Electronics & Communications Test

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Electronics & Communications Test - 1

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Weekly Quiz Competition

Question 1

2 / -0.66

Consider the logic circuit given below:

Q = ______ ?

A̅C + BC̅ + CD

ABC + C̅D

AB + BC̅ + BD̅

AB̅ + AC̅ + C̅D

Solution

The correct answer is "option 3".

Additional Information

Important Boolean algebra laws are:

BOOLEAN LAWS	FORMULA
De-Morgan’s law	(A+B)' = A'B', (AB)' = A'+B'
NAND law	(AB)’
NOR law	(A+B)’
XOR law	(A⊕B) = A’B+AB’
XNOR law	(A⊙B) = A’B’+AB

Question 2
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The following figure shows in block diagram form a scheme for controlling the direction of travel of road vehicle. Engine vehicle dynamics are represented by G(s) and the controller by G_c(s). Instrumentation of direction sensing and error determination is quite fast and its gain factor is included in k.

A
k < -45.69

B
k < 288

C
0< k < 201.6

D
No such k exists

Solution
Question 3
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It is given that carrier frequency (f_c) is 1000kHz. Then calculate the value of instantaneous frequency (f_i) at t = 1.5 sec and the peak frequency deviation (δf) of the above FM signal.

A
f_i =998.407 MHz , δf = 5000\π Hz

B
f_i =99.8407 kHz , δf = 5\π kHz

C
f_i =0.998407 MHz , δf = 5\π kHz

D
f_i =998.407 kHz , δf = 50\π kHz

Solution
Question 4
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For following Fig. if C(s) is Laplace Transform of output and R(s) is Laplace transform of input, the equivalent transfer function T(s) will be

A

B

C

D

Solution

Concept:

According to Mason’s gain formula, the transfer function is given by

Where, n = No of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

Given signal flow graph is
Question 5
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What will be the voltage reading of DC Voltmeter placed across the terminals of the Diode in the circuit below,

A
1.25 V

B
2.5 V

C
0 V

D
0.1 V

Solution

Concept:

The DC value is also the average value of a waveform and is calculated as:

The above voltage is applied to the circuit with a diode and resistance as shown:

When forward biased, the diode can be replaced with an equivalent forward resistance of r_a = 2 Ω as shown:

The waveform of V_D for only the positive half cycle ( till π ) will be:

Now, for negative half cycle, the diode will be reverse biased So, it will be open-circuited as shown.

The Waveform of V_D for the negative half cycle will therefore be:
Question 6
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The circuit shown in the figure uses an ideal op-amp working with +5 V and -5 V power supplies. The output voltage V₀ is equal to

A
-1 V

B
+5 V

C
-2 V

D
+1 V

Solution

For ideal op-amp, V_a = 0 V (by virtual short)

Due to the Infinite input impedance of ideal op-amp, the input current in op-amp will be zero.

Hence, I_in = 0 and I = 1 mA
Question 7
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Determine the approximate current through the diode assuming that the diode is ideal.

A
0.4 A

B
0.6 A

C
0.2 A

D
0.1 A

Solution

Concept:

The diode is a unidirectional device that allows current to flow only in one direction i.e. from anode to cathode.

Testing of a diode:

During the ON condition of the diode, I_PN > 0 and V_PN > 0.

During the OFF condition of the diode, IPN ≼ 0 and V_PN ≼ 0.

Calculation:

Let us assume the diode is OFF.

The current across R₂ is zero due to the minimum resistive path offered by the diode connected in parallel with it.
Question 8
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A
1 unit

B
√3 unit

C
√3/2

D

Solution
Question 9
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For the LTI system shown in the figure. Find impulse response of the system.

h₁(t) = h₂(t) = 5δ(t) ; h₃(t) = h₄(t) = u(t).

A
20u(t)

B
30u(t)

C
10u(t)

D
40u(t)
Solution
The correct answer is 30u(t)

Concept:
- When two blocks are in cascade form then take convolution of these two systems.
- When two blocks are in parallel to each other, then take the sum of these two systems.
Solution:

Here h₂(t) and h₃(t) are in cascade form.

∴ h'(t) = h₂(t) * h₃(t)
Question 10
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Which of the following statements with respect to the stability of a control system is true?

a. The roots of the characteristic equation are on the left half of the s-plane.

b. The system is bounded input bounded output (BIBO) stable.

c. Stability is independent of the input.

d. The roots of the characteristic equation are on the left half of the s-plane and on the imaginary axis.

A
Only (a) and (c) are true

B
Only (a), (b) are true

C
Only (a), (b) and (c) are true

D
Only (d) is true
Solution
Explanation:

The Characteristics equation is an essential tool for analyzing different system parameters
- It is obtained by equating the denominator of the closed-loop transfer function to 0
- It helps in the determination of system stability
- Also provides information about the system bandwidth, speed, response time, etc.
From solving the characteristics equation we could infer that
- The roots of the characteristic equation give the location of the Closed-loop poles of the system
- These roots if strictly lie on the left half of the S-plane, then the system is stable
- If these roots have at least one root on the imaginary axis or on the right half of S-plane then the system is unstable
- The system transfer function solely determines the characteristic equation
- It is independent of the input applied
- Therefore system stability is independent of input
Hence statement a and c are correct,

Statement d is incorrect

If the system is practically realizable (stable) then
- For bounded input, it must give bounded output
- If finite magnitude input is applied then the system must also produce finite magnitude output
- The above is known as (BIBO) stable
Hence statement b is correct

Therefore statements a b and c are correct

The correct answer is option 3.

NOTE:

It is to be noted here that stability or instability is the characteristic property of the control system and thus depends on the closed-loop poles of the system.

Therefore, we can say that stability is a factor of the system which is independent of the input of the system. However, the steady-state output of the system is dependent on the poles of the applied input.