.

Problem 3
Our control law is
(
)
*
,
*
,
*
*
1
1
r
fb
qs
r
ff
qs
v
rqs
rqs
v
m
r
rds
d
r
rqs
i
i
s
K
i
L
v


+
+
+
=

which yields a transfer function of

rqs
s
rqs
s
rqs
v
K
s
K
r
s
i
K
s
K
r
s
K
i +
+
+
+
+
+
=
)
(
)
(
)
1
(
*


In order to be able to neglect the low pass filter on the transfer function, we should set the
closed loop pole so as the transfer function has a cut-off frequency an order of magnitude
lower than the closed loop pole. Thus we have

(
)
1
200
2
=
+
K
K
r
s


However, we have two degrees of freedom,
K
and . From our control law and transfer
function observe two considerations. First, as we keep
K
small, it will keep current
ripple from entering our current command. Second, if we keep
K
large relative to
s
r
,
then in will reduce the sensitivity of the control performance on the machine parameters.
As a compromise, lets choose
s
r
K
10
=
which yields a value of 2 Ohms. Then we can
set

72
.
0
200
2
)
(
=
+
=
K
r
K
s
ms

Consider an indirect current control as shown in Fig. 3.5-1 and 3.5-2 of [2]. Suppose
the switching frequency is 20 kHz, and that the low pass filter has a cut-off frequency of
2 kHz. Further suppose that
2
.
0
=
s
r
Ohms. Select an appropriate value of
K
and .
Plot the magnitude and phase of the transfer function between
*
rqs
i
(input) and
rqs
i
(output)
as well as
rqs
v (input) and
rqs
i
(output).


Problem 4
I would be a wound rotor induction motor because I like well defined current paths.

Problem 5
N/A. See text for start/end points.

Problem 6
N/A. See text for start/end points.

Problem 7
N/A. See text for start/end points.

Problem 8
Barring an algebraic error, the result is:
2
2
2
2
'
2
'
2
2
3
dc
rr
r
r
M
r
r
e
i
L
r
L
r
P
T
+ =


Reference
[1]
Analysis of Electric Machinery and Drive Systems
[2]
Analysis and Design of Permanent Magnet Synchronous Machines