Signal To Quantisation Noise Power Ratio(SQNR) MCQ Quiz - Objective Question with Answer for Signal To Quantisation Noise Power Ratio(SQNR) - Download Free PDF

The SNR in a binary PCM system can be expressed as:

SNR = (6 × n) dB

where n is the number of bits used for quantization. This indicates that the SNR increases linearly with the number of bits used in the PCM system.

Effect of Increasing Number of Bits:

When the number of bits in the PCM system is increased from n to n + 2, the SNR also increases. The improvement in SNR can be calculated using the difference in SNR values before and after increasing the number of bits:

Improvement in SNR = SNR (n + 2) - SNR (n)

Substituting the values:

Improvement in SNR = (6 × (n + 2)) - (6 × n)

Improvement in SNR = 6 × n + 12 - 6 × n

Improvement in SNR = 12 dB

Hence the correct answer is 12 dB

Concept:

The quantization noise power P_q for a uniform quantizer is given by:

P_q = \(\frac{{\Delta^2}}{12}\)

where \(\Delta\) is the quantization step size.

Given:

Full-scale input voltage = 2B

Number of quantization levels = 2^N

Calculation:

Quantization step size:

\(\Delta = \frac{2B}{2^N} = \frac{B}{2^{N-1}}\)

Quantization noise power:

\(\frac{\Delta^2}{12} = \frac{1}{12} \cdot \left(\frac{B}{2^{N-1}}\right)^2 = \frac{B^2}{12 \cdot 2^{2(N-1)}} = \frac{B^2}{3 \cdot 2^{2N}}\)

Hence, the correct answer is option 1) \(\rm \frac{B^{2}}{3\left(2^{2 N}\right)}\)

Quantization:

• Quantization is the process of assigning voltage levels to the samples, is called quantization.
•  This quantization noise can be decreased by increasing the number of quantization levels, is called signal to quantization noise ratio.

Given that, 

\(s(t) = 5cos(200\pi t) \)

number of bits n =10.

\({s}\over{Nq} \)= \({3}\over{2} \)\(\times\)2²ⁿ

= \({3}\over{2}\)\(\times\)2^{2\(\times\)10}

= 1.5 x 2²⁰

= 1.57 \(\times\)10⁶

Here, option 3 is correct.

Concept:

A noise figure is a number by which the noise performance of a radio receiver, amplifier, mixer, or other circuit blocks can be specified.

The lower the value of the noise figure, the better the performance.

Essentially the noise figure defines the amount of noise an element adds to the overall system. It may be a pre-amplifier, mixer, or a complete receiver.

Noise figure(in dB) = SNR of input(dB) - SNR of output(dB) 

Calculation:

Given;

Input SNR =21.4 dB.

Output SNR =16 dB 

Noise figure(in dB) = SNR of input(dB) - SNR of output(dB) 

⇒ 21.4 - 16 = 5.4 dB

Concept:

The signal to noise ratio of a PCM system is given by:

SNR = 1.8 + 6n dB

n = number of encoded bits used.

As the number of encoded bits increases, the SNR is improved.

Application:

When the number of bits per sample is increased from n to n + 2, the SNR will be:

SNR2 = 1.8 + 6 (n+2)

SNR2 = 13.8 + 6n dB

We observe that:

SNR₂ = SNR + 12 dB

∴ The improvement in SNR is of 12 dB

Concept:

The signal to noise ratio of a PCM system is given by:

SNR = 1.8 + 6n dB

n = number of encoded bits used.

As the number of encoded bits increases, the SNR is improved.

Application:

When the number of bits per sample is increased from n to n + 2, the SNR will be:

SNR2 = 1.8 + 6 (n+2)

SNR2 = 13.8 + 6n dB

We observe that:

SNR₂ = SNR + 12 dB

∴ The improvement in SNR is of 12 dB

Concept:

A noise figure is a number by which the noise performance of a radio receiver, amplifier, mixer, or other circuit blocks can be specified.

The lower the value of the noise figure, the better the performance.

Essentially the noise figure defines the amount of noise an element adds to the overall system. It may be a pre-amplifier, mixer, or a complete receiver.

Noise figure(in dB) = SNR of input(dB) - SNR of output(dB) 

Calculation:

Given;

Input SNR =21.4 dB.

Output SNR =16 dB 

Noise figure(in dB) = SNR of input(dB) - SNR of output(dB) 

⇒ 21.4 - 16 = 5.4 dB

Concept:

SNR (Signal to Noise Ratio) for a sinusoidal message signal Quantized uniformly with ‘n’ bits in given by;  SNR_AB = 6n + 1.8

Calculation: Given, SNR = 40 dB, required at the output;

So, 40 ≤  6n + 1.8

6n ≥  40 – 1.8

n ≥  6.366

Concept:

• Power of sinusoidal signal is simply \(\frac{{{A^2}}}{2};\) where A is the amplitude of the sinusoid.
• Noise power is given by \(\frac{{{{\rm{\Delta }}^2}}}{{12}},\) where;

\({\rm{\Delta }} = step\;size = \frac{{{V_{p - p}}}}{{{2^n}}}\)  (Where V_p-p is the peak-to-peak voltage)

And n = number of bits.

Calculation:

Given sinusoid amplitude \(= \frac{1}{2}({V_{p - p}})\)    (∵ V_m = ½ of V_p-p )

So, Signal power \(= \frac{{1.V_{P - P}^2}}{{{{\left( 2 \right)}^2} \times 2}} = \frac{{V_{p - p}^2}}{8}\)

With n = 4 bits (number of encoded bits)

\({\rm{\Delta }} = \frac{{{V_{p - p}}}}{{{2^4}}}\)

\(So,\;Noise\;power = \frac{{{{\rm{\Delta }}^2}}}{{12}} = \frac{{4V_{pp}^2}}{{12 \times {2^8}}} = \frac{{V_{PP}^2}}{{12 \times {2^6}}}\) 

The signal to noise power ratio \(= \frac{{V_{pp}^2 \times 12 \times {2^6}}}{{8 \times V_{pp}^2}} = 12 \times {2^3}\)

The SNR in a binary PCM system can be expressed as:

SNR = (6 × n) dB

where n is the number of bits used for quantization. This indicates that the SNR increases linearly with the number of bits used in the PCM system.

Effect of Increasing Number of Bits:

When the number of bits in the PCM system is increased from n to n + 2, the SNR also increases. The improvement in SNR can be calculated using the difference in SNR values before and after increasing the number of bits:

Improvement in SNR = SNR (n + 2) - SNR (n)

Substituting the values:

Improvement in SNR = (6 × (n + 2)) - (6 × n)

Improvement in SNR = 6 × n + 12 - 6 × n

Improvement in SNR = 12 dB

Hence the correct answer is 12 dB

Concept:

The quantization noise power P_q for a uniform quantizer is given by:

P_q = \(\frac{{\Delta^2}}{12}\)

where \(\Delta\) is the quantization step size.

Given:

Full-scale input voltage = 2B

Number of quantization levels = 2^N

Calculation:

Quantization step size:

\(\Delta = \frac{2B}{2^N} = \frac{B}{2^{N-1}}\)

Quantization noise power:

\(\frac{\Delta^2}{12} = \frac{1}{12} \cdot \left(\frac{B}{2^{N-1}}\right)^2 = \frac{B^2}{12 \cdot 2^{2(N-1)}} = \frac{B^2}{3 \cdot 2^{2N}}\)

Hence, the correct answer is option 1) \(\rm \frac{B^{2}}{3\left(2^{2 N}\right)}\)

Quantization:

• Quantization is the process of assigning voltage levels to the samples, is called quantization.
•  This quantization noise can be decreased by increasing the number of quantization levels, is called signal to quantization noise ratio.

Given that, 

\(s(t) = 5cos(200\pi t) \)

number of bits n =10.

\({s}\over{Nq} \)= \({3}\over{2} \)\(\times\)2²ⁿ

= \({3}\over{2}\)\(\times\)2^{2\(\times\)10}

= 1.5 x 2²⁰

= 1.57 \(\times\)10⁶

Here, option 3 is correct.

For positive value

Step size \( = 0.05 = \frac{5}{{{2^n}}},\;n \cong 7\)

For negative values

Step size \( = 0.1\;V = \;\frac{5}{{{2^n}}},\;n \approx 6\)

SNR for Positive region \(= 1.8 + 6n = 43.8 dB\)

SNR for Negative region \(= 1.8 + 6n = 37.8 dB\)

So resulting SNR is best of the two values 43.8 dB

Concept:

The signal to noise ratio of a PCM system is given by:

SNR = 1.8 + 6n dB

n = number of encoded bits used.

As the number of encoded bits increases, the SNR is improved.

Application:

When the number of bits per sample is increased from n to n + 2, the SNR will be:

SNR2 = 1.8 + 6 (n+2)

SNR2 = 13.8 + 6n dB

We observe that:

SNR₂ = SNR + 12 dB

∴ The improvement in SNR is of 12 dB

Concept:

A noise figure is a number by which the noise performance of a radio receiver, amplifier, mixer, or other circuit blocks can be specified.

The lower the value of the noise figure, the better the performance.

Essentially the noise figure defines the amount of noise an element adds to the overall system. It may be a pre-amplifier, mixer, or a complete receiver.

Noise figure(in dB) = SNR of input(dB) - SNR of output(dB) 

Calculation:

Given;

Input SNR =21.4 dB.

Output SNR =16 dB 

Noise figure(in dB) = SNR of input(dB) - SNR of output(dB) 

⇒ 21.4 - 16 = 5.4 dB