Question 22.2: Consider the scalar system x(t) = ax(t) + u(t) (22.5.3) and ...
Consider the scalar system
\dot{x}(t)=ax(t)+u(t) (22.5.3)
and the cost function
J=\psi _{f}x(t_{f})^{2}+\int_{0}^{t_{f}}{(\psi x(t)^{2}+u(t)^{2})dt} (22.5.4)
22.2.1 Discuss this optimal control problem in the light of Lemma 22.1 and Lemma
22.2.
22.2.2 Discuss the convergence of the solutions of the CTDRE to P^{s}_{\infty } .
22.2.1 The associated CTDRE is
\dot{P}(t)=-2aP(t)+P(t)^{2}-\psi; P(t_{f})=\psi _{f} (22.5.5)
and the CTARE is
(P^{s}_{\infty })^{2}-2aP^{s}_{\infty }-\psi =0Case 1 ψ ≠ 0
Here (A, \Psi ^{\frac{1}{2} }) is completely observable (and thus detectable). From Lemma 22.1, part (a), there is only one non-negative solution of the CTARE. This solution coincides with the stabilizing solution. Making the calculations,we find that the only non-negative solution of the CTARE is
P^{s}_{\infty }=\frac{2a+\sqrt{4a^{2}+4\psi } }{2} (22.5.7)
leading to the following feedback gain:
K^{s}_{\infty }= a+\sqrt{a^{2}+\psi } (22.5.8)
The corresponding closed loop pole is at
p_{cl}=-\sqrt{a^{2}+\psi } (22.5.9)
This is clearly in the LHP verifying that the solution is indeed the stabilizing solution.
Case 2 ψ = 0
Here we will consider three positive values for a: one positive, one negative and one zero.
(i) a > 0. In this case, (A, \Psi ^{\frac{1}{2} }) has an observable pole outside the stability region. Then from part (c) of Lemma 22.1, in addition to the stabilizing solution, there exists at least one other non negative solution to the CTARE. Making the calculations, we find that there are two non negative solutions of the CTARE, namely: the stabilizing solution P^{s}_{\infty } =2a and one other solution P^{\prime }_{\infty }.
We see that P^{s}_{\infty } – P^{\prime }_{\infty } > 0 and that P^{s}_{\infty } gives a feedback loop gain of K^{s}_{\infty }=2a leading to closed loop poles at p_{cl}=-a. This is clearly in the LHP verifying that P^{s}_{\infty } is indeed the stabilizing solution.
(ii) a < 0. Here (A, \Psi ^{\frac{1}{2} }) is again detectable. Thus, from Lemma 22.1 on page 666, part (a), there is only one nonnegative solution of the CTARE. This solution coincides with the stabilizing solution.
Making the calculations, we find P^{s}_{\infty }=0, and the corresponding feedback gain is K^{s}_{\infty }=0, leading to closed loop poles at p_{cl}=a, which is stable, since a < 0, by hypothesis.
(iii) a = 0. Here (A, \Psi ^{\frac{1}{2} }) has an unobservable pole on the stability boundary. This is then not covered by Lemma 22.1 on page 666. However, it can be verified that the CTARE has only one solution, namely P_{∞}, which is not stabilizing.
22.2.2 To study the convergence of the solutions, we again consider two cases.
Case 1 ψ ≠ 0.
Here (A, \Psi ^{\frac{1}{2} }) is completely observable. Then P(t) converges to P^{s}_{\infty } given in (22.5.7) for any P(t_{f}) ≥ 0.
Case 2 ψ = 0
(i) a > 0. Here (A, \Psi ^{\frac{1}{2} }) has no unobservable poles on the jω axis, but it is not detectable. Thus, P(t) converges to P^{s}_{\infty } provided we choose P(t_f) >P^{s}_{\infty }.
(ii) a < 0. Here (A, \Psi ^{\frac{1}{2} }) is again detectable and P(t) converges to P^{s}_{\infty } for any P(t_{f}) ≥ 0.
(iii) a = 0. Here (A, \Psi ^{\frac{1}{2} }) has an unobservable pole on the jω axis. Thus, Lemma 22.2 on page 667 does not apply. Actually, P(t) converges to 0 for any P(t_{f}) ≥ 0. However,in this case, zero is not a stabilizing solution.