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Lecture 12 (revised October 2009)
The revision is minor.
The correction is in the –ve sign of j (in yellow)
The manipulations follow three steps:
Multiply the second and fourth row by –
j.
Thereafter in equation (6) the voltages are e
1
+je
2
, e
3
+je
4
, i
1
+ji
2
and i
3
+ji
4
as below:
12
34
()
(
)
SS
mR
m
R
ej
e
R
j
M
j
M
ij
i
e
j
nM
Rj
n
M
i
ωω
++
+
+
⎡⎤
⎡
⎤
⎡
⎤
=
⎢⎥
⎢
⎥
⎢
⎥
+−
+
−
+
+
⎣⎦
⎣
⎦
⎣
⎦
A
A
(6)
Just before equation (15),
Earlier the following definitions have been made:
S
I
i
=
+
±
,
R
I
i
=
+
±
.
End of Note on Revision
Derivation of Equivalent Circuit from Application of
γ

δ
Frame
From (20) in Lecture 10, the stator current vector in the
dq
frame is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
+
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
)
sin(
)
cos(
S
S
S
Sq
Sd
t
t
I
i
i
ϕ
ω
(
1
)
In (8) of Lecture 11, the vector is transformed to the
γ

δ
frame as
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
S
S
S
S
S
I
i
i
sin
cos
(
2
)
Equations (1) and (2) are related by the transformation matrix as:
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
=
⎥
⎦
⎤
⎢
⎣
⎡
S
S
Sq
Sd
i
i
t
t
t
t
i
i
cos
sin
sin
cos
(
3
)
One can define a complex number
S
I
i
=+
±
and express
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~
Re[
t
j
S
Sd
S
Sa
e
I
i
i
i
ω
α
=
=
=
(
4
)
In (4), one is reminded that the
a
axis, the
axis and the
d
axis of the stator remain in
the same position. The
γ

axis, however, rotates at the speed of the stator frequency
.
Electrical engineers are familiar with ac voltages and currents expressed as complex
numbers, such as
S
I
~
, associated with a synchronously rotating unit phasor
t
j
e
.
This lecture shows how the equivalent circuit of the induction machine is easily derived
from (5) of Lecture 11.
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 Spring '09
 BonTekOoi

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