Environmental Engineering Reference
In-Depth Information
(
I
rd
¼
/
M
L
M
I
sd
I
rq
¼
L
M
I
sq
Let us now define h
r
such as
0
1
r
2
3
I
ra
sin h
r
þ
2
I
rb
sin h
r
þ
3
p
2
@
A
I
rq
¼
I
rb
cos h
r
þ
2
I
rc
sin h
r
ð
2
:
35
Þ
3
p
2
I
rc
cos h
r
h
2
, I
rq
can be rewritten as
Using x
¼
tan
0
1
1
þ
x
2
þ
2
I
rb
2x
2x
1
þ
x
2
r
2
3
I
ra
þ
p
2
@
A
I
rq
¼
1
x
2
I
rb
1
þ
x
2
þ
2
I
rc
2x
1
þ
x
2
ð
2
:
36
Þ
p
2
1
x
2
1
þ
x
2
I
rc
After some algebraic manipulations, it comes that
2
3
I
rc
p
2
p
2
r
2
3
x
2
I
rb
4
5
1
1
þ
x
2
I
rq
¼
þ
x
2I
ra
þ
I
rb
þ
I
rc
ð
Þ
ð
2
:
37
Þ
þ
I
rc
þ
3
p
2
p
2
I
rb
This equation can further be rewritten in a compact form
I
rq
¼
x
2
a
þ
xb
þ
c
1
þ
x
2
ð
2
:
38
Þ
8
<
q
I
rq
þ
p
2
a
¼
ð
I
rb
I
rc
Þ
2
at
2
þ
bt
þ
c
¼
0
with
b
¼
2I
ra
I
rc
I
rc
c
¼
q
I
rq
:
p
2
ð
I
rb
I
rc
Þ
2
The solutions of this equation are the following.
x
1
¼
b
þ
(
p
2a
b
2
4ac
x
2
¼
b
ð
2
:
39
Þ
p
2a
b
2
4ac
Now we will estimate the derivative of the following value z
¼
2 arctan x
1
using
a second-order sliding mode. The same result is obtained with arctan x
2
.
As for the above-discussed problem, the proposed speed observer is designed
using the supertwisting algorithm [
19
].
