# Park's transform (dq0 transform)

Discussion in 'Homework Help' started by kokkie_d, Mar 8, 2011.

1. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
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0
Hi,

I am trying to understand Park's Transform also known as the dq0 transform. Although there are people who argue that they are not the same but in principle.

Anyway, Park's transform:

$
v_{d} = \frac{2}{3}\cdot (i_{a} \cdot cos(\theta) + i_{b}\cdot cos(\theta-120) + i_{c}\cdot cos(\theta+120))\\

v_{q} = -\frac{2}{3}\cdot (i_{a} \cdot sin(\theta) + i_{b}\cdot sin(\theta-120) + i_{c}\cdot sin(\theta+120))\\
v_{0} = \frac{1}{3}\cdot (i_{a} + i_{b} + i_{c})\\
$

I understand where the cosine and sines come from but not where the factors $\frac{2}{3}\$ and $\frac{1}{3}\$ come from.

Is there someone who knows?

The reason for my question is that I can not find the original explanation paper which is by Park out of 1928.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Have a look at the first 3 pages of the attached PDF.

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3. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
72
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Hi T_N_K,

I had seen that document before but it does not explain where the mentioned factors come from; it just shows they are there.

Any other ideas?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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782
Your original post appeared to ask how the '2/3' & '1/3' terms arose. Presumably these are scaling factors that ensure the correct relationships are maintained for the indicated vector quantities.

Was your question solely concerned with that matter or something more fundamental concerning the transform itself?

5. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
72
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Yes, I want to know where the scaling factors arise from; People are arquing these arise from the magneto motive force which indeed is a factor 3/2 higher than a single phase force generated by an individual current.
But the magnetomotive force in induction motors is generated by the way the coils and field are setup and its direction in relation (90 degrees angle from current and flux) as such in a phasor diagram the individual force enhance eachother resulting in the factor 3/2 higher than single phase, but this set of circumstances would not apply to the current, voltages or flux linkages. Yet, people include these factors in all their simulation and models.
So I am wondering: if what I wrote above is correct than the factor 2/3 would apply when one transforms the magneto motive force but not when one transforms the voltages, currents and flux linkages.
But if the above is not correct, then where does the factor of 2/3 come from?

So, I guess my question is more about the fundamentals concerning the transform.

What is the reason for including the correction factor?

Cheers,

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
My understanding is that the transformation maps all the (your) aforementioned electrical parameters onto the D & Q axes.

The undisputed consensus is that the transform is valid in all cases and that the parameter invariant or power invariant forms (take your pick) require the said scaling factors to ensure matching of values across the transform and inverse transform process.

Can you cite any supporting documentation concerning your discussion on MMF?

7. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
72
0
Hi T_N_K,

I just found the answer in a student manual (and with hindside I maybe should have thought of it myself).

The factor is there because of the perpendicular angle of the force to the torque and flux but because of the AC effects as explained in the attached screenshot from the student lecture notes.

Thanks anyway.

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8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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OK - But in essence this is what I was inferring in posts #4 & #6 with respect to the matter of scaling in relation to vector addition.

9. ### steveb Senior Member

Jul 3, 2008
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Perhaps this is just a typo, but the dq0 transformation is not from abc current components to dq0 voltage components. The transformation is either from voltage to voltage, or current to current.

Also, note that you can find a wide variety of transforms with different scale factors and different phase relations. One transform might reference the angle to the d-axis, and another to the q-axis. Also, one transform might scale the dq variables to the peak ABC frame values, and another version might correspond to the RMS value. It's very important to know the particular transform that is in use, or mistakes are very likely, particularly when calculating power.

Last edited: Mar 9, 2011
10. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
72
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Appologies, is indeed a typo.

Yes, but you can fairly easy recognise the similarities between dq or qd assumed conversion.

I am trying to verify some models of induction motors I found in various papers but all the results seemed to differ even from their published results.
That's why I am trying to find out where that factor came from since various books (as mentioned by steveb) use various approaches and some models seemingly seem to mix them up.

11. ### daveochs New Member

Dec 28, 2011
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I have come across the form of Park's transform that kokkie_d originally posted in various literature. However, MATLAB's SimPower Systems toolbox for simulink uses the following form:

vd=(2/3)(va*sin(wt) + vb*sin(wt-2∏/3) + vc*sin(wt+2∏/3))
vq=(2/3)(va*cos(wt) + vb*cos(wt-2∏/3) + vc*cos(wt+2∏/3))
v0=(1/3)(va + vb + vc)

Yet the MATLAB synchronous machine model as well as the literature that I've found the other definition in both have the same equations to model the machine. For example, everyone seems to agree that torque is controlled by q-axis current in a round-rotor, sinusoidal, permanent magnet synchronous machine. Depending on which definition of Park's transform you choose, q-axis current will be different though. Can anyone provide any insight? Thanks

12. ### steveb Senior Member

Jul 3, 2008
2,433
469
Personally, I use this transform exclusively. Coincidentally, I just ordered SimPower Systems toolbox and will be installing it later today. I'm happy to hear that it uses the one that I'm familiar with.

This particular definition is a bit of an oddball definition because the angle of the transform is referenced to the q-axis relative to the a-axis. This is odd because the dq-frame has the handedness orientation like the xy frame, and you would normally expect a rotation of an xy-frame to be referenced to the x-axis and not the y-axis.

As I mentioned in the previous post, there are many many different convensions for the Park transformation and it is critical to be consistant with all theory and use the same transformation exclusively in all derivations. This issue with torque being dependent on the values of current implies that the torque formula you use MUST be derived to be consistant with the particular Park's transform you are using. For exampe, this transform which I use has a corresponding torque formula with a 3/2 factor to compensate for the current scaling properly.

Last edited: Dec 28, 2011
13. ### dogcell New Member

May 26, 2014
1
0
you can search park's transform or dp0 transform on wikipedia. Through this genuine avenue,p transform may be easy.but if you're good at linear algebra(by Gilbert Strang ),well,that's an appllication of it.