GATE 2024 EC Q.57: solution and answer

GATE 2024 · ECQ.57 · 2 marks · NAT

Let $X (t) = A cos (2 π f_{0} t + θ)$ be a random process, where amplitude $A$ and phase $θ$ are independent of each other, and are uniformly distributed in the intervals $[- 2, 2]$ and $[0, 2 π]$ , respectively. $X (t)$ is fed to an 8-bit uniform mid-rise type quantizer. Given that the autocorrelation of $X (t)$ is $R_{X} (τ) = \frac{2}{3} cos (2 π f_{0} τ)$ , the signal to quantization noise ratio (in dB, rounded off to two decimal places) at the output of the quantizer is ______.

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Answer and solution

ANSWER

SQNR = 2^15 = 32768, which is about 45.15 dB.

Official answer: 45.00 to 45.30Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

Signal power is the autocorrelation at zero lag: $P_{X} = R_{X} (0) = \frac{2}{3}$ . Check: $A$ is uniform on $[- 2, 2]$ , so $E [A^{2}] = \frac{( 4 ) ^{2}}{12} = \frac{4}{3}$ , and $E [X^{2}] = \frac{E [ A ^{2} ]}{2} = \frac{2}{3}$ .
Since $∣ A ∣ \leq 2$ , we have $∣ X (t) ∣ \leq 2$ . The quantizer therefore covers $- 2$ to $+ 2$ , a total range of $4$ , and there is no overload.
An 8-bit quantizer has $L = 2^{8} = 256$ levels, so the step size is $Δ = \frac{4}{256} = \frac{1}{64}$ . A mid-rise quantizer has the same step size as a mid-tread one, so only the level positions differ.
Quantization noise power for a uniform quantizer with no overload is $P_{q} = \frac{Δ ^{2}}{12} = \frac{1}{12 \cdot 4096} = \frac{1}{49152} .$
$SQNR = \frac{P _{X}}{P _{q}} = \frac{2/3}{1/49152} = \frac{2}{3} \times 49152 = 32768 = 2^{15} .$
In dB: $10 lo g_{10} (2^{15}) = 15 \times 3.0103 \approx 45.15 dB$ .
The shortcut $6.02 n + 1.76 = 49.92$ dB assumes a full-scale sinusoid of fixed amplitude. Here $A$ is random, so the average signal power is only $\frac{2}{3}$ and not $\frac{A _{m a x}^{2}}{2} = 2$ . The shortcut therefore does not apply.

KEY IDEA

With a random amplitude, signal power is the average $R_{X} (0) = \frac{2}{3}$ , and quantization noise is $\frac{Δ ^{2}}{12}$ with $Δ = \frac{range}{2 ^{n}}$ taken over the full range of $\pm 2$ . Both are exact here, so SQNR $= 2^{15}$ .

WATCH OUT FOR

Using $6.02 n + 1.76 = 49.92$ dB. That formula is for a full-scale sinusoid, but here the amplitude is random and the average power is far below full-scale power. Another slip is taking the range as $2$ instead of $4$ when computing $Δ$ .

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