ISRO EC 2025 Q.26: solution and answer

ISRO 2025 · ECQ.26 · 1 mark · MCQ

For a 10-bit ADC, the signal to quantization noise ratio (SNR) is 62 dB. If the bit resolution is increased by 2 bits, what would be the new SNR value?

(A)
62 dB
(B)
74 dB
(C)
50 dB
(D)
31 dB

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Answer and solution

ANSWER

The new SNR is 74 dB, option (B).

Official answer: BSource: ISRO's final answer key of 15 May 2026

For an ideal $N$ -bit converter with a full-scale sinusoidal input, $SNR = 6.02 N + 1.76 dB$ .
Check the given case: $N = 10$ gives $6.02 (10) + 1.76 = 61.96 \approx 62 dB$ , which matches the value in the question.
Each extra bit halves the quantization step, so the quantization noise power falls by a factor of 4, which is $10 lo g_{10} 4 \approx 6.02 dB$ of improvement per bit.
Increasing the resolution by 2 bits gives $Δ SNR = 2 \times 6.02 = 12.04 dB$ .
New SNR $= 62 + 12 = 74 dB$ . Directly, $N = 12$ gives $6.02 (12) + 1.76 = 74.0 dB$ .
The other options do not fit: $62 dB$ means no change, $50 dB$ would be a drop, and $31 dB$ would be a halving. More bits can only improve the SNR, so only $74 dB$ is correct.

KEY IDEA

An ideal ADC gains about 6.02 dB of SNR for every extra bit ( $SNR = 6.02 N + 1.76 dB$ ), so adding 2 bits adds about 12 dB.

WATCH OUT FOR

Halving or keeping the SNR when bits are added, or treating the 6 dB per bit as 3 dB per bit. More bits always improve the SNR, by about 6 dB each.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO ADC & DAC questions, with notes

ISRO EC 2025 Q.26: solution and answer

Answer and solution

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