Does RMS(f1) +RMS(F2) = RMS(F1+F2)?

lennybogzy

If I hava non periodic waveform and I divide it into 3 cycles is taking the RMS of each cycle and averaging the same thing as taking the RMS of all 3 cycles at once?

es1

In general this is not true.

Let f1(t)=1 and and f2(t)=-1

rms(f1)=1
rms(f2)=1
rms(f1+f2)=0

es1

Just out of curiosity, if the signal is non-periodic, how did you determine how big a cycle is?

yungman

If it is non periodic, you need to find the average rates to do RMS calculation. That is if you have a random pulses, you need to use a frequency counter and find the pulse rate, then you can use the amplitude to get the RMS amplitude.

I did a lot of pulse counting work, they are random, just like current, it is not steady in microscopic level, we say it is 1A only because in average there are so many electrons part a point per second and we use that as the average current....Yes, current is a stream of electrons passing in a random rate.