Flux-Linkage State-Space Model
(Section 4.12)
One limitation of the current state-space model is that the effects of
saturation are difficult to represent.
This is because saturation occurs in the mutual fluxes, since mutual
flux exists in the core material, which is quite saturable (i.e., the
permeability decreases with high currents), whereas leakage flux
exists largely external to the core material, i.e., in the air, which
has permeability that is nearly constant with current.
So we will develop the flux-linkage state-space model, which uses
the λ’s as the state variables rather than the currents.
Our approach will follow four steps:
1. Develop d- and q- axis currents as function of λ’s and mutual
fluxes λAD and λAQ (we will define these shortly).
2. Develop state equations for λ’s by substituting currents (step 1)
into voltage equations.
3. Develop the torque equation in terms of flux linkages.
4. Approximate the effects of saturation.
Step 1: Develop d- and q- axis currents as function of λ’s and
mutual fluxes λAD and λAQ:
(This comes partly from Section 4.11 and partly from 4.12)
We will take three sub-steps here.
•
Step 1a: d-axis currents
•
Step 1b: q-axis currents
•
Step 1c: put them together
1
Step 1a: d-axis currents:
Recall that in per-unit, all d-axis mutual flux linkages are equal.
As we have seen, these d-axis mutual fluxes can be expressed as
the total flux less the leakage flux, and we call it λAD:
λAD=λd-ldid=λF-lFiF=λD-lDiD
(4.110)
This says the following:
•
If in each circuit on the d-axis,
•
the pu leakage flux linkage is subtracted from the pu main
flux linkage,
•
the remaining flux linkage is the same for all other circuits
coupled to it.
Solving (4.110) for each of the currents, we obtain:
id =
1
(λ d − λ AD )
ld
iF =
1
(λ F − λ AD )
lF
iD =
1
(λ D − λ AD )
lD
(4.118)
Now this appears to be good enough to allow us to go to step 2,
and substitute the currents into the voltage equations; however, we
have introduced an extra variable, λAD, which is actually dependent
on the other flux linkages.
So let’s see if we can express λAD in terms of the other fluxes:
λ AD = λd − l d id
2
Expand the total flux term, λd, as a function of the currents. From
eq. 4.20’ (pg 20 of “macheqts.doc”), we have that (in pu):
λ d = Ld id + kM F i F + kM D i D
Making this substitution results in
λ AD = [Ld id + kM F i F + kM D i D ] − l d id
Now combining the two terms containing id results in:
λ AD = [ Ld − l d ]id + [kM F ]i F + [kM D ]i D
(*)
where we recognize each term in brackets as LAD, which was
defined in the “perunitization” notes (pg 29) as the per-unit value
of any d-axis mutual inductance, i.e.,
LAD ≡ Lmdu = kM Fu = kM Du = M Ru
Substitution of LAD into (*) yields:
λ AD = [ L AD ]id + [ L AD ]i F + [ L AD ]i D = L AD [id + i F + i D ] (4.111)
Substitution of (4.118) into (4.111) eliminates the currents:
⎤
⎡1
1
1
λ AD = L AD ⎢ (λd − λ AD ) + (λ F − λ AD ) + (λ D − λ AD )⎥
lF
lD
⎦
⎣ ld
Now divide through by LAD and then collect terms in λAD:
⎡ 1
λ
λ
1 1 1⎤ λ
+ + + ⎥= d + F + D
λ AD ⎢
⎣ L AD
ld
lF
lD ⎦
ld
lF
lD
(**)
Recognizing that the term in the brackets is the inverse of a parallel
combination of the inductances, we define this term as:
⎡ 1
1
1 1 1⎤
=⎢
+ + + ⎥
LMD ⎣ L AD l d l F l D ⎦
Making this substitution into (**) results in:
λ AD λd λ F λ D
LMD
LMD
LMD
=
+
+
λ
=
λ
+
λ
+
λD
AD
d
F
LMD l d
lF
lD Î
ld
lF
lD
(4.120)
3
Step 1b: q-axis currents:
We repeat the above procedure for the q-axis currents. We will
actually do this, despite its similarity to step 1a, to be sure we
correctly deal with the G-circuit.
Recall that in per-unit, all q-axis mutual flux linkages are equal.
As we have seen, these d-axis mutual fluxes can be expressed as
the total flux less the leakage flux, and we call it λAQ:
λAQ=λq-lqiq=λQ-lQiQ=λG-lGiG
(4.110-Q)
This says the following:
If in each circuit on the q-axis,
•
•
the pu leakage flux linkage is subtracted from the pu main
flux linkage,
the remaining flux linkage is the same for all other circuits
•
coupled to it.
Solving (4.110-Q) for each of the currents, we obtain:
1
iq = (λ q − λ AQ )
lq
iQ =
1
(λQ − λ AQ )
lQ
iG =
1
(λG − λ AQ )
lG
(4.118-Q)
(similar to 4.123)
Now this appears to be good enough to allow us to go to step 2,
and substitute the currents into the voltage equations; however, we
have introduced an extra variable, λAQ, which is actually dependent
on the other flux linkages.
4
So let’s see if we can express λAQ in terms of the other fluxes.
λ AQ = λ q − l q iq
Expand the total flux term, λq, as a function of the currents. From
eq. 4.20’ (pg 20 of “macheqts.doc”), we have that (in pu):
λ q = Lq i q + kM Q iQ + kM G iG
Making this substitution results in
λ AQ = Lq iq + kM Q iQ + kM G iG − l q iq
[
]
Now combining the two terms containing iq results in:
λ AQ = [ Lq − l q ]iq + [kM Q ]iQ + [kM G ]iG
where we recognize each term in brackets as LAQ, which was
defined in the “perunitization” notes (pg 30) as the per-unit value
of any q-axis mutual inductance, i.e.,
LAQ ≡ Lmqu = kM Qu = kM Gu = M Yu
so that:
λAQ = [LAQ]iq +[LAQ]iQ +[LAQ]iG = LAQ[iq + iQ + iG ]
(4.111-Q)
(Similar to 4.112)
Substitution of (4.118-Q) into (4.111-Q) eliminates the currents:
⎡1
λ AQ = L AQ ⎢
⎢⎣ l q
(λ q − λ AQ ) +
⎤
1
1
(λQ − λ AQ ) + (λG − λ AQ )⎥
lQ
lG
⎥⎦
Now divide through by LAQ and then collect terms in λAQ, resulting
in:
⎡ 1
1 1 1 ⎤ λ q λQ λG
+ + + ⎥=
+
+
λ AQ ⎢
⎣⎢ L AQ
lq
lQ
l G ⎦⎥
5
lq
lQ
lG
Recognizing that the term in the brackets is the inverse of a parallel
combination of the inductances, we define it as:
⎡ 1
1
1 1 1⎤
=⎢
+ + + ⎥
LMQ ⎢⎣ L AQ l q l Q l G ⎥⎦
Making this substitution results in:
λ AQ λ q λQ λG
LMQ
LMQ
LMQ
=
+
+
λ AQ =
λq +
λQ +
λG
Î
L
l
l
l
MQ
q
Q
lq
G
lQ
lG
(4.120-Q)
(similar to 4.121)
We refer to eqs. (4.120) and (4.120-Q) as the auxiliary equations.
L
L
L
λ AD = MD λd + MD λ F + MD λ D
(4.120)
ld
lF
lD
Step 1c: Putting them together:
Now combine equations 4.118 with 4.118-Q, and write them all
into a single matrix expression of currents in terms of the fluxes:
⎡1
⎢l
⎢d
⎢
⎡ id ⎤ ⎢ 0
⎢i ⎥ ⎢
⎢ F⎥ ⎢0
⎢i D ⎥ ⎢
⎢ ⎥=⎢
⎢ iq ⎥ ⎢ 0
⎢iQ ⎥ ⎢
⎢ ⎥ ⎢
⎣⎢iG ⎦⎥ ⎢ 0
⎢
⎢0
⎢⎣
0
0
1
lF
0
0
1
lD
1
ld
1
−
ld
1
−
lD
0
0
0
0
−
0
0
0
0
0
0
0
0
0
0
1
lq
0
0
0
0
0
1
lQ
0
0
0
0
0
1
lG
⎤
0 ⎥
⎥ ⎡ λd ⎤
0 ⎥⎢ λ ⎥
⎥⎢ F ⎥
⎥ ⎢ λD ⎥
0 ⎥⎢
⎥
⎥ ⎢λ AD ⎥
1 ⎥⎢
⎥
− ⎥ ⎢ λq ⎥
lq ⎥ λ
⎢ Q⎥
1 ⎥⎢
− ⎥ λG ⎥
⎥
lQ ⎢
⎥ ⎢⎣λ AQ ⎥⎦
1
− ⎥
lG ⎥⎦
(4.124’)
where we know from 4.120 and 4.120-Q that λAD and λAQ are
“combination” states developed from the other flux-linkage states.
6
Step 2: Develop state equations for λ’s by substituting currents
(step 1) into voltage equations (see Section 4.12.1)
The procedure here is:
a. Start from a preliminary form of the voltage equation that
contains flux linkage derivatives instead of current derivatives.
This is convenient because it immediately provides us with the
derivatives on the states we desire. The form that we will use
is given by eq. 4.36 (where I have included the G-circuit
voltage equation) in the text, which is
⎡r
⎡ vd ⎤
⎢0
⎢ v ⎥
⎢
⎢ q ⎥
⎢0
⎢ − vF ⎥
=
−
⎢
⎥
⎢
⎢0
⎢v D = 0 ⎥
⎢0
⎢vQ = 0 ⎥
⎢
⎥
⎢
⎣⎢0
⎣⎢vG = 0⎦⎥
b.
0
0
0
0
r
0
0
0
0 rF
0
0
0
0
0
0
rD
0
0
rQ
0
0
0
0
0 ⎤ ⎡ id ⎤ ⎡− ωλ q ⎤ ⎡ λ&q ⎤
⎢ ⎥
0 ⎥⎥ ⎢⎢ iq ⎥⎥ ⎢ ωλd ⎥ ⎢ λ&d ⎥
⎥
⎢
0 ⎥ ⎢i F ⎥ ⎢ 0 ⎥ ⎢λ&F ⎥
⎥⎢ ⎥ + ⎢
⎥−⎢ ⎥
0 ⎥ ⎢i D ⎥ ⎢ 0 ⎥ ⎢λ&D ⎥
0 ⎥ ⎢iQ ⎥ ⎢ 0 ⎥ ⎢λ&Q ⎥
⎥⎢ ⎥ ⎢
⎥ ⎢ ⎥
rG ⎦⎥ ⎣⎢iG ⎦⎥ ⎣⎢ 0 ⎦⎥ ⎢⎣λ&G ⎥⎦
Substitute for the currents using 4.124’, and do some algebra.
Let’s do the procedure for the d-axis flux linkage (as in the text) as
well as the g-axis flux linkage.
d-axis flux linkage equation:
Step a:
v d = −rid − λ&d − ωλq
Step b: From eq. 4.124’, we have that:
7
id =
λd
ld
−
λ AD
ld
Substituting, we have:
λd λ AD
v d = −r (
ld
−
λ
λ
) − λ&d − ωλq = −r d + r AD − λ&d − ωλq
ld
ld
ld
λd
λ AD
&
Solving for dλd/dt results in: λd = −r l + r l − ωλq − v d
d
d
G-axis flux linkage equation:
vG = 0 = − rG iG − λ&G
Step a:
Step b: From eq. 4.124’, we have that:
iG =
λG
lG
Substituting, we have:
λ AQ
λ AQ &
λ
λ
0 = −rG ( G −
) − λ&G = −rG G + rG
− λG
lG
lG
lG
lG
λ
& = −r λG + r AQ
λ
G
G
Solving for dλG/dt results in: G
lG
lG
Summarize all equations
λ&d = −
r
r
λ d + λ AD − ωλ q − v d
ld
ld
(4.126)
λ&F = −
rF
r
λ F + F λ AD + v F
lF
lF
(4.128)
rD
r
λ D + D λ AD
lD
lD
(4.129)
λ&D = −
8
−
λ AQ
lG
λ&q = −
r
r
λq + λ AQ + ωλd − v q
lq
lq
λ&Q = −
λ&G = −
rQ
lQ
λQ +
rQ
lQ
(4.130)
λ AQ
(4.131)
rG
r
λG + G λ AQ
lG
lG
(4.131’)
Step 3: Develop the torque equation in terms of flux linkages.
We must also write the torque equation as a function of the flux
linkages.
Here, we return to the expression we developed for the torque
corresponding to the power crossing the air gap. This was given as:
Teφu = −λ q i d + λ d i q
When developing the current state-space equations, you recall that
at this point we substituted for the flux linkages. Now, when
developing the flux-linkage state-space equations, we substitute for
the currents. Use 4.124’ from above, which, for iq and id, is:
id =
λd − λ AD
iq =
ld
Substitution yields:
Teφu = −λ q
λ d − λ AD
ld
+ λd
λ q − λ AQ
lq
λ q − λ AQ
lq
Performing the multiplication, and then gathering terms, results in
⎛1 1⎞
1
1
Teφu = − λ d λ AQ + λ q λ AD + ⎜ − ⎟λ d λ q
(4.132a)
⎟
⎜l
lq
ld
l
q
d
⎠
⎝
A+F indicate that the leakage inductance is the same in the d- and
q-axis, that is, lq=ld. I believe this is true for round-rotor machines
but only approximate for salient-pole machines. Application of it
results in:
9
1
1
λd λ AQ + λq λ AD
(4.132b)
lq
ld
Recall the torque equation derived for the current-state-space
model:
⎡− D⎤
Tm
1
&
ω=
+
[−Teφ ] + ⎢
⎥ω
τ j 3τ j
τ
⎣⎢ j ⎦⎥
Teφu = −
Substitution of (4.132b) yields:
⎤ ⎡− D⎤
Tm ⎡ λ AQ
λ AD
+⎢
ω& =
λd −
λq ⎥ + ⎢
⎥ω
(4.133)
l d 3τ j ⎥⎦ ⎣⎢ τ j ⎦⎥
τ j ⎣⎢ l q 3τ j
And so we may gather our equations for the flux-linkage model as:
r
r
λ d + λ AD − ωλ q − v d
ld
ld
r
r
λ&F = − F λ F + F λ AD + v F
lF
lF
λ&d = −
λ&D = −
λ&q = −
(4.128)
rD
r
λ D + D λ AD
lD
lD
(4.129)
r
r
λq + λ AQ + ωλd − v q
lq
lq
λ&Q = −
λ&G = −
ω& =
(4.126)
rQ
lQ
λQ +
rQ
lQ
(4.130)
λ AQ
(4.131)
rG
r
λG + G λ AQ
lG
lG
(4.131’)
⎡ λ AQ
⎤ ⎡− D⎤
λ
+⎢
λ d − AD λ q ⎥ + ⎢
⎥ω
l d 3τ j ⎥⎦ ⎢⎣ τ j ⎥⎦
τ j ⎢⎣ l q 3τ j
Tm
δ& = ω − 1
(4.133)
(4.102)
where:
10
LMD
L
L
λd + MD λ F + MD λ D
ld
lF
lD
LMQ
LMQ
LMQ
=
λq +
λQ +
λG
lq
lQ
lG
λ AD =
λ AQ
(called the “auxiliary equations) and
⎡ 1
1
1 1 1⎤
=⎢
+ + + ⎥
LMD ⎣ L AD l d l F l D ⎦
⎡ 1
1
1 1 1⎤
=⎢
+ + + ⎥
LMQ ⎢⎣ L AQ l q l Q l G ⎥⎦
The state variables are λd, λF, λD, λq, λQ, λG, ω, and δ.
The forcing functions are vd, vq, vF, and Tm (vd and vq depend on
the currents, which in turn depend on the network loading – we
will deal with this issue later!).
Recall that λAD, λAQ represents the per-unit d-axis and q-axis
mutual flux linkages, respectively, i.e.,
(4.110)
λAD=λd-ldid=λF-lFiF=λD-lDiD
λAQ=λq-lqiq=λQ-lQiQ=λG-lGiG
(4.110-Q)
and that it is only in them, the mutual flux, where saturation
occurs. That is, of the total flux seen by each winding, saturation
does not affect the leakage, only the mutual. Since these mutual
fluxes occur in the same material, it is reasonable to assume the
saturation characteristics are the same for all. This allows us to
address saturation in a very simple way, through λAD, λAQ.
If we were to eliminate these two terms by making the appropriate
substitution, then the effects of saturation would be distributed
throughout our model, and its treatment would be very complex.
Î So how do we deal with saturation?
11