One of the simplest and often overlooked waveforms for RMS evaluation is the squarewave. If one "thinks" of a squarewave as a bi-phase modulated DC voltage of V volts, which is simply multiplying by +-1, then the result is and RMS value identical to the DC value V before bi-phase modulation since multiplying by + or - 1 doesn't change the magnitude of the DC value.
Of course one must consider an ideal squarewave as having zero rise and fall times, and thus possessing an infinite bandwidth, however the harmonics fall off as 1/n^2, so quickly become small. Selecting a low squarewave frequency is advantageous here as the measurement instruments have finite bandwidths. Also one can't actually generate zero rise and fall times, and these finite times affect the result, however if quick relative to the squarewave period they tend to have a small effect.
RMS is the heating value of the waveform by definition and thus must contain all waveform artifacts in the measurement which includes the frequency content and the amplitude content.
One interesting squarewave type waveform is generated by a simple CMOS flip-flop (FF), which can produce a very accurate duty cycle of 50% at low frequencies when configured as a divide by 2. The RMS and average value of the output are identical, and these are exactly 1/2 the VDD supply ideally since the waveform transverses 0 to VDD volts. The small effect of finite rise and fall times can be factored in also as:
VRMS = VDD(sqrt(1/2 -rise/(3*P))), where equal linear rise and fall times are assumed, P is the output waveform period, and zero output impedance is assumed for the FF.
Quite some time ago we built a DMM test and verification device based upon this concept (used discrete P and NMOS devices to lower the FF output Z), and VDD was from a Precision 5.000V reference.
Here's the link:
https://www.eevblog.com/forum/testgear/ac-rms-dmm-tests/msg3940957/#msg3940957Best,