Say I've got a motor drive system where two of the phase currents are measured, and the other is known because the motor is wired in star configuration and so the currents through each phase need to sum to zero.
So I setup my hardware to measure Iu and Iw, but maybe my Clarke and Park transform software asks for Iu and Iv, so I calculate Iw from my two measured currents and feed Iu(measured) and Iv(calculated) into the software, and the transform works.
But what happens when the measured signals are noisy (with both correlated noise appearing on Iu and Iw, and also random noise appearing on both channels)? Will the noise cancel out when applying the transform?
Iu(measured) = Iu(real current) + Iu(noise)
Iw(measured) = Iw(real current) + Iw(noise)
Iv(calculated) = Iu(measured) - Iw(measured) = Iu(real current) + Iu(noise) - (Iw(real current) + Iw(noise))
= Iu(real current) - Iw(real current) + Iu(noise) - Iw(noise)
Are there any considerations to take into account with this type of setup? Is there an issue with calculating one current from the other two?
My understanding of the subject was limited at best, so I consulted someone more knowledgeable.
Dave told me that there was no difference functionally between using a calculated Iw or a measured Iw:
Quoted from Dave (firstname.lastname@example.org):
Assuming equal amounts of noise on all the phase current measurements, there will be no difference from a performance point of view than if you had measured Iu and Iv, and calculated Iw. Whenever you have one phase which is reconstructed from the other two, that phase will see the noise from both other phases. For correlated noise, they must be added algebraically. For non-correlated noise, they must be added vectorally.
Also, your customer made a mistake in his equations. The calculated missing phase is always the algebraic sum of the NEGATIVE of the other two. So, here is what the equations should look like:
Iu(measured) = Iu + Iu(noise)
Iw(measured) = Iw + Iw(noise)
Iv(calculated) = -Iu(measured) – Iw(measured) = -Iu – Iu(noise) – Iw – Iw(noise)
As you can see, the two noises are of the same polarity, which means the correlated portions add algegraically.
Then we supply Iu(measured) and Iv(calculated) to our Clarke transform, which will attempt to recreate Iw. But watch what happens:
Iu(out) = Iu(measured) = Iu + Iu(noise)
Iv(out) = Iv(calculated) = -Iu – Iu(noise) – Iw – Iw(noise)
Iw(out) = -Iu(measured) – Iv(calculated) = -Iu – Iu(noise) +Iu +Iu(noise) +Iw +Iw(noise) = Iw + Iw(noise)
So, there is no difference. Two of the phases will have their measurement plus their noise, and the reconstructed phase will have the current plus the noise from both other phases.
I hope this helps!
Thanks for the correction. I didn't really check my equations for the calculated current, I was just trying to get the general point across.
Matt HeinSo, there is no difference. Two of the phases will have their measurement plus their noise, and the reconstructed phase will have the current plus the noise from both other phases.
So this means that the 'reconstructed phase' with have twice as much noise, which is bad, right? But you say that there is no difference functionally.
I think I may need to do some simulations to see exactly what happens.
Based on the Clarke and Park transform equations, I think Dave is getting at the point that even if you had simply measured Iu and Iv from the outset, the transform software still relies on the fact that the missing phase is the algebraic sum of the negative of the other two phases. In this case, Iw sees the noise from both phases.
So when you measure Iu and Iw to calculate Iv, Iv sees the noise from both phases. As long as you are not measuring all three, then one is going to be reconstructed.
I completely agree that simulations would help shed some light on this problem.
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