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

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SUMMARY

The discussion centers on the relationship between the root mean square (RMS) of non-periodic waveforms and the averaging of RMS values from individual cycles. It is established that for non-periodic signals, such as f1(t)=1 and f2(t)=-1, the RMS of the combined signal (f1+f2) does not equal the average of the individual RMS values. Specifically, rms(f1) and rms(f2) both equal 1, while rms(f1+f2) equals 0. The conversation highlights the necessity of determining average rates for RMS calculations in non-periodic signals.

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Electrical engineers, signal processing specialists, and researchers interested in waveform analysis and RMS calculations will benefit from this discussion.

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

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