Transcript d-qTransformation

d-q transformation
J. McCalley
Machine model
Consider the DFIG as two sets of abc windings, one on the stator and one on the rotor.
ωm
θm
2
Machine model
The voltage equation for each phase will have the form: v ( t )  ri ( t ) 
That is, we can write them all in the following form:
 v as   rs
  
v
0
 bs  
 v cs   0
  

 v ar   0
v   0
br
  
 v cr   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
0   i as 
  as 


 
 bs
0 ibs


 
0   i cs  d   cs 


  

ar

0   i ar
dt



  
0   ibr 
br


 


rr   i cr 
 cr 
d  (t )
dt
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.
3
Machine model
• 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.
4
Machine model
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.
  as 


 bs


  cs   L s

 
  ar   L rs
 
br


  cr 
 i as 
 
i
 bs 
L sr   i cs 
 
L r   i ar 
i 
br
 
 i cr 
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.
Each of the submatrices in the inductance matrix is a 3x3, as given on the next slide…
5
Machine model

 L s  L m
 1
L s    Lm
 2
 1
 L
 2 m

 L r  L m
 1
L r    Lm
 2
 1
 L
 2 m
L sr
L rs

1
2
Lm
L s  L m

1
2

1
2
Lm
Lm
L r  L m

1
2
cos  m


 L m cos  m  120

 cos  m  120
cos  m


 L m cos  m  120

 cos  m  120
Lm

Lm 
2

1
 Lm 
2


L s  L m


1

Lm 
2

1
 Lm 
2


L r  L m


1
cos  m  120




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.
ωm
θm

cos  m
cos  m  120

cos  m  120

cos  m
cos  m  120

cos  m  120 

cos  m  120 


cos  m
cos  m   120 
T

cos  m  120   L sr


cos  m
6
Machine model
Summarizing….
 v as   rs
  
v
0
 bs  
 v cs   0
  
 v ar   0
v   0
br
  
 v cr   0

 L r  L m
 1
L r    Lm
 2
 1
 L
 2 m
7

1
2
Lm
L r  L m

1
2
Lm
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0

 Lm 
2

1
 Lm 
2


L r  L m

1
  as 


 bs


  cs   L s

 
  ar   L rs
 
br


  cr 
0   i as 
  as 


 
 bs
0 ibs


 
0   i cs  d   cs 


  


0   i ar
dt

 ar 
 
0   ibr 
br


 

rr   i cr 
 cr 

 L s  L m
 1
L s    Lm
 2
 1
 L
 2 m

1
2
Lm
L s  L m

1
2
Lm

Lm 
2

1
 Lm 
2


L s  L m


1
L sr
L rs
cos  m


 L m cos  m  120

 cos  m  120
cos  m


 L m cos  m  120

 cos  m  120
cos  m  120



cos  m
cos  m  120
cos  m  120


 i as 
 
i
 bs 
L sr   i cs 
 
L r   i ar 
i 
br
 
 i cr 

cos  m
cos  m  120


cos  m  120 

cos  m  120 


cos  m
cos  m   120 
T

cos  m  120   L sr


cos  m
Machine model
Combining….
 v as   rs
  
v
0
 bs  
 v cs   0
  
 v ar   0
v   0
br
  
 v cr   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
0   i as 
 
0 ibs
 
0   i cs  d  L s
  

0   i ar  dt  L rs
0   ibr 
 
rr   i cr 
 i as 
 
i
 bs 
L sr   i cs 
 
L r   i ar 
i 
br
 
 i cr 
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 like d   dL i  di L
dt
dt
dt
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.

 L r  L m
 1
L r    Lm
 2
 1
 L
 2 m
8

1
2
Lm
L r  L m

1
2
Lm

 Lm 
2

1
 Lm 
2


L r  L m

1

 L s  L m
 1
L s    Lm
 2
 1
 L
 2 m

1
2
Lm
L s  L m

1
2
Lm

Lm 
2

1
 Lm 
2


L s  L m


1
L sr
L rs
cos  m


 L m cos  m  120

 cos  m  120
cos  m


 L m cos  m  120

 cos  m  120
cos  m  120


cos  m
cos  m  120
cos  m  120




cos  m
cos  m  120


cos  m  120 

cos  m  120 


cos  m
cos  m   120 
T

cos  m  120   L sr


cos  m
Transformation
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.
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.
θ
d-axis
Decomposing the b-phase currents and the c-phase currents
in the same way, and then adding them up, provides us with:
i q  k q i a cos   ib cos(   120  )  ic cos(   120  ) 
i d  k d i a sin   ib sin(   120  )  ic sin(   120  ) 
q-axis
ia
a
Constants kq and kd are chosen so as to simplify the numerical
coefficients in the generalized KVL equations we will get.
9
iq
id
a'
Transformation
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:
i 0  k 0 i a  ib  i c 
Recall our id and iq equations:
i q  k d i a cos   ib cos(   120  )  ic cos(   120  ) 
i d  k q i a sin   ib sin(   120  )  ic sin(   120  ) 
We can write our transformation more compactly as
 i q   k q cos 
  
i  k sin 
 d  d
 i 0   k 0
k q cos(   120 )
k d sin(   120 )
k0
k q cos(   120 )   i a 
 
k d sin(   120 ) ib
 
  i c 
k0
10
Transformation
 i q   k q cos 
  
i  k sin 
 d  d
 i 0   k 0
k q cos(   120 )
k d sin(   120 )
k0
k q cos(   120 )   i a 
 
k d sin(   120 ) ib
 
  i c 
k0
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.
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.
See
http://www.nap.edu/openbook.php
?record_id=5427&page=175 for
an interesting biography on Park,
written by Charles Concordia.
Robert H. Park,
1902-1994
11
Transformation
 i q   k q cos 
  
i  k sin 
 d  d
 i 0   k 0
k q cos(   120 )
k d sin(   120 )
k0
k q cos(   120 )   i a 
 
k d sin(   120 ) ib
 
  i c 
k0
Here, the angle θ is given by
 

t
 (  ) d   ( 0 )
0
where ɣ is a dummy variable of integration.
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].
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).
12
Transformation
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):
i q  k d i a cos   ib cos(   120  )  ic cos(   120  ) 
Let ia=Acos(ωt); ib=Acos(ωt-120); ic=Acos(ωt-240) and substitute into iq equation:
i q  k d  A cos  t cos   A cos(  t  120 ) cos(   120  )  A cos(  t  120 ) cos(   120  ) 
 k d A cos  t cos   cos(  t  120 ) cos(   120  )  cos(  t  120 ) cos(   120  ) 
Now use trig identity: cos(u)cos(v)=(1/2)[ cos(u-v)+cos(u+v) ]
iq 
kd A
2
cos(  t   )  cos(  t   )
iq 
 cos(  t  120    120 )  cos(  t  120    120 )
 cos(  t  120    120 )  cos(  t  120    120 )
kd A
2
cos(  t   )  cos(  t   )
 cos(  t   )  cos(  t    240 )
 cos(  t   )  cos(  t    240 )
Now collect terms in ωt-θ and place brackets around what is left:
iq 
kd A
2
3 cos(  t   )  cos(  t   )  cos(  t  
 240 )  cos(  t    240 ) 
Observe that what is in the brackets is zero! Therefore:
iq 
kd A
2
3 cos(  t   ) 
3k d A
2
3 cos(  t   )
Observe that for 3kdA/2=A,
we must have kd=2/3.
13
Transformation
Choosing constants k0, kq, and kd to be 1/3, 2/3, and 2/3, respectively, results in

i
 q
 cos 
  2
i 
sin 
 d 3
1
 i 0 

 2
cos(   120 )
sin(   120 )
1
2

cos(   120 )   i a 
 
sin(   120 )  ib
 
1
  i c 
2

The inverse transformation becomes:
cos 
ia  
  
i  cos(   120 )
 b 
 i c   cos(   120 )
sin 
sin(   120 )
sin(   120 )
1  i q 
 
1 id
 
1  i 0 
14
Example
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:
f as  cos t
f bs 
t
Note that these are not
balanced quantities!
2
f cs   sin t
The objective is to transform them into 0-d-q quantities, which he denotes as fqs, fds, f0s.

 f qs 
 cos 

 2
f ds   sin 

 3
1
 f 0 s 

 2

 cos 
2
  sin 
3 1

 2
cos(   120 )
sin(   120 )
1
2
cos(   120 )
sin(   120 )
1
2

cos(   120 )   f as 


sin(   120 )  f bs


1
  f cs 
2


cos(   120 )   cos t 


sin(   120 )  t / 2


1
   sin t 
2

15
Example
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:
 

t
0
 (  ) d   ( 0 ) 

 /3
0
1d   ( 

12
)

3


12


4
Now we can evaluate the above equations 3A-1, 3A-2, and 3A-3, as follows:
16
This results in
Example
17
Example

 f qs 
 cos 

 2
f ds   sin 

 3
1
 f 0 s 

 2
cos(   120 )
sin(   120 )
1
2

cos(   120 )   cos t 


sin(   120 )  t / 2


1
   sin t 
2

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).
Composite
of other 3
figures
Resolution of fcs=-sint into directions
of fqs and fds for t=π/3 (θ=π/4).
18
Inverse transformation
The d-q transformation and its inverse transformation is given below.


cos

cos(


120
)
cos(


120
)
 iq 

  ia 
  2
 
id 
sin 
sin(   120 ) sin(   120 )  ib
  3
 
1
1
1
 i 0 

  i c 
2    
2 
2 
 

K
K
s

 cos 
2
  sin 
3 1

 2
s
cos(   120 )
sin(   120 )
1
2

cos(   120 ) 
sin(   120 ) 

1

2

cos 
sin 
1  i q 
ia  
  
 
ib  cos(   120 ) sin(   120 ) 1 i d
  
 
 i c   cos(   120 ) sin(   120 ) 1  i 0 
          
K
K
1
s
1
s
cos 


 cos(   120 )

 cos(   120 )
sin 
sin(   120 )
sin(   120 )
1

1

1
It should be the case that Ks Ks-1=I, where I is the 3x3 identity matrix, i.e.,

 cos 
2
sin 
3 1

 2
19
cos(   120 )
sin(   120 )
1
2

cos(   120 )  
cos 

sin(   120 )   cos(   120 )
 
1
  cos(   120 )
2

sin 
sin(   120 )
sin(   120 )
1  1
 
1  0
 
1  0
0
1
0
0

0

1 
Balanced conditions
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

 iq 
 cos 
2
 
i   sin 
 d 3
1
 i 0 

 2
 i q  2  cos 
  
 i d  3  sin 
20
cos(   120 )
sin(   120 )
1
2
cos(   120 )
sin(   120 )

cos(   120 )   i a 
 
sin(   120 )  ib
 
1
  i c 
2

 ia 
cos(   120 )   
 ib
sin(   120 )   
 i c 
cos 
ia  
  
i  cos(   120 )
 b 
 i c   cos(   120 )
sin 
sin(   120 )
sin(   120 )
cos 
ia  
  
i  cos(   120 )
 b 
 i c   cos(   120 )
sin 
1  i q 
 
1 id
 
1  i 0 

  iq 
sin(   120 )  
 i
 d
sin(   120 ) 
Rotor circuit transformation
Our d-q transformation is as follows:
But, what, exactly, is θ?
K
s

 cos 
2
sin 

3 1

 2
cos(   120 )
sin(   120 )
1
2

cos(   120 ) 
sin(   120 ) 

1

2

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
t
    (  ) d   ( 0 )
0
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.
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.
21
Rotor circuit transformation
To 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.
We have also shown
• θm, which is the angle
θ
between the stator a-axis and
ωm
d-axis
q-axis
θm
the rotor a-axis, and
β
ω
ia
• β, which is the angle between
the rotor a-axis and the q-axis
of the synchronously rotating
a'
reference frame.
iq
id
a
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.
22
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 transformation
We compare our two transformations below.
Rotor winding transformation, Kr
Stator winding transformation, Ks
K
s

 cos 
2
  sin 
3 1

 2
 
cos(   120 )
sin(   120 )
1
2

t

cos(   120 ) 
sin(   120 ) 

1

2

 (  ) d   ( 0 )
0
K
r

 cos 
2

sin 

3
1

 2
 
cos(   120 )
sin(   120 )
1
2
t
 )   ( ) d 
 (
 

m
0
r

cos(   120 ) 
sin(   120 ) 

1

2

 ( 0 )   m ( 0 )
  
 (0)
We now augment our notation to distinguish between q-d-0 quantities from the
stator and q-d-0 quantities from the rotor:

 i qs 
 cos 
  2
i

sin 
 ds  3 
1
 i 0 s 

 2
23
cos(   120 )
sin(   120 )
1
2

cos(   120 )   i as 
 
sin(   120 )  ibs
 
1
  i cs 
2


 i qr 
 cos 
  2
i

sin 
 dr  3 
1
 i 0 r 

 2
cos(   120 )
sin(   120 )
1
2

cos(   120 )   i ar 
 
sin(   120 )  ibr
 
1
  i cr 
2

Transforming voltage equations
Recall our voltage equations:
 v as   rs
  
v
0
 bs  
 v cs   0
  
 v ar   0
v   0
br
  
 v cr   0

 L r  L m
 1
L r    Lm
 2
 1
 L
 2 m

1
2
Lm
L r  L m

1
2
Lm
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0

Lm 
2

1
 Lm 
2


L r  L m


1
0   i as   as 
    
0 ibs

    bs 
0   i cs   cs 

   
0   i ar   ar 
0   ibr   br 

  

rr   i cr    cr 

 L s  L m
 1
L s    Lm
 2
 1
 L
 2 m

1
2
Lm
L s  L m

1
2
Lm

Lm 
2

1
 Lm 
2


L s  L m


1
Let’s apply our d-q transformation to it….
24
  sa 


 sb


  sc   L s

 
  ra   L rs
 
rb


  rc 
L sr
L rs
cos  m


 L m cos  m  120

 cos  m  120
cos  m


 L m cos  m  120

 cos  m  120
cos  m  120



cos  m
cos  m  120
cos  m  120


 i sa 
 
i
 sb 
L sr   i sc 
 
L r   i ra 
i 
rb
 
 i rc 

cos  m
cos  m  120


cos  m  120 

cos  m  120 


cos  m
cos  m   120 
T

cos  m  120   L sr


cos  m
Transforming voltage equations
 v as   rs
  
v
0
 bs  
 v cs   0
  
 v ar   0
v   0
br
  
 v cr   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
0   i as   as 
    
0 ibs

    bs 
0   i cs   cs 
     
0   i ar    ar 
0   ibr   br 

  
rr   i cr   cr 
Let’s rewrite it in compact notation
 v abcs   r s


v
 abcr   0
0   i abcs    abcs


r r   i abcr    abcr



Now multiply through by our transformation matrices. Be careful with dimensionality.
0   v abcs   K s
0  r s
0   i abcs   K s
0    abcs 
K s

 





 
v
i
0
K
0
K
0
r
0
K
r 
   r 
abcr 
   r 
abcr 


abcr 
 
    r 
   
 


Term 1
25
Term 2
Term 3
Transforming voltage equations – Term 1
0   v abcs   K s v abcs   v qd 0 s 
K s

 
  v


v
K
v
0
K
qdor
   r 
abcr 


  r abcr  
Term 1
Therefore: the voltage equation becomes
 v qd 0 s   K s
0  r s
0   i abcs   K s
0    abcs 



 





 
v
i
0
K
0
r
0
K
r 
   r 
abcr 

 qdor     r 
abcr 
 
   
 


Term 2
26
Term 3
Transforming voltage equations – Term 2
0  r s
0   i abcs   K s r s
K s
 



0 r r   i abcr   0
0 Kr 
 
   
  i abcs 


K r r r   i abcr 
0
Term 2
What to do with the abc currents? We need q-d-0 currents!
Recall:
1
 i abcs   K s


i
 abcr   0
0   i qd 0 s 

1  
K r   i qd 0 r 
0  r s
0   i abcs   K s r s
K s
 



i
0
K
0
r
0
   r 
   r 
abcr 
 
 
and substitute into above.
1
K s

K r rr  0
0
0   i qd 0 s 

1  
K r   i qd 0 r 
Term 2
Perform the matrix multiplication:
0  r s
0   i abcs   K s r s K
K s
 



i
0
K
0
r
0
   r 
   r 
abcr 
 
 
1
s
  i qd 0 s 

1  
K r r r K r   i qd 0 r 
0
Term 2
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….
27
Transforming voltage equations – Term 2
0  r s
0   i abcs   K s r s K
K s
 



i
0
K
0
r
0
   r 
   r 
abcr 
 
 
1
s
  i qd 0 s   r s
 
1  
i
K r r r K r   qd 0 r   0
0
Term 2
Therefore: the voltage equation becomes
 v qd 0 s   r s

 
 v qdor   0
0   i qd 0 s   K s
0    abcs 


 i

r r   qd 0 r   0
K r    abcr 

  

Term 3
28
0   i qd 0 s 


r r   i qd 0 r 
Transforming voltage equations – Term 3
0   i qd 0 s   K s
0    abcs 


 i

r r   qd 0 r   0
K r    abcr 

  

Term
3
0    abcs   K s  abcs 
K s
 


 

0
K
  abcr   K r  abcr 
r 


 


 v qd 0 s   r s

 
 v qdor   0
Term 3 is:
Term 3
Focusing on just the stator quantities, consider:  qd 0 s  K s  abcs
Differentiate both sides
 qd 0 s  K s  abcs  K s  abcs
Solve for K s  abcs
K s  abcs   qd 0 s  K s  abcs
Use λabcs =K-1λqd0s:
1
K s  abcs   qd 0 s  K s K s  qd 0 s
A similar process for the rotor quantities results in K r  abcr   qd 0 r  K r K
1
r
 qd 0 r
Substituting these last two expressions into the term 3 expression above results in
0    abcs   K s  abcs
K s
 

 
K r    abcr   K r  abcr
 0

  

   qd 0 s   K s K
 
 
   qd 0 r   K r K
1
s
1
r
 qd 0 s 

 qd 0 r 
Term 3
29
Substitute this back into voltage equation…
Transforming voltage equations – Term 3
 v qd 0 s   r s

 
 v qdor   0
0   i qd 0 s   K s
0    abcs 


 i

r r   qd 0 r   0
K r    abcr 

  

0    abcs   K s  abcs
K s
 

 
0
K
 abcr   K r  abcr
r 



  

Term 3
   qd 0 s   K s K
 
 

  qd 0 r   K r K
1
s
1
r
 qd 0 s 
Term 3
 v qd 0 s   r s


 v qdor   0
30
0   i qd 0 s 

 i
r r   qd 0 r 
  qd 0 s 


  qd 0 r 

 qd 0 r 
 K s K s 1  qd 0 s 
 

1
K
K

 r r qd 0 r 
Transforming voltage equations – Term 3
Now let’s express the fluxes in terms of currents by recalling that
  qd 0 s   K s


  qd 0 r   0
0    abcs 


K r    abcr 
and the flux-current relations:
  as 


 bs


  cs   L s

 
  ar   L rs
 
br


  cr 
 i as 
 
i
 bs 
L sr   i cs 
  abcs   L s


 
 
L r   i ar 
  abcr   L rs
i 
br
 
 i cr 
Now write the abc currents in terms of the qd0 currents:
Substitute the third equation into the second:
L sr   i abcs 


L r   i abcr 
1
 i abcs   K s


i
 abcr   0
0   i qd 0 s 

1  
K r   i qd 0 r 
1
  abcs   L s



 abcr   L rs
L sr   K s

Lr  0
0   i qd 0 s 

1  
K r   i qd 0 r 
Substitute the fourth equation into the first:
31
  qd 0 s   K s


  qd 0 r   0
0  Ls

K r   L rs
1
L sr   K s

Lr  0
0   i qd 0 s 

1  
K r   i qd 0 r 
Transforming voltage equations – Term 3
  qd 0 s   K s



 qd 0 r   0
0  Ls

K r   L rs
1
L sr   K s

Lr  0
0   i qd 0 s 

1  
K r   i qd 0 r 
Perform the first matrix multiplication:
  qd 0 s   K s L s


  qd 0 r   K r L rs
and the next matrix multiplication:
  qd 0 s   K s L s K s 1


1
  qd 0 r   K r L rs K s
32
1
K s L sr   K s

K r Lr  0
0   i qd 0 s 

1  
i
K r   qd 0 r 
1
K s L sr K r   i qd 0 s 

1  
K r L r K r   i qd 0 r 
Transforming voltage equations – Term 3
  qd 0 s   K s L s K s 1


1
  qd 0 r   K r L rs K s
1
K s L sr K r   i qd 0 s 

1  
i
K r L r K r   qd 0 r 
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):
K s Ls K
1
s
1
K s L sr K
K r Lr K
33
r
1
r
3

L

Lm

s

2

 
0

0


 K r L rs K
1
s
3

L

Lm

r

2

 
0

0



0 

0   L sqd 0

L s 

0
L s 
3
2
Lm
0
3
 2 Lm

  0

 0

0
3
Lm
2
0
0
L r 
0
3
2
Lm

0

0   L mqd 0

0


0 

0   L rqd 0

L r 

  qd 0 s   L sqd 0


  qd 0 r   L mqd 0
L mqd 0   i qd 0 s 


L rqd 0   i qd 0 r 
And since our inductance matrix is
constant, we can write:
  qd 0 s   L sqd 0

 

  qd 0 r   L mqd 0
L mqd 0   iqd 0 s 


L rqd 0   i qd 0 r 
Substitute the above expression for flux
derivatives into our voltage equation:
Transforming voltage equations – Term 3
  qd 0 s   L sqd 0

 
  qd 0 r   L mqd 0
L mqd 0   iqd 0 s 


L rqd 0   i qd 0 r 
Substitute the above expressions for flux & flux derivatives into our voltage equation:
 v qd 0 s   r s


v
 qdor   0
 v qd 0 s   r s


v
 qdor   0
0   i qd 0 s 


r r   i qd 0 r 
0   i qd 0 s 

 i
r r   qd 0 r 
 L sqd 0

 L mqd 0
  qd 0 s 



 qd 0 r 
 K s K s 1  qd 0 s 
 

1
K
K

 r r qd 0 r 
L mqd 0   iqd 0 s 


L rqd 0   iqd 0 r 
 K s K s 1  qd 0 s 
 

1
K
K

 r r qd 0 r 
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 K K and K r K r 1
1
s
34
s
Transforming voltage equations – Term 3
1. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents:
  qd 0 s   L sqd 0


L
  qd 0 r   mqd 0
L mqd 0   i qd 0 s 


L rqd 0   i qd 0 r 
3

L

 s 2 Lm
  qs  

 
0
 ds

 
0s  
0

 
3
  qr  
Lm
   2
dr

 
0

 0 r  
0

0
3
0
Lm
2
3
L s 
3
2
2
Lm
0
0
3
0
Lm
2
0
L s
0
0
Lm
0
0
0
0
L r 
0
0
0
3
2
Lm
0
L r 
3
2
Lm
0
From the above, we observe:

 qr 
3

3

L m  i qs  L m i qr
2
2


3
3

L m  i ds  L m i dr
2
2

 dr 
3
 qs   L s 
3
 ds   L s 
35

3


L m i qs   L r  L m  i qr
2
2


3


L m i ds   L r  L m  i dr
2
2



0 
i
  qs 
 
0  i ds
 
0   i0 s 
 i 
0   qr 
  i dr 
 
0  i 
  0r 
L r 
Transforming voltage equations – Term 3
2. Obtain
K
K
r
K
36

 cos 
2
  sin 
3 1

 2
s
K
K s K
1
s
r
s
and
K r K
cos(   120 )
sin(   120 )
1
2
cos 


 cos(   120 )

 cos(   120 )

 cos 
2

sin 
3 1

 2
1
1

cos(   120 ) 
sin(   120 ) 

1

2

sin(   120 )
sin(   120 )
sin(   120 )
1
cos 


 cos(   120 )

 cos(   120 )
r
sin 
cos(   120 )
2
1
1

1

1

cos(   120 ) 
sin(   120 ) 

1

2

sin 
sin(   120 )
sin(   120 )
1

1

1
To get
 (t ) 
K
s

, we must consider:
t
 (  ) d   ( 0 )   ( t )  
0
Therefore:
K
  sin 

  cos 
3 
 0
2
s
 sin(   120 )
cos(   120 )
0
Likewise, to get
 
t
 )   ( ) d 
 (
 

m
0
K ,r
 sin(   120 ) 

cos(   120 )


0
we must consider:
  ( 0 )   m ( 0 )   ( t )     m
  
r
 (0)
Therefore:
K
  sin 

    m  cos 

3
 0
2
r
 sin(   120 )
cos(   120 )
0
 sin(   120 ) 

cos(   120 )


0
Transforming voltage equations – Term 3
2. Obtain
K s K
1
s
K r K
1
r

3
2
0
0
K r K

0
0
cos(   120 )
0

0
0
 sin(   120 )  
cos 

cos(   120 ) cos(   120 )

  cos(   120 )
0
sin 
sin(   120 )
sin(   120 )
1

1

1
0

0

0 
1
r
 (   m )
0
 sin(   120 )
0


0  


0   0

  sin 
2

    m  cos 

3
 0
 0

   m

 0
37
s
  sin 
2 
  cos 
3 
 0

0
2 3
 
3 2
0

Obtain
1
K s K
0

0

0 
 sin(   120 )
cos(   120 )
0
 sin(   120 )  
cos 

cos(   120 ) cos(   120 )

  cos(   120 )
0
sin 
sin(   120 )
sin(   120 )
1

1

1
Substitute into voltage equations…
Transforming voltage equations – Term 3
Substitute into voltage equations…
K s K
K r K
1
r
1
s

0

 

 0
0

0

0 
0
0
 (   m )
 0

   m

 0
0
0
 v qd 0 s   r s


v
 qdor   0
0

0

0 
0   i qd 0 s 

 i
r r   qd 0 r 
 L sqd 0

 L mqd 0
L mqd 0   iqd 0 s 


L rqd 0   iqd 0 r 
 K s K s 1  qd 0 s 
 

1
K
K

 r r qd 0 r 
This results in:
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
 v 0 r   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0

0
0
Lm
2
3
L s 
3
2
2
Lm
0
0
3
0
Lm
2
0
L s
0
0
Lm
0
0
0
Note the “Speed voltages” in the
first,
second,
fourth, and
fifth equations.
38
3
0
0
3
L r 
2
0
0
Lm
0
L r 
3
2
Lm
0
-ωλds
ωλqs
-(ω- ωm)λdr
(ω- ωm) λqs

0 
 i   0
 qs
  
0  ids

  

0   i0 s   0
 
 i
0
0   qr  
 i   0
 dr
  
0   i   0
  0r  
L r 

0
0
0
0
0
0
0
0
0
0
0
0
0
0
 (   m )
0
0
  m
0
0
0
0
0
0    qs 


0  ds


0  0 s 


0    qr 
0    dr 


0    0 r 
Transforming voltage equations – Term 3
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 daxis 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.
39
Transforming voltage equations – Term 3
K s K
K r K
1
r
1
s

0

 

 0
0

0

0 
0
0
 (   m )
 0

   m

 0
0
0
 v qd 0 s   r s


v
 qdor   0
0

0

0 
0   i qd 0 s 


r r   i qd 0 r 
 L sqd 0

 L mqd 0
L mqd 0   iqd 0 s 


L rqd 0   iqd 0 r 
 K s K s 1  qd 0 s 
 

1
K
K

 r r qd 0 r 
Substitute the matrices into voltage equation and then expand. This results in:
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
 v 0 r   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0

0
0
3
Lm
2
3
L s 
3
2
2
Lm
0
0
3
0
Lm
2
0
L s
0
0
Lm
0
0
0
0
L r 
0
0
0
3
2
Lm
0
L r 
0
3
2
Lm

0 
 i   0
 qs
  
0  ids

  


0   i0 s   0
 
 i
0
0   qr  
  0
 idr
  
0   i   0
  0r  
L r 

0
0
0
0
0
0
0
0
0
0
0
0
0
0
 (   m )
0
0
  m
0
0
0
0
0
Let’s collapse the last matrix-vector product by performing the multiplication….
40
0    qs 


0  ds


0  0 s 


0    qr 
0    dr 


0    0 r 
Transforming voltage equations – Term 3
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
 v 0 r   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
 v 0 r   0
Results
In 
From slide 35,
we have the
fluxes expressed
as a function of
currents 
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
 v 0 r   0
41
3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0

0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
Lm
2
3
L s 
2
Lm
0
0
2
3
0
Lm
2
L s
0
3
0
0
Lm
0
L r 
0
0
3
2
Lm
0
3
L r 
0
2
0
3
0
0
L s
0
0
L r 
0
0
3

L m  i ds  L m i dr
2
2

 dr 
3
3
L s 
2
0
0
3
2
Lm
0
0
3
Lm
0
0
0
0
L r 
0
0
L r 
0
0
0
Lm
0
3
2
Lm
0
3
2
Lm
0
0
0
0
0
0
0
0
0
0
0
0
0
0
 (   m )
0
0
  m
0
0
0
0
0
0    qs 


0  ds


0  0 s 


0    qr 
0    dr 


0    0 r 

0 
  ds
 i  

 qs
  

0  i ds
 qs
  



0
0   i0 s  
 

 i
 (   m )  dr 
0   qr  
   (   )  
 idr
m
qr
  

0   i  
0


0r 


L r 
And then substitute
these terms in:
3


L m i ds   L r  L m  i dr
2
2


2
L s
Lm

3


L m i qs   L r  L m  i qr
2
2


Lm
2
2
0
3
0
Lm
0
3
0

0
3
2

3
0
0
3
3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0

Lm

0 
 i   0
 qs
  
0  ids

  

0   i0 s   0
 
 i
0
0   qr  
 i   0
 dr
  
0   i   0
  0r  
L r 
Lm
2
 qr 
3

0
3
3

L m  i qs  L m i qr
2
2

 qs   L s 
 ds   L s 
0
0
0
0
0
3

0
 L s  2 L m
0   i qs 

3
  
0
L s  L m
0 i ds
  
2
0   i0 s  
0
0
  
0   i qr   3
Lm
0

0   i dr   2
3
 
0
Lm
rr   i 0 r  
2

0
0


0
0
0
L r 
0
3
2
Lm




3
3

    L s  L m  i ds  L m i dr 



0 
2
2




 iqs  




3
3



 
   L s  L m  i qs  L m i qr 
0  ids

  
2
2









i
0  0s
0



 
 i
3
3

 

qr


0 
 (   m )  L m i ds   L r  L m  i dr 

  
2

 
2
 idr


 
3
3

  
0   i  
  0 r   (   m )  L m i qs   L r  L m  i qr  
2

 
2
L r 


0


Transforming voltage equations – Term 3
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
v
 0 r   0
0
0
0
3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0

0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0
0
3
0
Lm
2
3
L s 
2
Lm
0
L s
0
0
3
Lm
0
0
0
2
3
0
Lm
2
0
0
L r 




3
3

    L s  L m  i ds  L m i dr 



0 
2
2






i 
 qs


3
3


  
   L s  L m  i qs  L m i qr 
0  i ds

  
2
2







0   i0 s  
0

 
 i


3
3




0   qr 
 (   m )  L m i ds   L r  L m  i dr 


 
2

 
2
 idr

  
3
3

  
0   i  
  0 r   (   m )  L m i qs   L r  L m  i qr  
2

 
2
L r 


0


0
0
3
2
Lm
0
L r 
0
0
3
2
Lm
0
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.
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
v
 0 r   0
42
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0

0


  L  3 L 


  s 2 m 

0


0

 3 (   m )
Lm

2

0

3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0

3


   L s  L m 
2


0
0
2
3
L s 
2
Lm
0
0
3
2
0
Lm
0
3
3
L s
0
0
0
0
0
L r 
0
3
2
0
0
0
0
0
0
0
0
3


(   m )  L r  L m 
2


0
Lm
0
L r 
0
0
3
2
Lm
0
 3
Lm
Lm
2
Lm
0
0
0
2
0
Lm
3
2
2
0
 3 (   m )
0
Lm
0
0
3


 (   m )  L r  L m 
2


0
0

0

0 
 i 
 qs
 
0  ids
 

0   i0 s 
 
 i
0   qr 
 
 idr
 
0   i 
  0r 
L r 
  i qs 
0   i ds 
 
0   i0 s 
 
i
0   qr 
 i 
dr
 
0   i 0 r 

0 
Now change the
sign on the last
matrix.
Transforming voltage equations – Term 3
 v qs   rs
  
v
0
 ds  
v0 s   0
  
 v qr   0
v   0
dr
  
 v 0 r   0
0
0
0
0
rs
0
0
0
0
rs
0
0
0
0
rr
0
0
0
0
rr
0
0
0
0

0


   L  3 L 
 s
m 

2



0


0

  3 (   m )
Lm

2

0

3

 L s  2 L m
0   i qs 

  
0
0 i ds
  
0   i0 s  
0
   
3
0   i qr 
Lm

0   i dr   2
 
0
rr   i 0 r  

0


  L s 


Lm 
2

3
0
0
2
3
L s 
2
Lm
0
3
0
0
Lm
0
0
0
0
0
3
L r 
Lm
2
0
0
0
0
0
0
0
0
0
Lm
3


 (   m )  L r  L m 
2


0
3
2
Lm
0
2
 3
0
L r 
0
3
0
Lm
2
L s
2
0
0
0
3
0
0
Lm
Lm
2
2
0
3 (   m )
3
0
Lm
0
0
3


(   m )  L r  L m 
2


0
0

0 
 i 
 qs
 
0  ids
 

0   i0 s 
 
 i
0   qr 
 
 idr
 
0   i 
  0r 
L r 

0
  i qs 
0   i ds 
 
0   i0 s 
 
i
0   qr 
 i 
dr
 
0   i 0 r 

0 
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….
43
Final Model

rs

 v qs  
      L  3 L 
v
s
m
 ds  
2



v0 s 
0
  
 v qr  
0
v  
dr
    3 (   )
m
Lm
 v 0 r  
2


0
3

L

 s 2 L m


0

0

 3
Lm

2


0

0

44
0

  L s 


Lm 
2

3
L s 
3
2
2
Lm
 3
0
rs
0
0
0
rr
3


(   m )  L r  L m 
2


2
Lm
2
0
2
0
0
0
0
3
3


 (   m )  L r  L m 
2


0
Lm
0
3
0
Lm
2
0
L s
0
0
Lm
0
0
0
0
L r 
0
0
0
3
2
Lm
Lm
Lm
0
2
3
3
0
rs
3 (   m )
0
0
0
L r 
0
3
2
Lm

0 
 i 
 qs
 
0  ids
 

0   i0 s 
 
 i
0   qr 
 
 idr
 
0   i 
  0r 
L r 
0
rr
0

0
  i qs 
0  i 
  ds 
0   i0 s 
 
i
0   qr 
 i 
  dr 
0   i 0 r 

rr 
This is the complete
transformed electric
machine state-space
model in “current form.”
Some comments about the transformation
• 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!
45
Torque in abc quantities
The electromagnetic torque of the DFIG may be evaluated according to
T em 
W c
 m
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:
W f
T em 
 m
We use electric rad/sec by substituting ϴm=θm/p where p is the number of pole pairs.
W f
T em  p
 m
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.
46
Consider the abc inductance matrices given in slide 6.
Torque in abc quantities

 L s  L m
 1
L s    Lm
 2
 1
 L
 2 m

 L r  L m
 1
L r    Lm
 2
 1
 L
 2 m
L sr
L rs
47

1
2

Lm 
2

1
 Lm 
2


L s  L m


Lm
L s  L m

1
2

1
2
Lm
L r  L m

2
cos  m


 L m cos  m  120

 cos  m  120
cos  m


 L m cos  m  120

 cos  m  120
1
Lm

cos  m
cos  m  120
cos  m  120


Wf 
1
2
T
T
i abcs ( L s  L s U ) i abcs  i abcs L sr i abcr 
1
T
2
i abcr ( L r  L r U ) i abcr
Applying the torque-energy relation
W f
1
cos  m  120


The stored energy is given by:

T em  p
Lm 
 m
2
 to the above, and observing that dependence
1
 L m  on θ only occurs in the middle term, we get
m
2

W f

 T
L r  L m

i abcs L sr i abcr


Lm
1

cos  m
cos  m  120


cos  m  120 

cos  m  120 


cos  m
cos  m   120 
T

cos  m  120   L sr


cos  m
 m
 m
So that
T em  p

 m
T
i abcs L sr i abcr
But only Lsr depend on θm, so
T
T em  p i abcs
 L sr
 m
i abcr
Torque in abc quantities
T em  p

T
 m
i abcs L sr i abcr
We may go through some analytical effort to show that the above evaluates to
T em
 
1
1
1
1
1
1





  pL m   i as  i ar  ibr  i cr   ibs  ibr  i ar  i cr   i cs  i cr  ibr  i ar   sin  m
2
2 
2
2 
2
2



 

3
2
i as ibr
 i cr   ibs i cr  i ar   i cs i ar

 ibr  cos  m 

Negative value for
generation
To complete our abc model we relate torque to rotor speed according to:
J is inertia of the rotor
in kg-m2 or joules-sec2
T em 
J d m
p
dt
Inertial
torque
48
 Tm
Mech
torque (has
negative
value for
generation)
Torque in qd0 quantities
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 quantities
using our inverse transformation, according to:
i abcs  K
1
i abcr  K
1
s
r
i qd 0 s
i qd 0 r
Substitute the above into our torque expression:
T
T em  p i abcs
49

 m

L sr i abcr  p K
1
s

T
i qd 0 s

 m
L sr K
1
r
i qd 0 r
Torque in qd0 quantities

T em  p K
K
1
s
cos 


 cos(   120 )

 cos(   120 )
L sr
T em

cos 

 p  cos(   120 )


  cos(   120 )
sin 
sin(   120 )
sin(   120 )
1
s
i qd 0 s
T
sin(   120 )
sin(   120 )
1  i qs  

  
1 i ds 
 
  m
1  i 0 s  
 m
1

1

1
cos  m


 L m cos  m  120

 cos  m  120
sin 


T
K
1
r
L sr K

cos  m
cos  m  120
 
cos  m
 
 L m  cos  m  120
  cos   120
m
 
cos  m  120


r
i qd 0 r
cos 


 cos(   120 )

 cos(   120 )
cos  m  120


1


cos  m
cos  m  120

sin 
sin(   120 )
sin(   120 )
cos  m  120 

cos  m  120 


cos  m
cos  m  120  
cos 

cos  m  120  cos(   120 )

  cos(   120 )
cos  m
I will not go through this differentiation but instead provide the result:
T em 
50
9
4
pL m i qs i dr  i ds i qr 
1

1

1
sin 
sin(   120 )
sin(   120 )
1  i qr  
 
1 i dr 
 
1  i 0 r  
Torque in qd0 quantities
T em 
9
4
pL m i qs i dr  i ds i qr 
Some other useful expressions may be derived from the above, as follows:
T em 
3
T em 
3
2
2
p  qr i dr   dr i qr 
p  ds i qs   qs i ds 
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.
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