V₀)/3R
On solving it, we get:
V_out = 11/2 V₂ - 3V₁

Concept:

• Norton’s Theorem states that “Any linear circuit containing several energy sources and resistances can be replaced by a single constant current source in parallel with a single resistor “.
• It is an analytical method used to change a complex circuit into a simple equivalent circuit consisting of a single resistance in parallel with a current source.

Steps to follow for Norton’s Theorem:

• Calculate Norton’s current source by removing the load resistor from the original circuit and replacing it with a short circuit.
• Calculating the current through a shorted wire.
• Calculating the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open).
• Calculating total resistance between the open connection points.
• Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance.
• The load resistor re-attaches between the two open points of the equivalent circuit.

Putting x = ∞ to the given limiting function, we get:

1^∞ is one of the indeterminant forms.

We can modify the given limits as:

The above can be written as:

e^f(x) --(1)

where:

---(2)

The above can be written as:

Putting on the limit of x = ∞, we get:

Applying L-Hospitals rule, we get:

Putting this in equation (1), we can write:

Concept:

A TEM wave traveling in the given direction must satisfy the Poynting theorem, as it describes the magnitude and direction of the flow of energy in electromagnetic waves.

Mathematically, the Poynting vector is the cross-product of the Electric field vector and the magnetic field vector, i.e.

Analysis:

Option 1:

With E = +8 ŷ and H = -4 ẑ

According to Poynting theorem, the direction of propagation will be:

P̅ = 8 ŷ × (-4ẑ)

P̅ = - 32 x̂

Since the direction of propagation given is +x direction, Option 1 is incorrect.

Option 2:

With E = -2ŷ and H = -3ẑ

P̅ = -2ŷ × (-3ẑ)

P̅ = 6 x̂

Since the direction of propagation given is +x direction, Option 2 is correct.

Option 3:

With E = +2ẑ and H = +2ŷ

P̅ = 2ẑ × (2ŷ)

P̅ = - 4 x̂

Since the direction of propagation given is +x direction, Option 3 is incorrect.

Option 4:

With E = -3ŷ and H = +4ẑ

P̅ = -3ŷ × (4ẑ)

P̅ = - 12 x̂

Since the direction of propagation given is +x direction, Option 4 is incorrect.

Concept:

For a given position of Fermi-level, the concentration of electrons and holes are given by:

N­c and Nv are the effective density of states.

We can now calculate the intrinsic Fermi level position.

For the intrinsic semiconductor, the electron and hole concentrations are equal.

∴ we can write:

Taking the natural log of both sides of the above equation, we get:

Since,

---(1)

Calculation:

Given V_T = 26 mV and N_V = 2 N_C

Using Equation (1), we can write:

= 9.01 meV

x[n] = 2 sin(π²n)
Let N be the period of the signal, then
x[n] = x[n + N]
⇒ 2 sin(π²n) = sin (π²n + π²N)
⇒ sin(π²n + 2πk) = sin(π²n + π²N)
Where k is an integer.\
⇒ 2πk = π²N
⇒ N = 2πk = π² = 2k/π
For any integer value of k, we cannot have an integer value of N.
So, the given signal is not periodic.

Closed loop transfer function

Characteristic equation,

(s + 5)³ + k = 0

⇒ s³ + 15s² + 75s + (125 + k) = 0

Characteristic equation for the dominant conjugate pole pair

---(1)

Given that ζ = 0.5

---(2)

By comparing both the equations,

α + ω_n = 15 ⇒ α = 15 - ω_n

⇒ ω_n = 5 rad/sec

⇒ α = 10

= (10) (5)² = 250

Some properties of impulse signal are shown below

• x(n) ⋆ δ(n) = x(n)
• x(n) ⋆ δ(n-a) = x(n-a)
• x(n-a) ⋆ δ(n-b) = x(n -a -b)
• x(n)δ(n-a) = x(a)δ(a)
• δ(an) = δ(n)

For a causal LTI system
h(n) = 0 ; n < 0
Calculation
If x(n) is the input to causal LTI system h(n) ,then the output is y(n)
y(n) = x(n) ⋆ h(n)
Calculation

h(n) = δ(n) + 2δ(n-1) + δ(n-2) + 3δ(n-3)
x(n) ⋆ h(n) = x(n) ⋆ [δ(n) + 2δ(n-1) + δ(n-2) + 3δ(n-3)]
y(n) = x(n) + 2x(n-1) + x(n-2) + 3x(n-3)
Hence the correct answer is option 2.

Concept:
The transfer function of the standard second-order system is:

ζ is the damping ratio
ωn is the natural frequency
Characteristic equation:


Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.

for a 5% tolerance band.
for 2% tolerance band
Calculation:

By comparing the above transfer function with the standard second-order system,
ωn² = 25 ⇒ ω_n = 5
2ζωn = 5 ⇒ ζ = 0.5
As the damping ratio is less than unity, the system is underdamped.
Settling time,

The denominator of any closed loop transfer function is ;
1 + G(s)H(s) = 0.
For a second order system;

where ζ is damping factor and ω_n is natural frequency of oscillations.
Note:
For 2% tolerance band:

For 5% tolerance band:

Concept:
Norton's Theorem:
In any linear, bidirectional circuit having more than one independent source, having more the active and passive element it can be replaced by a single equivalent current source IN in parallel with an equivalent resistance RN.

Where
IN = Norton or short circuit current
RN = Norton's resistance
Procedure in order to find Norton’s equivalent circuit, when only the sources of independent type are present.

• Consider the circuit diagram by opening the terminals with respect to which, Norton’s equivalent circuit is to be found.
• Find Norton’s current IN by shorting the two opened terminals of the circuit.
• Find Norton’s resistance RN across the open terminals of the circuit, eliminating the independent sources present in it.
• Norton’s resistance RN will be the same as that of Thevenin’s resistance RTh.
• Draw Norton’s equivalent circuit by connecting a Norton’s current IN in parallel with Norton’s resistance RN.

Explanation:
Given circuit is

To find Norton current through terminal ab, ab terminal is short circuited, so no current will flow through the 5 Ω resistor.

Now the circuit will look like


16 × 6 = 4(V_x - 16) + V_x
5 V_x = 10 × 16
V_x = 32 V
I_N = V_x / 16 = 32 / 16 = 2 A
To find Norton's resistance, source should be replaced with internal resistance, so

• Current source is open circuited.
• Voltage source is short circuited.

The circuit will become

R_N = 5 || (8 + 4+ 8) = 5 || 20 = (5 × 20) / (5 + 20) = 4 Ω
So the Norton equivalent circuit is

Given

This can be written as

Taking the inverse Laplace Transform of the above expression, we get:

x(t) = δ(t) + 2.e-3t u(t)

At t = 1 sec

x(1) = 0 + 2.e^-3 (∵ δ(t) = 0 for t ≠ 0)

Putting e^-3 = 0.05:

⇒ x(1) = 2 × 0.05 = 0.1

Concept:

Initial value theorem:

The initial value theorem is one of the basic properties of the Laplace transform used to find the response of the system at the initial state (t = 0) in the Laplace domain. Mathematically it is given by

Where

f(t) is system function

F(s) is Laplace transform of system function f(t)

f(0+) is the initial value of the system

NOTE:

• For the final value theorem to be applicable system should be stable in a steady-state and for that real part of the poles should lie on the left side of the s plane.
• In the given problem exactly the final value theorem is not applied but just X(0+) is calculated.
  

Calculation:

Given that,

= 3

Given that N₁ : N₂ = 4 : 1


But here

Concept:
In an N-type semiconductor, the doping done is of pentavalent atoms.
As the doping density increases the Fermi level moves towards the conduction band.
If ND is the doping density, then the upward shift in the Fermi level is given by

From the expression
As ND increases N_D/n_i increases and the shift increases.
Calculation:
The electrostatic potential of n type semiconductor


E_F - E_i = qν_n
E_F - E_i = 0.359 eV

Case – I:
Let us assume initially that both Zener and diode are off
i.e. V₀ = V_i and -0.7 < V­_I < 0.7
Case – II:
When V_i ≥ 0.7, the diode D₂ is on but Zener is still off


Case III:
For the Zener to be in breakdown,
V₀ = 2.7 V
To find the input at which Zener breaks down:
2.7 = 0.5 V_i + 0.35
V_i = 4.7 V
Case – IV
For V_i < 0.7 V, the Zener acts as a normal diode and the diode D₂ is off
V₀ = -0.7 V

∴ option 3 is correct answer.

• Microwave tubes used in satellite communication systems are mainly TWTs (Traveling Wave Tubes) and Klystrons.
• These are vacuum devices that amplify microwaves by converting the kinetic energy of an electron beam into microwave energy.
• In TWT, electrons emitted from the cathode are accelerated by a potential difference between the cathode and the anode forming an electron beam to be injected into the slow-wave circuit where it interacts with the propagating RF wave.
• The role of the RF attenuator placed in the center portion of the slow-wave circuit is to prevent feedback oscillation in the TWT.
• After passing over the slow-wave circuit, the electron beam reaches the collector, and the electron energy is converted to heat and dissipated.

Concept:
Bisection Method: This method is based on the intermediate value theorem, which states that if a function f(x) is continuous between ‘a and b’ and f(a) and f(b) are at opposite signs, then there exists at least one root between a and b for definiteness, i.e.
Let f(a) be negative and f(b) be positive.
Then the root will lie between a and b.

Application:
Given f(x) = x³ - 3x² + 1
f(-1) = (-1)³ - 3(-1)² + 1 = -3 (Negative)
f(0) = (0)³ - 3(0²) + 1 = 1 (positive)
f(0.5) = (.5)³ - 3(.5)² + 1 = .375 (Positive)
f(1) = (1)³ - 3(1)² + 1 = -3 (Negative)
f(2) = (2)² - 3(2)² + 1 = -3 (Negative)
Hence, Root must lie in between (-1, 0) & (0.5, 1)

Let the current through the load resistor R_L is I_L. We can find the current I_L by using superposition theorem.

When only voltage source is active:

When only current source is active.

Total current, I_L = I_L1 + I_L2

Power transfers through load resistance,


The above expression gets maximum for a minimum value of R.
So, the value of R = 0 Ω

Calculation:
The number of persons who have taken the insurance policy, excluding the year 2012 = 54 + 65 + 72 + 84 + 85
⇒ 360
Average = 360/5
⇒ 72
∴ The required answer is 72.

The position of the English alphabet letters is given below:

Logic: Addition of +3 in the first letter and +4 in the second letter.
Option 1) MPT
⇒ M(13) + 3 → P(16) + 4 → T(20)
Option 2) FIM
⇒ F(6) + 3 → I(9) + 4 → M(13)
Option 3) GJN
⇒ G(7) + 3 → J(10) + 4 → N(14
Option 4) DGI
⇒ D(4) + 3 → G(7) + 2 → I(9)
According to the logic, DGI is the odd one out.
Hence, the correct answer is "DGI".

Given:
The Data of Union Territory Area's
Solution:
The Area of Delhi in percentage = 44%
And The Area of Chandigarh in Percentage = 4%
Difference between Area of Delhi and Chandigarh = (44 - 4) %
⇒ 40%
Percentage bigger than Delhi form Chandigarh = (40 / 4) × 100
⇒ 10 × 100 = 1000 %
Delhi is 1000 % more Bigger than Chandigarh.

The logic followed here is:

Hence, "Option (4)" is the correct answer.Alternate MethodThe logic followed here is:
Given series is 6, 25, 62, 123, 214, ______
2³– 2 → 6
3³ – 2 → 25
4³ – 2 → 62
5³ – 2 → 123
6³ – 2 → 214
7³ – 2 → 341
Therefore, the missing number is '341'.
Hence, "Option (4)" is the correct answer.

The logic which followed here is :
Step I: Letter ' is getting shifted and replaced with letter '.

Step II: Letter ' is getting shifted and replaced with symbol .

Step III: Symbol '' is getting shifted and replaced with Symbol ' and letter '

So, the series of image will be:

Hence, the correct answer is "option 4".

Given:

Average age of A and B = 48 years

Average age of A, B and C = 62 years

Formula Used:

Average = Sum of all observations/total number of observations

Calculation:

Sum of ages of A and B,

(A + B) = Average × 2

⇒ 48 × 2 = 96 years

Sum of ages of A, B and C,

A + B + C = Average × 3

⇒ 62 × 3 = 186 years

Age of C = ( A + B + C) - ( A - B) = 186 - 96 = 90 years

∴ Age of professor C is 90 years