Signal Sampling & Nyquist Theorem

Derive the 6 dB-per-bit rule from scratch for an 8-bit converter — where does the stray 1.76 come from? — then say what it does not promise.

1. Rounding to the nearest level leaves an error no bigger than half a step, and over a busy signal that error is spread evenly across the interval. Its variance is the integral of e² over a uniform distribution.

   $$\Delta =\frac{V_{FS}}{2^N},e\sim \mathcal{U}\left(-\frac{\Delta }{2},\frac{\Delta }{2}\right),{\sigma }_e^2=\frac{1}{\Delta }{\int }_{-\Delta /2}^{\Delta /2}{e}^{2}de=\frac{{\Delta }^2}{12}$$

2. The reference signal is a sine filling the full input range, so its power is half the square of its amplitude.

   $$P_{\text{sig}}=\frac{(V_{FS}/2{)}^2}{2}=\frac{V_{FS}^2}{8}$$

3. Divide the two. The full-scale range cancels, which is the reason the result is a pure function of bit count and not of voltage.

   $$\mathrm{SNR}=\frac{V_{FS}^2/8}{{\Delta }^2/12}=\frac{12}{8}{\left(\frac{V_{FS}}{\Delta }\right)}^2=\frac{3}{2}\cdot 2^{2N}$$

4. Take logarithms. The 6.02 is 20 log₁₀2 — one bit doubles the number of levels — and the 1.76 is the leftover 3/2 from the previous line.

   $$\mathrm{SN}{\mathrm{R}}_{\mathrm{dB}}=10{\log }_{10}\frac{3}{2}+20N{\log }_{10}2=1.76+6.02N$$

5. Evaluate at 8 bits, then at 16. Every extra bit is worth just over 6 dB, so CD audio's eight extra bits buy 48 dB.

   $$N=8:6.02(8)+1.76=49.92\mathrm{dB};N=16:98.08\mathrm{dB}$$