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W
c
12.5
=
W
c2
10
=
W
c2
0
2
i
2
14
i
2
⋅
+
⌠
⌡
d
=
W
c1
5
2
=
W
c1
0
1
i
1
5i
1
⋅
⌠
⌡
d
=
Trajectory 1.
λ
2
i
1
4i
2
⋅
+
=
λ
1
1
⋅
3i
2
⋅
+
=
Problem 2
R
l
µ
A
⋅
=
F
ab
R
Φ
⋅
=
F
ab
l
Φ
µ
A
=
B
Φ
A
=
Φ
BA
⋅
=
Φ
S
B
⌠
⌡
d
=
F
ab
Bl
µ
=
H
B
µ
=
F
ab
Hl
⋅
=
F
ab
a
b
l
H
⌠
⌡
d
=
Problem 1
EE321  Spring 04  Final Exam  Solution
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i
f
1000
⋅
15.4
=
∆
i
f
v
dc
v
fsw
−
r
f
i
fbar
⋅
−
L
ff
d
⋅
1
f
sw
⋅
:=
i
fbar
7.325
=
i
fbar
v
fbar
r
f
:=
v
fbar
14.65
=
v
fbar
v
dc
v
fsw
−
()
d
⋅
v
fd
−
1d
−
⋅
+
:=
v
fd
0.75
:=
f
sw
10 10
3
⋅
:=
r
f
2
:=
v
fsw
1.5
:=
d
0.8
:=
L
ff
20 10
3
−
⋅
:=
v
dc
20
:=
Problem 3
The coenergy is a function of trajectory  therefore it is not conservative
W
c
14.5
=
W
c
11
2
9
+
=
W
c
0
1
t
11t
⌠
⌡
d
0
1
2t
9t
⌠
⌡
d
+
=
t will go from 0 to 1
i
2
=
i
1
t
=
Tractory 2
i
bs
α
sin
β
t
()
⋅
=
i
as
cos
β
t
=
Now we have that
W
bs
5
−
4
sin 4
φ
sm
⋅
⋅
=
W
as
5
4
cos 4
φ
sm
⋅
⋅
=
First we compute the winding functions.
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This note was uploaded on 02/19/2012 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University.
 Spring '08
 Staff

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