ISRO Scientist Electronics Mock Test - 1 - Electronics and Communication Engineering (ECE) MCQ

Concept:
The magnetic field intensity (H) of a circular coil is given by
H  = I/2R
Where I is the current flow through the coil
R is the radius of the circular coil

Calculation:
Given that, Current (I) = 2 A
Diameter = 1 m
Radius (R) = 0.5 m

Magnetic field intensity

Concept:
The sampling theorem states that “a signal can be exactly reproduced if it is sampled at the rate f_s which is greater than twice the maximum frequency W in the modulating signal."

If the sampled frequency is less than the Nyquist frequency, overlapping of lower and upper sidebands known as aliasing takes place.

The main reason for aliasing is under-sampling, i.e. sampling the signal at a frequency less than twice the Nyquist rate.
f_s < 2f_m
f_s = Sampling frequency
f_m = Modulating frequency

Application:
The minimum sampling rate to avoid aliasing will be:
f_s = 2 × 2f_m = 2 × 4 kHz
f_s = 8 kHz
∴ For 6 kHz, the Aliasing effect will be seen in the reconstructed signal as the sampling frequency is less than the Nyquist Rate.

Important Point

Aliasing is explained with the help of the spectrum as shown:

Concept:
If f(t) ↔ F(s)
Then, f(t - t₀) ↔ F(s).e^-sto

Calculation:
The given signal can be represented as the multiplication of the below two functions:

i.e. the given signal can be presented as:
= (sin t) [u(t) - u(t - π]
= sin t u(t) - sin t u(t - π)
Now, sin (t - π) = -sin t
So, the above can be represented as;
= sin t u(t) + sin (t - π) u(t - π)

So, the Laplace transform of sin t u(t) + sin (t - π) u(t - π) i.e. of the given signal is:

Both the power transistor and the diode are ideal.
When voltage V is positive, both transistor and diode will be ON and making voltage across them zero and current I will be flowing.
⇒ at V = 0, I = I_maximum (∞)
When voltage V is negative, then both will be OFF and offering infinite resistance and hence voltage is maximum, and the current is zero.
⇒ at V = V_max (∞), I = 0
Now, the V-I characteristics are as shown below.

Concept:

From the above circuit, the current flowing through the load, ‘I’ is given as

Power tranfered to the load,

In the above equation RL is a variable, therefore the condition for maximum power delivered to the load is determined by differentiating load power with respect to the load resistance and equating it to zero.

⇒ (RL + RTH)2 = 2RL (RTH + RL)
⇒ RL + RTH = 2RL
⇒ RL = RTH
This is the condition for maximum power transfer, which states that power delivered to the load is maximum, when the load resistance RL matches with Thevenin’s resistance RTH of the network.
Under this condition, power transfer to the load is
(by substituting RL = RTH)
Therefore, the power transfer to the load

Calculation:

By the above circuit diagram,
It obeys Wheatstone bridge i.e
6 × 9 = 6 × 9
The current across the load R_L is zero
V_th = 0 V (by KVL)
That also implies the power across load R_L is zero
Therefore, no power transfer across the load resistance R_L
Note:

R_th = (6 ∥ 6) + (9 ∥ 9)
= 3 + 4.5
= 7.5 Ω

Concept:
The gain bandwidth product of an op-amp is constant for both open loop as well as closed loop configuration.
A_OL × F_OL = A_CL × f_CL

Application:
For non-inverting amplifier


f_CL = 100 KHZ = 10⁵


R_F = 9R₁
R_F = 90 KΩ

Concept:

In a series RLC circuit, the conditions for different nature of damping are:

Over damped:

Under damped:

Critically damped:

Calculation:
For critically damped system:

i.e,




5 + C₂ = 10
C₂ = 5 mF

Balanced modulators and ring modulators are popularly used to generate DSB-SC signals.
The circuit diagram of a balanced modulator is as shown:

Balanced Modulator:-
1) It is used to generate a DSB-SC wave.
2) It consists of two AM modulators in a balanced configuration to suppress the carrier signal.
3) The message signal m(t) is applied at one modulator and – m(t) is applied at another modulator.

Concept:

• A decibel is a one-tenth unit of the Bel (B) which is used to describe a gain of either voltage, current, or power.
• It is used to describe the ratio of one magnitude of a power or field quantity to another, on a logarithmic scale, the logarithmic quantity being called a degree of power or a degree of field, respectively.

Let the voltage gain, current gain and power gain be X.
Voltage Gain (dB) = 20log(X)
Current Gain (dB) = 20log(X)
Power Gain (dB) = 10log(X)
log(10) = 1
log(Xⁿ) = nlog(X)

Calculation:
Given:
Voltage Gain = 10
Current Gain = 100
Voltage Gain (dB) = 20log(10) = 20 dB
Current Gain (dB) = 20log(100) = 40log(10) = 40 dB
Power Gain (dB) = 10log(V × I)
Power Gain (dB) = 10log(10 × 100)
Power Gain (dB) = 10log(10³) = 30 dB

Concept:

Frequency Hopping Spread Spectrum (FH-SS):

• In FH-SS, the spectrum of the transmitted signal is spread sequentially by randomly hopping the data modulated carrier from one frequency to the next.
• The synthesizer output at a given instant of time is the frequency hop.
• The output bits of the PN generator change randomly. Hence the synthesizer output frequency will also change randomly.
• If the number of successive bits at the output of the PN generator is n, then the total number of frequency hops will be 2ⁿ.

Calculation:

The number of register stages given is 5, i.e.

n = 5

∴ The number of slots available for frequency hopping will be:

Number of Slots = 2ⁿ = 2⁵

= 32

Concept:

Input offset voltage:

• Whenever both the input terminals of the op-amp are grounded, ideally the output voltage must be zero. However, in this condition, the practical op-amp always shows a small non-zero output voltage.
• The input offset voltage depends upon the temperature and given as:

V_id = V₁ – V₂

Input Bias current:

• For ideal op-amp, no current flows into the input terminals but in a practical case, op-amp does have some very small input currents of the order 10^-16 A to 10^-14 A.
• Therefore input bias current is defined as the current flowing into each of the two input terminals when they are biased to the same voltage level i.e. when the op-amp is balanced.

• Output offset voltage (V_o)_offset due to input offset is given by:

• Output offset voltage (V_o)_offset due to input bias current is given by:

(V_o)_offset = R_fI_B
R_f = Feedback resistor
R₁ = Input resistance of the Op-Amp

Calculation:
Given:
Gain = 100, V_id = 6 mV, I_b = 500 nA
Output offset voltage (V_o)_offset due to input offset is:

(V_O)_Offset = (1 + 100) 6mV
(V_O)_Offset = 606 mV
Output offset voltage (V_o)_offset due to input bias current is:
(V_o)_offset = R_fI_B
(V_O)_Offset = 47 kΩ × 500 nA
(V_O)_Offset = 23.5 mV

For the zenor diode to be on minimum voltage across it should be 20V.

Vi ≥ 20 x 1200/1000
Vi ≥ 24 V
Only option (B) satisfies this condition.
This objective approach solves time to calculate the upper bound on V_i

Transresistance or trans-impedance amplifier is normally used as a current to voltage converter which requires low input and low output impedances for its proper operation.
Requirements of input and output impedances for different amplifiers are listed below:

The distance between the plates of the capacitor is increased 10 times since:
C = εA/d
The capacitance decreases 10 times:
Cnew = C/10
The plates of the capacitor are isolated electrically ⇒ q, the charge is constant
q = CV
⇒ Voltage becomes 10 times previous value Vnew = 10 kV
Energy initially stored in capacitance is E = 1/2 CV2 = 1/2 x 100 x 10-12 x (104)2 
= 0.005 J
Energy after plates are moved to 10 times the original distance:
Enew = 1/2 Cnew ​​​​​​​ V2new
= 1/2 C/10 (10V)2
= 10 X 1/2CV2

= 0.05 J

Concept:
The capacity of a band-limited AWGN channel is given by the formula:
C = Blog₂( 1 + (S/N)
C = Maximum achievable data rate (in bits/sec)
B = channel bandwidth
S/N = signal Noise power (in W)
Note: In the expression of channel capacity, S/N is expressed not in dB.
Calculation:
Given B.W. = 4 kHz, S/N = 31 dB
Since the S/N ratio is in dB, we can write:
10log₁₀(S/N) = 31
10log₁₀(S/N) = 3.1
S/N = 10^3.1W
The channel capacity will be:
C = (4 x 103) log2 ( 1 + 10^3.1)
C = 41.19 kbps
The closest option is 32 kbps. Mistake Points Many of us will make a common mistake of selecting 20 kbps as the correct answer, which can't be right. Because, in the formula of channel capacity S/N is not in dB, it is in Watts.

​​

​The maximum frequency deviation is:

​​

= 12500 Hz

Concept:
The K-map is a graphical method that provides a systematic method for simplifying and manipulating the Boolean expressions or to convert a truth table to its corresponding logic circuit in a simple, orderly process.
In an 'n' variable K map, there are 2n cells
For 4 variables there will be 24 = 16 cells as shown:

Calculations:
F(a, b, c) = ∑(0, 2, 4, 5, 6)

F = c' + ab'

Concept:
Ampere's Circuital Law:

• It gives the relationship between the current and the magnetic field created by it.
• This law says that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

Where dl is a small element, μ0 is the permeability of free space and I is the electric current.

Nyquist plot:

Nyquist plots are an extension of polar plots for finding the stability of the closed-loop control systems. This is done by varying ω from −∞ to ∞, i.e. Nyquist plots are used to draw the complete frequency response of the open-loop transfer function.

Method of drawing Nyquist plot:

• Locate the poles and zeros of open-loop transfer function G(s)H(s) in ‘s’ plane.
• Draw the polar plot by varying ω from zero to infinity.
• Draw the mirror image of the above polar plot for values of ω ranging from −∞ to zero.
• The number of infinite radii half circles will be equal to the number of poles at the origin.
• The infinite radius half-circle will start at the point where the mirror image of the polar plot ends. And this infinite radius half-circle will end at the point where the polar plot starts.

CONCEPT:
Inductors: The coils of wire that are wound around any ferromagnetic material (iron cored) or wound around a hollow tube that increase their inductive value are called inductors.

• The inductance (L) is measured in Henry (H) and the instantaneous voltage in volts.
• Rate of instantaneous voltage is given by (v = L di/dt)

Analysis:
By integrating the voltage we will get the current,
the voltage is a rectangular function hence the current will be a ramp function.
Hence, option 1 is correct.

Concept:

Damped natural frequency of a second order system is given by

Where,

ωn = Natural frequency

ζ = Damping ratio

Calculation:

Given-

ωn = 3 rad/sec

Now, natural damped frequency can be calculated as

ζ² = 1 - (2.4/3)²

ζ = 0.6

For Nyquist plot of G(s), encirclements is
N = P – Z
Where P = right side poles of open loop transfer function
The characteristic equation is:
1 + (s-1)(s+2)(s+3) = 0
(s² + 5s + 6)(s - 1) + 1 = 0
s³ + 5s² + 6s - s² - 5s - 6 + 1 = 0
s³ + 4s² + s - 5 = 0

Z = right half-plane pole
Z = 1, P = 1.
⇒ N = 1 – 1 = 0
Therefore, the Nyquist plot never encircles (-1 + j0)

Concept :

The directivity for an antenna is mathematically calculated as:

η = Efficiency

a = Radius

Rearranging the above, we get:

Calculation:

Given:

f1 = 100 MHz, f2 = 200 MHz, η1 = 40%, η2 =80%

The ratio of the diameter of antenna 1 to antenna 2 will be:

Since √2 = 1.41, we the required ratio will be:

= 1.41 × 2 = 2.82

From the given Nyquist plot
At ω = 0, 1 ∠0°
ω = ∞, 0∠-180°
matches these conditions.

Concept:

For an n-channel MOSFET, the current in triode and saturation region is given by:

ID = Kn [2VDS(VGS - Vth) - VDS2)] when VDS < VGS - Vth

when VDS > VGS - Vth

VDS = Drain to Source Voltage

VGS = Gate to source voltage

Vth = Threshold Voltage

Kn = Conduction parameter

Analysis:

Given: Kn = 0.249 mA / V2

VGS = 2VTN

VTN = 0.75 V

∵ VGS – VTN = 0.75 V

And VDS is usually more than that

∴ VDS > VGS - VTN

And the MOSFET is in saturation.

I_D = 0.249 (2V_TN - V_TN)²

= 0.249 (V_TN)²

= 0.249 (0.75)²

= 0.14 mA

By redrawing the circuit, we get

As shown in the circuit diagram, given op-amp is inverted grounded and output is corresponding to the non-inverted terminal.
∴ V = +V_cc = 15 volt
Now by applying KCL at node ‘V’ we get
I₁ = I₂

1 – V = V – V₀
V₀ = 2V - 1
V₀ = 29 volt
But output voltage can't be greater than higher saturation voltage (V_cc)
∴ Output voltage (V₀) = +15 volt

Concept:
The shown bock diagram will be considered as the receiver in digital communication.

The impulse response of the filter is calculated by:

h(t) = s*(T_b - t)
Calculation:
In the given signal the duration is T. So, T_b = T
The shifted signal by T is given by:
s(t+T)

The time-reversed signal will be

Option 3 is correct.

Calculation:
Total production of X in 5 years = 13 + 15 + 12 + 14 + 11
⇒ 65
Average = 65/5
⇒ 13
Total production of Y in 5 years = 12 + 14 + 13 + 15 + 13
⇒ 67
Average = 67/5
⇒ 13.4
Total production of Z in 5 years = 14 + 13 + 15 + 12 + 15
⇒ 69
Average = 69/5
⇒ 13.8
∴ The requied answer is Z.

Persons: P, Q, R, S, T and U

With the help of the information given in the question, we can draw the arrangement,

Step 1: P sits second to the left of U and third to the right of Q.

Step 2: S sits second to the right of U.

Step 3: R sits second to the left of Q. Therefore the remaining spot is assigned to T.

From the diagram, we can see that the immediate neighbor of both Q and T is S.

Hence the correct answer is S.

The logic followed here is :-

• GTCHP : HSDGQ

Similarly,

• RVXIS : ?

Hence, the correct answer is "SUYHT".