A full scale sinusoidal signal is applied to a 10-bit ADC. The fundamental signal component in the ADC output has a normalized power of 1 W, and the total noise and distortion normalized power is 10 W. The effective number of bits (rounded off to the nearest integer) of the ADC is ________.
7
8
9
10
ANSWER
SINAD , so ENOB , which rounds to 8 bits. The answer is option (B).
Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)
The effective number of bits (ENOB) comes from the measured signal-to-noise-and-distortion ratio (SINAD), not from the nominal 10-bit resolution.
SINAD is the ratio of the fundamental's power to the total noise-plus-distortion power. With :
For a full-scale sine wave into an ideal -bit quantiser, the signal-to-quantisation-noise ratio is .
ENOB is the for which an ideal quantiser would have the same ratio as the measured SINAD:
Rounded to the nearest integer, ENOB bits, which is option (B). Noise and distortion cost this 10-bit ADC about 2 bits of resolution.
KEY IDEA
ENOB is found by putting the measured SINAD into the ideal-quantiser formula and solving for . The nominal bit count does not enter. Both quantities given are powers, so use .
WATCH OUT FOR
Three traps. First, answering 10 (option D) because the ADC is 10-bit: ENOB depends on the measured SINAD, not on the nominal resolution. Second, misreading as . That gives 60 dB and ENOB , which rounds to 10, also option D. The correct value is . Third, using on a power ratio, which gives 100 dB and an ENOB of about 16, not one of the options.
From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 18 ADC & DAC questions, with notes
