Baseband Transmission MCQ || Baseband Transmission Questions and Answers

21. 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.

 

22. Phase encoded group consists of

  1. Manchester coding
  2. Bi-phase-mark
  3. Miller coding
  4. All of the mentioned

Answer.4. All of the mentioned

Explanation:-

Different types of phase-encoded waveforms consist of
  • Biphase — mark or Manchester I code.
  • Biphase — level or Manchester II code.
  • Biphase — space 
  • Delay modulation (DM).
  • Miller coding

The phase-encoding schemes are used in magnetic recording systems and optical communications and in some satellite telemetry links.

With bi-φ-L, a one is represented by a half-bit-wide pulse positioned during the first half of the bit interval; a zero is represented by a half-bit-wide pulse positioned during the second half of the bit interval.

With bi-φ-M, a transition occurs at the beginning of every bit interval. A one is represented by a second transition one- half bit interval later; a zero is represented by no second transition.

 

23. 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:-

Baseband communication is given.

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

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

Where Rb = n fs

n = number of bits used to represent samples

fs = sampling rate.

Calculation:

In the Question, it is asked about minimum Bandwidth. ∴ we will consider that the transmission of data is taking place through sin pulses.

Given fs = 8 kHz

n = 8 bits/sample

M = 4

Rb = nfs

${\left( {BW} \right)_{min}} = \frac{{{R_b}}}{{2{{\log }_2}M}}$

$= \frac{{n\;{f_s}}}{{2{{\log }_2}M}} = \frac{{8 \times 8}}{{2{{\log }_2}4}}$

$= \frac{{64}}{{2 \times 2}} = 16\;kHz$

 

24. Comparing Delta Modulation (DM) with PCM systems, DM requires:

1. a lower sampling rate

2. a higher sampling rate

3. least bandwidth

4. simpler hardware

  1. 1, 2 and 4 only
  2. 1, 2 and 3 only
  3. 2, 3 and 4 only
  4. 1, 3 and 4 only

Answer.3. 2, 3 and 4 only

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.
  • However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ Hence there is a bandwidth saving in Delta modulation.
  • DM has a simple hardware requirement in comparison to PCM.

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

Parameter

PCM

Delta Modulation (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

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

Highest bandwidth is required since the number of bits are high

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

 

25. Which method is called as differential encoding?

  1. NRZ-L
  2. NRZ-M
  3. NRZ-S
  4. None of the mentioned

Answer.2. NRZ-M

Explanation:-

In Non-Return Zero- Mark or NRZ-M PCM waveform, the 1 or mark is represented by a change in level, and the 0 or space, is represented by no change in level. Referred to as differential encoding, NRZ-M PCM is used primarily in magnetic tape recording.

 

26. Which method is preferred in magnetic tape recording?

  1. NRZ-L
  2. NRZ-M
  3. NRZ-S
  4. None of the mentioned

Answer.2. NRZ-M

Explanation:-

In Non-Return Zero- Mark or NRZ-M PCM waveform, the 1 or mark is represented by a change in level, and the 0 or space, is represented by no change in level. Referred to as differential encoding, NRZ-M PCM is used primarily in magnetic tape recording.

 

27. Companding is used to

  1. Overcome quantizing noise in PCM
  2. Protect small signals in PCM from quantizing noise
  3. PCM receivers to reduce impulse noise
  4. Increase the power content of the modulated signal

Answer.2. Protect small signals in PCM from quantising noise

Explanation:-

The smaller the signal strength, the stronger it is hit by the quantization process. Companding is used to protect these small strength signals from quantization noise.

Companding:

  • Companding, also known as Companded PCM, is a non-uniform quantization technique.
  • It is implemented to improve the signal to quantization noise ratio of weak signals.
  • The compression of the signal at the transmitter and expansion at the receiver is combined to be called companding.
  • Companders like A-law and μ-law Companders are used in PCM to compress signals before input to the ADC in the transmitter stage and expand it after the input to DAC at the receiver.

 

28. Bandwidth of an N-bit binary-coded PCM signal for modulating a signal sampled at ‘f’ Hz is

  1. f/N Hz
  2. f/N2 Hz
  3. Nf Hz
  4. N2f Hz

Answer.3. Nf Hz

Explanation:-

  • PCM stands for Pulse Code Modulation.
  • With PCM, the amplitude of the analog signal is sampled at regular intervals and translated into a binary number.
  • The difference between the original signal and the translated digital signal is called the quantizing error.

The bandwidth of PCM is given by

BW = Nf

N = number of bits to encode

f = sampling frequency

Note:

The number of levels for an N-bit PCM system is given by:

L = 2N

We can also state that the number of bits for a given quantization level will be:

N = log2 L

 

29. In which waveform logic 1 is represented by half bit wide pulse and logic 0 is represented by the absence of pulse?

  1. Unipolar RZ
  2. Bipolar RZ
  3. RZ-AMI
  4. Manchester coding

Answer.1. Unipolar RZ

Explanation:-

Unipolar signal In positive unipolar logic signals, logic 1 is represented by a positive voltage and 0 by a zero. This is often called on-off keying.

In unipolar RZ waveform, logic 1 is represented by half bit wide pulse, and logic 0 is represented by the absence of a pulse.

 

30. Four voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signals using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be

  1. 8 kbps
  2. 64 kbps
  3. 256 kbps
  4. 512 kbps

Answer.3. 256 kbps

Explanation:-

The bitrate of a PCM system for an encoded signal sampled at a frequency of fs is given by:

Rb = m n fs

fs = Sampling frequency

n = number of bits used for encoding.

m = number of message signals

n is related to the number of quantization levels (L) as:

L = 2n

or n = log2L

Calculation:

Since the sampling frequency is not mentioned, we’ll assume it to be sampled at the Nyquist rate, i.e.

fs = 2fm

fm = Maximum frequency present at the modulating signal.

∴ For the given band-limited signal with a frequency 4 kHz, the sampling frequency will be:

fs = 2 × 4 = 8 kHz

With L = 256, the number of bits will be:

n = log2 256 = log2 28

n = 8 bits

Now for 4 Voice signals

Rb = 4 × n × fs = 4 × 8 × 8000

Rb = 256 kbps

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