advanced control theory - chibum lee · chibum lee -seoultech advanced control theory solution of...

Advanced Control Theory

State Space Solution and Realization

[email protected]

Advanced Control TheoryChibum Lee -Seoultech

Outline

State space solution

2


Solution of state-space equations

First, recall results for scalar equation:

Laplace transforming:

𝑠𝑋 𝑠 − 𝑥 0 = 𝑎 𝑋 𝑠 + 𝑏 𝑈 𝑠

(𝑠 − 𝑎)𝑋 𝑠 = 𝑥(0) + 𝑏 𝑈 𝑠

Solve for transformed variable:

𝑋 𝑠 =𝑥 0

𝑠 − 𝑎+

𝑏

𝑠 − 𝑎𝑈 𝑠

𝒙 𝒕 = 𝑨𝒙 𝒕 + 𝑩𝒖 𝒕

𝑥 𝑡 = 𝑎 𝑥 𝑡 + 𝑏 𝑢 𝑡

3


Recall Laplace transform

𝐿−11

𝑠 − 𝑎= 𝑒𝑎𝑡

𝐿−1 𝐹1 𝑠 𝐹2(𝑠) = 0

𝑡

𝑓1 𝑡 − 𝜏 𝑓_2 𝜏 𝑑𝜏

Taking inverse Laplace transform of

𝑥 𝑡 = 𝑒𝑎𝑡𝑥(0) + 0

𝑡

𝑒𝑎 𝑡−𝜏 𝑏𝑢 𝜏 𝑑𝜏


zero input solution zero state solution

4



Vector-matrix equation:

𝒙 𝒕 = 𝑨𝒙 𝒕 + 𝑩𝒖 𝒕

Laplace transforming:

𝒔𝑿 𝒔 − 𝒙 𝟎 = 𝑨 𝑿 𝒔 + 𝑩 𝑼 𝒔

(𝒔𝑰 − 𝑨)𝑿 𝒔 = 𝒙(𝟎) + 𝑩𝑼 𝒔

𝑿 𝒔 = 𝒔𝑰 − 𝑨 −𝟏𝒙(𝟎) + 𝒔𝑰 − 𝑨 −𝟏𝑩𝑼 𝒔

Introducing matrix exponential by analogy with scalar exponential

function

𝑒𝑎𝑡 = 1 +1

1𝑎𝑡 +

1

2!𝑎𝑡 2 +

1

3!𝑎𝑡 3 +

1

4!𝑎𝑡 4 +⋯

𝑒𝐴𝑡 = 𝐼 +1

1𝐴𝑡 +

1

2!𝐴𝑡 2 +

1

3!𝐴𝑡 3 +

1

4!𝐴𝑡 4 +⋯

5


Ref) Matrix exponential

𝑒𝐴𝑡 matrix exponential

If 𝐴 is an (𝑛 × 𝑛) square matrix, then 𝑒𝐴𝑡 is also an (𝑛 × 𝑛) matrix

From the following (verify!)

1

𝑠 − 𝑎=1

𝑠+

𝑎

𝑠2+𝑎2

𝑠3+𝑎3

𝑠4+⋯

𝑠𝐼 − 𝐴 −1 =𝐼

𝑠+𝐴

𝑠2+𝐴2

𝑠3+𝐴3

𝑠4+⋯

Laplace transform of 𝑒𝐴𝑡 can be found

𝐿−11

𝑠 − 𝑎= 1 +

1

1𝑎𝑡 +

1

2!𝑎𝑡 2 +

1

3!𝑎𝑡 3 +

1

4!𝑎𝑡 4 +⋯ = 𝑒𝑎𝑡

𝐿−1 𝑠𝐼 − 𝐴 −1 = 𝐼 +1

1𝐴𝑡 +

1

2!𝐴𝑡 2 +

1

3!𝐴𝑡 3 +

1

4!𝐴𝑡 4 +⋯ = 𝑒𝐴𝑡

6



Note that:

𝐿 𝑒𝐴𝑡 = 𝑠𝐼 − 𝐴 −1

From the previous solution in Laplace domain

𝑿 𝒔 = 𝒔𝑰 − 𝑨 −𝟏𝒙(𝟎) + 𝒔𝑰 − 𝑨 −𝟏𝑩𝑼 𝒔

Taking inverse Laplace transform

𝑥 𝑡 = 𝑒𝐴𝑡𝑥(0) + 0

𝑡

𝑒𝐴 𝑡−𝜏 𝑏𝑢 𝜏 𝑑𝜏

zero input solution zero state solution

7


State transition matrix

State transition matrix: It describes the transition of the states from

initial conditions 𝑥(0) to those at time 𝑡, when there is no input:

𝒙 𝒕 = 𝚽 𝒕 𝒙(𝟎)

State transition matrix is used to find the solution of linear time varying

system

𝒙 𝒕 = 𝑨(𝒕)𝒙 𝒕 + 𝑩(𝒕)𝒖 𝒕

The matrix exponential 𝑒𝐴𝑡 is the state transition matrix for LTI system

For LTI, Φ 𝑡 satisfies: 𝚽 𝟎 = 𝒆𝑨⋅𝟎 = 𝑰, 𝚽−𝟏 𝒕 = 𝒆−𝑨𝒕 = 𝚽 −𝒕

Other properties of Φ(𝑡): 𝚽 𝒕𝟏 + 𝒕𝟐 = 𝚽 𝒕𝟐 𝚽(𝒕𝟏)

8


Computation of state transition matrix

In principle, we can compute

• as a series, terminating when no significant change observed

Very simple example: double-integrator plant

• control orientation of satellite with thrusters

States:

S-S:

Φ 𝑡 = 𝑒𝐴𝑡 = 𝐼 +1

1𝐴𝑡 +

1

2!𝐴𝑡 2 +

1

3!𝐴𝑡 3 +⋯

𝐽 𝜃 = 𝐹 ⋅ 𝑑 = 𝑢,Φ 𝑠

𝑈 𝑠=

1

𝐽𝑠2

9


Computation of state transition matrix

In practice, more

sophisticated numerical

algorithms are employed to

compute the state transition

matrix

Here:

Hence:

Then, solution to homogeneous equation 𝑥 𝑡 = 𝐴𝑥(𝑡) with initial

conditions 𝑥0 = 𝜃0 𝜔0𝑇 is

10


Ex. State transition matrix via Laplace transform

State transition matrix can be obtained from Laplace transform solution

To illustrate method, consider example plant with transfer function

A state-space realization is

For this system

11

𝑠𝐼 − 𝐴 = 𝑠1 01 1

−−3 1−2 0

=𝑠 + 3 −12 𝑠

𝑌(𝑠)

𝑈(𝑠)=

1

𝑠2 + 3𝑠 + 2=

1

(𝑠 + 1)(𝑠 + 2)

𝑥1 𝑥2=

−3 1−2 0

𝑥1𝑥2

+01𝑢

y = 1 0𝑥1𝑥2



Hence

Thus

12

Φ 𝑠 = 𝑠𝐼 − 𝐴 −1 =

𝑠 1−2 𝑠 + 3𝑠2 + 3𝑠 + 2

=

𝑠

(𝑠 + 1)(𝑠 + 2)

1

(𝑠 + 1)(𝑠 + 2)−2

(𝑠 + 1)(𝑠 + 2)

𝑠 + 3

(𝑠 + 1)(𝑠 + 2)

=

−1

𝑠 + 1+

2

𝑠 + 1

1

𝑠 + 1+

−1

𝑠 + 1−2

𝑠 + 1+

2

𝑠 + 1

2

𝑠 + 1+

−1

𝑠 + 1

Φ 𝑡 = 𝐿−1 𝑠𝐼 − 𝐴 −1 = −𝑒−𝑡 + 2𝑒−2𝑡 𝑒−𝑡 − 2𝑒−2𝑡

−2𝑒−𝑡 + 2𝑒−2𝑡 2𝑒−𝑡 − 𝑒−2𝑡



Calculate with MATLAB (Symbolic math)

>>A = [-3 1; -2 0];

>>syms t

>>Phi = expm(A*t)

>>Phi1=subs(Phi,t,1)

Phi =

[ 2*exp(-2*t)-exp(-t), exp(-t)-exp(-2*t)]

[ -2*exp(-t)+2*exp(-2*t), -exp(-2*t)+2*exp(-t)]

can also compute numerical value with MATLAB fn

Phi1 =

-0.0972 0.2325

-0.4651 0.6004

13

Φ 𝑡 = 𝐿−1 𝑠𝐼 − 𝐴 −1

= −𝑒−𝑡 + 2𝑒−2𝑡 𝑒−𝑡 − 2𝑒−2𝑡

−2𝑒−𝑡 + 2𝑒−2𝑡 2𝑒−𝑡 − 𝑒−2𝑡



Total system response to a unit step input u(t) = 1(t) [U(s) = 1/s] is

applied to the system

Total response is:

14

𝑥1 𝑥2=

−3 1−2 0

𝑥1𝑥2

+01𝑢



>>A=[-3 1;-2 0]; B=[0;1]; C=[1 0]; D=0;

>>G = ss(A, B, C, D);

>>ltiview(G)

15



Zero input

response:syms s, syms x10 x20 real

xZI = Phi*[x10; x20]

xZI =

[ (2*exp(-2*t)-exp(-t))*x10+(exp(-t)-exp(-2*t))*x20]

[ (-2*exp(-t)+2*exp(-2*t))*x10+(-exp(-2*t)+2*exp(-t))*x20]

Zero state

Response:

U=1/s;

xZS = ilaplace(inv(s*eye(2)-A)*B*U)

xZS =

[ 1/2-exp(-t)+1/2*exp(-2*t)]

[ 3/2+1/2*exp(-2*t)-2*exp(-t)]

16


Outline

Simulation diagram

State space realization

• Controllable canonical form

• Observable canonical form

• Modal canonical form

17


Simulation diagrams

Consider the problem of constructing an analog system to produce a

given input-output behavior, using only integrators, gain elements and

summing elements (as in an ‘analog computer’)

For example, input-output behavior defined by:

Then:

18


Simulation diagrams

Define states as outputs of integrators

Now we can write the state-space equations:

A state-space realization

A matrix is in upper companion form

19


Simulation Diagram vs. Block Diagram

Simulation diagram, in which:

• signals are in the time domain; e.g. 𝑢(𝑡)

• the only system elements are integrators, gains and summers

Block diagram, in which:

• the signals are in the Laplace domain; e.g. 𝑈(𝑠)

• the system elements may be arbitrary transfer functions

20


Realization

Realization:

For a given LTI system, a state space model which is constructed from

a given transfer function 𝐺 𝑠 is called a realization of 𝐺.

We were able to realize an ODE with no derivatives on the RHS (i.e.,

no zeros in the TF) with a simple chain of integrators

(previous example)

Consider SISO 3rd order model

𝑦 + 𝑎2 𝑦 + 𝑎1 𝑦 + 𝑎0𝑦 = 𝑏2 𝑢 + 𝑏1 𝑢 + 𝑏0𝑢

where 𝑎𝑖 , 𝑏𝑖 are arbitrary real numbers

𝐺 𝑠 =𝑌(𝑠)

𝑈(𝑠)=

𝑏2𝑠2 + 𝑏1𝑠 + 𝑏0

𝑠3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0

21


Realization: Controllable canonical form

How to deal with 𝑦 + 𝑎2 𝑦 + 𝑎1 𝑦 + 𝑎0𝑦 = 𝑏2 𝑢 + 𝑏1 𝑢 + 𝑏0𝑢

i.e. 𝐺 𝑠 =𝑌(𝑠)

𝑈(𝑠)=

𝑏2𝑠2+𝑏1𝑠+𝑏0

𝑠3+𝑎2𝑠2+𝑎1𝑠+𝑎0

=𝐵(𝑠)

𝐴(𝑠)

Introduce a partial state 𝜉 as an auxiliary variable, such that:

𝜉 + 𝑎2 𝜉 + 𝑎1 𝜉 + 𝑎0𝜉 = 𝑢 ;Ξ(𝑠)

𝑈(𝑠)=

1

𝐴(𝑠)

• this can be realized with a chain of integrators as before

We can then construct the output thus:

𝑦 = 𝑏2 𝜉 + 𝑏1 𝜉 + 𝑏0𝜉; 𝑌 𝑠 =𝐵 𝑠

𝐴 𝑠𝑈(𝑠) = 𝐵 𝑠 Ξ(𝑠)

22


−𝑎2−𝑎1

−𝑎0

𝑏2

𝑏1

𝑏0

Realization: Controllable canonical form

23

𝑌(𝑠)

𝑈(𝑠)=

𝑏2𝑠2 + 𝑏1𝑠 + 𝑏0

𝑠3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0

𝑦 = 𝑏2 𝜉 + 𝑏1 𝜉 + 𝑏0𝜉

𝜉 = 𝑢 − 𝑎2 𝜉 − 𝑎1 𝜉 − 𝑎0𝜉

𝑥1 𝑥2 𝑥3

=−𝑎2 −𝑎1 −𝑎01 0 00 1 0

𝑥1𝑥2𝑥3

+100𝑢

y = 𝑏2 𝑏1 𝑏0

𝑥1𝑥2𝑥3


Realization: Observable canonical form

To get another canonical form, rewrite the transfer function

(𝑠3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0)𝑌 𝑠 = 𝑏2𝑠

2 + 𝑏1𝑠 + 𝑏0 𝑈(𝑠)

Divide throughout by 𝑠3 to obtain

(1 +𝑎2𝑠+𝑎1𝑠2

+𝑎0𝑠3)𝑌 𝑠 =

𝑏0𝑠3

+𝑏1𝑠2

+𝑏0𝑠

𝑈(𝑠)

Rearrange term then,

𝑌 =𝑏0𝑈 − 𝑎0𝑌

𝑠3+𝑏1𝑈 − 𝑎1𝑌

𝑠2+𝑏2𝑈 − 𝑎2𝑌

𝑠

24

𝑌(𝑠)

𝑈(𝑠)=

𝑏2𝑠2 + 𝑏1𝑠 + 𝑏0

𝑠3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0


−𝑎2−𝑎1−𝑎0

𝑏2𝑏1𝑏0

Realization: Observable canonical form

25

𝑌(𝑠)

𝑈(𝑠)=

𝑏2𝑠2 + 𝑏1𝑠 + 𝑏0

𝑠3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0

𝑥1 𝑥2 𝑥3

=

−𝑎2 1 0−𝑎1 0 1−𝑎0 0 0

𝑥1𝑥2𝑥3

+

𝑏2𝑏1𝑏0

𝑢

y = 1 0 0

𝑥1𝑥2𝑥3


Relationships between CCF and OCF

Observable

canonical

form

Controllable

canonical

form

26


Realization: Modal canonical form

If the poles of transfer function are all distinct, we can do the partial

fraction expansion:

Each term in the summation can be represented:

27


Realization: Modal canonical form

Hence, parallel or diagonal realization:

• The modal states are decoupled

• Output is a linear combination of the system modes

28


Realization: Natural modes

We saw previously that by performing a partial fraction expansion of the

transfer function of a dynamic system

• we could obtain a parallel state-space realization

for which the A matrix is diagonal

so that the state-space equations are

decoupled

This modal canonical form is

very fundamental:

• the 'natural modes' make independent

contributions to the system output

• each mode has a characteristic 'natural

frequency' (eigenvalue) and 'mode shape' (eigenvector)

29


Natural motions

For 'natural' (unforced) motions (i.e., u ≡ 0):

• We know from before that the unforced response is a combination of simple

exponential motions of the form

• The system is said to be moving in a natural mode when all the states have

the same form of motion:

• The vector 𝒑 (the mode shape) describes the relative amplitude of the

common motion for each state

30


Eigenvalue problem

If then

• But

• That is,

• We seek a solution for the eigenvalue (or mode shape) 𝒑.

The condition for a non-trivial solution (i.e., 𝒑 ≠ 0) is that

The roots of the characteristic equation are:

• the "eigenvalues" 𝝀𝒊(i = 1, 2, ..., n) of 𝐴

• the "natural frequencies" of the system

• the "poles" of the transfer function 𝑌(𝑠)/𝑈(𝑠)

31


Eigenvalue problem

The eigenvalues are the values of λ for which the assumed motion

x(t) = peλt is possible without any external input u

For each eigenvalue λi, we can find the initial conditions x(0) = p

which will excite this natural mode, as the solution of the auxiliary

equation:

The solution p(i) is:

• the "eigenvector" corresponding to the eigenvalue λi

• the "mode shape" of the i-th mode

Note that p(i) may be found easily as a vector proportional to any non-

zero column of the adjugate of the auxiliary matrix F =[λiI −A]

32


Example

Auxiliary matrix:

Characteristic equation:

Eigenvalues:

33


First mode: λ1 = −2

Auxiliary equation:

Note that the eigenvector, or mode shape, may be normalized in

various ways

• set the first or last element to 1, as above

• set the vector "magnitude" to 1 (unit norm vector):

Solution:

34


Rapid calculation of mode shape

Auxiliary matrix:

Adjoint:

• The columns of the adjoint of F are proportional to each other, and to the eigenvector

• It is thus only necessary to compute one column to find the mode shape

35


Motion in first natural mode

The first mode may be excited by setting up initial conditions

corresponding to the first mode shape:

Then

36


Motion in first natural mode

37


MATLAB calculation of responses

% Get controller canonical form

>>[A, B, C, D] = tf2ss(1, [1 5 6]);

>>G = ss(A, B, C, D);

% Check

>>G.a

ans =

-5 -6

1 0

% initial condition

>>x0 = [-2 1]';

% compute response

>>[y, t, x] = initial(G, x0);

% plot response

>>plot(t, x), grid

>>legend('x_1', 'x_2')

38


Double check with transition matrix solution

% Set up for symbolic computation

>>syms t

>>Phi = expm(A*t) % transition matrix

>>Phi =

[ 3*exp(-3*t)-2*exp(-2*t), -6*exp(-2*t)+6*exp(-3*t)]

[ exp(-2*t)-exp(-3*t), -2*exp(-3*t)+3*exp(-2*t)]

>>x0 = [-2 1]'; % initial condition = mode shape

% Compute zero-input response

>>X = Phi*x0

X =

[ -2*exp(-2*t)]

[ exp(-2*t)]

pure 1st mode

39


Second mode: λ2 = −3

Auxiliary equation:

40


Motion in second natural mode

41


Response to arbitrary initial conditions

Example: Displace 1 unit and release from rest

• Construct initial condition as superposition of mode shapes:

• Then:

Response is the sum of contributions from the independent natural

modes

42


Response to initial displacement

43


Modal contributions to output y

44


Outline

Similarity transform

45


Transformations between state variables

We have seen that there is no unique set of state variables to realize a

given input-output behaviour. That is, there are many s-s model for a

given input-output model!!

Given one realization,

We can form another by a (nonsingular) linear transformation of

variables:

T = transformation

matrix

46


Transformations between state variables

Then:

Hence where

This is called a similarity transformation

Two systems are called equivalent with each other

47


Transfer functions from state-space equations

Until now, we have seen that there are many different state space

realizations of a given transfer function

• The reverse process is also of interest; that is, SS TF

Consider a linear time invariant state dynamics

• Start by taking the Laplace Transform of these equations

48


Transfer functions from state-space equations

• which gives

• And

• By definition G(s) = C(sI - A)-1B + D is called the Transfer Function of the system.

• And C(sI - A)-1 x(0-) is the initial condition response. It is part of the response, but not part of the transfer function.

49


Transfer functions between equivalent models

Let’s get back to two equivalent models

where

Consider the two transfer functions:

50


Transfer functions between equivalent models

The transfer function is not changed by putting the state-space model

through a similarity transformation

51


Evaluation of transfer function matrix

To evaluate transfer function from s-s model, we should calculate

G(s) = C(sI - A)-1B + D

For a SISO system, G(s) may be evaluated using

For a SIMO system, G(s) may be evaluated using:

See Franklin et al. App C.6

The MATLAB function

ss2tf uses this formulation

52


• Choose

Then, state-space equations are:

• Transfer function:

Example

)3)(2(

1

65

1

1

65det

010

01

165

det

2

ssss

s

s

s

s

53


Let P [p (1) p (2) p (n) ]

And

We have:

Hence:

That is:

Transformation to modal state variables

modal matrix

assume eigenvalues are distinct

The modal matrix P is the similarity

transformation matrix between the original

state vector x and the modal

state vector xd :

54


Ex Modal matrix:

Similarity transformation:

modal

canonical

form

55

advanced control theory - chibum lee · chibum lee -seoultech advanced control theory solution of...

Documents