Hi David,

I looked at this a bit more after I posted the above correction yesterday.

If you have only a DC voltage, then calculating the integral from my previous post is trivial, and the result is that the rms value of the DC voltage is the value of the DC voltage.

Now consider calculating the rms value of a non-constant signal (e.g. a sine wave) offset by a DC bias. Let's use f(t) = A*sin(t) + B (where A is the amplitude of the sin wave and B is the DC offset). Now when you square f(t) you end up with a cross term.

f(t)^2 = A^2*sin^2(t) + 2AB*sin(t) + B^2

Now if you multiply that squared funtion by a factor (say delta-t divided by T from the previous post) and take the square root of the single term you get f(t1)*(D)^(1/2). | D = delta-t/T (which, if you read my previous posts you will see, in this case is 1)

{[A^2*sin^2(t1) + 2AB*sin(t1) + B^2]*D}^(1/2)

= [A^2*sin^2(t1) + 2AB*sin(t1) + B^2]^(1/2)*D^(1/2)
= [A*sin(t1) + B]*D(1/2) = f(t1)*D^(1/2)

But, when you integrate (or sum in the case of numeric approximation) you are adding many of these terms together before taking the square root.

{[A^2*sin^2(t1) + 2AB*sin(t1) + B^2 + A^2*sin^2(t2) + 2AB*sin(t2) + B^2 + A^2*sin^2(t3) + 2AB*sin(t3) + B^2 + ...]*D}^(1/2)

= [A^2*sin^2(t1) + 2AB*sin(t1) + B^2 + A^2*sin^2(t2) + 2AB*sin(t2) + B^2 + A^2*sin^2(t3) + 2AB*sin(t3) + B^2 + ...]^(1/2) * D^(1/2)

Notice that [A^2*sin^2(t1) + 2AB*sin(t1) + B^2 + A^2*sin^2(t2) + 2AB*sin(t2) + B^2 + A^2*sin^2(t3) + 2AB*sin(t3) + B^2 + ...]^(1/2) no longer nicely equals f(t), which means that in general it no longer equals a nice linear sum of A*sin(t) + B.

This means that the rms value of a signal with a DC offset, in general, is not equal to the linear sum of the rms value of the signal plus the rms value of the offset.

Note that for simplicity's sake I assumed that delta-t was constant.