Digital Communication Multiple Choice Questions

Digital Communication Multiple Choice Questions

1) What are the primary features of a transmitter?

Lower clock speed

Lower transmitting power

Higher clock speed

None of these 

Answer: d

Explanation: A transmitter refers to an electronic device that produces a radio signal with the help of an antenna. The primary features which make a transmitter complex are higher transmitting power, higher clock speed, directional antennas.

2) The window gives a number of

Bytes

Frames

Both option a and b

None of these 

Answer: c

Explanation: The window gives the number of frames or bytes.

3) A speech signal, band-limited to 8 kHz with peak to peak between +20 V to - 20 V, and the signal are sampled at Nyquist rate, and the bits 0 and 1 are transmitted using bipolar pulses. Find the minimum bandwidth for distortion-free transmission in KHz?

98

kHz

64 kHz

52 kHz

35 kHz 

Answer: b

Explanation:

While using polar pulses, the minimum bandwidth required is four times the theoretical bandwidth or Nyquist bandwidth.

So the required bandwidth = 4 * Nyquist bandwidth

= 4 × 2 FM = 4 × 2 × 8 = 64 KHz.

4) Which of the given device does the data compression?

Switches

Circuit breakers

Microcontroller

Source encoder 

Answer: d

Explanation: Data compression refers to a reduction in the number of bits required to present the data. The compressing data can save memory, increase the file transfer rate, and minimize the cost for storage hardware and network bandwidth. Source encoder is a type of encoder which converts the digital signal or analog signal to a binary digit sequence.

5) According to Federal communications commission regulations, the maximum allowable frequency deviation is 40 kHz, for a TV signal, given that the percentage modulation of the audio portion is 50%. Find the frequency deviation of the audio signal in KHz?

10

kHz

20 kHz

30 kHz

40 kHz 

Answer: b

Explanation:

Given;

Maximum allowable frequency deviation ?fmax = 40 KHz

Percentage modulation of the audio portion = 50%

We know that the actual frequency deviation for the TV signal is calculated as 

6) Which of the given filter has maximum flatness?

Bessel filter

Butterworth filter

Low pass filter

None of these 

Answer: b

Explanation: Butterworth filter refers to a type of filter whose frequency response is flat over the passband region. The low pass filter provides a constant output source from DC up to a cutoff frequency. If the signal is beyond the cutoff frequency, it simply rejects.

7) For an Amplitude modulation signal, the bandwidth is 20 kHz, and the highest frequency component present is 650 kHz. Find the carrier frequency used for this amplitude modulation signal?

640

kHz

900 kHz

440 kHz

260 kHz 

Answer: a

Explanation:

Given;

Bandwidth = 20 KHz

FC + FM = 650 KHz

We know that,

BW = 2 FM = 20 KHz

So, fm = 10 KHz

Therefore,

FC = 650 kHz - FM

FC = 650 - 10 = 640 KHz

8) Space loss occurs due to a decrease in

Phase shift

Momentum

Electric field strength

Power 

Answer: c

Explanation: Space loss occurs mainly due to a decrease in electric field strength. Due to a decrease in the electric field strength, there will be a reduction in the signal strength as a function of distance known as space loss.

9) Analog data with the highest harmonic at 40 kHz generated by a sensor has been digitized using 6 level PCM. Find the rate at which digital signal generated?

300 kbps

240 kbps

450 kbps

600 kbps 

Answer: b

Explanation:

We know that,

Nyquist rate = 2 × 40 kHz = 80 KHz

2n ≥ 6

Thus, n = 3

So the bit rate = 80 × 3 = 240 kbps

10) A satellite receiver with a noise figure of 5.6 dB has a bandwidth of 24 kHz and comprises a preamplifier with a noise temperature 150 k and a gain of 40 dB. If the reference temperature is 300 k, find the equivalent noise temperature of the receiver?

450 K

300 K

220 K

150 K 

Answer: d

Explanation:

Given;

F2 = 5.6 dB = 3.63

Teq = 150

T = 300

We know that, 

= 1.5 + ( 2.63 × 10 -3 ) = 1.50

Therefore, the equivalent noise temprature

Te = (F - 1) T = (1.50 - 1) 300 = 150 K

11) In frequency modulation broadcast, the maximum deviation is 80 kHz, and the maximum modulating frequency is 20 kHz. In reference to Carson's rule, find the maximum required bandwidth?

300 KHz

200 KHz

150 KHz

80 KHz 

Answer: b

Explanation:

Given;

Maximum deviation = 80 KHz

Maximum modulating frequency = 20 KHz

We know that,

According to Carson's rule,

BW = 2 (fm + ?f) = 2 (20 + 80) = 200 KHz

12) If noise figure of a receiver is 1.8 at 20 oC, find its equivalent noise temprature?

474.9 K

384.8 K

234.4 K

184.6 K 

Answer: c

Explanation:

Given;

Noise figure of a recevier (NF) = 1.8

TO = 273 + 20 = 293 K

We know that,

Equivalent noise temprature Teq = TO( NF - 1)

= 293 (1.8 - 1 ) = 234.4 K

13) If a 120 V carrier peak changes from 170 V to 50 V by a modulating signal, find the modulation factor?

0.5

1.5

2.5

3.5 

Answer: a

Explanation:

:

Modulation factor also refers in terms of the values referred to the modulated carrier wave and is given as 

Where,

Vmax ? = Maximum values of the amplitude of the modulated carrier wave.

Vmin ? = Minimum values of the amplitude of the modulated carrier wave.

Therefore, 

14) RF carrier 15 kV at 1 MHz is amplitude modulated and modulation index is 0.8. Find the peak voltage of the signal?

18 kV

22 kV

26 kV

12 kV 

Answer: d

Explanation:

Given;

Modulation index (m)= 0.8

RF carrier = 15 kV

We know that, 

Therefore the peak value of signal Vm = m × amplitude of carrier wave, Vc

= 0.8 × 15 kV

= 12 kV

15) A broadcast amplitude modulation radio transmitter radiates 140 kW when the modulation percentage is 75. Find the carrier power?

120 kW

142 kW

109 kW

172 kW 

Answer: c

Explanation:

Given;

Transmitted power Ptotal = 140 kW

Modulation percentage = 75 = 0.75

We know that, 

16) The power spectral density of a signal is

Even negative and complex

Odd, complex, and positive

Real, odd, and negative

Real, even, and non-negative 

Answer: d

Explanation: We know that the power of a signal is real and non-negative quantity; therefore, the power spectral density of a signal is real, even, and non-negative quantity.

17) A 1200 W carrier is amplitude modulated and has a sideband power of 400 W. Find the depth of the modulation?

1.245

0.775

3.689

1.059 

Answer: b

Explanation:

Given;

Pc = 1200 W

Psb = 400 W

We know that, power in sidebands is given as 

Or, m = √0.6 = 0.7746 = 0.775

18) A modulater is a device used to

Differenciates two frequencies

Amplify two radio frequency signal

Impress the information on to a radio frequency carrier

Reduce the modulating power requirement. 

Answer: d

Explanation: Modulation refers to a low-frequency signal process combined with a high-frequency signal to transmit it over a long distance. For example, the normal human voice signal constitutes low frequency to travel only to a short distance. Hence, to transfer it to a long-distance, usually, we need modulation. In other words, we can say that it is a device used to impress the information present in low-frequency signals onto a radio frequency carrier.

19) If the two signals modulate the same carrier with different modulation depths of 0.4 and 0.8. Find the resulting modulation signal?

0.89

0.66

0.54

0.16 

Answer: a

Explanation:

Given;

Two different modulation depths m1 = 0.4 and m2 = 0.8

We know that,

The total modulation index (depth) 

20) A balanced modulator is used in the generation of which of the following signal?

Frequency Modulation signal

DSB- SC signal

ISI signal

SSB -SC signal 

Answer: b

Explanation: A balanced modulator can be defined as a circuit in which two non-linear devices are linked in a balanced mode to generate a DSB-SC signal.

21) A given AM broadcast station transmits an average carrier power output of 50kW and uses a modulation index of 0.804 for sign wave modulation. Find the maximum amplitude of the output if a 60 ohm resistive load represents the antenna?

5.88 kV

7.66 kV

3.95 kV

1.05 kV 

Answer: c

Explanation:

Given;

Average carrier power output (Pc) = 50kW

Modualtion index (m)= 0.804

Resistive load (R) = 60 ohm

We know that, 

22) Which of the given modulator is an indirect way of generating FM?

Inductance FET modulator

Armstrong modulator

Reactance tube modulator

Zener diode modulator 

Answer: b

Explanation: If we are talking about the direct method of FM generation, the VCO output frequency is controlled by the instantaneous value of the baseband signal. Reactance tube modulator facilitates it. In Armstrong modulation, FM is determined from PM, so it is an indirect method.

23) In a frequency modulation system, the maximum frequency deviation is 2000, and modulating frequency is 2kHz. Find the modulation index ??

1

2

3

4 

Answer: a

Explanation:

Given;

?f = 2000

fm = 2 kHz = 2000 Hz

We know that, 

24) The maximum deviation allowed in a frequency modulation system is 120 kHz. If the modulating signal frequency is 20 kHz, find the bandwidth requirement as per Carson's rule?

120 kHz

240 kHz

280 kHz

320 kHz 

Answer: c

Explanation:

Given;

?f = 120 kHz

fm = 20 kHz

We know that,

The bandwidth requirement as per Carson's rule

= 2 (1 + δ) fm 

25) In FM modulation, when the modulation index increases, the transmitted power?

Half

Decreased

Doubled

Unchanged 

Answer: d

Explanation: Frequency modulation refers to the process of modulating a sinusoidal wave frequency as per the amplitude of the baseband bearing signal. The signal can be digital or analog; in frequency modulation, the FM wave's amplitude is always constant, and transmitted power Pt is equaled toremains unchanged.

26) In a frequency modulated system, when the audio frequency is 600 Hz, and audio frequency voltage is 2.8 V, the frequency deviation ? is 5.6 kHz. If the audio frequency voltage is now increased to 7.4 V, find the new value of deviation?

12. 7 kHz

14. 8 kHz

15. 2kHz

17. 6 kHz 

Answer: b

Explanation:

Given;

Modulation frequency , FM = 600 Hz

Audio frequency voltage Am= 2.8 V

Frequency deviation ?f = 5.6 kHz

We know that,

?f= Kf× Am 

When amplitude of message signal is increased to 7.4 V

That is, A'm = 7.4 V

So, new frequency deviation,

?f'= Kf × A'm = 2 × 7.4 = 14. 8 kHz

27) In phase modulation, phase deviation is proportional to which of the following.

The wavelength of the message signal

Message signal

The amplitude of the message signal

The phase shift of the message signal 

Answer: b

Explanation: In phase modulation, Phase deviation varies linearly with the message signal frequency so that phase deviation is proportional to the message signal.

28) A system has a receiver noise resistance of 60 ohms, and it is connected to an antenna with an output resistance of 60 ohms. Find the noise figure of the system?

1

2

3

4 

Answer: b

Explanation:

Given,

Req = 60 ohm

Ra = 60 ohm

We know that,

The noise figure is given as

NF = 1 + Req/Ra = 1 + 1/1 = 2

29) Which type of noise reduction by limiters in FM receivers

Jhonson noise

Cosmic noise

Impulse noise

Partition noise 

Answer: c

Explanation: Under the ordinary condition, limiters in FM receivers limit the impulse noise to the same extent as the random noise.

30) When a radio receiver is tuned to 600 kHz, its local oscillator provides the mixer with input at 1000 kHz. Find the frequency of other stations?

1400 kHz

1600 kHz

1800 kHz

2200 kHz 

Answer: a

Explanation:

Given;

Turned frequency, FC = 600 kHz

Local oscillator frequency, f = 1000

We know that,2

Local oscillator frequency, f = FC + fif

fif = f - FC

= 1000 - 600 = 400 kHz

Therefore, the image is given as

fc + 2 fif

= 600 + 2 × 400

= 600 + 800

= 1400 kHz

31) What is the automatic frequency control voltage of the FM transmitter VCO?

Ac voltage

Dc voltage

Ramp voltage

Sine wave voltage. 

Answer: b

Explanation: VCO stands for the voltage-controlled oscillator. An oscillator refers to a circuit that producers an AC signal, usually of high frequency. A voltage-controlled oscillator is an oscillator whose frequency is controlled by voltage. It ranges from 1 MHz with one-volt input to 1 MHz per volt to 10v/10MHz.

32) Without any filtering, a broadcast station at 1.800 kHz is heard together with another station at 2800 kHz on a superheterodyne receiver. Find the value of employed IF?

200 kHz

400 kHz

500 kHz

600 kHz 

Answer: c

Explanation:

Given

Image frequency = 2800 kHz

Signal frequency, fs = 1800 kHz

We know that the IF (Intermediate frequency), 

33) If the turn-on energy loss in a transformer is 55 MJ and turn-off energy loss is 78 mj. If the mean power loss is limited to 220 W, find the maximum switching rate that can be achieved?

1232.7 cycles/s

1654.1 cycles/s

1453.8 cycles/s

1862.9 cycles/s 

Answer: b

Explanation:

Given;

P = 220 W

We know that,

Energy loss = Mean power loss, P × time duration 

34) Analog data having the highest harmonic at 40 kHz generated by a sensor has been digitized using 6 level PCM; find the generated digital signal rate?

120 kbps

160 kbps

180 kbps

240 kbps 

Answer: d

Explanation:

Given

L = 6

FM = 40 kHz

Since the number of quantization levels,

L = 2n

Therefore, L= 6 , 2n = 6 or n = 3

So 3 bits are required to represent each sample,

So the rate of a digital signal generated is given as

Rb = n × 2 × FM = 3 × 2 × FM = 3 × 2 × 40 = 240 kbps

35) In a delta modulation scheme, the step height is 80 mV, and the step width is 2 ms. Find the maximum slope that the staircase can track?

50 V

60 V

40 V

30 V 

Answer: c

Explanation:

Given;

Step height = 80 Mv

Step width = 2 ms

We know that,

The maximum slope that the staircase can track in a delta modulation scheme is given as 

36) Frequency frogging is used in carrier system to

Reduce frequencies

Reduce cross-talk

Reduce error rate

Conserve narrowband signals. 

Answer: b

Explanation: Frequency frogging means the interchanging of frequencies. It performs specific tasks such as avoiding oscillations and feedback to reduce cross-talk and correct for a high-frequency response slope in the transmitter line.

37) A TDM link has 25 signal channels, and each channel is sampled at 7 kHz. If each sample is represented by 8 bits and contains an additional bit for synchronization, find the total bit rate for the TDM link?

695 kbps

888 kbps

1407 kbps

1209 kbps 

Answer: c

Explanation:

Given;

Number of channel N = 25

Sampling frequency, fs = 7 kHz

Number of bits, n = 8

Additional bit C = 1 (Smaplw contains one additional bit)

Bit rate, Rb = N × n × fs + Cfs

= 25 × 8 × 7 + 1 × 7

= 1407 kbps

38) Output data ratio of a 4 bit PCM - TDM system sampling 16 voice channels, comparing these using µ - law at the rate of 6 kHz and with an Iframe alignment word, is

6.04 × 104bit/sec

4.08 × 103bit/sec

8.03 × 102bit/sec

5.97 × 105bit/sec 

Answer: b

Explanation:

Given;

n =4

N = 16

a = 1

Fs = 6 × 103

We know that,

Each frame alignment consists of a synchronizing bit in addition to data, so the equation is given as 

As we know, Ts = 1/fs, therefore 

= n (N + a) × fs

= 4 (16 + 1) × 6 × 103

= 4.08 × 103bit/sec

39) 23 voice channels are sampled uniformly at a rate of 10 kHz and then time division multiplexed. The sampling process uses flat-top samples with a two µs duration. The multiplexing operation includes the provision of synchronization by adding an extra pulse of 2 µs duration. Find the spacing between successive pulses of the multiplexed signal?

One µs

Two µs

Three µs

Four µs 

Answer: b

Explanation:

Given;

Number of voice signals = 23

Synchronization pulses = 2

Total number of pulses in one rotation = 23 + 2 = 25

We knonw that,

The time duration for one time frame, 

So time duration utilized by 25 pulses = 100 µs

Therefore, the time utilised by each pulse 

Since actual duration of each pulse is 2 µs

So, the spacing between successive pulse = 4 - 2 = 2 µs

40) The unmodulated carrier power in an AM transmitter is 8 kW. A sinusoidal modulating signal modulates this carrier. The maximum percentage of modulation is 60 %. If it is reduced to 50 %, find the maximum unmounted carrier power in (kW) that can be used without overloading the transmitter?

37.76 kW

25.34 kW

18.55 kW

42.57 kW 

Answer: a

Explanation:

Unmodualted carrier power Pc = 8 kW

Maximum modulation index mmax = 0.6

We know that, total power is given as 

41) For an Amplitude modulation signal, the bandwidth is 10 kHz, and the highest frequency component present is 450 kHz. Find the carrier frequency used for this amplitude modulation signal?

50 KHz

930 KHz

440 KHz

280 KHz 

Answer: a

Explanation:

Given;

Bandwidth = 10 KHz

fc + fm = 450 KHz

We know that,

BW = 2 fm = 10 KHz

So, fm = 5 KHz

Therefore,

fc = 50 KHz - fm

fc = 450 - 10 = 440 KHz