Bukti:
Misalkan hipotesis teorema 5.7 berlaku. Maka berdasarkan syarat eksistensi dan
ketunggalan kontrol optimal masalah Ω
ଵ
seperti yang diberikan dalam bagian 2.2, Ω
ଶ
mempunyai kontrol optimal tunggal ݑ
∗
, dimana ݑ
∗
memenuhi persamaan 15, 16 dan 17. Dari transformasi 7 dan relasi 9 diperoleh
൬ݔ
∗
ݐ ݑ
∗
ݐ൰ =
൬ܰ ܫ
൰ ቌ
ݔ
1 ∗
ݐ ݔ
2 ∗
ݐ
ݑ
∗
ݐ ቍ
= ൬ܰ
ܫ
൰ ቌ
ݔ
1 ∗
ݐ −ܣ
22 −1
ܣ
21
ݔ
1 ∗
ݐ − ܣ
22 −1
ܤ
2
ݑ
∗
ݐ ݑ
∗
ݐ ቍ
= ൬ܰ
ܫ
൰ ቌ
ܫ
−ܣ
22 −1
ܣ
21
−ܣ
22 −1
ܤ
2
ܫ
ቍ ൬
ݔ
1 ∗
ݐ
−ܮ
ݔ
1 ∗
ݐ
൰
= ൬ܰ
ܫ
൰ ൮
ܫ
−ܣ
22 −1
ܣ
21
+ ܣ
22 −1
ܤ
2
ܮ
____________________ −ܮ
൲
ݔ
1 ∗
ݐ. ∎
VI. MATLAB CODE
This CODE is to find the optimal control-state pair of the LQ control problem subject to singular continuous time invariant
============================================================== Input the matrices E,A,B,C,D of appropriate dimension
Observe the rankE, the number of column of B and the number of column of C n = sizeE,1
p=rankE r=sizeB,2
q=sizeC,1 Determination of Singular value Decomposition of the Matrix E
[M,S,N] = svdE MEN=MEN
MAN=MAN A11 = MAN1:p,1:p
A12 = MAN1:p,p+1:n A21 = MANp+1:n,1:p
A22 = MANp+1:n,p+1:n MB=MB
B1=MB1:p,1:r B2=MBp+1:n,1:r
CN=CN C1=CN1:q,1:p
C2=CN1:q,p+1:n A22B2=[A22 B2]
Verify whether rank [A22 B2]=number of row of [A22 B2] If yes, to becontinued
If no, to be stop because the sistem uncontrollable impulse s1 = sizeA22B2,1
s=rank[A22 B2] Check singularitypositive definite of the transformed system
If test=test1, test2=test3 then continue test=rank[A22 B2;C2 D]
test1=n-p+r test2=rank[zerosn,n E zerosn,r;E A B;zerosq,n C D]
test3=n+p+r Choose the matrix W as follows:
W=null[A22 B2] Check=[A22 B2]W
W1=W1:n-p,1:r W2=Wn-p+1:n-p+r,1:r
Determine the generalized invers of [A22 B2] PseudoinversA22B2=pinv[A22 B2]
Abar=A11-[A12 B1]PseudoinversA22B2A21 Bbar=[A12 B1]W
Cbar=C1-[C2 D]PseudoinversA22B2A21 Dbar=[C2 D]W
Q11=CbarCbar detQ11=detQ11
Q12=CbarDbar Q22=DbarDbar
detQ22=detQ22 Q=[Q11 Q12;Q12 Q22]
detQ=detQ Test for imlementation of LQR tools
detQ11-Q12invQ22transposeQ12=0 detQ11-Q12invQ22Q12
c_stab=eigAbar stability1=p
stability2=n stability1=rank[Abar-11eyep Bbar]
stability2=rank[A-11E B] d_detect=eigAbar-BbarinvQ22Q12
detectability1=p detectability2=n+r
detectability1=rank[Abar-BbarinvQ22Q12-3.7096eyep;Q11- Q12invQ22Q12]
detectability2=rank[A-3.7096E B;C D] [L,P,e] = lqrAbar,Bbar,Q11,Q22,Q12
detP=detP ARE=AbarP+PAbar+Q11-PBbar+Q12invQ22PBbar+Q12
incondition=[eyep zerosp,n-p]M[1;2;0;0] koef=Abar-BbarL
lambda = eigkoef [x11,x12] = dsolveDx11=-9.3044x11-9.1776x12, Dx12 =4.7518x11+4.0852x12,
x110 = 1, x120 = 2 Simplify these results by the following way:
digits6 simplifyvpax11
simplifyvpax12 v=-L[x11;x12]
v1=transposev Objective=int[x11 x12 v1]Q[x11;x12;v],0,inf
Joptimum=evalObjective x1=N[eyep;gamma1-W1L][x11;x12]
u1=gamma2-W2L[x11;x12] Simplify these results by the following way:
digits6 x=vpax1
u=vpau1 =============Construction into Feedback form
K2=[1 0;3 1] r1=rankA22+B2K2
r2=detA22+B2K2 K1=gamma2-W2L-K2gamma1-W1L
controlfeedback=K1[x11;x12]+K2x2
VII. KESIMPULAN DAN SARAN
