Dq Transformation

5/21/2018 Dq Transformation

1/51

J. McCalley

d-q transformation


2/51

Machine model

2

Consider the DFIG as two sets of abc windings, one on the stator and one on the rotor.

m

m


3/51

Machine model

3

The voltage equation for each phase will have the form:

That is, we can write them all in the following form: dt

tdtritv )()()(

cr

br

ar

cs

bs

as

cr

br

ar

cs

bs

as

r

r

r

s

s

s

cr

br

ar

cs

bs

as

dt

d

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

00000

00000

00000

00000

00000

00000

All rotor terms are given on the

rotor side in these equations.

We can write the flux terms as functions of the currents, via an equation for each flux of

the form =Lkik, where the summation is over all six winding currents. However, wemust take note that there are four kinds of terms in each summation.


4/51

Machine model

4

Stator-stator terms: These are terms which relate a stator winding flux to a statorwinding current. Because the positional relationship between any pair of stator

windings does not change with rotor position, these inductances are not a function of

rotor position; they are constants. Rotor-rotor terms: These are terms which relate a rotor winding flux to a rotor

winding current. As in stator-stator-terms, these are constants.

Rotor-stator terms: These are terms which relate a rotor winding flux to a statorwinding current. As the rotor turns, the positional relationship between the rotor

winding and the stator winding will change, and so the inductance will change.

Therefore the inductance will be a function of rotor position, characterized by rotor

angle . Stator-rotor terms: These are terms which relate a stator winding flux to a rotor

winding current. As described for the rotor-stator terms, the inductance will be a

function of rotor position, characterized by rotor angle .


5/51

Machine model

5

There are two more comments to make about the flux-current relations:

Because the rotor motion is periodic, the functional dependence of each rotor-statoror stator-rotor inductance on is cosinusoidal.

Because changes with time as the rotor rotates, the inductances are functions oftime.

We may now write down the flux equations for the stator and the rotor windings.

cr

br

ar

cs

bs

as

rrs

srs

cr

br

ar

cs

bs

as

i

i

i

i

i

i

LL

LL

Each of the submatrices in the inductance matrix is a 3x3, as given on the next slide

Note here that all quantities are now referred to the

stator. The effect of referring is straight-forward,

given in the book by P. Krause, Analysis of ElectricMachinery, 1995, IEEE Press, pp. 167-168. I willnot go through it here.


6/51

Machine model

6

msmm

mmsm

mmms

s

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

mrmm

mmrm

mmmr

r

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

mmm

mmm

mmm

msr LL

cos120cos120cos

120coscos120cos

120cos120coscos

Diagonal elements are the self-inductance of

each winding and include leakage plus mutual.

Off-diagonal elements are mutual inductances

between windings and are negative because120 axis offset between any pair of windings

results in flux contributed by one winding to

have negative component along the main axis

of another winding.

T

sr

mmm

mmm

mmm

mrs LLL

cos120cos120cos

120coscos120cos

120cos120coscos

m

m


7/51

Machine model

7

msmm

mmsm

mmms

s

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

mrmm

mmrm

mmmr

r

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

Summarizing.

cr

br

ar

cs

bs

as

cr

br

ar

cs

bs

as

r

r

r

s

s

s

cr

br

ar

cs

bs

as

dt

d

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

00000

00000

00000

00000

00000

00000

cr

br

ar

cs

bs

as

rrs

srs

cr

br

ar

cs

bs

as

i

i

i

i

i

i

LL

LL

T

sr

mmm

mmm

mmm

mrs LLL

cos120cos120cos

120coscos120cos

120cos120coscos

mmm

mmm

mmm

msr LL

cos120cos120cos

120coscos120cos

120cos120coscos


8/51

Machine model

8

msmm

mmsm

mmms

s

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

mrmm

mmrm

mmmr

r

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

cr

br

ar

cs

bs

as

rrs

srs

cr

br

ar

cs

bs

as

r

r

r

s

s

s

cr

br

ar

cs

bs

as

i

i

i

i

i

i

LL

LL

dt

d

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

00000

00000

00000

00000

00000

00000

Combining.

It is here that we observe a difficultythat the stator-rotor and rotor-stator terms, LsrandLrs, because they are functions of r, and thus functions of time, will also need to bedifferentiated. Therefore differentiation of fluxes results in expressions like

The differentiation with respect to L, dL/dt, will result in time-varying

coefficients on the currents. This will make our set of state equations difficult to solve.

Ldt

dii

dt

dL

dt

d

T

sr

mmm

mmm

mmm

mrs LLL

cos120cos120cos

120coscos120cos

120cos120coscos

mmm

mmm

mmm

msr LL

cos120cos120cos

120coscos120cos

120cos120coscos


9/51

Transformation

9

This presents some significant difficulties, in terms of solution, that we would like to

avoid. We look for a different approach. The different approach is based on the

observation that our trouble comes from the inductances related to the stator-rotor

mutual inductances that have time-varying inductances.In order to alleviate the trouble, we project the a-b-c currents onto a pair of axes which

we will call the d and q axes or d-q axes. In making these projections, we want to obtain

expressions for the components of the stator currents in phase with the and q axes,

respectively. Although we may specify the speed of these axes to be any speed that is

convenient for us, we will generally specify it to be synchronous speed, s

.

ia

aa'idiq

d-axisq-axis

One can visualize the projection by thinking of the a-b-c currents as having sinusoidal

variation IN TIME along their respective axes (a space vector!). The picture below

illustrates for the a-phase.

Decomposing the b-phase currents and the c-phase currents

in the same way, and then adding them up, provides us with: )120cos()120cos(cos cbaqq iiiki

)120sin()120sin(sin cbadd iiiki

Constants kqand kdare chosen so as to simplify the numerical

coefficients in the generalized KVL equations we will get.


10/51

Transformation

10

We have transformed 3 variables ia, ib, and icinto two variables idand iq, as we did in

the -transformation. This yields an undetermined system, meaning We can uniquely transform ia, ib, and icto idand iq

We cannot uniquely transform idand iqto ia, ib, and ic.We will use as a third current the zero-sequence current:

Recall our idand iqequations:

cba iiiki 00

We can write our transformation more compactly as

)120cos()120cos(cos cbadq iiiki

)120sin()120sin(sin cbaqd iiiki

c

b

a

ddd

qqq

d

q

i

i

i

kkk

kkk

kkk

i

i

i

0000

)120sin()120sin(sin

)120cos()120cos(cos


11/51

Transformation

11

c

b

a

ddd

qqq

d

q

i

i

i

kkk

kkk

kkk

i

i

i

0000

)120sin()120sin(sin

)120cos()120cos(cos

A similar transformation resulted from the work done by Blondel (1923), Doherty and

Nickle (1926), and Robert Park (1929, 1933), which is referred to as Parkstransformation. In 2000, Parks 1929 paper was voted the second most importantpaper of the last 100 years (behind Fortescues paper on symmertical components).R, Park, Two reaction theory of synchronous machines, Transactions of the AIEE, v. 48, p. 716-730, 1929.G. Heydt, S. Venkata, and N. Balijepalli, High impact papers in power engineering, 1900-1999, NAPS, 2000.

Robert H. Park,

1902-1994

See

http://www.nap.edu/openbook.php

?record_id=5427&page=175foran interesting biography on Park,

written by Charles Concordia.

Parks transformation uses a frame ofreference on the rotor. In Parks case,

he derived this for a synchronous

machine and so it is the same as asynchronous frame of reference. For

induction motors, it is important to

distinguish between a synchronous

reference frame and a reference frame

on the rotor.
http://www.nap.edu/openbook.php?record_id=5427&page=175http://www.nap.edu/openbook.php?record_id=5427&page=175http://www.nap.edu/openbook.php?record_id=5427&page=175http://www.nap.edu/openbook.php?record_id=5427&page=175


12/51

Transformation

12

Here, the angle is given by

)0()(0

t

d

where is a dummy variable of integration.

The constants k0, kq, and kdare chosen differently by different authors. One popular

choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q

quantities to be equal to that of the three-phase quantities. However, it also causes a

3/2 multiplier in front of the power expression (Anderson & Fouad use k0=1/3,kd=kq=(2/3) to get a power invariant expression).

The angular velocity associated with the change of variables is unspecified. Itcharacterizes the frame of reference and may rotate at any constant or varying angular

velocity or it may remain stationary. You will often hear of the arbitrary referenceframe. The phrase arbitrary stems from the fact that the angular velocity of thetransformation is unspecified and can be selected arbitrarily to expedite the solution of

the equations or to satisfy the system constraints [Krause].

c

b

a

ddd

qqq

d

q

i

i

i

kkk

kkk

kkk

i

i

i

0000

)120sin()120sin(sin

)120cos()120cos(cos


13/51

Transformation

13

The constants k0, kq, and kdare chosen differently by different authors. One popular

choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q

quantities to be equal to that of the three-phase quantities. PROOF (iqequation only):

)120cos()120cos(cos cbadq iiikiLet ia=Acos(t); ib=Acos(t-120); ic=Acos(t-240) and substitute into iqequation:

)120cos()120cos()120cos()120cos(coscos

)120cos()120cos()120cos()120cos(coscos

tttAk

tAtAtAki

d

dq

Now use trig identity: cos(u)cos(v)=(1/2)[ cos(u-v)+cos(u+v) ]

)120120cos()120120cos(

)120120cos()120120cos(

)cos()cos(2

tt

tt

ttAk

i dq

)240cos()cos(

)240cos()cos(

)cos()cos(2

tt

tt

ttAk

i dq

Now collect terms in t-and place brackets around what is left:

)240cos()240cos()cos()cos(32

ttttAk

i dq

Observe that what is in the brackets is zero! Therefore:

)cos(32

3)cos(3

2

tAk

tAk

i ddqObserve that for 3kdA/2=A,

we must have kd=2/3.


14/51

Transformation

14

Choosing constants k0, kq, and kdto be 1/3, 2/3, and 2/3, respectively, results in

The inverse transformation becomes:

01)120sin()120cos(

1)120sin()120cos(

1sincos

i

i

i

i

i

i

d

q

c

b

a

c

b

a

d

q

i

i

i

i

i

i

2

1

2

1

2

1)120sin()120sin(sin

)120cos()120cos(cos

32

0


15/51

Example

15

Krause gives an insightful example in his book, where he specifies generic quantities

fas, fbs, fcsto be a-b-c quantities varying with time on the stator according to:

tf

tf

tf

cs

bs

as

sin

2

cos

The objective is to transform them into 0-d-q quantities, which he denotes as fqs, fds, f0s.

t

t

t

f

f

f

f

f

f

cs

bs

as

s

ds

qs

sin

2/

cos

2

1

2

1

2

1

)120sin()120sin(sin

)120cos()120cos(cos

3

2

2

1

2

1

2

1

)120sin()120sin(sin

)120cos()120cos(cos

3

2

0

Note that these are not

balanced quantities!


16/51

Example

16

This results in

Now assume that (0)=-/12 and =1 rad/sec. Evaluate the above for t= /3 seconds.First, we need to obtain the angle corresponding to this time. We do that as follows:

4123)

12(1)0()(

3/

00

ddt

Now we can evaluate the above equations 3A-1, 3A-2, and 3A-3, as follows:


17/51

Example

17

This results in


18/51

Example

18

t

t

t

f

f

f

s

ds

qs

sin

2/

cos

2

1

2

1

2


)120cos()120cos(cos

3

2

0

Resolution of fas=cost into directions

of fqsand fdsfor t=/3 (=/4).

Resolution of fbs=t/2 into directions


Resolution of fcs

=-sint into directions


Composite

of other 3

figures


19/51

Inverse transformation

19

The d-q transformation and its inverse transformation is given below.

c

b

a

K

d

q

i

ii

i

ii

s

2

1

2

1

2

1

)120sin()120sin(sin)120cos()120cos(cos

3

2

0

0

1

1)120sin()120cos(

1)120sin()120cos(1sincos

i

ii

i

ii

d

q

K

c

b

a

s

2

1

2

1

2


)120cos()120cos(cos

32

sK

1)120sin()120cos(

1)120sin()120cos(

1sincos1

sK

It should be the case that KsKs-1=I, where I is the 3x3 identity matrix, i.e.,

100

010

001

1)120sin()120cos(

1)120sin()120cos(

1sincos

2

1

2

1

2


)120cos()120cos(cos

3

2


20/51

Balanced conditions

20

Under balanced conditions, i0is zero, and therefore it produces no flux at all. Under

these conditions, we may write the d-q transformation as

c

b

a

d

q

i

i

i

ii

)120sin()120sin(sin)120cos()120cos(cos

32

c

b

a

d

q

i

i

i

i

i

i

2

1

2

1

2


)120cos()120cos(cos

3

2

0

01)120sin()120cos(

1)120sin()120cos(

1sincos

i

i

i

i

i

i

d

q

c

b

a

d

q

c

b

a

ii

i

i

i

)120sin()120cos(

)120sin()120cos(

sincos

f


21/51

Rotor circuit transformation

21

We now need to apply our transformation to the rotor a-b-c windings in order to obtain

the rotor circuit voltage equation in q-d-0 coordinates. However, we must notice one

thing: whereas the stator phase-a winding (and thus its a-axis) is fixed, the rotor

phase-a winding (and thus its a-axis) rotates. If we apply the same transformation tothe rotor, we will not account for its rotation, i.e., we will be treating it as if it were fixed.

2

1

2

1

2


)120cos()120cos(cos

3

2

sKOur d-q transformation is as follows:

But, what, exactly, is ?

)0()(0

t

d

can be observed in the below figure as the angle between the rotating d-q referenceframe and the a-axis, where the a-axis is fixed on the stator frame and is defined by the

location of the phase-a winding. We expressed this angle analytically using

where is the rotational speed of the d-q coordinate axes (and in our case, is

synchronous speed). This transformation will allow us to operate on the stator circuitvoltage equation and transform it to the q-d-0 coordinates.

i i f i


22/51

Rotor circuit transformationTo understand how to handle this, consider the below figure where we show our

familiar , the angle between the stator a-axis and the q-axis of the synchronouslyrotating reference frame.

22

ia

aa'idiq

d-axisq-axis

m

m

We have also shown

m, which is the anglebetween the stator a-axis and

the rotor a-axis, and

, which is the angle betweenthe rotor a-axis and the q-axis

of the synchronously rotating

reference frame.The stator a-axis is stationary,

the q-d axis rotates at , and therotor a-axis rotates at m.

Consider the iarspace vector, in blue,

which is coincident with the rotor a-axis.

Observe that we may decompose it

in the q-d reference frame only by

using instead of .

Conclusion: Use the exact same transformation, except substitute for , and.

account for the fact that to the rotor windings, the q-d coordinate system appears tobe moving at -m

i i f i


23/51

Rotor circuit transformationWe compare our two transformations below.

2

1

2

1

2

1

)120sin()120sin(sin

)120cos()120cos(cos

3

2

sK

)0()(0

t

d )0(

0)0()0()()(

mt

m d

r

2

1

2

1

2

1

)120sin()120sin(sin

)120cos()120cos(cos

3

2

rK

Stator winding transformation, Ks Rotor winding transformation, Kr

23

cs

bs

as

s

ds

qs

i

i

i

i

i

i

2

1

2

1

2


)120cos()120cos(cos

3

2

0

cr

br

ar

r

dr

qr

i

i

i

i

i

i

2

1

2

1

2


)120cos()120cos(cos

3

2

0

We now augment our notation to distinguish between q-d-0 quantities from the

stator and q-d-0 quantities from the rotor:

T f i l i


24/51

Transforming voltage equations

24

cr

br

ar

cs

bs

as

cr

br

ar

cs

bs

as

r

r

r

s

s

s

cr

br

ar

cs

bs

as

i

i

i

ii

i

r

r

r

rr

r

v

v

v

vv

v

00000

00000

00000

0000000000

00000

rc

rb

ra

sc

sb

sa

rrs

srs

rc

rb

ra

sc

sb

sa

i

i

i

i

i

i

LL

LL

mrmm

mmrm

mmmr

r

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

msmm

mmsm

mmms

s

LLLL

LLLL

LLLL

L

2

1

2

12

1

2

12

1

2

1

Recall our voltage equations:

Lets apply our d-q transformation to it.

T

sr

mmm

mmm

mmm

mrs LLL

cos120cos120cos

120coscos120cos

120cos120coscos

mmm

mmm

mmm

msr LL

cos120cos120cos

120coscos120cos

120cos120coscos

T f i l i


25/51

Transforming voltage equations

25

Lets rewrite it in compact notation

cr

br

ar

cs

bs

as

cr

br

ar

cs

bs

as

r

r

r

s

s

s

cr

br

ar

cs

bs

as

i

i

ii

i

i

r

r

rr

r

r

v

v

vv

v

v

00000

00000

0000000000

00000

00000

abcr

abcs

abcr

abcs

r

s

abcr

abcs

i

i

r

r

v

v

0

0

Now multiply through by our transformation matrices. Be careful with dimensionality.

321

00

00

00

00

Term

abcr

abcs

r

s

Term

abcr

abcs

r

s

r

s

Term

abcr

abcs

r

s

KK

ii

rr

KK

vv

KK

T f i lt ti T 1


26/51

Transforming voltage equationsTerm 1

26

Therefore: the voltage equation becomes

qdor

sqd

abcrr

abcss

Term

abcr

abcs

r

s

v

v

vK

vK

v

v

K

K 0

1

0

0

32

0

0

0

0

0

0

0

Term

abcr

abcs

r

s

Term

abcr

abcs

r

s

r

s

qdor

sqd

K

K

i

i

r

r

K

K

v

v

T f i lt ti T 2


27/51


27

What to do with the abc currents? We need q-d-0 currents!

abcr

abcs

rr

ss

Term

abcr

abcs

r

s

r

s

i

i

rK

rK

i

i

r

r

K

K

0

0

0

0

0

0

2

Recall:

rqd

sqd

r

s

abcr

abcs

i

i

K

K

i

i

0

0

1

1

0

0 and substitute into above.

rqd

sqd

r

s

rr

ss

Term

abcr

abcs

r

s

r

s

ii

KK

rKrK

ii

rr

KK

0

0

1

1

2

00

00

00

00

Perform the matrix multiplication:

rqd

sqd

rrr

sss

Term

abcr

abcs

r

s

r

s

i

i

KrK

KrK

i

i

r

r

K

K

0

0

1

1

2

0

0

0

0

0

0

Fact: KRK-1=R if R is diagonal having equal elements on the diagonal.

Proof: KRK-1=KrUK-1=rKUK-1=rKK-1=rU=R.

Therefore.

T f i lt ti T 2


28/51


28

rqd

sqd

r

s

rqd

sqd

rrr

sss

Term

abcr

abcs

r

s

r

s

i

i

r

r

i

i

KrK

KrK

i

i

r

r

K

K

0

0

0

0

1

1

2

0

0

0

0

0

0

0

0

Therefore: the voltage equation becomes

3

0

00

0

0

0

0

Term

abcr

abcs

r

s

rqd

sqd

r

s

qdor

sqd

K

K

i

i

r

r

v

v

T f i lt ti T 3


29/51


29

3

0

00

0

0

0

0

Term

abcr

abcs

r

s

rqd

sqd

r

s

qdor

sqd

K

K

i

i

r

r

v

v

Focusing on just the stator quantities, consider: abcsssqd K 0

Differentiate both sides abcssabcsssqd KK

0

Solve forabcssK

abcsssqdabcss KK

0Use abcs=K-1qd0s: sqdsssqdabcss KKK 0

1

0

abcrr

abcss

Term

abcr

abcs

r

s

K

K

K

K

3

0

0Term 3 is:

A similar process for the rotor quantities results inrqdrrrqdabcrr KKK 0

1

0

Substituting these last two expressions into the term 3 expression above results in

rqdrr

sqdss

rqd

sqd

abcrr

abcss

Term

abcr

abcs

r

s

KK

KK

K

K

K

K

0

1

0

1

0

0

3

0

0

Substitute this back into voltage equation

T f i lt ti T 3


30/51


30

3

0

00

0

0

0

0

Term

abcr

abcs

r

s

rqd

sqd

r

s

qdor

sqd

K

K

i

i

r

r

v

v

rqdrr

sqdss

rqd

sqd

abcrr

abcss

Term

abcr

abcs

r

s

KK

KK

K

K

K

K

0

1

0

1

0

0

3

0

0

rqdrr

sqdss

rqd

sqd

rqd

sqd

r

s

qdor

sqd

KK

KK

i

i

r

r

v

v

01

0

1

0

0

0

00

0

0

T f i lt ti T 3


31/51


31

Now lets express the fluxes in terms of currents by recalling that

abcr

abcs

r

s

rqd

sqd

K

K

0

0

0

0

abcr

abcs

rrs

srs

abcr

abcs

cr

br

ar

cs

bs

as

rrs

srs

cr

br

ar

cs

bs

as

i

i

LL

LL

ii

i

i

i

i

LL

LL

and the flux-current relations:

Now write the abc currents in terms of the qd0 currents:

rqd

sqd

r

s

abcr

abcs

i

i

K

K

i

i

0

0

1

1

0

0

Substitute the third equation into the second:

rqd

sqd

r

s

rrs

srs

abcr

abcs

i

i

KK

LLLL

0

01

1

00

Substitute the fourth equation into the first:

rqd

sqd

r

s

rrs

srs

r

s

rqd

sqd

i

i

K

K

LL

LL

K

K

0

0

1

1

0

0

0

0

0

0

Transforming voltage equations Term 3


32/51


32

rqd

sqd

r

s

rrs

srs

r

s

rqd

sqd

i

i

K

K

LL

LL

K

K

0

0

1

1

0

0

0

0

0

0

Perform the first matrix multiplication:

rqd

sqd

r

s

rrrsr

srsss

rqd

sqd

i

i

K

K

LKLK

LKLK

0

0

1

1

0

0

0

0

and the next matrix multiplication:

rqd

sqd

rrrsrsr

rsrssss

rqd

sqd

i

i

KLKKLK

KLKKLK

0

0

11

11

0

0



33/51


33

rqd

sqd

rrrsrsr

rsrssss

rqd

sqd

i

i

KLKKLK

KLKKLK

0

0

11

11

0

0

Now we need to go through each of these four matrix multiplications. I will here omit

the details and just give the results (note also in what follows the definition of

additional nomenclature for each of the four submatrices):

0

1

0

11

0

1

00

02

30

002

3

000

02

30

002

3

0002

3

0

002

3

rqd

r

mr

mr

rrr

mqdm

m

srsrrsrs

sqd

s

ms

ms

sss

L

L

LL

LL

KLK

LL

L

KLKKLK

LL

LL

LL

KLK

rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

i

i

LL

LL

0

0

00

00

0

0

And since our inductance matrix is

constant, we can write:

rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

i

i

LL

LL

0

0

00

00

0

0

Substitute the above expression for flux

derivatives into our voltage equation:



34/51


rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

i

i

LL

LL

0

0

00

00

0

0

rqdrr

sqdss

rqd

sqd

rqd

sqd

r

s

qdor

sqd

KK

KK

i

i

r

r

v

v

0

1

0

1

0

0

0

00

0

0

Substitute the above expressions for flux & flux derivatives into our voltage equation:

rqdrr

sqdss

rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

r

s

qdor

sqd

KK

KK

i

i

LL

LL

i

i

r

r

v

v

0

1

01

0

0

00

00

0

00

0

0

We still have the last term to obtain. To get this, we need to do two things.

1. Express individual q- and d- terms of qd0sand qd0rin terms of currents.2. Obtain and1ssKK 1rrKK

34



35/51

Transforming voltage equationsTerm 31. Express individual q- and d- terms of qd0sand qd0rin terms of currents:

35

rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

i

i

LL

LL

0

0

00

00

0

0

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

0

0

0

0

00000

02

300

2

30

002

300

2

3

00000

02

300

2

30

002

300

2

3

drmrdsmdrdrmdsmsds

qrmrqsmqrqrmqsmsqs

iLLiLiLiLL

iLLiLiLiLL

2

3

2

3

2

3

2

3

2

3

2

3

2

3

2

3

From the above, we observe:



36/51


2. Obtain and1ssKK

1

rrKK

36

2

1

2

1

2


)120cos()120cos(cos

32

sK

2

1

2

1

2

1

)120sin()120sin(sin

)120cos()120cos(cos

3

2

rK

1)120sin()120cos(

1)120sin()120cos(

1sincos1

s

K

1)120sin()120cos(

1)120sin()120cos(

1sincos1

rK

To get , we must consider:

)()0()()( 0 tdtt

mm

t

m td

r

)()0()0()()()0(

0

sK

Therefore:

000

)120cos()120cos(cos

)120sin()120sin(sin

3

2

sK

Likewise, to get , we must consider:rK

Therefore:

000

)120cos()120cos(cos

)120sin()120sin(sin

3

2

mrK



37/51


2. Obtain1

ssKK

1

rrKK

37

000

00

00

000

002

3

02

30

3

2

1)120sin()120cos(1)120sin()120cos(

1sincos

000)120cos()120cos(cos

)120sin()120sin(sin

3

21

ssKK

000

00

0)(0

1)120sin()120cos(

1)120sin()120cos(

1sincos

000

)120cos()120cos(cos

)120sin()120sin(sin

3

21

m

m

mrrKK

Obtain

Substitute into voltage equations



38/51


38

00000

001

ssKK

000

00

0)(01

m

m

rrKK

Substitute into voltage equations

rqdrr

sqdss

rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

r

s

qdor

sqd

KK

KK

i

i

LL

LL

i

i

r

r

v

v

0

1

0

1

0

0

00

00

0

00

0

0

r

dr

qr

s

ds

qs

m

m

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

i

i

i

i

i

i

LLLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

0

0

0

0

0

0

0

0

000000

00000

0)(0000

000000

00000

00000

0000002

3

002

3

0

002

300

2

3

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

This results in:

Note the Speed voltages in thefirst,

second,

fourth, and

fifth equations.

-dsqs-(- m)dr(-

m

) qs



39/51


39

Some comments on speed voltages: -ds, qs, -(- m)dr,(- m) qs:

These speed voltagesrepresent the fact that a rotating flux wave will createvoltages in windings that are stationary relative to that flux wave.

Speed voltages are so named to contrast them from what may be calledtransformer voltages, which are induced as a result of a time varying magnetic

field.

You may have run across the concept of speed voltages in Physics, where youcomputed a voltage induced in a coil of wire as it moved through a static

magnetic field, in which case, you may have used the equation Blv where B is flux

density, l is conductor length, and v is the component of the velocity of the moving

conductor (or moving field) that is normal with respect to the field flux direction (or

conductor).

The first speed voltage term, -ds, appears in the vqsequation. The second

speed voltage term, qs, appears in the vdsequation. Thus, we see that the d-axis flux causes a speed voltage in the q-axis winding, and the q-axis flux causesa speed voltage in the d-axis winding. A similar thing is true for the rotor winding.



40/51


40

000

00

001

ssKK

000

00

0)(01

m

m

rrKK

Substitute the matrices into voltage equation and then expand. This results in:

rqdrr

sqdss

rqd

sqd

rqdmqd

mqdsqd

rqd

sqd

r

s

qdor

sqd

KK

KK

i

i

LL

LL

i

i

r

r

v

v

0

1

0

1

0

0

00

00

0

00

0

0

r

dr

qr

s

ds

qs

m

m

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

i

i

i

ii

i

L

LLL

LLL

LLLL

LLL

i

i

i

ii

i

r

r

r

rr

r

v

v

v

vv

v

0

0

0

0

0

0

0

0

000000

00000

0)(0000

00000000000

00000

00000

02

300

2

30

002

300

2

3

0000002

3

002

3

0

002

300

2

3

00000

00000

00000

0000000000

00000

Lets collapse the last matrix-vector product by performing the multiplication.



41/51


41

r

dr

qr

s

ds

qs

m

m

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

0

0

0

0

0

0

0

0

000000

00000

0)(0000

000000

00000

00000

00000

02

300

2

30

002

3

002

3

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

drmrdsmdrdrmdsmsds

qrmrqsmqrqrmqsmsqs

iLLiLiLiLL

iLLiLiLiLL

2

3

2

3

2

3

2

3

2

3

2

3

2

3

2

3

0

)(

)(

0

00000

02

300

2

30

002

3

002

3

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

0

0

0

0

0

0

qrm

drm

qs

ds

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

02

3

2

3)(

2

3

2

3)(

0

2

3

2

3

2

3

2

3

00000

0

2

300

2

30

002

300

2

3

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

0

0

0

0

0

0

qrmrqsmm

drmrdsmm

qrmqsms

drmdsms

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

iLLiL

iLLiL

iLiLL

iLiLL

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

From slide 35,

we have the

fluxes expressed

as a function of

currents

And then substitute

these terms in:

Results

In



42/51


42

0

2

3

2

3)(

23

23)(

0

2

3

2

3

2

3

2

3

00000

02

300

2

30

002300

23

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

0

0

0

0

0

0

qrmrqsmm

drmrdsmm

qrmqsms

drmdsms

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

iLLiL

iLLiL

iLiLL

iLiLL

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

Observe that the four non-zero elements in the last vector are multiplied by two currents

from the current vector which multiplies the resistance matrix. So lets now expand back

out the last vector so that it is a product of a matrix and a current vector.

r

dr

qr

s

ds

qs

mrmmm

mrmmm

mms

mms

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

i

i

i

i

i

i

LLL

LLL

LLL

LLL

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

0

0

0

0

0

0

0

0

000000

00

2

3)(00

2

)(3

02

3)(00

2

)(30

000000

002

300

2

3

02

300

2

30

00000

02

300

2

30

002

300

2

3

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

Now change the

sign on the last

matrix.



43/51


43

r

dr

qr

s

ds

qs

mrmmm

mrmmm

mms

mms

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

r

r

s

s

s

r

dr

qr

s

ds

qs

i

i

i

i

i

i

LLL

LLL

LLL

LLL

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

r

r

r

r

r

v

v

v

v

v

v

0

0

0

0

0

0

0

0

000000

002

3)(00

2

)(3

02

3)(00

2

)(30

000000

00

2

300

2

3

02

300

2

30

00000

02

300

2

30

002300

23

00000

02

300

2

30

002

300

2

3

00000

00000

00000

00000

00000

00000

Notice that the resistance matrix and the last matrix multiply the same vector,

therefore, we can combine these two matrices. For example, element (1,2) in the

last matrix will go into element (1,2) of the resistance matrix, as shown. This results in

the expression on the next slide.

Final Model


44/51

Final Model

44

r

dr

qr

s

ds

qs

r

mrm

mrm

s

mms

mms

r

dr

qr

s

ds

qs

r

rmrmmm

mrmrmm

s

msms

mmss

r

dr

qr

s

ds

qs

i

i

i

i

i

i

L

LLL

LLL

L

LLL

LLL

i

i

i

i

i

i

r

rLLL

LLrL

r

LrLL

LLLr

v

v

v

v

v

v

0

0

0

0

0

0

00000

02

300

2

30

002

3

002

3

00000

02

300

2

30

002

300

2

3

00000

02

3)(00

2

)(3

02

3)(0

2

)(30

00000

00

2

30

2

3

02

300

2

3

This is the complete

transformed electric

machine state-space

model in current form.

Some comments about the transformation


45/51

Some comments about the transformation

45

idsand iqsare currents in a fictitious pair of windings fixed on a synchronouslyrotating reference frame.

These currents produce the same flux as do the stator a,b,c currents. For balanced steady-state operating conditions, we can use iqd0s = Ksiabcsto show

that the currents in the d and q windings are dc! The implication of this is that:

The a,b,c currents fixed in space (on the stator), varying in time produce thesame synchronously rotating magnetic field as

The ds,qs currents, varying in space at synchronous speed, fixed in time!

idrand iqrare currents in a fictitious pair of windings fixed on a synchronouslyrotating reference frame.

These currents produce the same flux as do the rotor a,b,c currents. For balanced steady-state operating conditions, we can use iqd0r = Kriabcrto show

that the currents in the d and q windings are dc! The implication of this is that: The a,b,c currents varying in space at slip speed ss=(s- m) fixed on the

rotor, varying in time produce the same synchronously rotating magnetic

field as

The dr,qr currents, varying in space at synchronous speed, fixed in time!

Torque in abc quantities


46/51


46

The electromagnetic torque of the DFIG may be evaluated according to

m

cem

WT

m

f

em

WT

The stored energy is the sum of

The self inductances (less leakage) of each winding times one-half the square ofits current and

All mutual inductances, each times the currents in the two windings coupled bythe mutual inductance

Observe that the energy stored in the leakage inductances is not a part of the

energy stored in the coupling field.

Consider the abc inductance matrices given in slide 6.

where Wcis the co-energy of the coupling fields associated with the various windings.We are not considering saturation here, assuming the flux-current relations are linear,

in which case the co-energy Wcof the coupling field equals its energy, Wf, so that:

We use electric rad/sec by substituting m=m/p where p is the number of pole pairs.

m

f

em

WpT



47/51


47

msmm

mmsm

mmms

s

LLLL

LLLL

LLLL

L

2

1

2

1 2

1

2

12

1

2

1

mrmm

mmrm

mmmr

r

LLLL

LLLL

LLLL

L

2

1

2

121

21

2

1

2

1

mmm

mmm

mmm

msr LL

cos120cos120cos

120coscos120cos

120cos120coscos

T

sr

mmm

mmm

mmm

mrs LLL

cos120cos120cos

120coscos120cos

120cos120coscos

The stored energy is given by:

abcrrr

T

abcrabcrsr

T

abcsabcsss

T

abcsf iULLiiLiiULLiW )(2

1

)(2

1

Applying the torque-energy relation

abcrsr

T

abcs

mm

fiLi

W

m

f

em

WpT

to the above, and observing that dependenceon monly occurs in the middle term, we get

abcrsr

T

abcs

m

em iLipT

So that

But only Lsrdepend on m, so

abcrm

srT

abcsem i

L

ipT



48/51


48

We may go through some analytical effort to show that the above evaluates to

abcrsr

T

abcs

m

em iLipT

mbrarcsarcrbscrbras

marbrcrcscrarbrbscrbrarasmem

iiiiiiiii

iiiiiiiiiiiipLT

cos

2

3

sin2

1

2

1

2

1

2

1

2

1

2

1

To complete our abc model we relate torque to rotor speed according to:

mm

em T

dt

d

p

JT

Inertial

torqueMech

torque (has

negative

value for

generation)

J is inertia of the rotor

in kg-m2or joules-sec2

Negative value for

generation

Torque in qd0 quantities


49/51


However, our real need is to express the torque in qd0 quantities so that we may

complete our qd0 model.

To this end, recall that we may write the abc quantities in terms of the qd0 quantitiesusing our inverse transformation, according to:

rqdrabcr

sqdsabcs

iKi

iKi

0

1

0

1

rqdrsrm

T

sqdsabcrsr

m

T

abcsem iKLiKpiLipT 01

0

1

Substitute the above into our torque expression:

49



50/51


50

1)120sin()120cos(

1)120sin()120cos(

1sincos1

sK

1)120sin()120cos(

1)120sin()120cos(

1sincos1

rK

mmm

mmm

mmm

msr LL

cos120cos120cos

120coscos120cos

120cos120coscos

r

dr

qr

mmm

mmm

mmm

m

m

T

s

ds

qs

em

i

i

i

L

i

i

i

pT

00 1)120sin()120cos(

1)120sin()120cos(

1sincos

cos120cos120cos

120coscos120cos

120cos120coscos

1)120sin()120cos(

1)120sin()120cos(

1sincos

rqdrsrm

T

sqdsem iKLiKpT 01

0

1

I will not go through this differentiation but instead provide the result:

qrdsdrqsmem iiiipLT 4

9



51/51


Some other useful expressions may be derived from the above, as follows:

qrdsdrqsmem iiiipLT

4

9

qrdrdrqrem iipT 2

3

dsqsqsdsem iipT 2

3

Final comment: We can work with these expressions to show that the

electromagnetic torque can be directly controlled by the rotor quadrature current iqr

At the same time, we can also show that the stator reactive power Qscan be directly

controlled by the rotor direct-axis current idr.

This will provide us the necessary means to control the wind turbine.

Documents

Dq Transformation