Lecture 12
Derivation of Equivalent Circuit from Application of
γ

δ
Frame
In 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
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
ϕ
ϕ
δ
γ
sin
cos
(2)
Equations (1) and (2) are related by the transformation matrix as:
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
−
=
⎥
⎦
⎤
⎢
⎣
⎡
δ
γ
ω
ω
ω
ω
S
S
Sq
Sd
i
i
t
t
t
t
i
i
cos
sin
sin
cos
(3)
One can define a complex number
δ
γ
S
S
S
j
i
I
+
=
~
and express
]
~
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
α
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
ω
.
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
ω
.
This lecture shows how the equivalent circuit of the induction machine is easily derived
from (5) of Lecture 11.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document