Presentazione di PowerPoint - uniroma1.itiacoviel/Materiale/Do... · Given the initial point x i the problem is to bring that point to the origin along a trajectory constituted by

Optimal Control

Prof. Daniela Iacoviello

Lecture

Double integrator (from Bruni et al.1993)

1)(

0)(,)()()(

)()(

2

21

===

=

tu

txxtxtutx

txtx

fii

=

=

1

0

00

10BA

0;0Eigenvalues:

The couple verifies the condition of controllability:)( BA

0101

10det)det( −=

=ABB

( )of

oo tux ,

Unique, non singular and

the optimal control is bang-bang

with a number of discontinuity points

11=− no

Goal: Minimizeif

t

t

ttdtJ

f

i

−==

(1)

Prof.Daniela Iacoviello- Optimal Control 16/11/2018

Description of the trajectories when 1=u

From system (1), it results:

( )

( ) ( )2211

22

2

1)(

)(

iiii

ii

ttttxxtx

xtttx

−−+=

+−=

( ) ii xtxtt 22 )( −=−

22

22

2222

22222

2

2222211

)(2

1

)()(2

1

2

1)(

2

1)(

)(2

1)()(

i

ii

iii

iiii

xtx

xtxxtx

xtxxxtx

xtxxtxxxtx

−=

+−=

−+−=

−−=−

Prof.Daniela Iacoviello- Optimal Control Pagina 3

Description of the trajectories when 1=u

From system (1), it results:

( )

( ) ( )2211

22

2

1)(

)(

iiii

ii

ttttxxtx

xtttx

−−+=

+−=

( ) ii xtxtt 22 )( −=−

22

22

2222

22222

2

2222211

)(2

1

)()(2

1

2

1)(

2

1)(

)(2

1)()(

i

ii

iii

iiii

xtx

xtxxtx

xtxxxtx

xtxxtxxxtx

−=

+−=

−+−=

−−=−


The optimal trajectory is described by parabolic arcs with equation:

22

2211 )(

2

1)( ii xtxxtx −=−

x1

x2 u=+1u=-1

Prof.Daniela Iacoviello- Optimal Control 16/11/2018 Pagina 5

Given the initial point xi the problem is to bring that point to the

origin along a trajectory constituted by parabolic arcs

with 1 or 0 switching points

-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-


Let’s define:

==+ 0,2

1: 2

221

2 xxxRx

−==− 0,2

1: 2

221

2 xxxRx

−=== −+ 0,2

1: 2221

2 xxxxRx

−=+221

2

2

1: xxxRx

−=−221

2

2

1: xxxRx

0\2R= −+−+

xi belongs to one

of these regions


Let’s define:

==+ 0,2

1: 2

221

2 xxxRx

−==− 0,2

1: 2

221

2 xxxRx

−=== −+ 0,2

1: 2221

2 xxxxRx

−=+221

2

2

1: xxxRx

−=−221

2

2

1: xxxRx

0\2R= −+−+

xi belongs to one

of these regions





-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-


Let’s define:

==+ 0,2

1: 2

221

2 xxxRx

−==− 0,2

1: 2

221

2 xxxRx

−=== −+ 0,2

1: 2221

2 xxxxRx

−=+221

2

2

1: xxxRx

−=−221

2

2

1: xxxRx

0\2R= −+−+

xi belongs to one

of these regions





-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-


Four cases:

== + 0,2

1: 2

221

2 xxxRxxi 1)

with a control u=1 and without

switching you reach the origin

0 10 20 30 40 50 60 70 80 90 100-10

-5

0

5

x1

x 2

+

xi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

−== − 0,2

1: 2

221

2 xxxRxxi 2)

with a control u= -1 and

without switching you

reach the origin

-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0-5

0

5

10

x1

x 2

-

xi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

3)

First use a control with u= +1

and you get curve γ-

then you switch to control u=- 1

and you get to the origin

−= +221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

3)





−= +221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

3)





−= +221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

3)





−= +221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

Prof.Daniela Iacoviello- Optimal Control

4)

First use a control with u= -1

and you get curve γ+

then you switch to control u=+ 1


−= −221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-


4)





−= −221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-


4)





−= −221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-


4)





−= −221

2

2

1: xxxRxxi


-100 -80 -60 -40 -20 0 20 40 60 80 100-10

-8

-6

-4

-2

0

2

4

6

8

10

x1

x2

-

+

+

-

( )

−

=

−−

++

)(1

)(1)(

txf

txiftxu

o

ooo


Calculus of the minimum time:

it depends on the location of the initial point xi

( )

−

=

−−

++

)(1

)(1)(

txf

txiftxu

o

ooo

A)+− = ix

The control doesn’t switch; from ( ) ii xtxtt 22 )( −=−

( ) iiii

of xxxtt 2220 ==−=−


B) +ix The control switches at an instant at the position t

( ) ( )ifif tttttt −+−=−

Time with

u=-1

Time with

u=+1

Time with

u=-1

Time with

u=+1

( ) ii xtxtt 22 )( −=−

From

iof xtt 2=−

ix2

iii xxtt 22 −=−

iii

of xxtt 222 −=−

16/11/2018 Pagina 24

B) +ix The control switches at an instant at the position t

( ) ( )ifif tttttt −+−=−

Time with

u=-1

Time with

u=+1

Time with

u=-1

Time with

u=+1

( ) ii xtxtt 22 )( −=−

From

iof xtt 2=−

ix2

iii xxtt 22 −=−

iii

of xxtt 222 −=−

16/11/2018 Pagina 25

The position must belong to the two parabolic arcs

Time with

u=-1

Time with

u=+1

ix2

22

2211 )(

2

1)( ii xtxxtx −=−

22

2211

2

1

2

1 ii xxxx −+=

221

2

1xx −=

22

22

221

2

1

2

1

2

1 ii xxxx +−=+

0,2

12

221

22 +−= xxxx ii

2212

2

1 ii xxx +−=

iii

of xxtt 222 −=−

By substituting in

iiii

of xxxtt 2

221 24 −+−=−Prof.Daniela Iacoviello- Optimal Control

C) −ix The same calculations yield:

iiii

of xxxtt 2

221 24 ++=−

The commutation curve is the γ-curve

2212

1)( xxxx +=

+−=−= 2212

1)()( xxxsignxsigntuo


Harmonic oscillator (from Bruni et al.1993)

0)()()(

)()(

12

21

+−=

=

tutxtx

txtx

−=

0

0

A

=

1

0b

jEigenvalues:

When u=0

0)()(

0)()(

22

2

12

1

=+

=+

txtx

txtx

The natural modes are oscillatory


Problem

1)(

0)()(

==

tu

txxtx fiiDetermine a control such that

That minimizes the cost index:

( ) singularnot01

0

=

Abb

There exists a unique, non singular solution

and the control is bang-bang for any initial condition

iff tttJ −=)(

Eigenvalues of A: j

( )of

oo tux ,,


)()()()()()(1 21221 tuttxttxtH +−+=

BUTSince the eigenvalues are complex we can’t apply the theorem about

the maximum number of commutation points

Alternative procedure:

Necessary conditions:

Introduce the Hamiltonian:

)()( tAt oTo −=

1)(:

,)()()()()()(1

)()()()()()(1

2122

2122

1

1

+−+

+−+

t

tttxttxt

tuttxttxt

ooooo

oooooo


)()(

)()(

1

1

2

2

tt

tt

oo

oo

−=

=

1)(:,)()()()(22

ttttut ooo

( )

+−−=

−=−=

)(sin

)()()(2

i

ooo

ttKsign

tsignbtsigntu

( ) +−=−= )(sin)()()( 22

2 2 iooo ttKttt

All the switching subintervals have length equal to

with the exception of the first and the last whose length is less or

equal than

16/11/2018 Pagina 31

)()( tAt oTo −=

1)(:

,)()()()()()(1

)()()()()()(1

2122

2122

1

1

+−+

+−+

t

tttxttxt

tuttxttxt

ooooo

oooooo

)()(

)()(

1

1

2

2

tt

tt

oo

oo

−=

=

1)(:,)()()()(22

ttttut ooo

( )

+−−=

−=−=

)(sin

)()()(2

i

ooo

ttKsign

tsignbtsigntu

( ) +−=−= )(sin)()()( 22

2 2 iooo ttKttt


with the exception of the first and the last whose length is less or

equal than

16/11/2018 Pagina 32

)()( tAt oTo −=

1)(:

,)()()()()()(1

)()()()()()(1

2122

2122

1

1

+−+

+−+

t

tttxttxt

tuttxttxt

ooooo

oooooo


With the exception of the first and the last

whose length is less or equal than

Figura 1 (Bruni et al.1993)Prof.Daniela Iacoviello- Optimal Control 16/11/2018 Pagina 33








Prof.Daniela Iacoviello- Optimal ControlFigura 1 (Bruni et al.1993)





Prof.Daniela Iacoviello- Optimal ControlFigura 1 (Bruni et al.1993)













Let’s assume

And integrate the system:

1)( =tu

0)()()(

)()(

12

21

+−=

=

tutxtx

txtx

1)(sin)(cos

1)( 211 −+−

= i

ii

i ttxttxtx

)(cos)(sin1

)( 212 ii

ii ttxttxtx −+−

=

( )22

2

12

2

1

1)(

1)(

2

ii xxtxtx +

=+


Figura 2. (Bruni et al.1993)

( )22

2

12

2

1

1)(

1)(

2

ii xxtxtx +

=+

The trajectories with control are circumferences with center

passing through the initial condition

The direction is clockwise.

1)( =tu

0,

1

ix

16/11/2018 Pagina 40

16/11/2018Prof.Daniela Iacoviello- Optimal Control Pagina 41

From Bruni et al. 2003

Only one trajectory for each of these families passes through the origin

1)(

1)( 2

2

1 2=+

txtx

Figura 3. (Bruni et al.1993)Prof.Daniela Iacoviello- Optimal Control 16/11/2018 Pagina 42

For a generic instant t:

= −

1

)(

)()(

1

1 2

tx

txtgt

2

1

1212

2

1

21

)(

)()(1

)()(

1)(

)(1

1)(

2

−

+

=

tx

txtxtxtx

tx

txt


For a generic instant t:

= −

1

)(

)()(

1

1 2

tx

txtgt

2

1

1212

2

1

21

)(

)()(1

)()(

1)(

)(1

1)(

2

−

+

=

tx

txtxtxtx

tx

txt

−== )(t

0

)()()(

)()(

12

21

+−=

=

tutxtx

txtx

The trajectories are traversed

with constant angular velocity


The trajectories are traversed with constant angular velocity

=t

1)( =tu

The trajectories are traversed with constant angular velocity

We must reach the origin along arcs with amplitude

There exists a unique way to do it!

16/11/2018

Definitions:

,....2,1,0,0,112

: 22

22

2

12 =

=+

+−=+ kxx

kxRxk

,...2,1,0,0,112

: 22

22

2

12 =

=+

++=− kxx

kxRxk

Figura 4. (from Bruni et al. 1993):

+k

−k

+k

−k

16/11/2018 Pagina 46

Let and , k=1,2,…., be the sets in Figure,

obtained rotating

and around

,

respectively

Define:

+k−k

=

++ =0k

k

=

−− =0k

k

=

++ =0k

k

=

−− =0k

k

−+ = Figura 5. (from Bruni et al. 1993)

16/11/2018 Pagina 47

+k

−k

0,

1

The initial states and may be moved to the origin with a

constant control equal to +1 and -1 respectively and the corresponding arc

is less or equal than π.

The points belonging to are the only ones that can be transfered to the

origin with a constant control, without switching

+ 0ix − 0ix

−+ 00

1a 1b


Figure from Bruni et al. 1993

Let’s consider initial states

They may be moved to the origin with the following path:

- First a constant control +1 up to the arc for an interval time

- Switching to control -1 to reach the origin.

1 switching instant

( )−++ 001 \ ix

−0

2a






1 switching instant

( )−++ 001 \ ix

−0

2a


1





1 switching instant

( )−++ 001 \ ix

−0

2a


1

2



- First a constant control -1 up to the arc for an interval time

- Switching to control +1 to reach the origin.

1 switching instant

+0

( )−+− 001 \ ix 2b






1 switching instant

+0

( )−+− 001 \ ix 2b


1





1 switching instant

+0

( )−+− 001 \ ix 2b


1

2




- Note that: apply strategy 2b

- Switch to control -1 to reach and then switch to control +1 to reach the

origin two switching instants

−+ 12 \ ix

−1




- Note that apply strategy 2a

- Switch to control +1 to reach and then switch to control -1 to reach the

origin: two switching instants

+

1

+− 12 \ ix

−− 11

+0

3a

3b

++ 11

−0

Figura 8. (from Bruni et al. 1993)Prof.Daniela Iacoviello- Optimal Control 16/11/2018 Pagina 56

3

1

2


1

2

Figura 6. (from Bruni et al. 1993)Prof.Daniela Iacoviello- Optimal Control 16/11/2018


The same happens for any intial condition: −−

+ 1\ kkix +−

− 1\ kkix

−+ \ixAt the generic state you must associate u=+1

At the generic state you must associate u=-1 +− \ix

( )

−

=

+−

−+

\)(1

\)(1)(

tx

txtxu

o

ooo

The number of switching points is given by the minimum index k among the ones

characterizing the sets and

Example: Consider the point

Then there is only one switching instant.

−k+k

+++ 321P


\ix + −

Time necessary to get to the origin

If is the number of switching instants:

( )

foii

of

o ttif +−+=− 11

If

where:

if

if

−= −

i

i

ix

xtg

1

21

1

o

Rotation of the optimal trajectory

to go from the initial point to the

first point of commutation

Rotation of the optimal trajectory

to go from the final point to the

origin

0=o

ii

of tt =−

+= −

i

i

ix

xtg

1

21

1

+ oix

− oix Pagina 61


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Presentazione di PowerPoint - uniroma1.itiacoviel/Materiale/Do... · Given the initial point x i the problem is to bring that point to the origin along a trajectory constituted by