dq transformation.ppt
DESCRIPTION
freeTRANSCRIPT
J. McCalley
d-q transformation
Machine model
2
Consider the DFIG as two sets of abc windings, one on the stator and one on the rotor.
θm
ωm
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 λ=ΣLkik, where the summation is over all six winding currents. However, we must take note that there are four kinds of terms in each summation.
Machine model
4
• Stator-stator terms: These are terms which relate a stator winding flux to a stator winding 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 stator winding 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 θ.
• 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 θ.
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-stator or stator-rotor inductance on θ is cosinusoidal. •Because θ changes with time as the rotor rotates, the inductances are functions of time. 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 Electric Machinery,” 1995, IEEE Press, pp. 167-168. I will not go through it here.
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 because 120° 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.
Tsr
mmm
mmm
mmm
mrs LLL
cos120cos120cos
120coscos120cos
120cos120coscos
θm
ωm
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
Tsr
mmm
mmm
mmm
mrs LLL
cos120cos120cos
120coscos120cos
120cos120coscos
mmm
mmm
mmm
msr LL
cos120cos120cos
120coscos120cos
120cos120coscos
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 difficulty – that the stator-rotor and rotor-stator terms, Lsr and Lrs, because they are functions of θr, and thus functions of time, will also need to be differentiated. Therefore differentiation of fluxes results in expressions likeThe differentiation with respect to L, dL/dt, will result in time-varyingcoefficients on the currents. This will make our set of state equations difficult to solve.
Ldt
dii
dt
dL
dt
d
Tsr
mmm
mmm
mmm
mrs LLL
cos120cos120cos
120coscos120cos
120cos120coscos
mmm
mmm
mmm
msr LL
cos120cos120cos
120coscos120cos
120cos120coscos
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, ωs.
ia
aa'idiq
d-axisq-axis
θ
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 α-β 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
i
i
i
kkk
kkk
kkk
i
i
i
0000
)120sin()120sin(sin
)120cos()120cos(cos
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 “Park’s transformation.” In 2000, Park’s 1929 paper was voted the second most important paper of the last 100 years (behind Fortescue’s 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.
Park’s 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 θ is given by
)0()(0
t
d
where ɣ 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 ω 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
i
i
i
kkk
kkk
kkk
i
i
i
0000
)120sin()120sin(sin
)120cos()120cos(cos
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 iiiki
Let ia=Acos(ωt); ib=Acos(ωt-120); ic=Acos(ωt-240) and substitute into iq equation:
)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 ωt-θ 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 t
Akt
Aki 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
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
3
2
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
tf
tf
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
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!
Example
16
This results in
Now assume that θ(0)=-π/12 and ω=1 rad/sec. Evaluate the above for t= π/3 seconds.First, we need to obtain the angle θ 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:
Example
17
This results in
Example
18
t
t
t
f
f
f
s
ds
qs
sin
2/
cos
2
1
2
1
2
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
0
Resolution of fas=cost into directions of fqs and fds for t=π/3 (θ=π/4).
Resolution of fbs=t/2 into directions of fqs and fds for t=π/3 (θ=π/4).
Resolution of fcs=-sint into directions of fqs and fds for t=π/3 (θ=π/4).
Compositeof other 3 figures
Inverse transformation
19
The d-q transformation and its inverse transformation is given below.
c
b
a
K
d
q
i
i
i
i
i
i
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
i
i
i
i
i
d
q
K
c
b
a
s
2
1
2
1
2
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
sK
1)120sin()120cos(
1)120sin()120cos(
1sincos1
sK
It should be the case that Ks Ks-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
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
Balanced conditions
20
Under balanced conditions, i0 is 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
i
i
)120sin()120sin(sin
)120cos()120cos(cos
3
2
c
b
a
d
q
i
i
i
i
i
i
2
1
2
1
2
1)120sin()120sin(sin
)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
i
i
i
i
i
)120sin()120cos(
)120sin()120cos(
sincos
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 it’s a-axis) is fixed, the rotor phase-a winding (and thus it’s a-axis) rotates. If we apply the same transformation to the 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
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
sKOur d-q transformation is as follows:But, what, exactly, is θ?
)0()(0
t
d
θ can be observed in the below figure as the angle between the rotating d-q reference frame 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 ω 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 circuit voltage equation and transform it to the q-d-0 coordinates.
Rotor circuit transformationTo understand how to handle this, consider the below figure where we show our familiar θ, the angle between the stator a-axis and the q-axis of the synchronously rotating reference frame.
22
ia
aa'idiq
d-axisq-axis
θ
θm
βω
ωm
We have also shown •θm, which is the angle between the stator a-axis and the rotor a-axis, and •β, which is the angle between the 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 ω, and the rotor a-axis rotates at ωm.
Consider the iar space 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 β instead of θ.
Conclusion: Use the exact same transformation, except substitute β for θ, and…. account for the fact that to the rotor windings, the q-d coordinate system appears to be moving at ω-ωm
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()()(
m
t
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
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
0
cr
br
ar
r
dr
qr
i
i
i
i
i
i
2
1
2
1
2
1)120sin()120sin(sin
)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:
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
i
i
i
r
r
r
r
r
r
v
v
v
v
v
v
00000
00000
00000
00000
00000
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:
Let’s apply our d-q transformation to it….
Tsr
mmm
mmm
mmm
mrs LLL
cos120cos120cos
120coscos120cos
120cos120coscos
mmm
mmm
mmm
msr LL
cos120cos120cos
120coscos120cos
120cos120coscos
Transforming voltage equations
25
Let’s 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
i
i
i
i
r
r
r
r
r
r
v
v
v
v
v
v
00000
00000
00000
00000
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
0
0
0
0
0
0
0
0
Term
abcr
abcs
r
s
Term
abcr
abcs
r
s
r
s
Term
abcr
abcs
r
s
K
K
i
i
r
r
K
K
v
v
K
K
Transforming voltage equations – Term 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
Transforming voltage equations – Term 2
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
i
i
K
K
rK
rK
i
i
r
r
K
K
0
0
1
1
2
0
0
0
0
0
0
0
0
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….
Transforming voltage equations – Term 2
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
Transforming voltage equations – Term 3
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 for abcssK abcsssqdabcss KK 0
Use λabcs =K-1λqd0s: sqdsssqdabcss KKK 01
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 in rqdrrrqdabcrr KKK 01
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
01
01
0
0
3
0
0
Substitute this back into voltage equation…
Transforming voltage equations – Term 3
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
01
01
0
0
3
0
0
rqdrr
sqdss
rqd
sqd
rqd
sqd
r
s
qdor
sqd
KK
KK
i
i
r
r
v
v
01
01
0
0
0
00
0
0
Transforming voltage equations – Term 3
31
Now let’s 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
i
i
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
K
K
LL
LL
0
0
1
1
0
0
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
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
Transforming voltage equations – Term 3
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):
01
011
01
00
02
30
002
3
000
02
30
002
3
00
02
30
002
3
rqd
r
mr
mr
rrr
mqdm
m
srsrrsrs
sqd
s
ms
ms
sss
L
L
LL
LL
KLK
LL
L
KLKKLK
L
L
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:
Transforming voltage equations – Term 3
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
01
01
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
01
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 λqd0s and λqd0r in terms of currents.2.Obtain and
1ssKK 1
rrKK
34
Transforming voltage equations – Term 31. Express individual q- and d- terms of λqd0s and λqd0r in 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
300000
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:
Transforming voltage equations – Term 32. Obtain and 1
ssKK 1rrKK
36
2
1
2
1
2
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
sK
2
1
2
1
2
1)120sin()120sin(sin
)120cos()120cos(cos
3
2
rK
1)120sin()120cos(
1)120sin()120cos(
1sincos1
sK
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
Transforming voltage equations – Term 32. Obtain
1ssKK
1rrKK
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…
Transforming voltage equations – Term 3
38
000
00
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
01
01
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
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
300
2
300000
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.
-ωλds
ωλqs
-(ω- ωm)λdr
(ω- ωm) λqs
Transforming voltage equations – Term 3
39
Some comments on speed voltages: -ωλds, ωλqs, -(ω- ωm)λdr, (ω- ωm) λqs:
•These speed voltages represent the fact that a rotating flux wave will create voltages in windings that are stationary relative to that flux wave. •Speed voltages are so named to contrast them from what may be called transformer 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 you computed 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, -ωλds, appears in the vqs equation. The second speed voltage term, ωλqs, appears in the vds equation. Thus, we see that the d-axis flux causes a speed voltage in the q-axis winding, and the q-axis flux causes a speed voltage in the d-axis winding. A similar thing is true for the rotor winding.
Transforming voltage equations – Term 3
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
01
01
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
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
300
2
300000
02
300
2
30
002
300
2
3
00000
00000
00000
00000
00000
00000
Let’s collapse the last matrix-vector product by performing the multiplication….
Transforming voltage equations – Term 3
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
300
2
300000
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
300
2
300000
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)(
02
3
2
32
3
2
3
00000
02
300
2
30
002
300
2
300000
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
Transforming voltage equations – Term 3
42
02
3
2
3)(
2
3
2
3)(
02
3
2
32
3
2
3
00000
02
300
2
30
002
300
2
300000
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 let’s 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
002
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
300000
02
300
2
30
002
300
2
3
00000
00000
00000
00000
00000
00000
Now change the sign on the last matrix.
Transforming voltage equations – Term 3
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
002
300
2
3
02
300
2
30
00000
02
300
2
30
002
300
2
300000
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 inthe expression on the next slide….
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
300
2
300000
02
300
2
30
002
300
2
3
00000
02
3)(00
2
)(3
02
3)(0
2
)(30
00000
002
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
• ids and iqs are currents in a fictitious pair of windings fixed on a synchronously rotating 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 = Ksiabcs to 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 the
same synchronously rotating magnetic field as• The ds,qs currents, varying in space at synchronous speed, fixed in time!
• idr and iqr are currents in a fictitious pair of windings fixed on a synchronously rotating 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 = Kriabcr to 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 sωs=(ωs- ωm) 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
The electromagnetic torque of the DFIG may be evaluated according to
m
cem
WT
m
fem
WT
The stored energy is the sum of•The self inductances (less leakage) of each winding times one-half the square of its current and•All mutual inductances, each times the currents in the two windings coupled by the 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 Wc is 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 Wc of the coupling field equals its energy, Wf, so that:
We use electric rad/sec by substituting ϴm=θm/p where p is the number of pole pairs.
m
fem
WpT
Torque in abc quantities
47
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
Tsr
mmm
mmm
mmm
mrs LLL
cos120cos120cos
120coscos120cos
120cos120coscos
The stored energy is given by:
abcrrrTabcrabcrsr
Tabcsabcsss
Tabcsf iULLiiLiiULLiW )(
2
1)(
2
1
Applying the torque-energy relation
abcrsrTabcs
mm
f iLiW
m
fem
WpT
to the above, and observing that dependence on θm only occurs in the middle term, we get
abcrsrTabcs
mem iLipT
So that
But only Lsr depend on θm, so
abcrm
srTabcsem i
LipT
Torque in abc quantities
48
We may go through some analytical effort to show that the above evaluates to
abcrsrTabcs
mem iLipT
mbrarcsarcrbscrbras
marbrcrcscrarbrbscrbrarasmem
iiiiiiiii
iiiiiiiiiiiipLT
cos2
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 Tdt
d
p
JT
Inertial torque
Mech torque (has negative value for generation)
J is inertia of the rotor in kg-m2 or joules-sec2
Negative value for generation
Torque in qd0 quantitiesHowever, 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 quantities using our inverse transformation, according to:
rqdrabcr
sqdsabcs
iKi
iKi
01
01
rqdrsrm
T
sqdsabcrsrm
Tabcsem iKLiKpiLipT 0
10
1
Substitute the above into our torque expression:
49
Torque in qd0 quantities
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
mm
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
01
I will not go through this differentiation but instead provide the result:
qrdsdrqsmem iiiipLT 4
9
Torque in qd0 quantities
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 Qs can be directly controlled by the rotor direct-axis current idr.
This will provide us the necessary means to control the wind turbine.