J. McCalley

Machine Transformations

Space vectors

2

sin2a

a

Nn

Consider a 2-pole induction machine with the a-phase stator winding, having Na turns, is distributed about its axis as shown below.

The conductor density of the phase a winding is

where a negative density is interpreted as the return path of the turns.

Space vectors

3

If the coil carries a current ia, then the magneto-motive force (MMF, the magnetic circuit analogue to voltage in an electric circuit) is given by an application of Ampere’s circuital law, which relates the intensity of a magnetic field to the current that produces it.

A

dAJdlHF

Apply to the closed path shown in the figure:

cos

sin2

aa

aa

aaa

iN

diN

dinF

Space vectors

4

Symmetry allows us to associate half of this path mmf with each side of the machine. Thus, the mmf per airgap is

Because this and other variables are distributed around the air-gap periphery, it is convenient to represent them by space vectors. A space vector is represented by a complex number having•a magnitude equal to the peak amplitude of the variable and •an angle equal to the angular position of this positive peak value. Each space vector is considered to have a cosinusoidal distribution around the machine periphery.

cos2as

a

iNF

The MMF space vector is denoted aF

We may also denote the current as a space vector: aiThus, a

aa iN

F2

Space vectors

5

Symmetry allows us to associate half of this path mmf with each side of the machine. Thus, the mmf per airgap is

Because this and other variables are distributed around the air-gap periphery, it is convenient to represent them by space vectors. A space vector is represented by a complex number having•a magnitude equal to the peak amplitude of the variable and •an angle equal to the angular position of this positive peak value. Each space vector is considered to have a cosinusoidal distribution around the machine periphery.

cos2as

a

iNF

The MMF space vector is denoted aF

We may also denote the current as a space vector: aiThus, a

aa iN

F2

Space vectors

6

The two space vectors corresponding to F and i are illustrated below.

Space vectors

7

We may go through the same procedure for the phase currents b and c, in whichcase we represent them with space vectors

bi ciDefine a positive rotation of 2π/3 as:

2

3

2

1)3/2( jea j

Because the b and c phase windings are spatially displaced from the a phase winding by 120 and 240 degrees, respectively, then:

cc

bb

iai

aii2

The total effect of a set of instantaneous currents in all three phase windings is then found by vector addition according to:

cba iiii This is not the same as the sum of the currents: ia+ib+ic, which is zero!

Then a positive rotation of 4π/3 is:2

3

2

1)3/4(2 jea j

Space vectors

8

Time-domain waveforms Space vectors and their summation for the “instant” indicated on the other plot

cbacba iaaiiiiii 2

Observe: at the “instant,” ia is positive and just after its peak, ib and ic are negative.

Space vectors

9

cbacba iaaiiiiii 2Let’s add them mathematically:

2

3

2

1)3/2( jea j

2

3

2

1)3/4(2 jea j

Now use:

And we obtain:

cbcba

cba

iijiii

ijijii

2

3

2

1

2

3

2

1

2

3

2

1

But ia+ib+ic=0ia=-(ib+ic). Making this substitution in the above results in:

cbacbaa iijiiijiii 2

3

2

3

2

3

2

1

Space vectors

10

It will be convenient for us later on to scale this vector by 2/3 resulting in :

cba iijii 2

3

2

3

cba

cbacbas

iiji

iijiiijii

3

1

3

3

2

3

2

3

3

2

The above relation concentrates the effects of the three phase currents into a single complex variable. Very nice!

Now, let’s consider the converse process….

Space vectors

11

The converse 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?

si

We will project the vector onto the respective a, b, and c axes, as shown below. si

The analytic equivalent of this projection is:

sc

sb

sa

iai

iai

ii

Re

Re

Re

2

Just take the real part of si

Rotate the is vector forward by 240°. This places +b-axis where the +a-axis was. Then take real part.

Rotate the is vector forward by 120°. This places +c-axis where the +a-axis was. Then take real part.

Space vectors

12

We have represented the phase currents as space vectors. In doing so, however, the only thing required was they were balanced three-phase quantities. 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 sx

α-β transformation

13

The space vector can also be represented by two “phase magnitudes,” called xα and xβ in the real-imaginary complex plane, as illustrated below.

sx

We can express this relationship mathematically according to:

cbas xaaxxjxxx 2

3

2

The α-β components of the space vector can be calculated from the abc magnitudes according to:

cbas xxxxx

2

1

2

1

3

2Re

cbs xxxx2

3

2

3

3

2Im

Real partof axb

Real partof a2xc

Im partof axb

Im partof a2xc

α-β transformation

14

These two relations can be represented in matrix form as follows:

We define the matrix as the T-matrix:

cbas xxxxx

2

1

2

1

3

2Re

cbs xxxx2

3

2

3

3

2Im

c

b

a

x

x

x

x

x

2

3

2

30

2

1

2

11

3

2

2

3

2

30

2

1

2

11

3

2T

This transformation is also called the “Clarke transformation” for the person who developed it, Edith Clarke.

February 10, 1883 - October 29, 1959

AT&T, MIT, GE, U-Texas

α-β transformation

15

We will often represent the Clarke transformation in one of the two equivalent ways:

cbas xxxxx

2

1

2

1

3

2Re

cbs xxxx2

3

2

3

3

2Im

α-β transformation

16

It is interesting to observe, in a balanced system, so that xa+xb+xc=0,

cbas xxxxx

2

1

2

1

3

2Re

cbs xxxx2

3

2

3

3

2Im

aaaa

cbas

xxxx

xxxxx

2

3

3

2

2

1

3

2

2

1

3

2Re

We can solve for xb and xc in terms of xα and xβ, in which case we obtain the inverse Clarke transformation as

x

x

x

x

x

c

b

a

2

3

2

12

3

2

101

α-β transformation

17

We often represent the inverse Clark transformation in two equivalent ways:

x

x

x

x

x

c

b

a

2

3

2

12

3

2

101

Other transformations

18

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.

The space vector referred to it rotates at the synchronous speed ωs. We denote the corresponding space vector with a superscript “s” (stationary) according to:

jxxxs

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 frame.

QD

rjxxx

qd

ajxxx

Other transformations

19

Reference frame transformations

20

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

Revolving rotor with constant flux

21

Consider the rotating magnetic field (or a sinusoidal traveling wave) of an air gap, i.e., the plot on the left of the below figure and how it “moves” with time for a revolving rotor with constant flux (like that of a synchronous machine rotor). We see that, for fixed time (just one of the plots), there is sinusoidal variation of flux density with space. Also, if we stand on a single point on the stator (e.g., θ=90°) and measure flux (or flux density) as a function of time, we see that for fixed space (the vertical dotted line at 90°, and the red eye on the pictures to the right), there is sinusoidal variation of flux density w/time.

0

θ

N

N

N

θ

θ

θ

θ

θ

θ=90°

We can show that 3-phase, balanced currents of 3 stator windings separated by 120 degrees achieve the same thing. Let’s look at that….

Rotating magnetic field

22

Consider the 3-phase currents developed in the stator windings concentrated at points along the stator circumference 120 degrees apart. The currents are given by:

)240cos(

)120cos(

cos

tIi

tIi

tIi

sc

sb

sa

Whenever we have a current carrying coil, it will produce a magnetomotive force (MMF) equal to Ni. (MMF is the magnetic circuit analogue to voltage in an electric circuit.) And so each of the above three currents produce a time varying MMF around the stator. Each MMF has a maximum in space, occurring on the axis of the phase, of Fam, Fbm, Fcm, expressed as

)240cos()240cos()(

)120cos()120cos()(

coscos)(

tNItFtF

tNItFtF

tNItFtF

ssmcm

ssmbm

ssmam

Define the angle θ as measured from the a-phase axis, and consider points in the airgap. At any time t, the spatial maximums expressed above occur on the axes of the corresponding phases and vary sinusoidally with θ around the air gap. We can combine the time variation with the spatial variation in the following way:

)240cos()(),(

)120cos()(),(

cos)(),(

tFtF

tFtF

tFtF

cmc

bmb

ama

Now substitute (*) into these equations….

(*)

Each individual phase MMF here varies with θ around the air gap and has an amplitude that varies with time.

Rotating magnetic field

23

Now add the three MMFs above:

)240cos()240cos(),(

)120cos()120cos(),(

coscos),(

tFtF

tFtF

tFtF

smc

smb

sma

)240cos()240cos(

)120cos()120cos(

coscos

),(),(),(),(

tF

tF

tF

tFtFtFtF

sm

sm

sm

cba

Use cosαcosβ=0.5[cos(α-β)+cos(α+β)] and then simplify, and you will obtain:

)cos(2

3),( tFtF sm

If we plot the above along the airgap, we will see exactly the same MMF distribution that was created by the revolving rotor with constant flux.

Space vectors

24

)cos(2

3),( tFtF sm

0

θ

θ

θ

θ=90°