English CL 2

8/12/2019 English CL 2

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2. State equations

State equations

Solution of the state equations

Assumption: We assume that all the Laplace transforms involved in the

following reasonings exist.

x(t) = A x(t) + Bu(t)

y(t) = C x(t) + Du(t)

L

x(s) = (sIn A)

1x(0) + (sIn A)1Bu(s)

y(s) = C(sIn A)1x(0) + [C(sIn A)1B + D]u(s)

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2. State equations

Back to the time domain

x(s) = (sIn A)1x(0) + (sIn A)1Bu(s)

y(s) = C(sIn A)1x(0) + [C(sIn A)1B + D]u(s)

L1

x(t) = L1{(sIn A)1} x(0) + L1{(sIn A)1}B u(t)

y(t) = CL1{(sIn A)1} x(0 ) + [CL1{(sIn A)1}B + D] u(t)

L1{(sIn A)1} = ?

(sInA)1 is a rational matrix function that is (strictly) proper. Its

Laplace transform can be computed componentwise.

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2. State equations

Example:

ComputeL1{(sIn A)1} for A=

1 2

2 1

.

(sIA)1 = s1(s1)2+4 2(s1)2+4

2(s1)2+4

s1(s1)2+4

L1{(sIA)1}= et

cos 2t sin 2t

sin 2t cos 2t

(check this!)

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2. State equations

General expression for L1{(sIn A)1}

(sIA)1 = s1I+k=1

Aksk1

L1{(sIn A)1} = L1{s1}I+

k=1

AkL1{sk1}

= I+k=1

AkL1{dk1s

dsk(1)k

k!}

=k=0

Aktk

k!=: eAt

this notation is chosen by analogy with the scalar case

L1{ 1sa

}= eat =

k=0aktk

k!

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2. State equations

General form of x(t) and y(t)

x(t) = L1{(sIn A)1} x(0) + L1{(sIn A)1}B u(t)

y(t) = CL1{(sIn A)1} x(0 ) + [CL1{(sIn A)1}B + D] u(t)

x(t) = e

At

x(0) + eAt

B u(t)y(t) = CeAt x(0 ) + [CeAtB + D] u(t)

x(t) = eAtx(0) +

t

0 eA(t)B u()d

y(t) = CeAtx(0) +t0

CeA(t)B u()d+Du(t)

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2. State equations

x(t) = eAtx0 +t0

eA(t)B u()d is a solution ofx= Ax + Bu.

It is the only solution ofx= Ax+ Bu such that x(0) = x0, for a fixed

given u. [Prove this fact using the following theorem.]

Theorem - existence and uniqueness of solution

Consider a first order differential equation x= F(x) with initial condition x(t0) = x0 (IVP

- initial value problem). IfF is a lipschitzian, then the (IVP) has a unique solution. This

solution is of class C1, i.e. is continuously differentiable.

x(t) is defined for all inputs u(t) that guarantee the existence of theintegral

t0

eA(t)B u()d. Here we take as admissible the inputs u(t)

which are piecewise continuous.

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2. State equations

The solutions of the state eqautions can also be written as follows (if

the initial conditions are given at time t0):

x(t) = eA(tt0)x(t0) +tt0

eA(t)B u()d

y(t) = CeA(tt0)x(t0) +tt0

CeA(t)B u()d +Du(t)

Check this!

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2. State equations

Zero-input evolution/response- state and output evolution for zero input

Zero-state evolution/response- state and output evolution for zero initial

state

x(t) = eAtx(0) xl(t)

+ t

0eA(t)B u()d

xf(t)

y(t) = CeAtx(0) yl(t)

+ t

0

CeA(t)B u()d+Du(t)

yf(t)

zero-input evolution zero-state evolution

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2. State equations

Impulse response and transfer function

yf(t) =t0

CeA(t)B u()d+Du(t)

Impulse response

yf(t) = [CeAtB + D] u(t)L1 L

yf(s) = [C(sInA)1B+D] u(s)

Transfer function

Impulse = Dirac -function; ui = yf = CeAtB + Di.

ui - i-th component of u Di - i-th colunm of D

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2. State equations

Discretization

Discretization starting at time t0, with discretization interval .

xd(k) := x(t0+k); analogous definitions for ud e yd.

Process 1

Approximate x(t) x(t+)x(t)

.

This leads to:

xd(k+ 1) =(I+A)xd(k) +Bud(k)yd(k) = Cxd(k) +Dud(k) Check!

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2. State equations

Process 2

Suppose u(t) constant in each interval [t0+k t0+ (k+ 1))

Compute the state trajectories of the continuous system at the

discretization instants:

x(t0+ (k+ 1)) =

eA((t0+(

k+1)

(t0+

k))x(t0 + k) +

t0+(k+1)t0+k e

A(t0+(

k+1)

)Bu()d

This leads to:

xd(k+ 1) = eAxd(k) +

0 eABd

ud(k)

yd(k) = Cxd(k) +Dud(k) Check this!

Exercise: Compare the discrete systems obtained by the two different

processes.

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2. State equations

DiscretizaoExacta

EXEMPLO:

Considere-se o sistema compartimental contnuo,

1 0 1 0

0 0 0 1

0 1 1 0

x x u

= +

& (1)

( ) ( ) ( )

0.05 0.05 0.05 0.05

0.05 0.05 0.05

1 1.05 0.05 1.95 2.05

1 0 1 0 0.05

0 1 0.95

e e e e

x k x k u k

e e e

+

+ = + +

E o sistema compartimental discretizadocorrespondente, para h=0.05seg:

(2)

Assim

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2. State equations

0 50

2

4

6

Tempo (seg)

Amplitude

0 50

2

4

6

Tempo (seg)

Amplitude

0 50

2

4

6RESPOSTA AO DEGRAU UNITRIO

Tempo (seg)

Amplitude

DiscretizaoExacta

Respostas ao degrau do sistema contnuo e da sua discretizaoexacta

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2. State equations

DiscretizaoExacta

Respostas foradas do sistema contnuo e da sua discretizao exacta

nos intervalos de discretizao

0 50

5

10

15

Tempo (seg)

Amplitude

0 50

2

4

6

Tempo (seg)

Amplitude

0 50

5

10RESPOSTA FORADA

Tempo (seg)

Amplitude

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S


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2. State equations

DiscretizaoAproximada

EXEMPLO:

Considere-se novamente o sistema compartimental contnuo (1):

( ) ( ) ( )

0.95 0 0.05 0

1 0 1 0 0.05

0 0.05 0.95 0

x k x k u k

+ = +

E o sistema compartimental discretizado aproximadamente correspondente, para

h=0.05seg:

Assim

(3)

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2 St t ti


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2. State equations

DiscretizaoAproximada

Respostas ao degrau do sistema contnuo e da sua discretizaoaproximada

0 50

2

4

6

Tempo (seg)

Am

plitude

0 50

2

4

6

Tempo (seg)

Am

plitude

0 50

2

4

6RESPOSTA AO DEGRAU UNITRIO

Tempo (seg)

Am

plitude

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2 St t ti


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2. State equations

General solution for discrete state equations

x(k+ 1) = Ax(k) +Bu(k)

y(k) = Cx(k) +Du(k)

x(k) = Akx(0) +k1

l=0 Ak1lBu(l)

y(k) = CAkx(0) +l=0CAk1Ak1lBu(l) + Du(k)

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2 State equations


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2. State equations

Invertible transformations (isomorphisms) in the state space

State transformation: x x= Sx

S invertible(i.e., x x= Sx is an isomorphism)

Question: What are the evolution equations for x(t)?

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2 State equations


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2. State equations

x = Ax + Buy = Cx + Du

Sx = SAS1Sx + SBuy = CS1Sx + Du

Replacing Sx by x, yields:

x =

A SAS1 x +

BSB u

y = CS1 Cx + Du

x = Ax + Bu

y = Cx + Du

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2 State equations


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2. State equations

Thus:

x = Ax + Bu

y = Cx + Du

x = Ax + Bu

y = Cx + DuTransformation S

(A , B , C , D) (A, B, C, D) =

= (SAS1

, SB, CS

1

, D)

(A , B , C , D) Algebraically equivalent

e systems

(A, B, C, D) invertible matrix Ssuch that

A= SAS1, B = SB

C= CS1, D = D

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2. State equations


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2. State equations

Prove That:

Proposition: If two systems are algebraically equivalent then they have the

same transfer function.

Remark: Two systems with the same transfer function are called

zero-state equivalent.

So, the previous proposition states thatwhenever two systems are

algebraically equivalent systems, they are also zero-state equivalent.

However: there are systems that are zero-state equivalent, but not

algebraically equivalent.

Give an example!

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English CL 2