STATE-SPACE DYNAMIC SYSTEMS (CONTINUOUS-TIME)mocha-java.uccs.edu/ECE5520/ECE5520-CH02.pdf ·...

ECE4520/5520: Multivariable Control Systems I. 2–1

STATE-SPACE DYNAMIC SYSTEMS

(CONTINUOUS-TIME)

2.1: Introduction to LTI state-space models

1. What are they?

2. Why use them?

3. How are they related to the transfer

functions we have used already?

4. How do we formulate them?

What are they?

■ Representation of dynamics of an

nth-order system as a first-order

differential equation in an n-vector called

the state ➠ n first-order equations.

“Nice artwork kiddo. I have a feeling

that a great many people will make

a living off that third line someday!”

(Out of Control , IEEE Control Sys-

tems Magazine)

■ Classic example: 2nd-order equation of motion (EOM).

m u.t/

y.t/

k

b

m Ry.t/ D u.t/ ! b Py.t/ ! ky.t/

➠ Ry.t/ Du.t/ ! b Py.t/ ! ky.t/

m:

■ Define a state vector:

x.t/ D

"

y.t/

Py.t/

#

then; Px.t/ D

"

Py.t/

Ry.t/

#

D

2

4

Py.t/

!k

my.t/ !

b

mPy.t/ C

1

mu.t/

3

5 :

Lecture notes prepared by Dr. Gregory L. Plett. Copyright c" 2015, 2011, 2009, 2007, 2005, 2003, 2001, 2000, Gregory L. Plett

ECE4520/5520, STATE-SPACE DYNAMIC SYSTEMS—CONTINUOUS-TIME 2–2

■ We can write this in the form Px.t/ D Ax.t/ C Bu.t/, where A and B

are constant matrices:

Px.t/ D

"

Py.t/

Ry.t/

#

D

2

4

3

5

„ ƒ‚ …

A

"

y.t/

Py.t/

#

C

2

4

3

5

„ ƒ‚ …

B

u.t/:

■ Complete picture by finding y.t/ as function of x.t/: General form is

y.t/ D Cx.t/ C Du.t/;

where C and D are constant matrices:

C Dh i

; D Dh i

:

■ Fundamental form for LTI state-space model:

Px.t/ D Ax.t/ C Bu.t/

y.t/ D Cx.t/ C Du.t/;

where u.t/ is the input, x.t/ is the “state”, A, B, C , D are constant

matrices.

DEFINITION: The state of a system at time t0 is the minimum amount of

information at t0 that, together with the input u.t/, t # t0, uniquely

determines the behavior of the system for all t # t0.

■ Contrast with impulse-response (convolution) representation which

requires all past history of u.t/

y.t/ DZ t

0

h.!/u.t ! !/ d!:

Why use them?

■ Transfer functions provide input-output mapping: u ! G.s/ ! y.

State variables provide access to what is going on inside the system.



■ Convenient way to express EOM. Matrix format great for computers.

■ Allows new analysis and synthesis tools.

■ Great for multi-input, multi-output systems. These are very hard to

work with transfer functions.

Converting state-space to transfer function

■ Start with the state equations


y.t/ D Cx.t/ C Du.t/:

■ Laplace transform

sX.s/ ! x.0/ D AX.s/ C BU.s/

Y.s/ D CX.s/ C DU.s/;

or

.sI ! A/X.s/ D BU.s/ C x.0/

X.s/ D .sI ! A/!1BU.s/ C .sI ! A/!1x.0/

and

Y.s/ D ŒC.sI ! A/!1B C D"„ ƒ‚ …

transfer function of system

U.s/ C C.sI ! A/!1x.0/„ ƒ‚ …

response to initial conditions

:

■ So,Y.s/

U.s/D C.sI ! A/!1B C D;

but

.sI ! A/!1 Dadj.sI ! A/

det.sI ! A/D

Œcofactor array of .sI ! A/"T

det.sI ! A/:



■ Slightly easier to compute (for SISO systems)

Y.s/

U.s/D C.sI ! A/!1B C D D

det

"

sI ! A B

!C D

#

det.sI ! A/:

$ We will develop this result later in this chapter when defining

transmission zeros of a state-space system.

EXAMPLE: Our mass-spring-damper example:

A D

2

4

0 1

!k

m!

b

m

3

5 B D

2

4

01

m

3

5 C Dh

1 0i

■ The transfer function is:

G.s/ D C.sI ! A/!1B C 0 Dh

1 0i

2

4

s !1k

ms C

b

m

3

5

!12

4

01

m

3

5

D

h

1 0i

2

64

s Cb

m1

!k

ms

3

75

2

4

01

m

3

5

s2 C .b=m/s C .k=m/D

1m

s2 C bm

s C km

D1

ms2 C bs C k:

■ This is exactly what we expect from the first example in this section.

EXAMPLE: Using the special SISO formula,

G.s/ D

det

2

6664

s !1 0k

ms C

b

m

1

m!1 0 0

3

7775

det

2

64

s !1

k

ms C

b

m

3

75

D1=m

s2 C .b=m/s C .k=m/D

1

ms2 C bs C k:



■ Same result.

■ Example shows that the characteristic equation for the system is

#.s/ D det.sI ! A/ D 0:

■ Poles of system are roots of det.sI ! A/ D 0 (eigenvalues).

■ In transfer function matrix form, G.s/ D C.sI ! A/!1B C D, a pole of

any entry in G.s/ is a pole of the system.

SIMULATING SYSTEMS IN SIMULINK: To investigate how state-space

systems work, we can simulate them in Simulink. We could use the

“State Space” block from the “Continuous” library, or we can make our

own. The following method has advantages because it gives us

explicit access to the state and other internal signals. It is a direct

implementation of the transfer function above, and the initial state

may be set by setting the initial integrator values.

Note: All (square) gain blocks are MATRIX GAIN blocks

from the Math Library.

1y

s1

K

D

K

C

K

BK

A

1u

xdot x



2.2: Four canonical forms for LTI state-space models

■ We can make state-space forms from EOM, as we have seen.

■ Also from transfer functions: there are four main standardized forms,

plus a couple of other forms we will look at later.

Controller canonical form

■ Three cases:

1] Transfer function is made up only of poles.

G.s/ D1

s3 C a1s2 C a2s C a3

DY.s/

U.s/

➠ «y.t/ C a1 Ry.t/ C a2 Py.t/ C a3y.t/ D u.t/:

$ Choose output and derivatives as the state.

x.t/ Dh

Ry.t/ Py.t/ y.t/iT

: Then

Px.t/ D

2

64

«y.t/

Ry.t/

Py.t/

3

75 D

2

64

!a1 !a2 !a3

1 0 0

0 1 0

3

75

2

64

Ry.t/

Py.t/

y.t/

3

75C

2

64

1

0

0

3

75u.t/

y.t/ Dh

0 0 1i

x.t/ Ch

0i

u.t/:

$ Note the special form of A (top-companion matrix).

2] Transfer function has poles and zeros, but is strictly proper.

G.s/ Db1s

2 C b2s C b3

s3 C a1s2 C a2s C a3D

Y.s/

U.s/:

$ Break up transfer function into two parts.V.s/

U.s/contains all of the

poles ofY.s/

U.s/. Then,



Y.s/ D Œb1s2 C b2s C b3"V.s/:

$ Or,

y.t/ D b1 Rv.t/ C b2 Pv.t/ C b3v.t/:

$ But,

V.s/Œs3 C a1s2 C a2s C a3" D U.s/;

or,

«v.t/ C a1 Rv.t/ C a2 Pv.t/ C a3v.t/ D u.t/:

$ The representation for this is the same as in Case 1. Let

x.t/ Dh

Rv.t/ Pv.t/ v.t/iT

:

$ Then

Px.t/ D

2

64

«v.t/

Rv.t/

Pv.t/

3

75 D

2

64

!a1 !a2 !a3

1 0 0

0 1 0

3

75

2

64

Rv.t/

Pv.t/

v.t/

3

75C

2

64

1

0

0

3

75u.t/

represents the dynamics of v.t/. All that remains is to couple in the

zeros of the system.

Y.s/ D Œb1s2 C b2s C b3"V.s/

y.t/ Dh

b1 b2 b3

i

x.t/ Ch

0i

u.t/:

3] Non-proper transfer function.

G.s/ Db0s

3 C b1s2 C b2s C b1


Dˇ1s

2 C ˇ2s C ˇ3

s3 C a1s2 C a2s C a3C D;

where the ˇi terms are computed via long division. The remainder

D is the feedthrough term.



■ This particular method of implementing a system in state-space form

is called controller canonical form.

■ MATLAB command tf2ss(num,den) converts a transfer-function

form to state-space form.

■ Analog computer implementation:

y.t/

u.t/x1c x2c x3cR R R

b1

b2

b3

!a1

!a2

!a3

Observer canonical form

■ Now, using the same transfer function,

.s3 C a1s2 C a2s C a3/Y.s/ D .b1s

2 C b2s C b3/U.s/;

divide both sides by s3

Y.s/ D!

!a1

s!

a2

s2!

a3

s3

"

Y.s/ C#

b1

sC

b2

s2C

b3

s3

$

U.s/

D1

s

b1U.s/ ! a1Y.s/ C1

s

b2U.s/ ! a2Y.s/ C1

s

b3U.s/ ! a3Y.s/

!!!

:

■ This has block-diagram:



y.t/

u.t/

x1ox2ox3o R RR

b1b2b3

!a1!a2!a3

■ This is called observer canonical form:

Px.t/ D

2

64

!a1 1 0

!a2 0 1

!a3 0 0

3

75x.t/ C

2

64

b1

b2

b3

3

75u.t/

y.t/ Dh

1 0 0i

x.t/:

■ A is a left-companion matrix.

Controllability canonical form

■ Third, consider the block diagram:

y.t/

u.t/

x1co x2cox3coRR R

ˇ1

ˇ2

ˇ3

!a1!a2!a3



x3 D1

s.x2 ! a1x3/ X3.s/ D

1


U.s/

x2 D1

s.x1 ! a2x3/ ➠ X2.s/ D

s C a1


U.s/

x1 D1

s.u ! a3x3/ X1.s/ D

s2 C a1s C a2


U.s/:

■ Thus,

Y.s/ Dˇ3 C ˇ2.s C a1/ C ˇ1.s

2 C a1s C a2/


U.s/:

■ In order to get the correct transfer function, we must compute the fˇigvalues to get the desired numerator:

2

64

1 0 0

a1 1 0

a2 a1 1

3

75

2

64

ˇ1

ˇ2

ˇ3

3

75 D

2

64

b1

b2

b3

3

75

2

64

ˇ1

ˇ2

ˇ3

3

75 D

2

64

1 0 0

a1 1 0

a2 a1 1

3

75

!12

64

b1

b2

b3

3

75 D

2

64

b1

b2 ! a1b1

b3 ! a1b2 ! a2b1 C a21b1

3

75 :

■ Or,

Px.t/ D

2

64

0 0 !a3

1 0 !a2

0 1 !a1

3

75x.t/ C

2

64

1

0

0

3

75u.t/

y.t/ Dh

ˇ1 ˇ2 ˇ3

i

x.t/:

■ A is a right-companion matrix.

Observability canonical form

■ Note that H.s/ is a scalar. So, H T .s/ D H.s/.



H.s/ D C.sI ! A/!1B C D

D BT .sI ! A/!T C T C DT

D BT .sI ! AT /!1C T C DT :

■ So, C $ BT , A $ AT , B $ C T and D $ DT are dual forms.

■ We have already seen this(!). Controller and observer are dual forms.

Likewise, we can come up with

Px.t/ D

2

64

0 1 0

0 0 1

!a3 !a2 !a1

3

75 x.t/ C

2

64

ˇ1

ˇ2

ˇ3

3

75u.t/

y.t/ Dh

1 0 0i

x.t/

as a dual form with the controllability form.

y.t/

u.t/

x1obx2obx3ob RRR

ˇ1ˇ2ˇ3

!a1

!a2

!a3

■ A is a bottom-companion matrix.

■ We will see that we have a lot of freedom when making our

state-space models (i.e., in choosing the components of x.t/).



2.3: One more canonical form, transformations

Modal (diagonal) form

■ Yet another “canonical” form. Very useful. . .

■ Assume G.s/ DN.s/

D.s/; D.s/ has distinct roots pi (real).

G.s/ DN.s/

.s ! p1/.s ! p2/ % % % .s ! pn/

Dr1

s ! p1

Cr2

s ! p2

C % % % Crn

s ! pn

:

■ Now, let

X1.s/

U.s/D

r1

s ! p1

➠ Px1.t/ D p1x1.t/ C r1u.t/

:::

Xn.s/

U.s/D

rn

s ! pn

➠ Pxn.t/ D pnxn.t/ C rnu.t/:

■ Or,


y.t/ D Cx.t/ C Du.t/

A D

2

66664

p1 0

p2

: : :

0 pn

3

77775

; B D

2

66664

r1

r2:::

rn

3

77775

C Dh

1 1 % % % 1i

; D Dh

0i

:

■ Easily extends to handle complex poles $i D %i C j!i .



$ If A and B may have complex elements, no change is necessary.

$ Otherwise, use “real modal form” which is made via partial-fraction

expansion where complex pole-pairs are represented as

Gi.s/ D˛is C ˇi

.s ! %i /2 C !2i

:

$ The real-modal form has an A matrix which is block diagonal, and

of the form

A D diag

ƒr;

"

%rC1 !rC1

!!rC1 %rC1

#

; : : : ;

"

%n !n

!!n %n

#!

where ƒr is a diagonal matrix containing the real poles, and

$i D %i C j!i; i D r C 1; : : : ; n

are the complex poles.

$ The B matrix has corresponding entries:"

bi;1

bi;2

#

D

"

1 1

!!i ! %i !i ! %i

#!1 "

˛i

ˇi

#

:

■ Modal form is convenient for keeping track of system poles: : : they

are right on the diagonal!

■ Good representation to use: : : numerical robustness.

■ All canonical forms related by linear algebra—change of basis.

■ Diagonal is very useful, but we cannot always put a system in

diagonal form. (What was our assumption above?)

■ We will see one more canonical (Jordan) form in a little while that is

very similar to diagonal. All systems can be put in Jordan form.



Transformations

■ We have seen that state-space representations are not unique.

Selection of vector components in x is quite arbitrary.

■ Can we convert from one representation to another and get

equivalent systems?

■ Analyze the transformation of


y.t/ D Cx.t/ C Du.t/:

■ Let x.t/ D T ´.t/, where T is an invertible (similarity) transformation

matrix.

P.t/ D T !1 Px.t/

D T !1ŒAx.t/ C Bu.t/"

D T !1ŒAT ´.t/ C Bu.t/"

D T !1AT„ ƒ‚ …

NA

´.t/ C T !1B„ƒ‚…

NB

u.t/

y.t/ D C T„ƒ‚…

NC

´.t/ C D„ƒ‚…

ND

u.t/

so P.t/ D NA´.t/ C NBu.t/

y.t/ D NC ´.t/ C NDu.t/:

■ Argue that we should be able to use either model.

■ Are they going to give the same transfer function?

H1.s/ D C.sI ! A/!1B C D

H2.s/ D NC .sI ! NA/!1 NB C ND:



■ Need H1.s/ D H2.s/.

H1.s/ D C.sI ! A/!1B C D

D C T T !1.sI ! A/!1T T !1B C D

D .C T /ŒT !1.sI ! A/T "!1.T !1B/ C D

D NC .sI ! NA/!1 NB C ND D H2.s/:

Transfer function not changed by similarity transform.

OBSERVATION: Consider

H.s/ Db1s

2 C b2s C b3

s3 C a1s2 C a2s C a3:

■ Only six parameters in transfer function. But, A has 3 & 3, B has 3 & 1,

C has 1 & 3: a total of 15 parameters.

■ Appears that we have 9 degrees of freedom in state-space model.

Contradiction?

9 D sizeh i

:

■ We will see (Chapter 5) how to design T to put a system into the

various canonical forms.

EXAMPLE: Controller canonical form for

2s C 3

.s C 1/.s C 2/D

2s C 3

s2 C 3s C 2

is

Pxc.t/ D

"

!3 !2

1 0

#

xc.t/ C

"

1

0

#

u.t/

y.t/ Dh

2 3i

xc.t/:



■ Suppose we have the transformation matrix (note: det.T / D 1 so T is

invertible).

T D

"

2 !1

!1 1

#

:

■ Let xc D T Nx, where Nx is a new state. Then,

PNx.t/ D .T !1AT / Nx.t/ C .T !1B/u.t/

y.t/ D .C T / Nx.t/:

■ Plugging in A, B, C and T :

PNx.t/ D

"

!2 0

0 !1

#

Nx.t/ C

"

1

1

#

u.t/

y.t/ Dh

1 1i

Nx.t/;

which gives the diagonal realization of the transfer function!

■ We’ll often change coordinates in a system, for example to solve a

particular problem more easily.

EXAMPLE: Consider the system in the above example, implemented in

the four main canonical forms. Let the initial state for each form be

x.0/ D Œ1 1"T . Simulate response of each system.

■ The systems have the same

transfer function, but different

responses to initial states since

the states have different

interpretations.

0 1 2 3 4 5 6−1

0

1

2

3

4

5

Time

Am

plit

ude

Initial responses for same initial state



2.4: Time (dynamic) response

■ Develop more insight into the system response by looking at

time-domain solution for x.t/.

■ Scalar case first, then many states and MIMO.

Homogeneous part (scalar)

■ Px.t/ D ax.t/; x.0/.

■ Take Laplace. X.s/ D .s ! a/!1x.0/.

■ Inverse Laplace. x.t/ D eatx.0/.

Homogeneous part (full solution)

■ Px.t/ D Ax.t/; x.0/.

■ Take Laplace. X.s/ D .sI ! A/!1x.0/.

■ x.t/ D L!1Œ.sI ! A/!1"x.0/.

■ But,

.sI ! A/!1 DI

sC

A

s2C

A2

s3C % % %

so,

L!1Œ.sI ! A/!1" D I C At CA2t2

2ŠC

A3t3

3ŠC % % %

4D eAt matrix exponential

x.t/ D eAtx.0/:

■ eAt is called the transition matrix or state-transition matrix.

■ Matrix exponential expm.m



■ e.ACB/t D eAteBt iff AB D BA: (i.e., not in general).

■ Will say more about eAt when we discuss the structure of A.

■ Computation of eAt D L!1Œ.sI ! A/!1" straightforward for 2 & 2.

EXAMPLE:

Px D Ax; A D

"

0 1

!2 !3

#

.sI ! A/!1 D

"

s !1

2 s C 3

#!1

D

"

s C 3 1

!2 s

#

1

.s C 2/.s C 1/

D

2

664

2

s C 1!

1

s C 2

1

s C 1!

1

s C 2!2

s C 1C

2

s C 2

!1

s C 1C

2

s C 2

3

775

eAt D

"

2e!t ! e!2t e!t ! e!2t

!2e!t C 2e!2t !e!t C 2e!2t

#

1.t/

■ This is the best way to find eAt if A 2 & 2.

Forced solution (scalar)

Px.t/ D ax.t/ C bu.t/; x.0/

x.t/ D eatx.0/ CZ t

0

ea.t!!/bu.!/ d!„ ƒ‚ …

convolution

:

■ Where did this come from?

1. Px.t/ ! ax.t/ D bu.t/

2. e!at Œ Px.t/ ! ax.t/" Dd

dtŒe!atx.t/" D e!atbu.t/:



3.

Z t

0

d

d!Œe!a!x.!/" d! D e!atx.t/ ! x.0/ D

Z t

0

e!a!bu.!/ d!:

Forced solution (full solution)

■ Now, let Px.t/ D Ax.t/ C Bu.t/; x 2 Rn&1; u 2 R

m&1:

■ Follow three steps above to get

x.t/ D eAtx.0/ CZ t

0

eA.t!!/Bu.!/ d!

■ Clearly, if y.t/ D Cx.t/ C Du.t/;

y.t/ D CeAtx.0/„ ƒ‚ …

initial resp:

CZ t

0

CeA.t!!/Bu.!/ d!„ ƒ‚ …

convolution

C Du.t/„ƒ‚…

feedthrough

:

Thought: Maybe we can simplify the matrix exponential. . .

■ Have seen the key role of eAt in the solution for x.t/. Impacts the

system response, but need more insight.

■ Consider what happens if the matrix A is diagonalizable, that is, if

there exists a matrix T such that T !1AT D ƒ Ddiagonal.

■ Then, eAt D TeƒtT !1; and

eƒt D

2

66664

e$1t 0

e$2t

: : :

0 e$nt

3

77775

:

■ Much simpler form for the exponential, but how to find T; ƒ?

■ Eigenvalues/eigenvectors.



2.5: Diagonalizing the A matrix

■ “Eigen” is a German word meaning “self” or “characteristic”.

■ The eigenvectors and eigenvalues of A characterize its behavior.

■ An eigenvector is a vector satisfying

Av D $v;

where $ is a (possibly complex) constant, and v ¤ 0.

■ Multiplying by A does nothing to the vector except change its length!

■ This is a very unusual vector. There are usually only n of them if A

has size n & n.

$ Note that if v is an eigenvector, kv is also an eigenvector—so

eigenvectors are often normalized to have unit length: kvk2 D 1.

■ The constant $ is an eigenvalue. Specifically, it is the eigenvalue

associated with eigenvector v.

■ Since there are (usually) n eigenvectors with n corresponding

eigenvalues, we label the eigenvectors and eigenvalues vi and $i

where 1 ' i ' n.

■ To find eigenvalues, consider that .$I ! A/v D 0.

■ Since v ¤ 0, $I ! A must drop rank for some value of $ associated

with v. A matrix which is not full rank has zero determinant. So, we

can solve for the eigenvalues by solving

det.$I ! A/ D 0:

■ We have seen already that this is how we solve for the poles of a

state-space system, so the eigenvalues of the A matrix are the poles

of the dynamic system.



■ There are very efficient and numerically robust methods of finding

eigenvectors/values. These methods do not use the determinant rule,

above. The determinant rule is useful for mathematical analysis.

■ In MATLAB, [V,Lambda]=eig(A);

Example: Finding eigenvalues/eigenvectors of 2 & 2 matrix

■ Let

A D

"

3 3

!5 !5

#

:

■ To find eigenvalues, we solve:

det.$I ! A/ D 0

det

"

$ 0

0 $

#

!

"

3 3

!5 !5

#!

D 0

det

"

$ ! 3 !3

5 $ C 5

#!

D 0

.$ ! 3/.$ C 5/ C 15 D 0

$2 C 2$ D 0:

■ We see that there are eigenvalues at $1 D 0 and $2 D !2.

■ To find eigenvector corresponding to $1 D 0, we solve

Av1 D $1v1"

3 3

!5 !5

#"

v1a

v1b

#

D 0

"

v1a

v1b

#

:

■ This gives us the equation that v1a D !v1b. We can arbitrarily choose

one component, so let v1a D 1 giving v1 Dh

1 !1iT

.



■ We should further normalize this to have unit length by dividing byp

.1/2 C .!1/2. This gives

vnormalized1 D

"

1=p

2

!1=p

2

#

:

■ We find the second eigenvector similarly:"

3 3

!5 !5

#"

v2a

v2b

#

D !2

"

v2a

v2b

#

:

■ Again, arbitrarily choose v2a D 1. Then, we find that v2b D !5=3.

■ We can normalize by dividing byp

.1/2 C .!5=3/2. We get the answer

vnormalized2 D

"

0:5145

0:8575

#

:

The diagonal form

■ Assume that eigenvectors v1; v2; : : : vn are linearly independent.

Stack together in a matrix equation:

Avi D $ivi i D 1; 2; : : : ; n

Ah

v1 v2 : : : vn

i

„ ƒ‚ …

T

Dh

v1 v2 : : : vn

i

2

64

$1 0: : :

0 $n

3

75

„ ƒ‚ …

ƒ

■ AT D Tƒ ➠ T !1AT D V !1AV D ƒ.

■ We have found the transformation matrix T that diagonalizes A!

■ However, not all matrices are diagonalizable. For example,

A D

"

0 1

0 0

#

det.$I ! A/ D $2:



■ One eigenvalue $ D 0. Solve for the eigenvectors"

0 1

0 0

#"

va

vb

#

D 0 ➠ all vectors of the form

"

va

0

#

¤ 0:

Dynamic interpretation using diagonal form

■ Diagonal form makes it easier to understand what is happening in a

state-space dynamic system.

$ Assume A is diagonalizable by T D V .

$ Define new coordinates by x.t/ D T Qx.t/ so

PQx.t/ D T !1Ax.t/ D T !1AT Qx.t/ D ƒ Qx.t/:

$ In new coordinate system, system is diagonal (decoupled).

■ Trajectories consist of n independent modes; that is,

Qxi.t/ D e$i t Qxi.0/

hence the name, modal form.

■ To understand in original coordinate system, write T !1AT D ƒ as

T !1A D ƒT !1 with

T !1 D

2

66664

wT1

wT2:::

wTn

3

77775

; i:e:; rows of T !1:

wTi A D $iw

Ti ; so wi is a left eigenvector of A and note that

wTi vj D ıi;j .

■ How does this help?

eAt D TeƒtT !1



Dh

v1 v2 : : : vn

i

2

66664

e$1t 0

e$2t

: : :

0 e$nt

3

77775

2

66664

wT1

wT2:::

wnT

3

77775

DnX

iD1

e$i t viwTi :

1=s

1=s

$1

$n

Qx1

Qxn

■ Can write

x.t/ D eAtx.0/

D TeƒtT !1x.0/

DnX

iD1

e$i tvi .wTi x.0//:

Thus, trajectory can be expressed as linear

combination of modes.

Interpretation.

■ Left eigenvectors decompose initial state x.0/ into modal components

wTi x.0/:

■ e$i t term propagates i th mode forward t seconds. Stability?

■ Reconstruct state as linear combination of right eigenvectors, where

vi corresponds to “relative phasing” of state contribution to the modal

response.

EXAMPLE: Let’s consider a specific system

Px.t/ D Ax.t/

y.t/ D Cx.t/

with x.t/ 2 R16&1; y.t/ 2 R. (16-state, single output).



■ A lightly damped system.

■ Typical output to initial

conditions shown.

■ Output waveform is very

complicated. Looks almost

random.0 50 100 150 200 250 300

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Impulse Response

Am

plit

ude

Time (sec.)

■ However, such a solution can be decomposed into much simpler

modal components (see next page).

BOTTOM LINE: While responses of state-space systems can appear very

complicated, they are fundamentally simply summations of very

simple responses. Poles (eigenvalues of) A determine character of

the response.



0 50 100 150 200 250−1

0

1

0 50 100 150 200 250

−0.5

0

0.5

0 50 100 150 200 250−1

0

1

0 50 100 150 200 250

−0.5

0

0.5

0 50 100 150 200 250

−0.5

0

0.5

0 50 100 150 200 250−1

0

1

0 50 100 150 200 250−1

0

1

0 50 100 150 200 250

−0.5

0

0.5



2.6: The Jordan canonical form; MIMO canonical forms

■ What if A cannot be diagonalized?

■ Any matrix A 2 Rn&n can be put in Jordan canonical form by a

similarity transformation.

■ That is, we can find a transformation matrix T such that T !1AT D J ,

where J is a matrix in Jordan form.

■ The Jordan form of a matrix is block-diagonal and looks like

J D

2

64

J1 0: : :

0 Jq

3

75

where

Ji D

2

66664

$i 1 0

$i: : :: : : 1

0 $i

3

77775

2 Rni&ni

is called a Jordan block of size ni with eigenvalue $i (so n DqX

iD1

ni).

■ J is block-diagonal and upper bidiagonal.

■ J is diagonal is the special case of n Jordan blocks of size ni D 1.

■ Jordan form is unique (up to permutations of the blocks).

■ Can have multiple blocks with the same eigenvalue.

NOTE: The Jordan form is a conceptual tool, never used in numerical

computations!



■ #.$/ D det.$I ! A/ D det.$I ! J / D .$ ! $1/n1 % % % .$ ! $q/nq D 0 hence

distinct eigenvalues ➠ ni D 1 ➠ A diagonalizable.

■ dimN .$iI ! A/ D dimN .$iI ! J / is the number of Jordan blocks with

eigenvalue $i .

■ The sizes of each Jordan block may also be computed, but this is

complicated. i.e., leave it to MATLAB! ➠ jordan(A)

EXAMPLE: Consider

A D

2

6666666664

3 !1 1 1 0 0

1 1 !1 !1 0 0

0 0 2 0 1 1

0 0 0 2 !1 !1

0 0 0 0 1 1

0 0 0 0 1 1

3

7777777775

:

■ From MATLAB, we find that A has eigenvalue 2 with multiplicity 5 and

eigenvalue 0 with multiplicity 1:

det.$I ! A/ D .$ ! 2/5$:

■ rank.2I ! A/ D 4, so dim.N .2I ! A// D 6 ! 4 D 2 so there are two

Jordan blocks with eigenvalue 2.

■ We can check this in MATLAB: jordan(A)

J D

2

6666666664

0 0 0 0 0 0

0 2 1 0 0 0

0 0 2 1 0 0

0 0 0 2 0 0

0 0 0 0 2 1

0 0 0 0 0 2

3

7777777775



■ Note that without further information (computation) the following form

might also be the Jordan form for A (but it isn’t)

J D

2

6666666664

0 0 0 0 0 0

0 2 0 0 0 0

0 0 2 1 0 0

0 0 0 2 1 0

0 0 0 0 2 1

0 0 0 0 0 2

3

7777777775

■ System written in Jordan form is decomposed into independent

Jordan block systems PQxi.t/ D Ji Qxi.t/

y.t/u.t/1

s ! $

1

s ! $

1

s ! $

c1

c2

cn

■ Jordan blocks sometimes called Jordan chains (diagram shows why).

■ What does this mean in the time domain?

.sI ! J$/!1 D

2

66664

s ! $ !1 0

s ! $ : : :: : : !1

0 s ! $

3

77775

!1

D

2

66664

.s ! $/!1 .s ! $/!2 % % % .s ! $/!k

.s ! $/!1 % % % .s ! $/!kC1

: : : :::

0 .s ! $/!1

3

77775



D .s ! $/!1I C .s ! $/!2F1 C % % % C .s ! $/!kFk

where Fk is the matrix with ones on the kth upper diagonal.

■ Hence, the matrix exponential is

eJ$t D e$t

2

66664

1 t % % % t k!1=.k ! 1/Š

1 % % % t k!2=.k ! 2/Š: : : :::

0 1

3

77775

D e$t .I C tF1 C % % % C t k!1=.k ! 1/ŠFk/:

■ Thus, Jordan blocks expanded in eAt yield repeated poles and terms

of the form tpe$t .

Canonical forms for MIMO systems

■ Consider

ŒG.s/" D C.sI ! A/!1B C D

DC Œadj.sI ! A/"B

det.sI ! A/C D:

■ Now,

det.sI ! A/ D sn C ˛1sn!1 C ˛2s

n!2 C % % % C ˛n

and

C adj.sI ! A/B D N.s/ D ŒN1sn!1 C N2s

n!2 C % % % C Nn":

■ We can then write



Px.t/ D

2

6666664

!˛1I !˛2I % % % !˛n!1I !˛nI

I 0 0 0

0 I 0 0::: ::: ::: :::

0 0 % % % I 0

3

7777775

x.t/ C

2

6666664

I

0

0:::

0

3

7777775

u.t/

y.t/ Dh

N1 N2 % % % Nn!1 Nn

i

x.t/ C G.1/u.t/;

which is multivariable controller canonical form.

■ We can also write

Px.t/ D

2

6666664

!˛1I I 0 % % % 0

!˛2I 0 I % % % 0::: ::: ::: :::

!˛n!1I 0 0 % % % I

!˛nI 0 0 % % % 0

3

7777775

x.t/ C

2

6666664

N1

N2:::

Nn!1

Nn

3

7777775

u.t/

y.t/ Dh

I 0 0 % % % 0i

x.t/ C G.1/u.t/;

which is multivariable observer canonical form.

■ We generally find that SISO canonical forms are more useful than

MIMO canonical forms.



2.7: Zeros of a state-space system

■ Have seen eigenvalues of A, or poles of the entries of G.s/ are the

poles. Zeros of transfer function?

■ What is a zero? Two types of zero in a MIMO system: Blocking zeros

and transmission zeros.

■ Consider a system with input U.s/ DK

s ! $, K ¤ 0 a constant vector.

■ If $ is a blocking zero, e$t will not appear at the output for any K, x.0/.

(Not considered a very useful definition for MIMO zero).

■ If $ is a transmission zero, e$t will not appear at the output for some

specific K, x.0/.

fblocking zerosg ( ftransmission zerosg:

IDEA: (but not the entire story) Consider a two-input two-output system

u1.t/

u2.t/

y1.t/

y2.t/

G11.s/

G12.s/

G21.s/

G22.s/

■ A blocking zero will show up in G11.s/, G12.s/, G21.s/ and G22.s/. No

matter what K is, if U.s/ D K=.s ! $/, and $ is a blocking zero, the

output does not have an e$t term.



■ A transmission zero may not show up as a zero in any of the

individual transfer functions, but will in combinations thereof (with

specific initial states).

■ To find transmission zeros, put in u.t/ D u0eí t and you get a zero

output at “frequency” eí t .

■ State space: Have input and state contributions (consider first the

SISO case)

u.t/ D u0eí t ; x.t/ D x0e

í t : : : y.t/ D 0:

Px.t/ D Ax.t/ C Bu.t/ ➠ íeí tx0 D Ax0e

í t C Bu0eí t

➠

h

íI ! A Bi

2

4

x0

!u0

3

5 D 0

y.t/ D Cx.t/ C Du.t/ ➠ Cx0eí t C Du0e

í t D 0

➠

h

!C Di

2

4

x0

!u0

3

5 D 0:

■ Put the two together

2

4íI ! A B

!C D

3

5

2

4

x0

!u0

3

5 D 0:

■ Zero at frequency í if there exists a nontrivial solution of

det

2

4íI ! A B

!C D

3

5 D 0:



■ Recall

G.s/ D

det

2

4sI ! A B

!C D

3

5

det.sI ! A/:

Ahah! (The !u0 before gave us the correct sign in G.s/).

■ In the MIMO case, with n state variables, p inputs and q outputs, a

transmission zero is any value í for which

rank

2

4íI ! A B

!C D

3

5 < n C minfp; qg

EXAMPLE:

G.s/ D

2

64

s.s C 1/

s2 C 1

s C 1

s C 2

0.s C 2/.s C 1/

s2 C 2s C 2

3

75 :

■ We can find that G.s/ has a blocking zero at s D !1 and has

transmission zeros at s D 0, s D !1 and s D !2.

Y1.s/ Ds.s C 1/

s2 C 1U1.s/ C

s C 1

s C 2U2.s/

Y2.s/ D.s C 2/.s C 1/

s2 C 2s C 2U2.s/:

Let U D K=.s ! $/

Y1.s/ Ds C 1

s ! $

%

s.s C 2/k1 C .s2 C 1/k2

.s2 C 1/.s C 2/

&

Y2.s/ Ds C 1

s ! $

%

.s C 2/k2

s2 C 2s C 2

&

:



■ For all k1, k2, s D !1 is a zero. Therefore, both blocking and

transmission.

■ For k2 D 0, k1 2 R, s D 0 is a zero. Therefore, transmission.

■ Not so obvious, but s D !2 is also a zero. Therefore, transmission. [In

a MIMO system, we can have a zero and pole at the same frequency!]

■ Recall from before,2

4íI ! A B

!C D

3

5

2

4

x0

!u0

3

5 D 0:

gives the initial state x0 and K D u0 if í is a transmission zero.

■ In MATLAB: tzero(sys)



2.8: Linear time-varying systems

■ With relatively small changes in what we’ve seen, we can also model

linear time-varying (LTV) systems

Px.t/ D A.t/x.t/ C B.t/u.t/

y.t/ D C.t/x.t/ C D.t/u.t/:

■ LTV systems do not have transfer functions (so we cannot analyze in

the Laplace domain).

■ However, they do have (time varying) impulse responses.

■ Let g.t; !/ be the output signal from the system corresponding to an

input signal ı.t ! !/.

■ Then, the general output from a causal LTV system can be written as

y.t/ DZ 1

0

g.t; !/u.!/ d!:

Homogeneous response

■ To see responses of LTV state-space systems, consider first the

homogeneous case

Px.t/ D A.t/x.t/; x.t0/ D x0; t # 0:

FACT: The unique solution to this equation can be given by

x.t/ D ˆ.t; t0/x0; t # 0;

where the n & n state-transition matrix ˆ.t; t0/ is given by

ˆ.t; t0/ D I CZ t

t0

A.s1/ ds1 CZ t

t0

A.s1/

Z s1

t0

A.s2/ ds2 ds1

CZ t

t0

A.s1/

Z s1

t0

A.s2/

Z s2

t0

A.s3/ ds3 ds2 ds1 C % % %



■ Some important properties of the state-transition matrix are:

1. P .t; t0/ D A.t/ˆ.t; t0/; ˆ.t0; t0/ D I; t # 0.

2. ˆ.t; s/ˆ.s; !/ D ˆ.t; !/.

3. ˆ.t; !/!1 D ˆ.!; t /.

■ To prove the first property, we start with

x.t/ D ˆ.t; t0/x0

Px.t/ D P .t; t0/x0 C ˆ.t; t0/ Px0„ ƒ‚ …

0

D A.t/x.t/:

■ So, we have

A.t/x.t/ D P .t; t0/x0

A.t/ˆ.t; t0/x0 D P .t; t0/x0:

■ Since this must be true for arbitrary x0, we have shown that

P .t; t0/ D A.t/ˆ.t; t0/:

■ The second property is easily shown by breaking up the time line into

the intervals from ! to s, and then from s to t .

x.s/ D ˆ.s; !/x.!/

x.t/ D ˆ.t; s/x.s/

x.t/ D ˆ.t; s/ˆ.s; !/„ ƒ‚ …

ˆ.t;!/

x.!/:

■ The third property relates to going backward in time from a final state

to an initial state.

x.t/ D ˆ.t; t0/x.t0/

x.t0/ D ˆ!1.t; t0/x.t/:



■ For an LTI system, we have ˆ.t; t0/ D eA.t!t0/. So, this new

state-transition matrix is simply a generalization of what we have

already seen.

Forced response

■ Generalizing the prior result, we can find that the nonhomogeneous

state solution is

x.t/ D ˆ.t; t0/x0 CZ t

t0

ˆ.t; !/B.!/u.!/ d!:

■ To show this, start with the assumed answer and differentiate with

respect to time t

Px.t/ D P .t; t0/x0 CZ t

t0

P .t; !/B.!/u.!/ d! C ˆ.t; t/B.t/u.t/;

where we have needed Liebnitz’ rule for differentiating an integral,

which is

d

d'

Z b.'/

a.'/

f .x; '/ dx DZ b.'/

a.'/

@f .x; '/

@'dx C f .b.'/; '/

@b.'/

@'

! f .a.'/; '/@a.'/

@';

assuming that f .x; '/ is continuous and that the integral exists.

■ Using our known result for P .t; t0/ and the fact that ˆ.t; t/ D I ,

Px.t/ D A.t/ˆ.t; t0/x0 CZ t

t0

A.t/ˆ.t; !/B.!/u.!/ d! C B.t/u.t/:

■ Since the integral is with respect to !; we can factor out A.t/

Px.t/ D A.t/

#

ˆ.t; t0/x0 CZ t

t0

ˆ.t; !/B.!/u.!/ d!

$

C B.t/u.t/

D A.t/x.t/ C B.t/u.t/:



■ Substituting into y.t/ D C.t/x.t/ C D.t/u.t/, we get

y.t/ D C.t/ˆ.t; t0/x0 CZ t

t0

C.t/ˆ.t; !/B.!/u.!/ d! C D.t/u.t/:

■ We can rewrite this as

y.t/ D C.t/ˆ.t; t0/x0 CZ t

t0

ŒC.t/ˆ.t; !/B.!/ C D.!/ı.t ! !/" u.!/ d!

D C.t/ˆ.t; t0/x0 CZ t

t0

h.t; !/u.!/ d!

which allows us to recognize the impulse response of a LTV system

to be

h.t; !/ D C.t/ˆ.t; !/B.!/ C D.!/ı.t ! !/:

Equivalent transformations for LTV systems

■ Transformations of LTV systems are usually time-varying.

■ Let T .t/ be nonsingular for all t and define

x.t/ D T .t/´.t/

such that

Px.t/ D PT .t/´.t/ C T .t/ P.t/ D A.t/T .t/´.t/ C B.t/u.t/:

■ Then, we have

T .t/ P.t/ D'

A.t/T .t/ ! PT .t/(

´.t/ C B.t/u.t/;

which gives the equivalent LTV system

P.t/ D T !1.t/'

A.t/T .t/ ! PT .t/(

´.t/ C T !1.t/B.t/u.t/

y.t/ D C.t/T .t/´.t/ C D.t/u.t/:



■ A transformation matrix is called a fundamental matrix when it

annihilates the term in square brackets

PT .t/ D A.t/T .t/; T .t0/ ¤ 0:

■ In this case, then

P.t/ D T !1.t/B.t/u.t/

or

´.t/ D ´.t0/ CZ t

t0

T !1.!/B.!/u.!/ d!

and

x.t/ D T .t/´.t0/ CZ t

t0

T .t/T !1.!/B.!/u.!/ d!

D T .t/T !1.t0/x0 CZ t

t0

T .t/T !1.!/B.!/u.!/ d!

D ˆ.t; t0/x0 CZ t

t0

ˆ.t; !/B.!/u.!/ d!:

■ This gives us a practical way to compute the state-transition matrix

ˆ.t; !/ D T .t/T !1.!/:

EXAMPLE: Consider the LTV system

Px.t/ D

"

0 0

t 0

#

x:

■ This is equivalent to stating

Px1.t/ D 0

Px2.t/ D tx1.t/

orx1.t/ D x1.t0/

x2.t/ D x2.t0/ C1

2.t2 ! t 2

0 /x1.t0/:



■ We wish to find a valid fundamental matrix T .t/, and from it find the

state-transition matrix ˆ.t; t0/.

■ So, we need to find any matrix T .t/ that satisfies

PT .t/ D A.t/T .t/; T .t0/ ¤ 0:

■ We choose to initialize the fundamental matrix T .0/ D I . This then

implies that if"

T11.0/

T21.0/

#

D

"

1

0

#

; then

"

T11.t/

T21.t/

#

D

2

4

11

2t2

3

5

and if"

T12.0/

T22.0/

#

D

"

0

1

#

; then

"

T12.t/

T22.t/

#

D

"

0

1

#

:

■ Combining these results as columns of T , we have

T .t/ D

2

4

1 01

2t2 1

3

5 :

■ You can verify that this matrix satisfies PT .t/ D A.t/T .t/.

■ Using this to find the state-transition matrix,

ˆ.t; t0/ D T .t/T !1.t0/

D

2

4

1 01

2t2 1

3

5

2

4

1 01

2t20 1

3

5

!1

D

2

4

1 01

2t2 1

3

5

2

4

1 0

!1

2t20 1

3

5 D

2

4

1 0

!1

2.t2 ! t 2

0 / 1

3

5 :



2.9: What about nonlinear systems?

■ Nonlinear continuous-time state-space systems are modeled as

Px.t/ D f .x.t/; u.t//

y.t/ D g.x.t/; u.t//:

Linearizing around an equilibrium point

■ To work with nonlinear systems, using small input signals, we can

linearize these equations.

■ Suppose that there is an equilibrium constant solution

u.t/ D ueq; x.t/ D xeq; and y.t/ D yeq D g.xeq; ueq/;

that satisfies the nonlinear form.

■ Then, let the actual input signal and initial state be written as

u.t/ D ueq C ıu.t/; x.t0/ D xeq C ıxeq.t0/

for small ıu.t/.

■ Then, we can write actual state and output as

x.t/ D xeq C ıx.t/; y.t/ D yeq C ıy.t/:

■ Using the known form for y.t/, we have

ıy.t/ D g.x.t/; u.t// ! yeq

D g.xeq C ıx.t/; ueq C ıu.t// ! g.xeq; ueq/:

■ We expand the function g.%/ using Taylor-series around the

equilibrium point .xeq; ueq/



g.xeq C ıx.t/; ueq C ıu.t// D g.xeq; ueq/ C#

dg.xeq; ueq/

dx

$

ıx.t/

C#

dg.xeq; ueq/

du

$

ıu.t/ C h:o:t:

■ Substituting into our result for ıy.t/,

ıy.t/ )#

dg.xeq; ueq/

dx

$

„ ƒ‚ …

C

ıx.t/ C#

dg.xeq; ueq/

du

$

„ ƒ‚ …

D

ıu.t/:

■ To determine the evolution of ıx, we first take its derivative

ı Px.t/ D Px.t/ D f .xeq C ıx; ueq C ıu/:

■ By Taylor series,

ı Px.t/ D#

df .xeq; ueq/

dx

$

„ ƒ‚ …

A

ıx.t/ C#

df .xeq; ueq/

du

$

„ ƒ‚ …

B

ıu.t/:

■ To summarize, the linearized perturbation system is

ı Px.t/ D Aıx.t/ C Bıu.t/

ıy.t/ D Cıx.t/ C Dıu.t/

and the overall state and output can be computed as

x.t/ D xeq C ıx.t/ and y.t/ D yeq C ıy.t/:

Linearizing around a trajectory

■ Instead of linearizing around an equilibrium point, which requires

small ıu.t/, we can also choose to linearize around a known solution

trajectory.



■ Suppose it is known that

usol.t/; xsol.t/; ysol.t/

form a time-varying solution to the nonlinear dynamics of the system.

■ This means that

Pxsol.t/ D f .xsol.t/; usol.t//

ysol.t/ D g.xsol.t/; usol.t//:

■ Then, let general input, state, and output be

u.t/ D usol.t/ C ıu.t/

x.t/ D xsol.t/ C ıx.t/

y.t/ D ysol.t/ C ıy.t/:

■ Proceeding as before,

Px.t/ D Pxsol.t/ C ı Px.t/

ı Px.t/ D Px.t/ ! Pxsol.t/

D f .xsol.t/ C ıx.t/; usol.t/ C ıu.t// ! f .xsol.t/; usol.t//:

■ Using Taylor series,

ı Px.t/ )#

df .xsol; usol/

dx.t/

$

„ ƒ‚ …

A.t/

ıx.t/ C#

df .xsol; usol/

du.t/

$

„ ƒ‚ …

B.t/

ıu.t/

ıy.t/ )#

dg.xsol; usol/

dx.t/

$

„ ƒ‚ …

C.t/

ıx.t/ C#

dg.xsol; usol/

du.t/

$

„ ƒ‚ …

D.t/

ıu.t/:



■ To summarize, the linearized perturbation system is

ı Px.t/ D A.t/ıx.t/ C B.t/ıu.t/

ıy.t/ D C.t/ıx.t/ C D.t/ıu.t/

and the overall state and output can be computed as

x.t/ D xsol C ıx.t/ and y.t/ D ysol C ıy.t/:

■ Notice that even if the nonlinear system is time-invariant, the

linearized system will be time-varying, in general.

Where to from here?

■ We have now seen how to model continuous-time LTI, LTV, and

nonlinear systems in state-space form.

■ We have seen some ways to analyze and predict system behaviors.

■ Our next step is to repeat this approach with discrete-time systems.

■ We will see that the results are very similar, which will allow us to use

common frameworks to develop deeper insights into state-space

systems and to synthesize controllers.


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STATE-SPACE DYNAMIC SYSTEMS (CONTINUOUS-TIME)mocha-java.uccs.edu/ECE5520/ECE5520-CH02.pdf ·...

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