Pulse Code Modulation (PCM) System MCQ Quiz - Objective Question with Answer for Pulse Code Modulation (PCM) System - Download Free PDF

Explanation:

Improvement in Signal to Quantization Noise Ratio in PCM System

Definition: In a Pulse Code Modulation (PCM) system, the signal to quantization noise ratio (SQNR) is a measure of the quality of the digitized signal. It compares the power of the original signal to the power of the quantization noise, which is introduced during the digitization process.

Quantization Noise: Quantization noise is the error introduced when an analog signal is quantized to a finite number of levels. This noise is inherent in the process of converting an analog signal to a digital signal and is a function of the number of bits used in the quantization process.

Key Formula: The SQNR in a PCM system can be approximated using the following formula:

SQNR (dB) ≈ 6.02N + 1.76

where N is the number of bits per sample.

Analysis:

Given the initial number of bits per sample (N₁) is 16, the initial SQNR (SQNR₁) can be calculated as follows:

SQNR₁ (dB) ≈ 6.02 × 16 + 1.76 = 96.32 dB

When the number of bits per sample is increased to 17 (N₂), the new SQNR (SQNR₂) can be calculated as:

SQNR₂ (dB) ≈ 6.02 × 17 + 1.76 = 102.34 dB

The improvement in SQNR is the difference between the new SQNR and the initial SQNR:

Improvement in SQNR (dB) = SQNR₂ - SQNR₁

Improvement in SQNR (dB) = 102.34 dB - 96.32 dB = 6.02 dB

However, since we are looking for an approximate improvement, we can round this value to the nearest whole number, which gives us:

Improvement in SQNR ≈ 6 dB

Correct Option:

The correct option is:

Option 3: 6 dB

This option reflects the fact that increasing the number of bits per sample by 1 in a PCM system results in an improvement of approximately 6 dB in the signal to quantization noise ratio.

The correct option is 3

Concept:

The bit transmission rate in the PCM system is given by:

Rb = m × n × fs

where Rb = Bit transmission rate

fs = Sampling frequency

m = No. of message signal

n = No. of bits required for encoding

n is related to no. of quantization level by:

L = 2n

Calculation:

Given, L = 16

16 = 2n

n = 4

fm = 4 kHz

fs = 2fm

fs = 8 kHz

m = 1

Rb = 1 × 4 × 8

Rb = 32 K bits/sec

The correct option is 2

Concept:

• Quantizing noise is the noise or error introduced by the quantization process in the analog-to-digital conversion (ADC) process. It is the difference between the actual analog value and the quantized digital value.
• The level of quantization noise can be reduced by increasing the number of standard quantum levels, or in other words, by using more bits in the digital representation of the signal. This increases the resolution of the digital signal, allowing it to more accurately represent the original analog signal and thus reducing the amount of quantization noise.
• Increasing the sampling rate or the bandwidth will not directly reduce the quantization noise. In fact, sampling at a higher rate could potentially introduce more quantization noise because there are more opportunities for the quantization error to occur. Similarly, increasing the bandwidth means capturing more frequencies which again means more opportunities for the quantization error.

Concept:

In Pulse code modulation, 

R_b =  n ×  f_s  

where Rb is the bit rate that is in bits per second

n is no of bits per sample

fs is sample per second 

Analysis:

number of reference patterns is N;

bits = log₂(N);

number of samples = K;

the transmission rate in bits per sample (r) = \(\frac {bits} {samples}\)

r = \(\rm\frac{\log _2(N)}{K}\)

hence correct option is 4.

Concept:

In a PCM system with uniform quantization, the signal-to-noise ratio is directly proportional to 22n

S/N ratio ∝ 22n

Where n is the number of bits

Calculation:

The number of bits increases from 8 to 9.

The increase in the signal to quantization noise ratio by the factor is,

Concept:

The concept of converting an analog signal to its digital counterpart is explained with the help of the following diagram:

∴ The maximum quantization is given as:

\({Q_{e\left( {max} \right)}} = \frac{{\rm{\Delta }}}{2}\)

Δ = step size given by:

\({\rm{\Delta }} = \frac{{{V_{max}} - {V_{min}}}}{L}\)

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:

\({\rm{\Delta }} = \frac{{8 - 0}}{{8}} = 1\)

Now, the maximum quantization error will be:

\({Q_{e\left( {max} \right)}} = \frac{{1}}{2} = 0.5\)

Concept: 

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

Bit rate = n fS

fS = Sampling frequency

n = number of bits used for encoding.

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

L = 2n

n = log2 L

Calculation:

the maximum frequency for the signal will be 4 kHz.

n = 10 bits, R_b = 60 Kbits/sec

Bit rate = n fS

f_s = 6 kHz

And f_m is given as 4 kHz i.e

f_s ≥ 2f_m

f_s ≥ 8 kHz

So, f_s = 6kHz leads to undersampling.

Hence, option 1 and 2 can't be correct.

Option 4: f_s = 9 kHz leads to oversampling

Hence, option 3 is correct f_s = 8 kHz 

∴ The appropriate value of the sampling frequency (fs) will be 8 kHz

Concept:

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

L = 2n

Also, the bandwidth of PCM is given by:

\(BW=n{f_s}\)

n = number of bits to encode

fs = sampling frequency

Calculation:

For n = 8, the bandwidth will be:

B.W. = 8 fs

Similarly, For n = 16, the bandwidth will be:

B.W. = 16 fs

We observe that the Bandwidth is increased by 2 times.

Concept:

The Signal to Noise (SNR) Ratio for a PCM system is given by:

SNR = (1.8 + 6n) dB

Where ‘n’ is the number of bits per sample.

Calculation:

For n number of bits, the Signal to Noise Ratio will be:

SNR1 = (1.8 + 6n) dB

With the increase in two-bit ( "n+2" bits), the SNR becomes:

SNR2 = (1.8 + 6(n+2)) dB

SNR2 - SNR1 = 12 dB

i.e if the number of bits increased from 10 bits to 12 bits, then the SNR increases by 12 dB. 

Some salient features of a PCM system are:

• Immunity to transmission noise and interference.
• It is possible to regenerate the coded signal along the transmission path.
• The Quantization Noise depends on the number of quantization levels and not on the number of samples produced per second.

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.

• Companding refers to a technique for compressing and then expanding (or decompressing) an analogue or digital signal.
• For digital audio signals, companding is used in pulse code modulation (PCM)
• The process involves decreasing the number of bits used to record the strongest (loudest) signals
• In the digital file format, companding improves the signal-to-noise ratio at reduced bit rates, it gets the almost uniform signal to noise ratio

Concept:

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

R_b = 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

R_b = 4 × n × fs = 4 × 8 × 8000

R_b = 256 kbps

In the digital modulation techniques, pulse code modulation (PCM) has higher bandwidth compare to DPCM, DM, and ADM.

Comparison between the PCM, DPCM, DM, and ADM is as shown:

PCM:

• 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:

\(BW=N{f}\)

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

Concept:

Pulse code modulation is a technique to change an analog signal to digital data. A large bandwidth is required in PCM system.

Explanation:

A PCM encode contain three components:

1) The analog signal is sampled

2) The sampled signal is quantized

3) The quantized values are encoded as stream of bits

Diagram: