1. Machine Transformations
J. McCalley

2. Space vectors
We have already considered this concept, but without calling it a “space vector,” when we discussed the rotating magnetic field in the slide deck called “Wind GeneratorsSteady-StateBasics” (see slides 8-12). We repeat slide 12 from that slide deck in the next slide. 2

3. Rotating magnetic field
Now let’s consider the magnetic field from all three windings simultaneously.
(1)
(2)
(3)
Add them up, then perform trig manipulation to obtain:
(4)
Notice that each of (1), (2), and (3) has a spatial maximum which is fixed in space, as dictated by the second term of each the 3 expressions. But the spatial maximum of (4) rotates. This is a characteristic of a rotating magnetic field. One can observe this using the following:
The shape of the individual winding fields throughout the airgap are spatially fixed, but their amplitudes pulsate up and down. In contrast, the amplitude of the composite is fixed in time, but it rotates in space. What you will see in the video are just the variation of the maximum field point.

4. Space vectors
Consider the current ia=Iacosωt. Let’s represent it as a vector having
a magnitude equal to the peak amplitude of the variable and
an angle equal to the angular position of the peak MMF produced by this current around the circumference of a machine.
The vector is then given by:
We may think of it as oscillating up and down in a direction as shown below. 4

5. Space vectors
We may do the same for phase currents b and c, in which case we represent them as
Identify a positive rotation of 2π/3 as:
Then a positive rotation of 4π/3 is:
Because the b and c phase windings are spatially displaced from the a phase winding by 120 and 240 degrees, respectively, then:
The total effect of a set of instantaneous currents in all three phase windings is then found by vector addition according to:
This is not the same as the sum of the currents: ia+ib+ic, which is zero! 5

6. Time-domain waveforms
Space vectors
Time-domain waveforms
Current vectors and their summation for the “instant” indicated on the other plot
+b axis
+a axis
+c axis
Note: Need to check these plots.
Perspective 1: Assume that the time sequence of the phase currents is consistent with the spatial sequence. In this case, the above plots are correct if we consider phase rotation on the right-hand plot to be in the CW direction. If we consider phase rotation in the CCW direction, then the above plots need to be changed, because the time domain waveform indicates a phase sequence of a-b-c but the right-hand plot indicates a phase sequence of a-c-b. To fix this (for phase rotation in the CCW direction), we can exchange the ib and ic labels on the time-domain waveforms (giving us a-c-b phase sequence), but then we also need to exchange the lengths of the ib and ic vectors on the right-hand-plot.
Perspective 2: Assume that the time sequence of the phase currents is different than the spatial sequence. In this case, the above is OK.
Observe: at the “instant,” ia is positive and just after its peak, ib and ic are negative.
The current vector for each phase oscillates up and down its axis, but the composite (space vector) i rotates.
(Eqt 1) 6

7. Space vectors Let’s add them mathematically: (Eqt 1) Now use:
And we obtain:
Let’s assume that the three quantities ia, ib and ic are balanced 3-phase quantities.
Then ia+ib+ic=0ia=-(ib+ic). Making this substitution in the above results in: 7

8. Space vectors
It will be convenient for us later on to scale this vector by 2/3 resulting in :
The above relation concentrates the effects of the three phase currents into a single complex variable. Very nice!
Note, because
and
that
(Eqt 2)
So the above is how to obtain a space vector given values of the balanced currents.
Now, let’s consider the reverse process…. 8

9. Space vectors The reverse process….
Consider that we know the space vector and that we want to find the instantaneous values of the individual phase currents. How to do this?
We project the vector onto the respective a, b, and c axes, as shown below.
+b-axis
The analytic equivalent of this projection is:
Take the real part of the is vector.
Rotate the is vector forward by 240°. The projection of this rotated vector on the a-axis (via taking real part) is the same as the projection of the original vector on the b-axis.
+a-axis
Rotate the is vector forward by 120°. The projection of this rotated vector on the a-axis (via taking real part) is the same as the projection of the original vector on the c-axis.
+c-axis
is may be decomposed to other sets of 3 vectors (e.g., 3 vectors collinear with is having 1/3 the magnitude), but if we impose that the 3 vectors be on the a, b, and c axes, then the projection is unique.
Recall (Eqt 1):
Recall (Eqt 2): 9

10. Space vectors
We have represented the phase currents as space vectors. In doing so, however, the only thing required was they were balanced three-phase quantities (see slide 7). Any other variables can be similarly represented as long as they are balanced three-phase quantities, e.g., currents, voltages, and fluxes.
Let’s generically refer to any such variables as xa, xb, and xc and the corresponding space vector as 10

11. α-β transformation
The space vector can also be represented by two phase magnitudes, called xα and xβ in the real-imaginary complex plane, as illustrated below.
Whereas vectors corresponding to xa, xb, and xc oscillate up and down the a, b, and c axes, respectively, the vectors corresponding to xα and xβ oscillate up and down the α and β axes, respectively. xc xb xa
We can express this relationship mathematically according to:
We also know (from Eqt 2, slide 8) that :
The α-β components of the space vector can be calculated from the abc magnitudes according to: 11

12. α-β transformation
These two relations can be represented in matrix form as follows:
We define the matrix as the T-matrix:
This transformation is also called the “Clarke transformation” for the person who developed it, Edith Clarke. 12

13. α-β transformation
E. Clarke, Circuit Analysis of AC Power Systems. New York: Wiley,
1943, vols. I, II.
February 10, October 29, 1959
AT&T, MIT, GE, U-Texas 13

14. α-β transformation
We will often represent the Clarke transformation in one of the two equivalent ways:
Note that 14

15. α-β transformation
It is affirming to observe, in a balanced system (which implies that xa+xb+xc=0),
Using xα=xa, we can solve for xb and xc in terms of xα and xβ, in which case we obtain the inverse Clarke transformation as 15

16. α-β transformation
We often represent the inverse Clark transformation in two equivalent ways: 16

17. Some observations on xα and xβ
Let’s expand
using
Let’s apply the trig identity:
Recalling cos120=-1/2, sin120=√3/2, cos240=-1/2, sin240=-√3/2, the above becomes
From this we learn that:
xα=xa (we knew that from slide 14)
xα is a sinusoidally varying quantity.
The amplitude of xα is equal to the amplitudes of xa, xb, and xc.
Similar work on xβ leads to the following conclusions:
xβ is a sinusoidally varying quantity.
The amplitude of xβ is equal to the amplitudes of xa, xb, and xc.
xα and xβ are 90 degrees phase-shifted (we also knew this already)
This is because of our choice of 2/3 as the coefficient in the transformation. 17

18. Other transformations
When we transform variables to the α-β transformation we have just established, we are said to be working in the Stator Reference Frame.
This reference frame is aligned with the stator, and the rotational speed of this reference frame, since it is aligned with the stator, is 0. It is a fixed reference frame.
The space vector referred to it rotates at the synchronous speed ωs. We will denote the corresponding space vector with a superscript “s” (stator) according to:
We can also define a space vector aligned with the rotor. The reference frame in this case is called the D-Q reference frame and it rotates with angular speed of ωm. Therefore the space vector referred to it rotates at the slip speed of ωr.
Finally, we can also define a space vector aligned with the synchronous reference frame, at a speed of ωs. The space vector referred to it does not rotate, that is, it presents constant real and imaginary parts. This is called the d-q reference frame. 18

19. Other transformations
The below figure shows the three different reference frames discussed on the previous slide: stator (α-β); rotor (D-Q); and synchronous (d-q).
xβcos(90-θm)
xαcosθm
We illustrate just one of the transformations, to the D-Q coordinates:
xD=xαcosθm+xβcos(90-θm)=xαcosθm+xβsinθm
We can make a similar projection for xQ to find… 19

20. Other transformations
We can make a similar projection for xQ to find… 20

21. Other transformations
You can find a book on electric machine theory at This book contains ch 3, titled “Reference frame theory.” Below are some cut-outs from that chapter.
P. Krause, O. Wasynczuk, and S. Sudhoff, “Analysis of electric machinery and drive systems,” 2nd edition, IEEE Press, 2002. 21

22. Reference frame transformations
Our objective now is to represent balanced, but time-varying (dynamic) behavior of a DFIG. There will be three types of equations that we will need:
Voltage equations
Flux linkage equations
Motion equations
We will do this by transforming to the d-q coordinate reference frame
We desire to achieve this objective because the control becomes much easier.
One can see, for example, on the more detailed block diagram on the right, when in the d-q frame, we may specify Te by idr and Qs by iqr. 22

23. Reference frame transformations
We will come back, later, to the Clarke (α-β) transformation, because we will find it very useful in designing the switching sequence for the RSC and the GSC. 23