Multiple Choice Questions and Answers on Analog Communication (Part-4)

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1)   Notch filter is a

DRex Electronics

a. Band pass filter
b. Band stop filter
c. Low pass filter
d. High pass filter

ANSWER: (b)  Band stop filter

2)   Noise is added to a signal in a communication system

a. At the receiving end
b. At transmitting antenna
c. In the channel
d. During regeneration of the information

ANSWER:  (c) In the channel

3)   Noise power at the resistor is affected by the value of the resistor as

a. Directly proportional to the value of the resistor
b. Inversely proportional to the value of the resistor
c. Unaffected by the value of the resistor
d. Becomes half as the resistance value is doubled

ANSWER:  (c) Unaffected by the value of the resistor

4)   Low frequency noise is

a. Transit time noise
b. Flicker noise
c. Shot noise
d. None of the above

ANSWER:  (b) Flicker noise

5)   Hilbert transform may be used in

a. Generation of SSB signals
b. Representation of band pass signals
c. Designing of minimum phase type filters
d. All of the above

ANSWER:  (d) All of the above

6)   At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 20 KΩ when the bandwidth is 10 KHz.

a. 4.071 * 10-6 V, 5.757 * 10-6 V
b. 6.08 * 10-6 V, 15.77 * 10-6 V
c. 16.66 * 10-6 V, 2.356 * 10-6 V
d. 1.66 * 10-6 V, 0.23 * 10-6 V

ANSWER:  (a) 4.071 * 10-6 V, 5.757 * 10-6 V

Explanation:
Noise voltage Vn = √(4R KTB)
Where, K = 1.381×10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by individual resistors
Vn1 = √(4R1 KTB)
= √(4 * 10 * 103 * 1.381 * 10-23 * 3000 * 10 * 103)
= √16.572 * 10-12
= 4.071 * 10-6 V
Vn2 = √(4R2 KTB)
= √(4 * 20 * 103 * 1.381 * 10-23 * 3000 * 10 * 103)
= √33.144 * 10-12
= 5.757 * 10-6 V

7)   At a room temperature of 293K, calculate the thermal noise generated by two resistors of 20KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in series.

a. 300.66 * 10-7
b. 284.48 * 10-7
c. 684.51 * 10-15
d. 106.22 * 10-7

ANSWER:  (b) 284.48 * 10-7

Explanation:
Noise voltage Vn = √(4R KTB)
Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in series is
Vn = √{4(R1 + R2) KTB}
= √{4(20 * 103 + 30 * 103) * 1.381 × 10-23 * 293 * 10 * 103 }
= 284.48 * 10-7

8)   At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in parallel.

a. 30.15 * 10-3
b. 8.23 * 10-23
c. 11.15 * 10-7
d. 26.85 * 10-7

ANSWER: (c)  11.15 * 10-7

Explanation:
Noise voltage Vn = √(4R KTB)
Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in parallel is
Vn = √{4R KTB}
Here for resistors to be in parallel,
1/R = 1/R1 + 1/R2
= 1/10K + 1/30K
= 0.1333
R = 7.502KΩ
Vn = √{4 * 7.502 * 103 * 1.381×10-23 * 300 * 10 * 103}
= √124.323 * 10-14
= 11.15 * 10-7

9)   A periodic signal is

a. May be represented by g(t) = g(t + T0)
b. Value may be determined at any point
c. Repeats itself at regular intervals
d. All of the above

ANSWER: (d) All of the above

10)   Sine wave is a

a. Periodic signal
b. Aperiodic signal
c. Deterministic signal
d. Both a and c

ANSWER: (a)  Periodic signal

11)   Properties of Hilbert transform are:

a. The signal and its Hilbert transform have same energy density spectrum
b. The signal and its Hilbert transform are mutually diagonal
c. Both a and b are correct
d. None of the above

ANSWER:  (c) Both a and b are correct

12)   An even function f(x) for all values of x and x holds

a. f(x) = f(-x)
b. f(x) = -f(x)
c. f(x) = f(x)f(-x)
d. None of the above

ANSWER:  (a) f(x) = f(-x)

13)   Random signals is

a. May be specified in time
b. Occurrence is random
c. Repeat over a period
d. None of the above

ANSWER: (b)  Occurrence is random

14)   Unit step function is

a. Exists only for positive side
b. Is zero for negative side
c. Discontinuous at time t=0
d. All of the above

ANSWER:  (d) All of the above

15)   In Unit impulse function

a. Pulse width is zero
b. Area of pulse curve is unity
c. Height of pulse goes to infinity
d. All of the above

ANSWER: (d) All of the above

16)   For a Unit ramp function area of pulse curve is unity

a. Discontinuous at time t=0
b. Starts at time t=0 and linearly increases with t
c. Both a and b
d. None of the above

ANSWER:  (b) Starts at time t=0 and linearly increases with t

17)   Thermal noise is also known as

a. Johnson noise
b. Partition noise
c. Flicker noise
d. Solar noise

ANSWER:  (a) Johnson noise

18)   Threshold effect is:

a. Reduction in output signal to noise ratio
b. Large noise as compared to input signal to envelope detector
c. Detection of message signal is difficult
d. All of the above

ANSWER:  (d) All of the above

19)   The rms value of thermal noise voltage is related to Boltzmann’s constant k as

a. Vn is Directly proportional to k2
b. Vn is Directly proportional to k
c. Vn is Directly proportional to √k
d. Vn is Directly proportional to k3

ANSWER:  (c) Vn is Directly proportional to √k

20)   The spectrum of the sampled signal may be obtained without overlapping only if

a. fs ≥ 2fm
b. fs < 2fm
c. fs > fm
d. fs < fm

ANSWER: (a) fs ≥ 2fm

21)   The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if

a. The sampled signal is passed through low pass filter
b. Filter has the cut off frequency wm
c. Both a and b
d. None of the above

ANSWER: (c)  Both a and b

22)   A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is

a. fs > 2fm
b. fs < 2fm
c. fs = 2fm
d. fs ≥ 2fm

ANSWER: (b) fs < 2fm

23)   Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt

a. 100 Hz
b. 200 Hz
c. 400 Hz
d. 250 Hz

ANSWER: (c) 400 Hz

Explanation:
In the given signal, the highest frequency is given by f = 400 π/ 2π
= 200 Hz

The minimum sampling rate required to avoid aliasing is given by Nyquist rate. The nyquist rate is = 2 * f
= 2 * 200
= 400 Hz.

24)   Calculate the Nyquist rate for sampling when a continuous time signal is given by
x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt

a. 300Hz
b. 600Hz
c. 150Hz
d. 200Hz

ANSWER: (a)  300Hz

Explanation:
For the given signal,
f1 = 100π/2π = 50Hz
f2 = 200π/2π = 100Hz
f3= 300π/2π = 150Hz

The highest frequency is 150Hz. Therefore fmax = 150Hz
Nyquist rate = 2 fmax
= 2 * 150
= 300Hz.

25)   A low pass filter is

a. Passes the frequencies lower than the specified cut off frequency
b. Rejects higher frequencies
c. Is used to recover signal from sampled signal
d. All of the above

ANSWER:  (d) All of the above

26)   The techniques used for sampling are

a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above

ANSWER: (d) All of the above

27)   The instantaneous sampling

a. Has a train of impulses
b. Has the pulse width approaching zero value
c. Has the negligible power content
d. All of the above

ANSWER: (d) All of the above

28)   The sampling technique having the minimum noise interference is

a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above

ANSWER: (b) Natural sampling

29)   Types of analog pulse modulation systems are

a. Pulse amplitude modulation
b. Pulse time modulation
c. Frequency modulation
d. Both a and b

ANSWER: (d) Both a and b

30)   In pulse amplitude modulation,

a. Amplitude of the pulse train is varied
b. Width of the pulse train is varied
c. Frequency of the pulse train is varied
d. None of the above

ANSWER:  (a) Amplitude of the pulse train is varied

31)   Pulse time modulation (PTM) includes

a. Pulse width modulation
b. Pulse position modulation
c. Pulse amplitude modulation
d. Both a and b

ANSWER: (d)  Both a and b

32)   Drawback of using PAM method is

a. Bandwidth is very large as compared to modulating signal
b. Varying amplitude of carrier varies the peak power required for transmission
c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver
d. All of the above

ANSWER: (d)  All of the above

33)   In Pulse time modulation (PTM),

a. Amplitude of the carrier is constant
b. Position or width of the carrier varies with modulating signal
c. Pulse width modulation and pulse position modulation are the types of PTM
d. All of the above

ANSWER: (d)  All of the above

34)   In different types of Pulse Width Modulation,

a. Leading edge of the pulse is kept constant
b. Tail edge of the pulse is kept constant
c. Centre of the pulse is kept constant
d. All of the above

ANSWER: (d) All of the above

35)   In pulse width modulation,

a. Synchronization is not required between transmitter and receiver
b. Amplitude of the carrier pulse is varied
c. Instantaneous power at the transmitter is constant
d. None of the above

ANSWER:  (a) Synchronization is not required between transmitter and receiver

36)   In PWM signal reception, the Schmitt trigger circuit is used

a. To remove noise
b. To produce ramp signal
c. For synchronization
d. None of the above

ANSWER:  (a) To remove noise

37)   In Pulse Position Modulation, the drawbacks are

a. Synchronization is required between transmitter and receiver
b. Large bandwidth is required as compared to PAM
c. None of the above
d. Both a and b

ANSWER:  (d) Both a and b