For an ideal N-bit ADC with a full-scale sine input, the signal-to-noise ratio in dB is given by which of the following?

• A. SNRdB = 6.02N - 1.76
• B. SNRdB = 6.02N + 1.76
• C. SNRdB = 6.02N + 3.01
• D. SNRdB = 6.02N + 1.76 dB

Answer: (D)

The key idea is how quantization noise behaves in an ideal N-bit ADC and how it compares to the signal when the input is a full-scale sine.

Quantization error in an ideal ADC is modeled as a uniform noise with power Δ^2/12, where the step size Δ is the full-scale range V_FS divided by 2^N. For a full-scale sine input, the peak is V_FS/2, so its RMS value is V_FS/(2√2), giving signal power V_FS^2/8. The noise power is Δ^2/12 = V_FS^2/(12·2^{2N}). The SNR is the ratio of these powers:

SNR = (V_FS^2/8) / (V_FS^2/(12·2^{2N})) = (12·2^{2N})/8 = (3/2)·2^{2N}.

Convert to dB: SNRdB = 10 log10((3/2)·2^{2N}) = 10 log10(3/2) + 20N log10(2) ≈ 1.76 + 6.02N dB. So the correct expression is SNRdB = 6.02N + 1.76 dB.

This also clarifies why the offset is +1.76 dB and why the per-bit gain is about 6.02 dB, reflecting the 20 log10(2) increase per additional bit.