Error analysis for uniform quantization of uniform input

Question

I was told that

error energy $$\delta^2_e=\Delta^2/12,\quad \Delta=(B-A)/2^R$$ signal energy $$\delta^2_x = (B-A)^2/12$$ _{Here signal bounded:$A\le x[n]\le B.$ $[A,B]$ is split into $2^R$ subintervals.}

But i don't understand how the signal energy was derived. More precisely, i don't understand why signal energy equals to variance.

EDIT: Ah, probably because we center signal around $\frac{B-A}{2}$.

Answer 1

This assumes that both quantization error and original signal are uniformly distributed and have zero mean.

That's a good assumption for quantization noise, but it feels very arbitrary for the signal.

The variance will be proportional to the power (not energy), ONLY if the signal is mean free. Power is defined roughly as the "mean of the squares" and variance as the "mean of the squares after subtracting the signal mean".