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ECSE 462
PRACTICE SESSION I (2009)
The exercises are for practice only. Do NOT submit.
Trigonometric Identities
cos(A+B)=cosAcosBsinAsinB
cos(AB)=cosAcosB+sinAsinB
sin(A+B)=sinAcosB+cosAsinB
sin(AB)=sinAcosBcosAsinB
cos
sin
jx
ex
j
x
=+
1.
A voltage
cos
vV
t
ω
ϕ
can be expressed as
Re
jt
e
=
. What is the phasor
(or complexor)
V
? Answer
j
VV
e
=
.
2.
A balanced 3 phase voltage vector in the abc frame is:
0
0
cos
cos
120
cos
240
a
b
c
vt
t
ωϕ
+
⎡⎤
⎡
⎤
⎢⎥
⎢
⎥
−
⎢
⎥
⎢
⎥
+−
⎣⎦
⎣
⎦
The voltages can be written in the complexor form:
Re
a
a
bb
c
c
V
v
e
v
V
=
. What is the
phasor (complexor) vector
a
b
c
V
V
V
?
3. The voltages in Q.2 is transformed to the 0
α

β
frame using the equations:
11
1
22
2
21
1
1
32
2
33
0
oa
b
c
vv
α
β
⎡
⎤
⎢
⎥
=−
−
⎢
⎥
⎢
⎥
⎣
⎦
−
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View Full DocumentIn the new frame, the voltages are expressed as:
0
0
Re
jt
V
v
vV
e
v
V
ω
αα
β
⎡⎤
⎢⎥
=
⎣⎦
What is the phasor (complexor) vector
0
V
V
V
α
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
?
As the transformation from the 0
α

β
frame to the 0dq frame is by the identity
matrix, the stator quantities remain unchanged. The transformation from the 0dq
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 Spring '09
 BonTekOoi

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