Lecture 04 OCLS

7/30/2019 Lecture 04 OCLS

1/48

Lecture 4: Well-Posedness and Internal Stability

Well-Posedness of Feedback Systems

c A.Shiriaev/L. Freidovich. Januar 29, 2010. O timal Control for Linear S stems: Lecture 4 . 1/17


2/48

Lecture 4: Well-Posedness and Internal Stability

Well-Posedness of Feedback Systems

Internal Stability of Feedback Systems



3/48

One Degree of Freedom Feedback Control System

K and P represent the controller and the plant,

r and y are the reference and the output,

up and yp the input and output of the plant,

di and d the input and output disturbances,

n is the measurement noise.



4/48

Example of a not realizable system

Suppose that

P(s) = s 1

s + 2, K(s) = 1



5/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)up + d, up = u + di

u = K(s) r n y



6/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n y



7/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu = K(s)

r n

P(s)

u + di

+ d



8/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu = r n

P(s)

u + di

+ d



9/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu + P(s)u = r n P(s)di d



10/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu =

1 + P(s)

1

[r n P(s)di d]



11/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu =

1 + P(s)

1

[r n P(s)di d]

u =

1

s 1

s + 2

1

r n +

s 1

s + 2di d



12/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu =

1 + P(s)

1

[r n P(s)di d]

u =

s + 2 (s 1)

s + 2

1

r n +

s 1

s + 2di d



13/48


Suppose that

P(s) = s 1

s + 2, K(s) = 1

y = P(s)

u + di

+ d

u = K(s) r n yu =

1 + P(s)

1

[r n P(s)di d]

u =

s + 2 (s 1)

s + 2

1

r n +

s 1

s + 2di d

u =

s + 2

3

r n d

+

s 1

3 di



14/48

Concept of Well-Posedness:

A closed-loop system is said to be well-posed, if all transferfunctions (matrices) from all the external signals

r, n, di, d

to the input and output signals

y and u

are proper (the degree of the numerator is not bigger than thedegree of the denominator).

If this is not true, the system is not physically realizable.



15/48

Combining external signals



16/48

Combining external signals

K(s) = K(s), w1 = di, w2 = n + d r



17/48




18/48


e1 = w1 + K(s)e2, e2 = w2 + P(s)e1



19/48


e1 = w1 + K(s)e2, e2 = w2 + P(s)e1

e1 = w1+Kw2 + P e1 e1 = I K(s)P(s)1

w1 + K(s)w2



20/48


e1 = w1 + K(s)e2, e2 = w2 + P(s)e1

e1 = I K(s)P(s)1

w1 + K(s)w2e2 =

I P(s)K(s)

1

w2 + P(s)w1



21/48

Criteria for Well-Posedness

The feedback system with proper K(s) and P(s)

e1 = I K(s)P(s)1

w1 + K(s)w2e2 =

I P(s)K(s)

1

P(s)w1 + w2

is well-posed if and only if



22/48



e1 = I K(s)P(s)1

w1 + K(s)w2e2 =

I P(s)K(s)

1

P(s)w1 + w2

is well-posed if and only if the matrix transfer function

I K(s)P(s)

1

exists and is proper.



23/48



e1 = I K(s)P(s)1

w1 + K(s)w2e2 =

I P(s)K(s)

1

P(s)w1 + w2

is well-posed if and only if the matrix

I K(s)P(s)

s=+is invertible.



24/48



e1 = I K(s)P(s)1

w1 + K(s)w2e2 =

I P(s)K(s)

1

P(s)w1 + w2

is well-posed if and only if the matrix

I K(s)P(s)

s=+is invertible.

It is sufficient to have either K(s) or P(s) strictly proper.



25/48

Lecture 4: Concepts of Well-Posedness and Internal

Stability

Well-Posedness of Feedback System

Internal Stability of Feedback System



26/48

Concept of Internal Stability of Feedback System:

The well-posed feedback system

e1 = w1 + K(s)e2, e2 = w2 + P(s)e1

or the well-posed feedback system

e1 =

I K(s)P(s)

1

w1 + K(s)w2

e2 = I P(s)K(s)1

P(s)w1 + w2is said to be internally stable if all 4 transfer functions

from (w1, w2) to (e1, e2)

have no poles in the closed right-half plane, i.e.

(I KP)1, (I KP)1K, (I PK)1P, (I PK)1

belong to RH .



27/48

Example

Suppose that

P(s) = s 1

s + 1, K(s) =

1

s 1



28/48

Example

Suppose that

P(s) = s 1

s + 1, K(s) =

1

s 1

Then

e1e2

=

I K(s)P(s)1

I

K(s)P(s)1

K(s)I P(s)K(s)

1

P(s)

I P(s)K(s)1

w1w2



29/48

Example

Suppose that

P(s) = s 1

s + 1, K(s) =

1

s 1

Then

e1e2

=

I K(s)P(s)1

I

K(s)P(s)1

K(s)I P(s)K(s)

1

P(s)

I P(s)K(s)1

w1w2

=

s + 1s + 2

(s + 1)(s 1)(s + 2)

s 1

s + 2

s + 1

s + 2

w1

w2



30/48

Example

Suppose that

P(s) = s 1

s + 1, K(s) =

1

s 1

Then

e1e2

=

I K(s)P(s)1

I

K(s)P(s)1

K(s)I P(s)K(s)

1

P(s)

I P(s)K(s)1

w1w2

=

s + 1s + 2

(s + 1)(s 1)(s + 2)

s 1

s + 2

s + 1

s + 2

w1

w2



31/48

Co-Prime Factorization

Two polynomials

p(s) = sk + p1 sk1 + + pk1 s + pk

q(s) = sm + q1 sm1 + + qm1 s + qm

are called co-prime if they have no common roots.



32/48


Two polynomials

p(s) = sk + p1 sk1 + + pk1 s + pk

q(s) = sm + q1 sm1 + + qm1 s + qm

are called co-prime if they have no common roots.

If p(s) and q(s) are co-prime, then there are two polynomialsx(s) and y(s) such that

x(s)p(s) + y(s)q(s) = 1

and the pair ( x(s) , y(s) ) can be found from the (reversed)Euclids algorithm.



33/48


Two transfer functions p(s) RH , q(s) RH arecalled co-prime if there are two new transfer functionsx(s) RH and y(s) RH such that

x(s)p(s) + y(s)q(s) = 1



34/48


Two transfer functions p(s) RH , q(s) RH arecalled co-prime if there are two new transfer functionsx(s) RH and y(s) RH such that

x(s)p(s) + y(s)q(s) = 1

Meaning: if p(s) and q(s) have a common divisor h RH :

p(s) = h(s)p1(s) and q(s) = h(s)q1(s)

such that

p1(s) RH and q1(s) RH

then

h1(s) RH

i.e. h(s) has neither zeros no poles in the closed right-halfplane (such transfer functions are called minimum-phase).



35/48


Two matrix transfer functions M(s), N(s) RH are calledright co-prime over RH if

they have the same number of columns

there are matrices Xr(s) , Yr(s) RH such that

Xr(s), Yr(s)

M(s)

N(s)

= Xr(s)M(s)+Yr(s)N(s) = I



36/48


Two matrix transfer functions M(s), N(s) RH are calledright co-prime over RH if

they have the same number of columns

there are matrices Xr(s) , Yr(s) RH such that

Xr(s), Yr(s)

M(s)

N(s)

= Xr(s)M(s)+Yr(s)N(s) = I

Two matrix transfer functions M(s), N(s) RH are calledleft co-prime over RH if

they have the same number of rows there are matrices Xl(s) , Yl(s) RH such that

M(s), N(s) Xl(s)Yl(s)

= M(s)Xl(s)+N(s)Yl(s) = Ic A.Shiriaev/L. Freidovich. Januar 29, 2010. O timal Control for Linear S stems: Lecture 4 . 13/17


37/48

Defining a Co-Prime Factorization

Given a matrix transfer function G(s) , then

G(s) = N(s)M1(s) is a right co-prime factorization, if

M(s), N(s) are right co-prime over RH ;



38/48





G(s) = M1(s)N(s) is a left co-prime factorization, if

M(s),

N(s) are left co-prime over

RH;



39/48





G(s) = M1(s)N(s) is a left co-prime factorization, if

M(

s)

, N(

s) are left co-prime over

RH;

G(s) is said to have double co-prime factorization, if thereare right and left co-prime factorizations such

Xr(s) Yr(s)

N(s) M(s)

M(s) Yl(s)

N(s) Xl(s)

= I



40/48

Computing a Co-Prime Factorization

Given a matrix transfer function G(s) = A B

C D

with

(A, B) stabilizable, i.e. F : (A + BF) is stable;

(A, C) detectable, i.e. L : (A + LC) is stable;


C i C P i F i i


41/48

Computing a Co-Prime Factorization

Given a matrix transfer function G(s) = A B

C D

with

(A, B) stabilizable, i.e. F : (A + BF) is stable;

(A, C) detectable, i.e. L : (A + LC) is stable;

The left and right co-prime factorizations can be computed as

M(s) Yl(s)N(s) Xl(s)

:=

A + BF B LF I 0C + DF D I

Xr(s) Yr(s)

N(s) M(s)

:=

A + LC (B + LD) LF I 0C D I

That is, G(s) = N(s)M

1(s) = M

1(s)N(s) , and theseare coefficients of the double co-prime factorization.


U i C P i F t i ti


42/48

Using a Co-Prime Factorization



U i C P i F t i ti


43/48


Given plant and controller matrix transfer functions

P(s) = N(s)M1(s) = M1(s)N(s)

K(s) = U(s)V

1

(s) =V

1

(s)

U(s)


Using a Co Prime Factorization


44/48



P(s) = N(s)M1(s) = M1(s)N(s)

K(s) = U(s)V

1

(s) =V

1

(s)

U(s)

The following conditions are equivalent:

the closed-loop system is internally stable; (I KP)1 , (I KP)1K , (I PK)1P , and

(I PK)1 RH ;




45/48



P(s) = N(s)M1(s) = M1(s)N(s)

K(

s) =

U(

s)

V1

(s

) =V1

(s

)U

(s

)



(I PK)1 RH ;

the matrix function

M U

N V

is invertible in RH ;




46/48



P(s) = N(s)M1(s) = M1(s)N(s)

K(

s) =

U(

s)

V1

(s

) =V1

(s

)U

(s

)



(I PK)1 RH ;

the matrix function

M U

N V

is invertible in RH ;

the matrix function V U

N M

is invertible in RH


Next Lecture / Assignments:


47/48


Next lecture (February 2, 13:15-15:00, in A208):Performance Specifications and Limitations.

The first assignment is due to hand in.

Next practice: February 3, 10:15-12:00, in A205,

You are invited to attend Ph.D. defense of Uwe Metin:February 5, 9:00-12:00, in N200, Naturvetarhuset;Title: Principles for Planning and Analyzing Motions of Underactu-

ated Mechanical Systems and Redundant Manipulators.




48/48


Next lecture (February 2, 13:15-15:00, in A208):Performance Specifications and Limitations.

The first assignment is due to hand in.

Next practice: February 3, 10:15-12:00, in A205,

You are invited to attend Ph.D. defense of Uwe Metin:February 5, 9:00-12:00, in N200, Naturvetarhuset;Title: Principles for Planning and Analyzing Motions of Underactu-

ated Mechanical Systems and Redundant Manipulators.

Problem: Let G(s) = (s1)(s+2)(s3) . Find a stable co-prime

factorization G(s) = n(s)m(s)

and x, y RH such that

xn + ym = 1 .


Documents

Lecture 04 OCLS