... jωψ
1
(3. 34a)
i
1
=
1
L
σ
ψ
1
− ψ
2
(3. 34b)
u
2
= jωψ
2
− i
2
R
2
(3. 34c)
u
2
= i
2
R
L
(3. 34d)
i
2
= i
1
− i
M
(3. 34e)
i
M
=
ψ
2
L
M
(3. 34f)
ψ
2
= kψ
2
(3. 34g)
i
2
= ki
2
(3. 34h)
The ... −
L
short−circuit
L
open−circuit
(3. 20)
66 FUNDAMENTALS OF ELECTRICAL DRIVES
m-file Tutorial 1, chapter 3
%Tutorial 1, chapter 3
plot(datout(: ,3) ,datout(:,1)/5)...
... were open
%%%%%%define variables
A_c=10e -3 * 10e -3 ; % cross-section of core and armature 10mm x10 mm
A_a=10e -3 * 10e -3 ; % assumed cross-section of airgap
g=10e -3 ; % distance (airgap) between armature ... parameters
depth=20e -3 ; % thickness E and I core
g=10e -3 ; % gap
w_A=45e -3 ; % E core center leg width
w_B=22.5e -3 ; % E core side leg width
n=500; % number of turns coil
i...
... those given in figure 2.17.
Simple Electro-Magnetic Circuits 39
SCOPE1
SCOPE2
SCOPE3
u
L
u
iR
@
i
-2 16 .32 3m
0
216 .32 3m
94.101m
108.162m
Figure 2. 13. Caspoc simulation: linear inductance ... ‘vector of input values:’ set to tanh( [-5 :0.1:5]), and ‘vector
38 FUNDAMENTALS OF ELECTRICAL DRIVES
0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.5
1
1.5
(a) time (s)
voltag...
... input power
of the circuit is equal to the real power, hence
P = p
in
(5 .33 )
144 FUNDAMENTALS OF ELECTRICAL DRIVES
m-file Tutorial 3, chapter 5
%Tutorial 3, chapter 5
%we set E=100; eta =-7 *pi/6 assume ... SCOPE5
SCOPE4
SCOPE1
SCOPE3
B_S
+
-
i
p
c
u
c
i
p
L
u
L
i
u
c
u
L
u
s
i
p
s
u
s
i
u
c
u
L
10
ground
-9 89.810u -1 581.294n 1597.573u
-9 89.810u
-9 896.523u...
... results
NOTE1
PI
@
1
i
m
u
1
1.25
@
1
@
s
U
2
u
2
u
1
U
1
i
2
i
1
I1
I
2
P
1
Q
1
@
s
@
s
@
2
1.250
833 .33 3m
39 2.699
1.250 31 4.159
45 .34 5
78.540
39 2.699 226.725
154.418
25.642 18. 132 89.1 53
9. 538 k
7.818k
31 4.159
31 4.159
250.000m
Figure 6.15. Caspoc ... =0.79Wb
respectively.
The m-file to plot the results shown in figure 6.18 is as follows
m-file Tuto...