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EE321 Spring 2008 HW#10
Problem 42
The winding functions are
w
as
50
−
sin 2
φ
s
⋅
()
⋅
=
w
bs
50
−
cos 2
φ
s
⋅
⋅
=
but now the currents are
i
as
5 cos 400t
⋅
=
i
bs
4 sin 400t
⋅
=
So the total MMF is given by
F
250
−
cos 400 t
⋅
⋅
sin 2
φ
s
⋅
⋅
200 sin 400 t
⋅
(
) cos 2
φ
s
⋅
⋅
−
=
F
125
−
sin 2
φ
s
⋅
400t
+
sin 2
φ
s
⋅
400t
−
+
⋅
100 sin 400t
2
φ
s
⋅
+
sin 400t
2
φ
s
⋅
−
+
−
=
F
225
−
sin 2
φ
s
⋅
400t
+
25 sin 2
φ
s
⋅
400t
−
−
=
The first, larger term of the MMF is moving at a speed of 200 rad/s in the CW direction, while
the second, smaller term is moving at the same speed in the CCW direction.
Problem 43
Starting with 4.516
n
as
φ
sm
N
s
sin
P
φ
sm
2
⎛
⎜
⎝
⎞
⎟
⎠
⋅
N
s3
sin 3
P
φ
sm
2
⎛
⎜
⎝
⎞
⎟
⎠
⋅
+
=
w
as
φ
sm
1
2
0
2
π
P
φ
sm
n
as
φ
sm
⌠
⎮
⎮
⌡
d
⋅
0
φ
sm
φ
sm
n
as
φ
sm
⌠
⎮
⌡
d
−
=
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as
φ
sm
()
1
2
2
−
N
s
P
cos
P
φ
sm
2
⎛
⎜
⎝
⎞
⎟
⎠
⋅
2N
s3
3P
cos 3
P
φ
sm
2
⎛
⎜
⎝
⎞
⎟
⎠
⋅
−
⎛
⎜
⎝
⎞
⎟
⎠
φ
sm
0
=
2
π
P
,
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This note was uploaded on 09/22/2011 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University.
 Spring '08
 Staff

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