Environmental Engineering Reference
In-Depth Information
i
d
¼
i
deq
þ
i
dn
ð
3
:
25
Þ
p
u
i
L
þ
C
2E
s
c
1
e
q
1
=
p
1
i
deq
¼
ð
3
:
26
Þ
1
E
s
i
dn
þ
T
1
i
dn
¼
v
ð
3
:
27
Þ
v
¼ð
k
1
þ
g
1
Þ
sgn
ð
s
1
Þ
ð
3
:
28
Þ
Þ;
T
1
[ 0
;
k
1
[ 0.
where g
1
¼
max
ð
T
1
i
dn
Proof
Consider the following Lyapunov function:
V
¼
1
2
s
1
ð
3
:
29
Þ
Differentiating V with respect to time gives:
V
¼
s
1
s
1
ð
3
:
30
Þ
Based on Eqs. (
3.22
) and (
3.23
), it can be obtained:
p
u
s
1
¼
2
C
E
s
i
d
þ
2
i
L
þ
c
1
e
q
1
=
p
1
1
C
Substituting Eqs. (
3.25
) and (
3.26
) into the above equation gives:
s
1
¼
2
C
E
s
i
dn
Differentiating s
1
with respect to time, it can be obtained:
s
1
¼
2
C
E
s
i
dn
Substituting the above equation into Eq. (
3.30
) gives:
s
1
s
1
¼
2
C
E
s
s
1
i
dn
Substituting Eqs. (
3.27
) and (
3.28
) into the above equation gives: