Hi Neil,

The rms (Root Mean Square) value of a signal, over a definite period, does not depend on it being sinusoidal, or even periodic for that matter. This would be easier to write if I could put mathematical symbols here, but since I can't I will just state the following and trust you to transcribe it appropriately.

The rms value of a signal, from a time t0 to a time t0 + T, is equal to the square root of {{the definite integral of the square of the signal from the limits t0 to t0 + T} divided by the constant T}.

Note that if the signal is a DC voltage, it works out that the rms value of the DC voltage is the DC voltage.

Now, if the signal is sinusoidal, then there are lots of nice people who have done the math and can tell you to multiply by a judicious square-root-of-two here or there (or its reciprocal), and that will give you the rms of the sine wave. If you are using such a formula, then and only then it is imperative that you know the signal is sinusoidal.

To address the question of how to perform the above integral, and to look at how clock speed will affect the result, we must think about what an integral is. Then we can approximate it quite closely. If you already know this then you can skim over it.

The integral of a function of t, f(t), is the summation of a series of products, each being the product of the value of the function at t multiplied by the infinitesimally small step along t, dt. This is why you've heard the integral described as the area under the curve. It is the sum of a series of infinitessimaly narrow rectangles.

In the case of a sampled voltage you are measuring the value of f(t) at discrete steps along t, delta-t, such that delta-t is not infinitessimaly small. When you multiply the function value at each step by the step size, f(t) * delta-t, you are calculating the area of a rectangle that is approximately the same value as the integral of f(t) over that same step delta-t. By adding the areas of all of these discrete rectangles, you are calculating a sum that is approximately equal to the actual integral of the function over the whole interval. Now notice that the smaller you make the discrete step, delta-t, the more closely the small rectangles will approximate the actual value of f(t)dt. Consequently, the closer the sum of the rectangles will be to the actual integral over the same interval. This is why you will get a more accurate result with smaller step sizes.

So, if you sample the signal at intervals of delta-t, the rms value of the signal from a time t0 to a time t0 + T, where T = n*delta-t, in pseudo code is

for i = 1 to n
{
sum = sum + (f(t_i)^2 * delta-t)
rms = sqrt(sum/(i * delta-t))
}

Now I would further caution you that the rms value of a signal can be greatly dependant upon the interval over which the rms value is calculated. The rms value of a DC voltage will be equal to the DC voltage over any interval. However, if the signal is changing with time, then the rms value will change as you calculate it over different intervals. If a signal is monotonically increasing, the rms value will do likewise over increasing intervals. And if the signal is periodic, then the rms value will vary depending on how much of the period you calculate the rms value over. The rms value of a sine wave, calculated over half its period, will not be equal to the rms value calculated over 2/3 of a period. And, of course, in electronics people are interested in full period values.

I hope this helps. Good luck.