Page 142 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 133

where the current function t à x 7 (t) is uniquely determined in solving the vari-
ational inequality VI(−A,−Du(t), ).

4.15 A RECTIFIER–STABILIZER CIRCUIT

Let us here again consider the rectifier–stabilizer circuit of Section 3.4.Let
u ∈ R be a given supplied voltage. Let us first consider the first variational in-
equality described in Section 3.4 of Chapter 3, that is, VI(M, ,Fu) with u ∈ R,
4
F ∈ R ,
4
(∀x ∈ R ) (x) = ϕ D 4 (x 1 ) + θ D 3 (x 2 ) + ϕ D 1 (x 3 ) + ϕ D 2 (x 4 ),
and
⎛ ⎞
R −1 R 0
⎜ 1 0 1 ⎟
⎜
⎟.
−1 ⎟
M = ⎜
⎝ R −1 R 0 ⎠
0 1 0 0
The matrix has rank 3 and is positive semidefinite since
4
2
(∀x ∈ R ) : x,Mx = R(x 1 + x 3 ) ≥ 0.
We have
T 4
ker{M + M }={x ∈ R : x 1 =−x 3 }.
Note also that
⎛ ⎞
−x 2
T ⎜ −x 4 ⎟
x ∈ ker{M + M }=⇒ Mx = ⎜ ⎟ .
−x 2
⎝ ⎠
+x 2
We may consider the variational inequality VI(M, ,Fu) for different types of
diodes.

Example 64. Suppose that all diodes are ideal, that is,
(x) (1 ≤ i ≤ 4).
(∀x ∈ R) : ϕ D i (x) = R +
4
Then ≡ 4 , D( ) ∞ = R , and clearly
R +
+
T
D( ) ∞ ∩ ker{M + M }∩ K(M, ) ={0}.
T
4
Indeed, x ∈ ker{M + M } yields x 1 =−x 3 and thus x 1 = x 3 = 0 since x ∈ R .
+
Using then x ∈ K(M, ), we get also −x 2 ≥ 0 and −x 4 ≥ 0, and thus x 2 =