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 -

egsi-AntBu@O07SSys.A

( 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

tqeCt3@oIEIzyI-EsutKtDt2Iutfects2Iu.e fe④ Equality constraints on fas of control

andst_ate@Cz.u , ⇒ = 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. * ..