]> 8.3 Abstract dynamical model

8.3 Abstract dynamical model

Let
x(t)(θ) = z(t + θ),t 0,θ [r, 0] .

Then

tx(t)(θ) = tz(t + θ) = θz(t + θ) = θx(t)(θ) .

In the Hilbert space H = L2(r, 0) L2(r, 0) with standard scalar product, the system–dynamics equations (8.5) can be written in the abstract form proposed in Section 7.6:

(t) =σx(t) τx(t) =u(t) y(t) =c#x(t) (8.6)

where

σx = x,D(σ) = x = x1 x2 W1,2(r, 0) W1,2(r, 0) : x 1(0) = x2(r) ,

τx = x2(0) + bx1(r),D(τ) C[r, 0] C[r, 0] D(σ) ,

c#x = c 0x(r) = ax 2(r),D(c#) C[r, 0] C[r, 0] ,

and σ is a closed linear operator. Notice that the first equation of the system in (8.5) was encountered into the definition of σ while the second one defines τ.

Now we seek the factor control vector d D(σ) such that σd = 0 and τd = 1. Elementary calculations show that

d = 1 1 + b 1 1

where 1 denotes the constant function taking the value 1 on [r, 0]. The knowledge of the factor control vector allows us to derive, as in Section 7.6, a final form of an abstract model of the system:

(t) =𝒜[x(t) u(t)d] y(t) =c#x(t) (8.7)

where 𝒜 = σker τ, i.e.

𝒜x = x,D(𝒜) = {x W1,2(r, 0) W1,2(r, 0) : x(0) = Ax(r)} (8.8)

From the theory of neutral delay systems, 𝒜 generates a C0–semigroup on H (or even a C0–group if detA0). This semigroup is EXS iff λ(A) < 1 [30, p. 148 - 154] or equivalently b < 1 . Since ρ > 1 and κ [1, 1], these inequalities are always satisfied. Observe that the condition κ = 1 corresponds to R1 = (unloaded line), while κ = 0 means that R1 = Z (the line loaded by the wave impedance), and κ = 1 means R1 = 0 (shunted line). To get the transfer function of (8.7), we rewrite the first equation of the system (8.7) in an equivalent form

𝒜1(t) = d dt 𝒜1x(t) = x(t) u(t)d .

Next, assuming null initial conditions and applying the Laplace transform we get

x̂(s) = s𝒜1 I1dû(s) = (sI 𝒜)𝒜1 1û(s) = =𝒜(sI 𝒜)1dû(s) = d (sI 𝒜)1dû(s) .

Hence,

ĝ(λ) = c#𝒜(𝒜 λI)1d = c#d λc#(λI 𝒜)1d (8.9)

because d D(c#), i.e., the compatibility assumption (7.5) holds. Now, by (8.8),

𝒜(𝒜 λI)1d = = d dθ eλθ I I eλrA1 d̂ F (λ) θ0eλ(θτ)ddτ = = eλθ 1 + b (eλr 1)A(I eλrA)1 + I 1 1 = = eλθ 1 + b(I A)(I eλrA)1 1 1 = eλθ 1 + be2λr eλr 1 λσ(𝒜) (8.10)

where d̂F (λ) =r0eλθddθ is the finite Laplace transform of d. Substituting (8.10) into (8.9), we obtain

ĝ(λ) = aeλr 1 + be2λr (8.11)

Exercise 8.3.1. Prove that ĝ H(Π+) but 1ĝH(Π+). Hint: aesr = aer Re s and 1 be2r Re s 1 + be2sr 1 + be2r Re s for all s , provided that b < 1.