ISRO EC 2023 Q.24: solution and answer

ISRO 2023 · ECQ.24 · 1 mark · MCQ

What is the quantization noise power for N bit uniform quantizer with full scale input voltage 2B?

(A)
$\frac{B ^{2}}{3 ( 2 ^{2 N} )}$
(B)
$\frac{2 B}{3 ( 2 ^{N} )}$
(C)
$\frac{B ^{2}}{12 ( 2 ^{2 N} )}$
(D)
$\frac{B}{3 ( 2 ^{N} )}$

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Answer and solution

ANSWER

The quantization noise power is $\frac{B ^{2}}{3 ( 2 ^{2 N} )}$ , option (a).

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

An $N$ -bit quantizer has $2^{N}$ levels spanning the full-scale range $2 B$ , so the step size is $Δ = \frac{2 B}{2 ^{N}}$ .
For a uniform quantizer the rounding error is uniformly distributed over $(- \frac{Δ}{2}, \frac{Δ}{2})$ with density $\frac{1}{Δ}$ , so the noise power is $\int_{- Δ/2}^{Δ/2} e^{2} \frac{d e}{Δ} = \frac{Δ ^{2}}{12} .$
Substitute the step size: $\frac{Δ ^{2}}{12} = \frac{1}{12} \cdot \frac{4 B ^{2}}{2 ^{2 N}} = \frac{B ^{2}}{3 ( 2 ^{2 N} )} .$
This is option (a). Options (b) and (d) vary as $2^{- N}$ , so they are voltages rather than powers. Option (c) is what you get from a step of $B / 2^{N}$ , which is the step for a range of $B$ , not $2 B$ .

KEY IDEA

Quantization noise power for a uniform quantizer is $Δ^{2} /12$ , where the step $Δ$ equals the full-scale range divided by $2^{N}$ .

WATCH OUT FOR

Taking the step as $B / 2^{N}$ instead of $2 B / 2^{N}$ (the full-scale range is $2 B$ ), which gives $B^{2} / (12 \cdot 2^{2 N})$ . Another slip is picking an expression that varies as $2^{- N}$ , which is an rms voltage, not a power.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO ADC & DAC questions, with notes

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