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Lecture
Infinite horizon LER#
Continuous Time Discrete Time
##
- p.
= Atp t PA - PBR' ' BTPTQ Pk=QtAP Gtp BE
' Bt Priti)tp÷A
Allow T→a,
01=0 ⇒ M = O Allow ,N -70
,Pk = Pkt ,
= : Pa,
M=0
Be = O. DARE ( Discrete-time Algebraic
@ ARE ( continuous-time Algeb }Ri coati '
Eq 's. ) Rica ate Eat ) :
O = At Pat % A - Pa BR- ' BTPata Pg =QtAP£( It Pok BR
- 'BTpjIjPg¥A
Only make sense for LTI Only makes sense for LTI
Proposition
Existence,
Uniqueness ) proposition Existence,
Uniqueness )
-Let CA , B) be a controllable ditto,
i.e . Pato solves DARE.
( actually , just need stabilize ble ) /pain . Then 3 ! Pa Yo thatsowesCARE.LI#PrTnositione**--CzcoDTPaECo
) → ditwfe.cn?I?IaaetodoueindM )
Propositioned
Yogasooptasfabie,
Proposition
toYID.io/p-ustabiegT T
Suppose , O &Q= @ C Suppose of @
=p. c .
= '
Suppose ( A , B ) controllableAlso,
let CA , B) controllable
( actually ,need and CA ,
observable,
stabilizable ) f-and CA ,
observable,
Hnk@
=(RtBTPaBJ¥5pgAthem
Cactually tfefdctable) makes the closed - loop
figure optional closed -b)system stable
system I = CA - Bka )E &
HA - BR' ' BT Pg) z Int ,
= CA - B Kafka is @gmp)
is ( asymptotically ) stable stable ④⇒ taermatnix CA -BkaThe matrix CA - Bka ) is
is Hurwitz-
Schurr - Cohn stable.
ii )
pgyo.CI) ¥711121 )
Protoss '± ILet Ace : = CA- BR
- '
Bt Po ),
audlQ-a.cc?z-
- theyre C Thos is alwaysclosed - loop " A
" possible since Orfeo.
Now by CARE ,we have i
.: By spectral
AtP÷A=¥Bp⇒decompose . :
Q = U DV- '
¥aE¥WE Would like to investigate ⇒ @ = v DayUnder What condition,
X E @-
-
@pen lefthalf plane )which is equivalent to closed - loop stability .
Let ( It,
net) be the complex conjugate pair for G,
I )Pne - multiplying the above boxed eat by I
*
,and post -
- multiplying the same by I, yields .
.
( At 7) zo* pard = - I. care - z*pBR-'BT Pg I
Since the RHS of the last eat is the negative of a
pos.se?idef. quadratic form ,
hence RHS ← o.
On the Lets,
I-
Paid So since Pato .
Therefone,taeonlywagCHS=RHScanhapbe-nis
that @to ) E 0.
However , if Atx * =o,
then we get
O = - WI @ Ctw - I- PgBR-
'
BIBI.
Since the RHS above is sum of two quadratics ,
hence X td.
= O mandates
c The-_ o and 12-4255 Pak =
Em× ,
→ Xl I ( -
.
. Ace : = A - BR - ' BT Pa )
Are = Ace I = did .
⇒ fcTE=0n×,andAE⇒w#
Recall from linear systemstheorytheaterare
( A, C) detectable# And = And
,Ctw = o
,
I # 0
implies Rely ) LO .
However,
Red ) = STIL = o intros case,
which
is a contradiction. Therefore ,
We
camnotkave.ttX *= O
.
At this point ,we know that d THE O and
that ytx. to .
.
'
. Atd * Lo ⇒ REG ) C O.
EAce is Hurwitz .
This proof is from
Anderson femme ,Optimal Control : Linear Quadratic
Methods,
Eh 3£
Stabilize able ( in continuous - time ) sunkBaeagroydiy#*
Given CA , B),
taupe exist some matrix K,
s .t.
( A,- BK ) is Hurwitz
,i. e.
, ICA - BKD Lotti .
1€ Ended
7%7:" am" :Kind:c.in?IIi7gsta.ef...@/t.ne::I....:-:
Ez .
• stabilized e rank LIE- A) I BJ = n
Here A is nxntf E ④t
B is nxm,
MF n-
MATLAB command to compute LQR infinite horizon.
#.
( computing Pa )⇒ Efa,
Pa,
⇒ =
lq.ncsss.QLR.sn
)matrices appearing
Kalman ↳Eof\ in the fainter . n
gain CARE or
D ARE eigenvaluesOf the
closed - loopmatrix
( A - Bka )⇒ sys = SS ( AB -
( will print the
Amatrix )
( You could create"
Sgs"
as discrete time system by
passing extra argument (Tss : sampling time ) , e. g. says ⇒ s CAs Be , Did
Handling AdditionalConstraints in the Ocp ( Continuous
time )÷
④/ Integral ) Iso perimetric constraints :
-
To NCE ,I ,t)dt must be conserved
.
I C- IRN
iced Sjw CE , IF ) dt = £ given-
To handle this : introduce extra state Knt ,
S.t. Int ,
= NEE, Est )
and sent , ( 07=0, sent ,
CT) = ④ Given it
- -
Now apply necessary conditions to Hamiltonian :
H = Lt LILI,
t Anti N-In particular , y ;+,=
-
ZI-n.io#u+FEtagt.
④ Contrivealityco#nt.
#
-
@ Crest ) = O,
A- ERM,
m > 2 .
( For me ',
this doesn't makeL scalar function sense since thenptkeesigw.no
¥687Itamietonian : H = Lt If
[email protected] fe④ Equality constraints on fas of control
[email protected] , ⇒ = O Where
2¥ to
Again , forarmy I.
H = Lt dtftpe G) @
I = -3¥ =- Ez - REITs-Keeffe
.
¥90 Equality constraint on functions of state-
Variables .
SEE,
t ) = O . - f* , for all to Et ET
Its = 2£ + zIa⑧/f " ⇒= o - - - * * )
Now,
CAA ) may on may not have explicit
dependence on I.
→ IL it does,
them Eta plays the mole ofCEE
,
Est7=0as in # 3However
, we must either eliminate one
Component of I in terms of the remaining( n - a ) components using # ) . OR
add @ ) as a B. C.
@ t = to or t If.
→ If * * ) still does not have explicit dependence
on U, them do dfdt again
Teen doing until u
appearsexplicitly
Suppose this happens @ qth orders £
Then sake, a ,⇒=o . where 59=117
.
-
plays the mole of ECI, at ) = o.
In addition,
we must eliminate q
Components of z in terms of the remaining
( n - q ) components using q algebraic eats .
.
(s
:÷¥¥¥ ! - l a. * ..