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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:
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