dq transformation 2

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 kq and kd are chosen so as to simplify the numerical coefficients in the generalized KVL equations we will get.

Transformation

10

We have transformed 3 variables ia, ib, and ic into two variables id and iq, as we did in the - transformation. This yields an undetermined system, meaning We can uniquely transform ia, ib, and ic to id and iq We cannot uniquely transform id and iq to ia, ib, and ic.We will use as a third current the zero-sequence current:

Recall our id and iq equations:( )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

iii

kkkkkkkkk

iii

0000

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

Transformation

11

+

+

=

c

b

a

ddd

qqq

d

q

iii

kkkkkkkkk

iii

0000


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 Parks transformation. In 2000, Parks 1929 paper was voted the second most important paper 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=175 for an interesting biography on Park, written by Charles Concordia.

Parks transformation uses a frame of reference on the rotor. In Parks case, he derived this for a synchronous machine and so it is the same as a synchronous frame of reference. For induction motors, it is important to distinguish between a synchronous reference frame and a reference frame on the rotor.

Transformation

12

Here, the angle is given by)0()(

0 +=

td

where is a dummy variable of integration.

The constants k0, kq, and kd are 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. It characterizes 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 reference frame. The phrase arbitrary stems from the fact that the angular velocity of the transformation 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

iii

kkkkkkkkk

iii

0000


Transformation

13

The constants k0, kq, and kd are 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 (iq equation only): ( ))120cos()120cos(cos +++= cbadq iiikiLet ia=Acos(t); ib=Acos(t-120); ic=Acos(t-240) and substitute into iq equation:

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

)120cos()120cos()120cos()120cos(coscos++++=

++++=

tttAktAtAtAki

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

ttAki dq {

})240cos()cos()240cos()cos(

)cos()cos(2

++++

+++

++=

tt

tt

ttAki dq

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

[ ]{ })240cos()240cos()cos()cos(32

+++++++= ttttAki dqObserve that what is in the brackets is zero! Therefore:

{ } )cos(32

3)cos(32

== tAktAki ddq Observe that for 3kdA/2=A, we must have kd=2/3.

Transformation

14

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

The inverse transformation becomes:

++

=

01)120sin()120cos(1)120sin()120cos(1sincos

iii

iii

d

q

c

b

a

+

+

=

c

b

a

d

q

iii

iii

21

21

21


32

0

Example

15

Krause gives an insightful example in his book, where he specifies generic quantities fas, fbs, fcs to be a-b-c quantities varying with time on the stator according to:

tf

tftf

cs

bs

as

sin2

cos

=

=

=

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

+

+

=

+

+

=

t

t

t

fff

fff

cs

bs

as

s

ds

qs

sin2/

cos

21

21

21


32

21

21

21


32

0

Note that these are not balanced quantities!

dq transformation 2

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