# How to find the variance of a noise signal distorting an ADC measurement

Say we have a 12-bit ADC measuring a signal with a voltage range of 0-10V, but the signal is corrupted by uniformly distributed 3-bit white noise. What's the correct way to get the variance of the noise?

I would have thought that you calculate the maximum voltage of the noise signal by doing

$$\frac{2^{\text{noise bits}} \cdot V_{fs}}{2^n-1} = \frac{2^3\cdot 10\text{V}}{2^{12}-1} \approx 0.0195\text{V}$$

and then assume the distribution is centered at zero (since, as far as I can tell, the information given isn't sufficient to determine the actual center and it shouldn't affect the variance anyway) and then have the endpoints of the distribution be $$\pm\frac{0.0195}{2}$$, which can then just be plugged into the formula for the variance of a uniform distribution. However, the solution from my professor has the endpoints as $$a=\frac{0}{2^{12-3}}, b=\frac{10}{2^{12-3}}$$ (he didn't include units; I'm not sure if that's a mistake or if he's expressing $$a$$ and $$b$$ as multiples of the voltage resolution or something). The only explanation he gives for this is that "if [the] noise had more bits, its strength would increase". I can see that subtracting the noise bits from the total ADC bits in the exponent results in a larger value for $$b$$, since it makes the denominator smaller, but that doesn't seem like sufficient justification for why those particular expressions give the endpoints of the distribution.

I've been searching online for explanations and example problems of how to get the variance of a noise signal, both generally and in the context of an ADC measurement, but I everything I found was too general to be of use. Could someone explain where those expressions for $$a$$ and $$b$$ come from?

Why do you have the $$-1$$ in your denominator? With $$n$$ bits, you always have $$2^n$$ possible ADC output values. The highest numerical value is $$2^n-1$$ but the number of values is higher by one as $$0$$ is also an output value.
So, if you ditch the $$-1$$ in your expression, you get: $$\frac{2^3\cdot10V}{2^{12}}$$ which can be rewritten as $$\frac{10V}{2^9}$$ which is exactly your professor's expression for $$b$$. Including the unit or not is irrelevant, just choose one and stick to it.