794 Chapter 10 / Control Systems Design in State SpaceNow let us solve the optimization problem. Substituting Equation (10–113) into
Equation (10–112), we obtain
In the following derivations, we assume that the matrix A-BKis stable, or that the
eigenvalues of A-BKhave negative real parts.
Substituting Equation (10–113) into Equation (10–114) yields
Let us set
wherePis a positive-definite Hermitian or real symmetric matrix. Then we obtain
Comparing both sides of this last equation and noting that this equation must hold true
for any x, we require that
(10–115)
It can be proved that if A-BKis a stable matrix, there exists a positive-definite ma-
trixPthat satisfies Equation (10–115). (See Problem A–10–15.)
Hence our procedure is to determine the elements of Pfrom Equation (10–115) and
see if it is positive definite. (Note that more than one matrix Pmay satisfy this equation.
If the system is stable, there always exists one positive-definite matrix Pto satisfy this
equation. This means that, if we solve this equation and find one positive-definite matrix
P, the system is stable. Other Pmatrices that satisfy this equation are not positive definite
and must be discarded.)
The performance index Jcan be evaluated as
Since all eigenvalues of A-BKare assumed to have negative real parts, we have
Therefore, we obtain
(10–116)
Thus, the performance index Jcan be obtained in terms of the initial condition x(0)
andP.
To obtain the solution to the quadratic optimal control problem, we proceed as
follows: Since Rhas been assumed to be a positive-definite Hermitian or real symmetric
matrix, we can write
R=T* T
J=x*( 0 ) Px( 0 )
x(q)S 0.
J=
3
q0x(Q+K RK)xdt=-x* Px^2
q0=-x(q) Px(q)+x( 0 ) Px( 0 )
(A-BK) P+P(A-BK)=-(Q+K RK)
x(Q+K RK) x=-x
Px-xPx
=-xC(A-BK) P+P(A-BK)D x
x(Q+KRK) x=-
d
dt
(x*Px)
=
3
q0x(Q+KRK) xdt
J=
3
q0(xQx+xK*RKx)dt
x
=Ax-BKx=(A-BK) x
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