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Quiz Electronics and Communication Engineering 22 July 2020

Quiz Electronics and Communication Engineering
Exam: GATE
Topic: Miscellaneous
Date: 22/07/2020

Each Question carries 2 Mark
Negative Marking: 1/3
Time: 20 Minutes

Q1. The addressing mode used in an instruction of the form ADD A, B is _____
(a) Direct
(b) Absolute
(c) Indirect
(d) Indexed

Q2. Ring shift and Johnson counters are:
(a) Synchronous counters
(b) Asynchronous counters
(c) True binary counters
(d) Synchronous and true binary counters

Q3. Two 100 W, 200 V lamps are connected in series across a 200 V supply. The total power consumed by each lamp will be(watts).
(a) 25
(b) 50
(c) 100
(d) 200

Q4. The circuit shown in the figure is that of

(a) a non-inverting amplifier
(b) a Schmitt triggers
(c) an oscillator
(d) an inverting amplifier

Q5. What is the value of the damping ratio of a second order system when the value of the resonant peak is unity?
(a) √2
(b) 1
(c) 1/√2
(d) 0

Q6. If the potential, V = 4x + 2 volts, then electric field is
(a) 6 V/m
(b) 2 V/m
(c) 4 V/m
(d) -4 (a_x ) ⃗ V/m

Q7. The transfer function G(s) = 1/((3s+1)) has a corner frequency
(a) 3 rad/s
(b) 0.33 rad/s
(c) 1 rad/s
(d) 30 rad/s

Q8. The wavelength beyond which photo-electric emission cannot take place is called
(a) critical wavelength
(b) optical wavelength
(c) photoelectric wavelength
(d) long wavelength

Q9. Compression in PCM refers to relative compression of
(a) higher signal amplitudes
(b) lower signal amplitudes
(c) lower signal frequencies
(d) higher signal frequencies

Q10. If ( (27s + 97)/(s^2 +33s) ) is the Laplace transform of f(t), then f(0+) is
(a) 0
(b) 97/33
(c) 27
(d) ∞


S1. Ans.(a)
Sol. In the given instruction data or operand from register B is added to the data present in the register A and the result is finally stored in register ‘A’. This instruction is in “Direct Addressing Mode”.

S2. Ans (a)
Sol. Since, all of the clock inputs are connected through a single clock pulse in ring shift and Johnson counters. So, both are Synchronous counters.

S3. Ans.(c)
Sol. Both bulbs have same voltage and power rating so both will have same resistance.
⇒ R = V^2/P
⇒ R = (200)^2/100
∴ R = 400Ω
As both the bulbs are connected in series so total resistance will be added.
⇒ R_T = 400 Ω + 400 Ω = 800 Ω
Current flowing through bulb is
⇒ I = P/V
∴ I =100/200 = 0.5A
Total power consumed by both bulbs are
∴ I^2 R_T=(0.5)^2 × 800 = 0.25 × 800 = 200 Watt
Hence, Total power consumed by each bulb = 200/2 = 100 watt.

S4. Ans. (b)
Sol. In the given figure, the feedback is connected to positive terminal of Op-amp so it behaves as a Positive feedback and input is applied to inverting terminal. So, the given figure is of Schmitt Trigger.

S5. Ans.(c)
Sol. Given:
|M_r| = 1
We know that in second order system,
⇒ |M_r| = 1/(2ζ√(1-ζ^2 ))
⇒ 1 = 1/(2ζ√(1-ζ^2 ))
⇒ 4ζ^2(1-ζ^2) = 1
Solving we get,
⇒ ζ = 1/√2

S6. Ans.(d)
Sol. Given:
V = 4x + 2 volts
We know that,
E ⃗ = -dV/dx (a_x ) ⃗
⇒ E ⃗ = -(d(4x + 2 ))/dx (a_x ) ⃗
So, E ⃗ = -4 (a_x ) ⃗ V/m

S7. Ans.(b)
Sol. Given:
G(s) = 1/((3s+1))
We know that in transfer function
G(s) = 1/((1+sT))
The corner frequency is ω = 1/T
Comparing we get Corner frequency as,
⇒ ω = 1/3
So, ω = 0.33 rad/sec

S8. Ans.(a)

S9. Ans.(a)
Sol. In PCM, Companding is a technique which compresses the data at the transmitter and expands the same data at the receiver. It is a combination of the words “Compressing” and “Expanding.”
So, Compression in PCM refers to relative compression of higher signal amplitudes.

S10. Ans.(c)
Sol. Given:
F(s) = ( (27s + 97)/(s^2 +33s) )
We know that, Initial Value Theorem is
f(0+) = lim┬(s → ∞)⁡〖sF(s)〗
Now, f(0+) = lim┬(s → ∞)⁡s ( (27s + 97)/(s^2 +33s) ) = lim┬(s → ∞) (27 + 97/s)/(1 + 33/s ) = 27

Therefore, f(0+) = 27

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