Why is finding the RMS current from a superimposed DC and AC signal the same as the pythagorean theorem?

Question

Suppose two currents of value $a$ (from dc source) and $b\sin(\omega t)$ (from ac source) are superimposed. We want to know the effective RMS current. By taking the average of the square of $a + b\sin(wt)$ I get the value as $\sqrt{a^2+b^2/2}$. However, that exactly matches the value if we square and add the RMS value individually like $a^2+(b/\sqrt{2})^2$ and then take its square root! Why do these two methods give the same answer? Its just like applying Pythagoras theorem!

Answer 1

It can be analytically easily shown. Let $T$ be the period of the ac signal ($T=1/2\pi\omega$). The square of the RMS current will then be given by:

$I_{rms}^2 = \frac{1}{T} \int_0^T (a+b \sin \omega t )^2dt=\frac{1}{T}[ \int_0^T a^2 dt + \int_0^T 2ab \sin \omega tdt +\int_0^T b^2 \sin^2 \omega tdt] $

Given that the second integral vanishes we have:

$I_{rms}^2 = \frac{1}{T}\int_0^T a^2 dt + \frac{1}{T}\int_0^T b^2 \sin^2 \omega tdt = I_{rms,a}^2 + I_{rms,b}^2$

which is just adding the squares of the RMS values of the two currents individually. Would this 2nd integral of the cross products be non zero (which could well be the case in other scenarios) one can not simply add the squares. Basically, the cross term between the instantaneous currents contributes at any given moment to the square of the instantaneous currents but it vanishes when one takes the time integral and does not contribute to the total RMS current.

Edits Since the main title was changed to why this looks similar to the Pythagorean theorem: I do not think there is any geometrical connection with a triangle. You can make some connection with generalizing the Pythagorean theorem to functional analysis in which you define inner products of functions by integrals. There you have a similar relation if the functions are orthogonal.

Answer 2

The Parseval and Planquerel are the key facts to note here. They are equivalent to the notion that the Fourier transform is a unitary map and that sinusoids of different frequency are orthogonal. You need to define your notion of orthogonality carefully and this leads to the rather technical idea of the tempered distributions, but here you can make things easier for yourself and consider only functions that are periodic over the period of the sinusoid you're looking at. Then you can resolve all such periodic functions into Fourier series and the sinusoids with periods that are integer fractions of the period of consideration can be made in to an orthonormal basis for such periodic functions. Crucially here we must include a sinusoid of zero frequency to make our basis complete.

You can take the fact that the sinusoids are orthogonal to be a given, or you can think of them as resulting from a certain Sturm-Liouville problem, namely one with $p(x)=w(x) = const$ and $q(x)=0$ in the Wikipedia page notation. Sturm-Liouville systems alway yields a complete set of orthogonal eigenfunctions, but the inner product in question is not always the simple integral of the product of the functions in question as it is for Fourier series.