- #1
hadoque
- 43
- 1
I'm calculating current through a resistor by measuring a single current pulse, integrating it and multiplying it with its frequency. This would correspond to calculating an arithmetic mean.
I also tried calculating RMS of this waveform, using GNU Octave, and was a bit surprised by the difference.
My interpretation of RMS current is the DC-equivalent of the AC. Should be more or less the same, or should it? I started looking into this by looking at the definition of the two means. For the arithmetic mean, I use the absolute of the function, since I don't want negative charge to cancel positive.
Ia = ∫T2T1 √f(t)2dt/(T2-T1)
IRMS = √( ∫T2T1 f(t)2dt/(T2-T1) )
If we use f(t)=Sin(t) as an example, we get:
Ia=2/π ≈ 0.64
IRMS = 1/√2 ≈ 0.71
Using a square pulse wave with height 1, period 1 and time on T, we get
Ia= T
IRMS = √T
Quite a difference.
So, I think the arithmetic mean is the most correct in this case, since it is equal to the actual definition of current. So why do we use RMS? Am I missing something?
I have searched the net, but haven't found a discussion about this.
I also tried calculating RMS of this waveform, using GNU Octave, and was a bit surprised by the difference.
My interpretation of RMS current is the DC-equivalent of the AC. Should be more or less the same, or should it? I started looking into this by looking at the definition of the two means. For the arithmetic mean, I use the absolute of the function, since I don't want negative charge to cancel positive.
Ia = ∫T2T1 √f(t)2dt/(T2-T1)
IRMS = √( ∫T2T1 f(t)2dt/(T2-T1) )
If we use f(t)=Sin(t) as an example, we get:
Ia=2/π ≈ 0.64
IRMS = 1/√2 ≈ 0.71
Using a square pulse wave with height 1, period 1 and time on T, we get
Ia= T
IRMS = √T
Quite a difference.
So, I think the arithmetic mean is the most correct in this case, since it is equal to the actual definition of current. So why do we use RMS? Am I missing something?
I have searched the net, but haven't found a discussion about this.