Digital Modulation MCQ || Introduction To Digital Modulation Questions and Answers

Ques.11. A speech signal is sampled at 8 kHz and encoded in PCM format using 8-bit/sample PCM data is transmitted through a baseband channel via 4-level PAM. The minimum Bandwidth required for transmission is

  1. 16 kHz
  2. 8 kHz
  3. 24 kHz
  4. 10 kHz

Answer.1. 16 kHz

Explanation:-

Given

fs = 8 kHz

n = 8 bits/sample

M = 4

Rb = nfs

Minimum Bandwidth will be required if transmission of data takes place using sinc pulses and the bandwidth is given by:

$B{W_{{\text{min}}}} = \frac{{{R_b}}}{{2{{\log }_2}M}}$

Where Rb = n  fs

n = number of bits used to represent samples

fs = sampling rate.

$\begin{gathered} {\left( {BW} \right)_{min}} = \frac{{{R_b}}}{{2{{\log }_2}M}} \hfill \\ \hfill \\ = \frac{{n\:{f_s}}}{{2{{\log }_2}M}} = \frac{{8 \times 8}}{{2{{\log }_2}4}} \hfill \\ \end{gathered}$

64/(2 × 2) = 16 KHz

 

Ques.12. The bandwidth of a video signal is 4.5 MHz. The signal is to be transmitted using PCM with 1024 quantization levels. The minimum bit rate required for transmission is

  1. 90 Mbps
  2. 45 Mbps
  3. 4.5 Mbps
  4. 10 Mbps

Answer.1. 90 Mbps

Explanation:-

For a minimum bit rate, the modulating signal must be sampled at the minimum sampling rate, i.e. at the Nyquist Rate.

fs = 2fm

fm = Maximum frequency present at the modulating signal.

∴ For the given bandlimited signal with a maximum frequency of 4.5 MHz, the sampling frequency will be:

fs = 2 × 4.5 M = 9 MHz

With L = 1024, the number of bits will be:

n = log2 1024 = log2 210

n = 10 bits

Now, the required minimum bit rate will be:

Rb = n fs = 10 × 9 Mbps

Rb = 90 Mbps

 

Ques.13. What are the advantages of digital circuits?

  1. Less noise
  2. Less interference
  3. More flexible
  4. All of the above

Answer.4. All of the above

Explanation:-

  1. Digital circuits are more reliable.
  2. Digital circuits are easy to design and cheaper than analog circuits.
  3. The hardware implementation in digital circuits is more flexible than analog.
  4. The occurrence of cross-talk is very rare in digital communication.
  5. Digital circuits are less subject to noise, distortion, and interference as it works on digital pulses, and also the pulses can be regenerated.

 

Ques.14. In digital transmission, the modulation technique that requires the minimum bandwidth is:

  1. PCM
  2. PAM
  3. DPCM
  4. Delta modulation

Answer.4. Delta modulation

Explanation:-

  • In PCM an analog signal is sampled and encoded into different levels before transmission
  • The bandwidth of PCM depends on the number of levels
  • If each sample is encoded into n bits, then the bandwidth of PCM is nfs
  • The bandwidth of DPCM is almost the same as that of PCM signal, the only difference between PCM and DPCM is that the dynamic range is reduced in the DPCM signal
  • However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ
  • Hence there is bandwidth saving in Delta modulation

A comparison of different modulation schemes is as shown in the table below:

Parameter

PCM

DM

 DPCM

Number of bits

It can use 4, 8

or 16 bits per sample

It uses only

one bit for one sample

Bits can be more than one but are less than PCM

Level/Step size

Step size

is fixed

Step size is fixed

and

cannot be varied

A fixed number of levels are used.

Quantization error or Distortion

Quantization error depends

on the number

of levels used

Slope

overload distortion and granular

noise is

present

Slope overload distortion

and quantization noise is

present

Bandwidth

of the transmission channel

 The highest bandwidth is required since the number of bits is high

The lowest bandwidth is required

The bandwidth required is lower than PCM

Signal to Noise ratio

Good

Poor

Fair

Area of Application

Audio and

Video

Telephony

Speech and images

Speech and video

 

Ques.15. Multiplexing scheme which uses carrier phase shifting and synchronous detection to permit two DSB signals to occupy the same frequency band is called

  1. NBFM
  2. CDMA
  3. QAM
  4. FDMA

Answer.3. QAM

Explanation:-

QAM (quadrature amplitude modulation) is a method of combining two amplitude modulation (AM) signals into a single channel. QAM is used to achieve high levels of spectrum usage efficiency. This is accomplished by utilizing both the amplitude and phase components to provide a form of modulation.

1) QAM scheme uses carrier phase shifting and synchronous detection to permit two DSB signals to occupy the same frequency band.

2) QAM uses two carrier signals with the same frequency but with which are in quadrature. Quadrature here means out phase by 90°.

3) Since the two signals have the same frequency, they are detected using synchronous detection.

4) Hence, amplitude and the phase of the carrier frequency both vary with the message signal.

A constellation diagram of a QAM signal with 2 different amplitude levels and 8 different phases is shown:

 

Ques.16. In QAM, both ________ of a carrier frequency vary.

  1. frequency and amplitude
  2. phase and amplitude
  3. phase and frequency
  4. Phase and frequency

Answer.2. phase and amplitude

Explanation:-

QAM (quadrature amplitude modulation) is a method of combining two amplitude modulation (AM) signals into a single channel. QAM is used to achieve high levels of spectrum usage efficiency. This is accomplished by utilizing both the amplitude and phase components to provide a form of modulation.

QAM is a mixture of ASK and PSK. Hence, amplitude and the phase of the carrier frequency both vary with the message signal.

 

Ques.17. The main advantage of PCM system is lower

  1. Bandwidth
  2. Power
  3. Noise
  4. None of these

Answer.3. Noise

Explanation:-

PCM (Pulse Code Modulation):

  • PCM (Pulse Code Modulation) is a digital scheme for transmitting analog data.
  • The amplitude of an analog signal can take any value over a continuous range, i.e. it can take on infinite values.
  • But, digital signal amplitude can take on finite values.
  • Analog signals can be converted into digital by sampling and quantizing.

Advantages of PCM:

  • Encoding is possible in PCM.
  • Very high noise immunity, i.e. better performance in the presence of noise.
  • Convenient for long-distance communication.
  • Good signal-to-noise ratio.

Disadvantage of PCM:

  • The circuitry is complex.
  • It requires a large bandwidth.
  • Synchronization is required between transmitter and receiver.

 

Ques.18. An analog voltage is in the range of 0 to 8 V is divided in eight equal intervals for conversion to 3-bit digital output. The maximum quantization error is

  1. 0 V
  2. 0.5 V
  3. 1 V
  4. 2 V

Answer.2. 0.5 V

Explanation:-

The maximum quantization is given as:

Qe(max) = Δ/2

Δ = (Vmax − Vmin)/L

L = Number of levels

Calculation:

With n = 3, the number of levels will be:

L = 23 = 8

With analog input in the range 0 to 8 V and L = 8, the step size will be:

Δ = (8 − 0)/8 = 1

Now, the maximum quantization error will be:

Qe(max) = 1/2 = 0.5

 

Ques.19. To transmit N signals each band limited to fm Hz by time-division multiplexing will require a minimum bandwidth of

  1. fm
  2. 2fm
  3. N fm
  4. fm/N

Answer.3. N fm

Explanation:-

Given:

N signal with fm Bandwidth.

TDM minimum B.W = Nnfm

Assume n = 1 Bit

BWmin = Nfm

 

Ques.20. Which block or device does the data compression?

  1. Channel encoder
  2. Source encoder
  3. Modulator
  4. None of the mentioned

Answer.2. Source encoder

Explanation:-

Source encoding aims to convert information waveforms (text, audio, image, video, etc.) into bits, the universal currency of information in the digital world. Source encoder converts the digital or analog signal to a sequence of binary digits. This process is called source encoding or compression.

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