LINEAR SYSTEM THEORY

LECTURE NOTES

Prof. Dr. Mario Edgardo Magaña

School of Electrical Engineering

and

Computer Science

Oregon State University

The course deals with the theoretical and practical aspects of linear dynamic systems as

they apply to engineering modeling, analysis and design. The mathematical concepts of

time and complex frequency domain representation of linear dynamic systems are covered

in detail. Furthermore, the theoretical foundations and application of dynamic system

stability are discussed thoroughly. Finally, the properties of controllability and

observability are studied in order to apply them to both feedback controller and state

estimator design.

MATLAB and Simulink are heavily used in the homework and term project to emphasize

the practical aspects of the course material.

Fundamental knowledge of linear algebra (matrices, determinants, vectors, eigenvalues,

eigenvectors, similarity transformations, etc.), differential equations, and signal and

system analysis is required. Also, knowledge of computer simulation will be helpful, as

you will often be performing dynamic system simulation in order to verify theoretical

results.

2

The behavior of physical systems can be characterized using heuristic or empirical methods

by applying a variety of input signals and observing their outputs. Furthermore, if such

behavior is not satisfactory, a compensating mechanism based on heuristic knowledge can

be introduced to modify the system behavior to meet design criteria. Such an approach is

based on experience and requires the use of trial and error. However, this methodology

may take an inordinate amount of time to achieve the desired behavior of the system. This

is further complicated when the complexity of the system increases to a point where trial

and error is no longer feasible.

This shortcoming may be addressed by using mathematical models of the system

components that take into account the limitations of the physical systems. When doing so,

the process of modifying the behavior of complex physical systems becomes manageable

because a large number of formal analysis and design approaches have been developed

over the years and the need for trial and error is practically removed from the design

process.

3

4

Mathematical System Description:

A system N is a device that maps a set of admissible inputs U to a set of

admissible output responses Y.

Mathematically, N: U Y or y() = N [u()].

Alternatively, a system can be described either by differential or difference

equations or by the impulse or unit-sample response in the time domain, or by

algebraic equations in the complex frequency domain.

Example: Let us describe the relationship between the input voltage and the

output voltage of the following active filter system:

where vi(t) is the input voltage and v0(t) is the output voltage.

5

Using nodal analysis and assuming ideal amplifiers, we get

Adding the currents at the nodes of the second amplifier yields

Assuming infinite input impedance (ideal amplifiers) , we get . Hence,

Furthermore, the two capacitor currents are described by

021

6

01

2

1

1

1

cc

i iiR

vv

R

v

R

vv

03

21

1

R

veic

0

5

02

4

22

R

ve

R

ve

1 2 0e e

1

52 3 0 2

4

= and c

Rv R i v v

R

6

Thus,

1

1

1

1 1 1 1 1

c

c

dv dvdi c c v e c

dt dt dt

2

2

1 2

2 2 1 2 2 2

c

c

dv dv dvdi c c v v c c

dt dt dt dt

12 3 1

dvv R c

dt dv

cRv 2

13

1

1

5 3 51 10 3 1 1

4 4

R R Rdv dvv R c c

R dt R dt

02 4

5

dvdv R

dt R dt

1 4 40 1 0

3 5 1 3 5 1

dv R Rv v v d

dt R R c R R c

7

In terms of the input and output voltages,

0 1 1 21 1 2

1 2 6 1 6

1 1 10iv v dv dv dv

v c cR R R R R dt dt dt

0 04 4 4 2 40 0 0 2

3 5 1 1 2 6 1 6 3 5 3 5 1 5

1 1 10iv v dvR R R c R

v d v v cR R c R R R R R R R R R c R dt

2

0 0 54 2 4 4 20 2

3 5 1 1 2 6 1 1 3 5 6 5 4

1 1 1 1 11 0idv dv d v RR c R R c

vR R c R R R R dt c R R R dt R dt R

2

5 5 0 020 2 2

3 1 1 2 6 1 4 1 3 4 6 2

1 1 1 1 1 11 0iR dv R dv d vc

v cR c R R R R R dt c R R R dt dt c

or

2

0 5 0 502

2 1 3 4 6 2 3 1 2 1 2 6 1 4 2

1 1 1 1 1 1 1 id v R dv R dvv

dt c c R R R c dt R c c R R R R R c dt

8

The last differential equation represents a time domain model of the active filter.

In the complex frequency domain, assuming zero initial conditions, the algebraic

relationship between the input and the output voltage is

Since both models assume linear behavior of the active amplifier circuits, we

could also obtain an input-output model in terms of the convolution relationship in

the time domain, namely,

where

5

1 4 20

2 5

2 1 3 4 6 2 3 1 2 1 2 6

( ) ( )1 1 1 1 1 1 1

( ) ( )

i

i

Rs

R R cV s V s

Rs s

c c R R R c R c c R R R

H s V s

0 ( ) ( ) ( )iv t h t v t

1( ) ( )h t H s L

9

Input-Output Linearity:

A system N is said to be linear if whenever the input u1 yields the output N[u1],

the input u2 yields the output N[u2], and

for arbitrary real or complex constants c1 and c2

Example:

Let the spring force be described by fk(x) = Kx, then

is an external force, and are the first and second derivatives of x

with respect to time.

22112211 uNcuNcucucN

)(tfKxxBxM a

)(tf a

M

B

kf x

x

af t

x x

10

Let x1(t) be the solution when fa(t) is replaced by fa1(t) and x2(t) be the solution

when fa(t) is replaced be fa2(t) .

Then, if , the total response is given by

x(t) =c1x1(t) + c2x2(t) .

Let the spring force be now described by fk(x) = Kx2, then

This time, however, x(t) ≠ c1x1(t) + c2x2(t) , i.e., the linearity property does not

hold because the system is now nonlinear.

Time Invariance and Causality

Let N represent a system and y() be the response of such system to the input

stimulus u(), i.e., y() = N[u()]. If for any real T, y( - T) = N[u( - T)], then the

system is said to be time invariant . In other words, a system is time invariant if

delaying the input by T seconds merely delays the response by T seconds.

)(2 tfKxxBxM a

1 1 2 2( ) ( ) ( )a a af t c f t c f t

11

Let the system be linear and time invariant with impulse response h(t), then

If the same system is also causal, then for t ≥ ≥ 0, (h(t) = 0, t < 0)

Example: Let a system be described by the ordinary, constant coefficients

differential equation

then the system is said to be a lumped-parameter system.

Systems that are described by either partial differential equations or linear

differential equations that contain time delays are said to be distributed-

parameter systems.

( ) ( ) ( ) ( ) ( )

t t

y t u h t d h u t d

0 0

( ) ( ) ( ) ( ) ( )

t t

y t h u t d u h t d

)()()('...)()( 1

)1(

1

)( tutyatyatyaty nn

nn

12

Example: Consider the dynamic system described by

According to the previous definition, this equation describes a distributed-

parameter system because of the presence of the time delays.

Definition: The state of a system at time t0 is the minimum (set of internal

variables) information needed to uniquely specify the system response given

the input variables over the time interval [t0, ).

Example:

vi(t): Input voltage (external stimulus or excitation)

i(t): Current flowing through circuit

y(t): Output (measured) variable (current flowing through inductor)

( ) 2 ( 1) 5 ( ) 4 ( 2)y t y t u t u t

13

Let the state be i(t) and the output be y(t) = i(t), then for t ≥ t0, i(t0) = i0,

and the solution is given by

Hence, regardless of what vi(t) is for t < t0, all we need to know to predict the

future of the output y(t) = i(t) is the initial state i(t0) and the input voltage vi(t),

for t ≥ t0.

State Models

They are elegant, though practical mathematical representations of the

behavior of dynamical systems. Moreover,

• A rich theory has already been developed

• Real physical systems can be cast into such a representation

( ) ( ) 1( ) ( ) 0 ( ) ( )i i

di t di t Rv t Ri t L i t v t

dt dt L L

0

0

( ) ( )

0 0

1( ) ( ) ( ) ,

t R Rt t t

L Li

t

i t e v d e i t t tL

14

Example:

By KVL, for t t0,

After taking the time derivative of the last equation and dividing by L, we get

0R L C

0

0

( ) ( ) 1( ) ( ) ( ) ( ) ( ) 0

t

C C

t

di t di tRi t L v t Ri t L i d t

dt dt C

2

2

( ) ( ) 1( ) 0

d i t R di ti t

dt L dt LC

15

To solve this second-order homogeneous differential equation, we may proceed

as follows:

Let 1 and 2 (1 ≠ 2) be the roots of the auxiliary equation

then, for t ≥ t0,

C1 and C2 can be uniquely obtained as follows:

From the knowledge of i(t0) and vc(t0) we can compute

and therefore C1 and C2.

012

LCL

R

)(

2

)(

10201)(

tttteCeCti

210 )( CCti

22110

)( CC

dt

tditt

0

)(tt

dt

tdi

16

Using a state variable approach, let x1(t) = vc(t) and x2(t) = i(t), then for t ≥ t0

or

or

This is a first-order linear, constant coefficient vector differential equation! In

principle, its solution should be easy to find.

)(1

)(1)()(

21 tx

Cti

Cdt

tdv

dt

tdx c

)()(1

)()(11

)()()(

21

)(

02

0

txL

Rtx

Ltvdi

CLti

L

R

dt

tdi

dt

tdx

tv

C

t

t

C

,)(

)(

1

10

)(

)(

2

1

2

1

tx

tx

L

R

L

Ctx

tx

dt

d

)(

)(

)(

)(

0

0

02

01

ti

tv

tx

tx c

0( ) ( ) , ( ).t A t tx x x

17

Specifically, for t ≥ t0 ,

The solution to the vector state equation is more elegant, easier to obtain

(provided there is an algorithm to compute eAt) and it specially makes the role

of the initial conditions (initial state) clear.

Linear State Models for Lumped-Parameter Systems

Consider the system described by the following block diagram

Mathematically, for t t0,

0( )0( ) ( )

A t tt e t

x x

0( ) ( ) ( ) ( ) ( ) , ( )t A t t B t t t x x u x

( ) ( ) ( ) ( ) ( )t C t t D t t y x u

B(t) C(t)

A(t)

D(t)

u(t) y(t)

x(t)x(t)

18

where x(t) Rn is the state vector, u(t) Rm is the input vector, y(t) Rr is the

output vector, A(t) Rnxn is the feedback (system) matrix, B(t) Rnxm is the

input distribution matrix, C(t) Rrxn is the output matrix and D(t) Rrxm is the

feed-forward matrix. Also, A(), B(), C() and D() are matrices whose entries

are piecewise continuous functions of time.

Definitions:

The zero-input state response is the response x() given x(t0) 0 and u() 0.

The zero-input system response is the response y() given x(t0) 0 and u() 0.

The zero-state state response is the response x() to an input u() 0 and

x(t0)=0.

The zero-state system response is the response y() to an input u() 0 and

x(t0)=0.

Let yzi() be the zero-input system response and yzs() be the zero-state system

response, then the total system response is given by y() = yzi() + yzs().

19

Example:

Now,

or

or

Let and , then for t ≥ t0, the state model is

3 ( ) 0.5 ( ) ( )v t v t u t

)(2)(6)( tutvtv

.)(,)(,)(2)(6)( 00 tytytutyty

)()(1 tytx )()()(2 tvtytx

1 01 1

0

2 02 2

( )( ) ( )0 1 0( ) ( ) ( ) ( ) , ( )

( )( ) ( )0 6 2

x tx t x tt u t A t B t t

x tx t x t

x x u x

1

2

( )( ) 1 0 ( )

( )

x ty t C t

x t

x

0.5m Kg

input force u t

3friction force v t

output position y t

( )velocity v t y t

20

Now, assuming we know eAt, the solution of the state equation is given by

where for t ≥ t0 = 0, the matrix exponential eAt is described by the 2x2 matrix

(how this was obtained will be explained later)

1. Zero-input state response: u(t) = 0, t ≥ 0 and x(0) ≠0.

0

0

( ) ( )

0( ) ( ) ( )

t

A t t A t

t

t e t e B d x x u

t

tAt

e

ee

6

6

06

1

6

11

6610 2010

6 20 6

20

1 11 11

( ) (0) 6 66 6

0

tt

At

t t

x e xxet e

xe e x

x x

21

2. Zero-input system response: u(t) = 0, t ≥ 0 and x(0) ≠0.

3. Zero-state state response: x(0) = 0. Let the input be the unit

step function, then.

6

10 20 610 20

620

1 11 1

6 6( ) ( ) (0) 1 06 6

t

At t

t

x e xt C t Ce x e x

e x

y x x

6( )

( ) ( )

6( )0 0 0

1 101

( ) ( ) 6 62

0

tt t t

A t A t

s

t

et e Bu d e Bd d

e

x

t

t

t

t

t

t

t

t

t

e

et

de

de

d

e

e

6

6

0

)(6

0

)(6

0)(6

)(6

13

1

118

1

3

1

2

3

1

3

1

23

1

3

1

( ) ( ),su t u t

22

4. Zero-state system response: x(0) = 0.

For t ≥ 0, the complete state response is then given by

and the complete system response by

6

6

6

1 11

1 13 18( ) ( ) 1 0 1

1 3 181

3

t

t

t

t e

y t C t t e

e

x

66

10 20( )

66020

1 11 1 1

3 18( ) (0) ( ) 6 6

11

3

ttt

At A t

tt

t ex e x

t e e Bu d

ee x

x x

1 6 6

1 10 20

2

( ) 1 1 1 1( ) ( ) 1 0 ( ) 1

( ) 6 6 3 18

t tx t

y t C t x t x e x t ex t

x

23

State Models from Ordinary Differential Equations

Let a dynamic system be described by the nth order scalar differential equation

with constant coefficients, i.e.,

where m ≤ n (because of physical realizability).

Let the initial energy of the system be zero, then with n = 3 and m = 2,

Let us implicitly solve this equation, namely,

( ) ( 1) ( )

1 1 1 1( ) ( ) ... ( ) ( ) ( ) ( ) ... ( )n n m

n n m my t a y t a y t a y t b u t b u t b u t

2

2

123322

2

13

3 )()()()(

)()()(

dt

tudb

dt

tdubtubtya

dt

tdya

dt

tyda

dt

tyd

2

2

123322

2

13

3 )()()()(

)()()(

dt

tudb

dt

tdubtubtya

dt

tdya

dt

tyda

dt

tyd

)()()()()()( 3322112

2

tyatubtyatubdt

dtyatub

dt

d

24

Integrating both sides of the last equation one step at a time, we get

This is the implicit solution of the original differential equation. This solution is

obtained via nested integration.

To obtain a state variable representation, we need to represent this implicit

solution in block diagram form (traditional analog simulation diagram).

t

dyaubtyatubtyatubdt

d

dt

tyd )()()()()()(

)(3322112

2

t

ddyaubyaubtyatubdt

tdy

)()()()()()()(

332211

dddyaubyaubyaubtyt

)()()()()()()( 332211

25

Block diagram representation:

If we select the output of the integrators as the state variables. Then

3

13123

23212

3331

xy

ubxaxx

ubxaxx

ubxax

b3

a1

b2 b1

a3 a2

Σ Σ Σx1 x3x21x 3x2x

u

y

+ ++++

_ __

26

In matrix form,

This is the so-called observable canonical form representation.

Alternative state variable representation:

Let us apply the Laplace transform to the original scalar ordinary differential

equation with constant coefficients, assuming zero initial conditions, i.e.,

or

3 3

2 2 0 0

1 1

0 0

1 0

0 1

a b

a b u A B u

a b

x x x

00 0 1y C x x

3 2 2

1 2 3 3 2 13 2 2

( ) ( ) ( ) ( ) ( )( ) ( )

d y t d y t dy t du t d u ta a a y t b u t b b

dt dt dt dt dt

L

)()( 2

12332

2

1

3 sUsbsbbsYasasas

11

0

0

( ) ( )( )

n knn n k

tn kk

d f t d f ts F s s

dt dt

L

27

In transfer function form,

Let us rewrite the last equation as follows:

where

and

Observation:

The overall transfer function is a cascade of two transfer functions.

2 2 1 2 33

3 2 1 3 2 1 1 2 3

3 2 3 2 3 1 2 3

1 2 3 1 2 3 1 2 3

( )

( ) 1

b b s b s b b s b s b s b s b sY s s

U s s a s a s a s a s a s a s a s a s a s

1 2 3

1 2 31 2 3

1 2 3

ˆ( ) ( ) ( ) 1,

ˆ( ) ( ) 1( )

Y s Y s Y sb s b s b s

U s U s a s a s a sY s

3

3

2

2

1

11

1

)(

)(ˆ

sasasasU

sY

3

3

2

2

1

1)(ˆ

)( sbsbsbsY

sY

28

Each of the transfer functions can be expressed in block diagram form, i.e.,

Observation: the term s-1 in the complex frequency domain corresponds to an

ideal integrator in the time domain.

( )U s ˆ( )Y s1 ˆ( )s Y s 2 ˆ( )s Y s 3 ˆ( )s Y s

1

s

1

s

1

s

1a

2a

3a

_ _ _

ˆ( )Y s

1 ˆ( )s Y s 2 ˆ( )s Y s 3 ˆ( )s Y s

1

s

1

s

1

s

1b 2b 3b

( )Y s

29

Putting the two diagrams together, yields

Again, choosing the outputs of the integrators as the state variables, we get

312213

3122133

32

21

xbxbxby

uxaxaxax

xx

xx

( )u t 3x 3x 2x 1x1

s

1

s

1

s

1a

2a

3a

_ _ _

1b 2b 3b

( )y t

2x 1x

30

In matrix form,

This form of the state equation is the so-called controllable canonical form.

Observation:

Both canonical forms are the dual of each other.

Consider the controllable canonical form of some linear time invariant dynamic

system, i.e.,

3 2 1

0 1 0 0

0 0 1 0

1

c cu A B u

a a a

x x x

3 2 1 cy b b b C x x

c c c c

c c

A B u

C

x x

y x

31

Then the observable canonical form is given by

Controllability and Observability (a conceptual introduction):

Suppose now that the initial conditions of an nth order scalar ordinary differential

equation are not equal to zero. How do we build state models such that their

responses will be the same as that of the original scalar model?

Method 1: Given the nth order scalar differential equation

with state model

0 0 0 0 0

0 0 0

0

,

T T

T T Tc c

c c cT

c

A B A CA A B C and C B

B

x x u x u

y x

( ) ( 1) ( 1)

1 1 1 1( ) ( ) ... ( ) ( ) ( ) ( ) ... ( )n n n

n n n ny t a y t a y t a y t b u t b u t b u t

( ) ( ) ( )

( ) ( ) ( )

t A t Bu t

y t C t Du t

x x

x

32

where x(t) Rn, u(t), y(t) R, A,B,C and D are constant matrices of appropriate

dimensions.

Observability Concept: Determine the initial state vector x(0)=[x1(0) … xn(0)]T

from the initial conditions and the initial input values

.

Remark: In the derivation of both observable and controllable canonical forms

from an ordinary linear differential equation with scalar constant coefficients and

m<n, we found out that D = 0, hence, for m = n-1,

)0(),...,0(),0( )1( nyyy

)0(),...,0(),0( )1( nuuu

2

( 1) 1 2 3 ( 3) ( 2)

(0) (0)

(0) (0) (0) (0)

(0) (0) (0) (0) (0) (0) (0)

(0) (0) (0) (0) (0) (0)n n n n n n

y C

y C CA CBu

y C CA CBu CA CABu CBu

y CA CA Bu CA Bu CABu CBu

x

x x

x x x

x

33

In matrix form,

where Rnxn, T Rnxn.

To get a unique solution x(0) for the last algebraic equation, it will be necessary

that the matrix be non-singular ( has full rank), i.e., its inverse exists and

x(0) = -1[Y(0) – TU(0)].

The existence of -1 is directly related to the property of observability of a system.

Hence, to uniquely reconstruct the initial state vector x(0) from input and

output measurements, the system must be observable, i.e., -1 must exist.

In fact, is called the observability matrix.

2

( 2)

( 1) 1 2 3 ( 1)

(0) 0 0 0 (0)

(0) 0 0 (0)

(0) (0)(0)

(0)

(0) 0 (0)

(0) (0)

n

n n n n n

y C u

y CA CB u

y CA CAB CB

u

y CA CA B CA B CB u

T

Y x

x U

34

Controllability Concept: Suppose now that instead of using the input-output

measurements to reconstruct the state at time t = 0 we use impulsive inputs to

change the value of the state instantaneously,

Let

with x(0-) = x0, A Rnxn, B Rn, describe an nth order scalar differential

equation and let the input be described by

Clearly, the input u(t) is described by a linear combination of impulsive inputs

where the scalars are unknown. Now, we know that for t 0-

( ) ( ) ( ), 0t A t Bu t t x x

( 1)

0 1 1( ) ( ) ( ) ( ), , 0, , 1n

n iu t t t t i n

( )

0

( ) (1) ( 1)

0 1 1

0

( ) ( ) ( ) ( ) ( 1)

0 1

0 0 0

( ) (0 ) ( )

(0 ) ( ) ( ) ( )

(0 ) ( ) ( ) ( )

t

At A t

t

At A t n

n

t t t

At A t A t i A t n

i n

t e e Bu d

e e B d

e e B d e B d e B d

x x

x

x

i

35

Integrating the ith term by parts recursively, we get

Therefore, x(t) is given by

where is an unknown vector.

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

00 0

( ) ( 2) 2 ( 2)

00

( ) 1

0

( ) ( ) ( )

( ) ( )

( )

t tt

A t i A t i At A ii i

tt

A t i At A ii

tA t i At A

e B d e B e e AB d

e AB e e A B d

e A B e e

0

0

00

( )

t

i At ii iA B d e A B

1

0 1 1

1

( ) (0 )

(0 )

At At At At n

n

At n

t e e B e AB e A B

e B AB A B

x x

x

0 1 1

T

n

36

At time t = 0+, we get

where Q is the so-called controllability matrix.

Clearly, an impulsive input that will take the state from x(0-) to x(0+) will exist if

and only if the inverse of Q exists, namely, we can uniquely compute the vector

Digital Simulation of State Models

Dynamic systems are nonlinear in general, therefore, let us begin with the

following nonlinear time-varying dynamic system which is described by

1(0 ) (0 ) (0 )n

Q

B AB A B Q

x x x

1 (0 ) (0 ) 0Q x x

0 0 0( ) ( , ( ), ( )), ( ) ,

( ) ( , ( ), ( ))

t t t t t t t

t t t t

x f x u x x

y g x u

1 2 , namely,T

n

37

Objective: We would like to know the behavior of the system over the time

interval t [t0, tn] for a given initial state x(t0) and input u(t), t [t0, tn].

In principle, for t [t0, tn],

However, to compute the integral analytically is very difficult in most cases.

Let’s examine the following numerical approximations to the integral. Let n = 10.

Case 1 (forward Euler formula, scalar case):

0

0( ) ( ) ( , ( ), ( ))

t

t

t t d x x f x u

38

Over the kth time interval,

Hence, the integral under the curve over the whole interval is given by

Now, at time tk, starting at tk-1, we get

Case 2 (backward Euler formula, scalar case):

9910112001 )()()())(),(,(10

0

fttfttfttduxf

t

t

11)())(),(,(

1

kkk

t

t

fttduxfk

k

))(),(,()()()( 11111 kkkkkkk tutxtftttxtx

39

For the kth time interval,

Therefore,

and the approximate solution at tk, starting at tk-1, is given by

Case 3 (trapezoidal rule, scalar case):

10910212101 )()()())(),(,(10

0

fttfttfttduxf

t

t

kkk

t

t

fttduxfk

k

)())(),(,( 1

1

))(),(,()()()( 11 kkkkkkk tutxtftttxtx

40

In this case, the integral over the entire interval is approximately equal to

Therefore, the solution at time tk , starting at tk-1 is approximately equal to

Example: Obtain an approximate solution over one time interval of the following

linearized pendulum state model at equally spaced time instants, tk – tk-1 = 0.5.

The linear model with l=g/4 is . Let

2

)()(

2

)()(

2

)()())(),(,( 109

91021

12

10

01

10

0

fftt

fftt

ffttduxf

t

t

))(),(,())(),(,()(2

1)()( 11111 kkkkkkkkkk tutxtftutxtftttxtx

1 2

2 1

( ) ( )

( ) 4 ( )

x t x t

x t x t

1

2

(0)40

(0)0

x

x

sin( ) sin( )g

mg ms mll

1 2

1 2 2 1

Let and , then

and sin( ) sin( )

x x

g gx x x x

l l

41

Forward Euler Method:

Backward Euler Method:

Trapezoidal Rule Method:

1 1 1 2 1 1 1 2 1

2 2 1 1 1 2 1 1 1

( ) ( ) ( ) ( ) 0.5 ( )0.5

( ) ( ) 4 ( ) ( ) 2 ( )

k k k k k

k k k k k

x t x t x t x t x t

x t x t x t x t x t

1 1 1 2 1 1 2

2 2 1 1 2 1 1

( ) ( ) ( ) ( ) 0.5 ( )0.5

( ) ( ) 4 ( ) ( ) 2 ( )

k k k k k

k k k k k

x t x t x t x t x t

x t x t x t x t x t

1 1 1 2 2 1

1

2 2 1 1 1 1

1 1 2 2 1

2 1 1 1 1

( ) ( ) ( ) ( )0.5( )

( ) ( ) 4 ( ) 4 ( )

( ) 0.25 ( ) ( )

( ) ( ) ( )

k k k k

k k

k k k k

k k k

k k k

x t x t x t x tt t

x t x t x t x t

x t x t x t

x t x t x t

42

Linear Discrete-Time Systems

Implementation of dynamic systems is actually done using digital devices like

computers and/or DSPs. Moreover, there are some naturally occurring processes

which are discrete-time. Hence, it is convenient to model such systems as

discrete-time systems.

In most cases, a continuous-time linear time-invariant dynamic system is

discretized at time t = tk, where tk+1 - tk may be fixed or variable. When such

system is described by its impulse response, the discretization process is

illustrated in the figure below (a sampled-data system).

t=tk t=tk

u(t) u(tk) y(t) y(tk)

h(t)

43

The equivalent single-input, single-output, discrete-time system can be depicted

by

Since the system is linear, its behavior is described by the convolution relation.

Let tk = kT, then

Let u(kT) = (kT), the unit sample, i.e.,

then,

is called the unit sample response.

n

nTunTkThkTy )()()(

otherwise

kkT

,0

0,1)(

)()()()( kThnTnTkThkTyn

u(tk) y(tk)h(tk)

44

Let the system be causal, i.e., h(kT) = 0, k < 0 then

In addition, if u(kT) = 0 for k < 0, then

k

n

nTunTkThkTy )()()(

k

n

nTunTkThkTy0

)()()(

Consider now a continuous-time linear time-invariant multiple-input, multiple-

output dynamic system in state-space form, i.e.

0 0 0( ) ( ) ( ), ( ) ,

( ) ( ) ( )

t A t B t x t x t t

t C t D t

x x u

y x u

We know that the solution of the state equation is

0

0

( ) ( )

0( ) ( ) ( )

t

A t t A t

t

t e t e B d x x u .

45

Let 0t kT and t kT T , then the last equation becomes

( )( ) ( ) ( )

kT T

AT A kT T

kT

kT T e kT e B d

x x u .

Suppose ( ) ( ),u t u kT kT t kT T , namely, the input does not change in the interval

[ , )kT kT T , (equivalent to the presence of a sample-and-hold device) then

( )( ) ( ) ( )

kT T

AT A kT T

kT

kT T e kT e d B kT

x x u .

Let kT T , then d d and

0

( )

0

( )

kT T T

A kT T A A

kT T

e d e d e d

.

The discretized system now becomes

0

0

( ) ( ) ( ), (0) , 0,1,2,

T

AT AkT T e kT e d B kT k x x u x x .

46

State representation of discrete-time dynamic systems:

Consider a linear discrete-time dynamic system described by the difference equation

where the sampling interval has been normalized, i.e., T = 1 sec.

)()1()2()1()( 121 kyakyankyankyanky nn

)()1()( 11 mkubkubkub mm

Define the discrete-time equivalent system and input distribution matrices as AT

dA e

and 0

T

A

dB e d B , then the discrete-time equivalent system is described by

0( ) ( ) ( ), (0) , 0,1,2,

( ) ( ) ( )

d dkT T A kT B kT k

kT C kT D kT

x x u x x

y x u

Note that this discretization of the original continuous-time system is exact when a

sample-and-hold device processes the input u(t) and the input and output samplers (D/A

and A/D) clocks are in synchronism.

47

If we now replace differentiations with forward shift operators and integrators with

backward shift operators then we can construct the same type of canonical

realizations that we built for continuous-time systems.

Example: Let n = 3, m = 2 and y(0) = y(1) = y(2) = u(0) = u(1) = u(2) = 0, then

Let us apply the backwards shift operator to this equation recursively, i.e.

The solution y(k) can now be computed implicitly using a simulation block

diagram.

)()()1()1()2()2()3( 332211 kyakubkyakubkyakubky

)()()()()1()1()2()3( 33

1

2211

1 kyakubqkyakubkyakubkykyq

)()()()()()()1()2( 33

1

22

1

11

1 kyakubqkyakubqkyakubkykyq

)()()()()()()()1( 33

1

22

1

11

11 kyakubqkyakubqkyakubqkykyq

48

Simulation diagram implementation:

Using the outputs of the shift operators as the state variables, we get

)()()()1(

)()()()1(

)()()1(

13123

23212

3331

kubkxakxkx

kubkxakxkx

kubkxakx

)()( 3 kxky

b3

a1

b2 b1

a3 a2

Σ Σ Σx1 x3x2

u

y

+ ++++

_ __q-1 q-1 q-1

49

In matrix form, we obtain the observable canonical form, i.e.

In general, a multiple-input, multiple-output linear shift-invariant discrete-time

dynamic system can be represented by (assuming T = 1)

where x(k) Rn, u(k) Rm, y(k) Rr, A, B, C and D are constants matrices of

appropriate dimensions.

As in the continuous-time case, we may reconstruct the state from input-output

measurements when the system is observable.

3 3

2 2

1 1

0 0

( 1) 1 0 ( ) ( )

0 1

a b

k a k b u k

a b

x x

( ) 0 0 1 ( )y k k x

( 1) ( ) ( )

( ) ( ) ( )

k A k B k

k C k D k

x x u

y x u

50

Iteratively,

In matrix form,

2

1 2 3

( ) ( ) ( )

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

( 2) ( 2) ( 2) ( ) ( ) ( 1) ( 2)

( 1) ( ) ( ) ( 1) ( 2) ( 1)n n n

k C k D k

k C k D k CA k CB k D k

k C k D k CA k CAB k CB k D k

k n CA k CA B k CA B k CB k n D k n

y x u

y x u x u u

y x u x u u u

y x u u u u

2

1 2 3

( ) 0 0 0 ( )

( 1) 0 0 ( 1)

( ) ( )( 2) ( 2)

0

( 1) ( 1)

( ) ( )

n n n

k C D k

k CA CB D k

k kk CA CAB CB D k

k n CA CA B CA B CB D u k n

k T k

y u

y u

Y xy u

y

x U

51

If has full rank, then x(k) = -# [Y(k)-TU(k)], i.e., if the system is observable then

we can reconstruct the state at time k using input, output measurements up to time

k+n-1, where -# is the pseudoinverse of .

Solution of the discrete-time state equation

Iteratively,

and

2

3 2

4 3 2

11

0

(1) (0) (0)

(2) (1) (1) (0) (0) (1)

(3) (2) (2) (0) (0) (1) (2)

(4) (3) (3) (0) (0) (1) (2) (3)

( ) (0) ( ), 0k

k k l

l

A B

A B A AB B

A B A A B AB B

A B A A B A B AB B

k A A B l k

x x u

x x u x u u

x x u x u u u

x x u x u u u u

x x u

11

0

( ) ( ) ( ) (0) ( ) ( )k

k k l

l

k C k D k CA C A B l D k

y x u x u u

52

From the solution of the discrete-time state equation, starting at time index k, the

value of the state at time index k+n is

where the controllability matrix Q and the vector are given by

To assure the existence of an input such that the state of the system can reach a

desired state at time k+n given the value of the state at time k, the matrix Q must

have rank n. Thus, the following relationship is satisfied

, where is the pseudo inverse of Q

In other words, the system must be controllable.

( ) ( ) ( ), hence, ( ) ( ) ( )n n

r rk n A k Q k Q k k n A k x x U U x x

1

( 1)

( 2)[ ] and ( )

( )

n

r

k n

k nQ B AB A B k

k

u

uU

u

#( ) ( ) ( )n

r k Q k n A k U x x#Q

( )r kU

53

Linearization of Nonlinear Systems

Consider the following scalar (single-input, single-output) nonlinear

memoryless system

where g(·):.

Let the nominal operating point be and let N be a neighborhood of it,

i.e., N = {x : a < x < b} and a < x0 < b. If the function g(·) is analytic

on N, i.e., it is infinitely differentiable on N, then for x such that

x0+xN, we get the following Taylor series expansion

For small x,

0x

0 0

22

0 0 2

1( ) ( )

2!x x x x

dg d gg x x g x x x

dx dx

0 00 0 0( ) ( ) x x x x

dg dgy g x x g x x y x

dx dx

)x(t) y(t)g(·)

54

Therefore, the linear approximation of a nonlinear system y = g(x) near

the operating point x0 has the form

or

where

Example: Consider a semiconductor diode described by

where vo=vt.

In this case,

00 x x

dgy y x

dx

0y y y m x

0x x

dgm

dx

0 ln 1 ( )s

iv v g i

i

0 0 0

0

0

( ) 1 1

1i i i i o i d

s s

s

vdg i dvm v r

idi di i i i

i

55

The linearized model is then given by

Using the parameters = 35.4, vt = 0.026 V, is = 1 nA, i0 = 0.05 A, v0 = 0.92 V,

rd = 18.44 , we get the following linear approximation:

0

0d

s

vv m i i r i

i i

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

i in Amps

v in V

olts

56

If, on the other hand, y=g(x1,x2,,xn) , then if g(·) is analytic on the

set N={xn:a<||x||<b}, x0N and x0+xN, x0=[x10,,xn0]T,

where and

Systems with memory:

I. Scalar Case. Consider a dynamic system described by

Suppose xn(t) is the system response resulting from a nominal

operating input un(t), assuming a nominal initial state xn(t0) = xn0

0 0

0

10 0 11

0

( ) ( )( , , ) higher order terms

( ) ( ) higher order terms

n nn

g gy g x x x x

x x

g g

x x x x

x x

x x

x x x

1 2

( ) ( ) ( )( )

n

g g gg

x x x

x x xx

0 0 0( ) ( ( ), ( ), ), ( ) , ( ), ( ) forx t f x t u t t x t x x t u t t t

1 2

T

nx x x x

57

In other words,

Assume that we know xn(t). Perturb the state and the input by taking

x(t) = xn(t) + x(t) and u(t) = un(t) + u(t).

We want to find the solution to the differential equation

For fixed values of t, and f(·) an analytic function on some neighborhood

of xn(t) and un(t), using a Taylor series expansion, we get

where

0 0 0( ) ( ( ), ( ), ) , ( ) , forn n n n nx t f x t u t t x t x t t

0 0 0 0( ) ( ( ) ( ), ( ) ( ), ) , ( ) ( ) , forn n nx t f x t x t u t u t t x t x t x t t

( ( ) ( ), ( ) ( ), ) ( ( ), ( ), )

( )( , , ) higher order terms

( )n

n

n n n n

x xu u

f x t x t u t u t t f x t u t t

x tf x u t

u t

( , , )f f

f x u tx u

58

Let and

Then, if we neglect the higher order terms in the expansion, we get

Therefore,

Example: Let

Clearly,

Now, if the nominal operating state and input are xn(t) and un(t) = 0, t 1,

( ( ), ( ), )( )

n

n

x xu u

f x t u t ta t

x

( ( ), ( ), )( )

n

n

x xu u

f x t u t tb t

u

( )( ) ( )( ( ), ( ), ) ( ) ( ) ( ) ( )n

n n

dx tdx t d x tf x t u t t a t x t b t u t

dt dt dt

0 0

( )( ) ( ) ( ) ( ), ( )

d x ta t x t b t u t x t x

dt

20 0( ) ( ) ( ) , ( )x t x t u t x t x

2( ( ), ( ), ) ( ) ( )f x t u t t x t u t

0 0

( ( ), ( ), )( ) 2 ( ) 2 ( )

n n

n n

x x x x nu u u u

f x t u t ta t x t x t

x

59

and

So,

With

Finally,

0 0

( )2 ( ) ( ) ( ), ( )n

d x tx t x t u t x t x

dt

0 0( ) 0, 1 and ( ) (1) 1, thenn n nu t t t x t x

2

1 2 1 22

( )( ) , (1) 1

( ) 1 1and ( ) , 1

( )( )

nn n

nn

nn

dx tx t x

dt

dx tdt c t c c c x t t

x t tx t

0

( ) 2( ) ( ) , (1) , 1

d x tx t u t x x t

dt t

0

( ( ), ( ), )( ) 1

n

n

x xu u

f x t u t tb t

u

60

II. Vector Case. Consider now the case of a system described by the following

nonlinear vector differential equation:

The ith element of the vector differential equation is described by

Moreover,

where

0 0 0( ) ( ( ), ( ), ) , ( ) , ( ) , ( ) forn mt t t t t t t t t x f x u x x x u

0 0 0( ) ( ( ), ( ), ) , ( ) , ( ) , ( ) forn mi ix t f t t t t t t t t x u x x x u

( ( ) ( ), ( ) ( ), ) ( ( ), ( ), )

( )( , , ) higher order terms

( )n

n

i n n i n n

i

f t t t t t f t t t

tf t

t

x xu u

x x u u x u

xx u

u

( , , ) ( , , ) ( , , )i i if t f t f t x ux u x u x u

61

and

or

1 2

( , , ) i i ii

n

f f ff t

x x x

x x u

1 2

( , , ) i i ii

m

f f ff t

u u u

u x u

1 1

1 1

2 2 2 2

( , , ) ( , , )

( , , ) ( , , )

( , , ) ( , , )

n n

n n

n n

n n

n n

n n

n n

n n

f t f t

x x

x f t x f td

dt

x x

f t f t

x x x u x xu u u u

x x x u x xu u u u

x x x u x xu u u u

x u x u

x u x u

x u x u

1

2

m

u

u

u

62

or

where

are real nxn and nxm matrices, respectively.

0 0

( )( ) ( ) , ( ),

d tA t B t x t t t

dt

xx u

1 1

2 2

( , , ) ( , , )

( , , ) ( , , ) and

( , , ) ( , , )

n n

n n

n n

n n

n n

n n

n n

f t f t

f t f tA B

f t f t

x x x u x xu u u u

x x x u x xu u u u

x x x u x xu u u u

x u x u

x u x u

x u x u

63

Finally, if the outputs of the nonlinear system are of the form

Then

( ) ( ( ), ( ), ) , ( ) pt t t t t y g x u y

1 1

1 1

2 2 2 2

( , , ) ( , , )

( , , ) ( , , )( )

( , , ) ( , , )

n n

n n

n n

n n

n n

n n

p n

p p

g t g t

y x

y g t x g tt

y x

g t g t

x x x u x xu u u u

x x x u x xu u u u

x x x u x xu u u u

x u x u

x u x uy

x u x u

1

2

m

u

u

u

64

or

where

are real pxn and pxm matrices, respectively.

( ) ( ) ( ) , ( ) pt C t D t t y x u y

1 1

2 2

( , , ) ( , , )

( , , ) ( , , ) and

( , , ) ( , , )

n n

n n

n n

n n

n n

n n

p p

g t g t

g t g tC D

g t g t

x x x u x xu u u u

x x x u x xu u u u

x x x u x xu u u u

x u x u

x u x u

x u x u

65

Example: Suppose we have an orbiting micro satellite modeled as a point mass

in an inverse square law force field, e.g., a gravity field as shown below

where r(t) is the radius of the orbit at time t

(t) is the angle relative to the horizontal axis at time t

u1(t) is the thrust in the radial direction at time t

u2(t) is the thrust in the tangential direction at time t

m is the mass of the orbiting body

m

u1(t)u2(t)

(t)r(t)

orbit

66

From the laws of mechanics and assuming m = 1kg, the total force in the radial

direction is described by

and the total force in the tangential direction is

Select the states as follows:

Then,

22

12 2

( ) ( )( ) ( )

( )

d r t d t Kr t u t

dtdt r t

2

22

( ) 2 ( ) ( ) 1( )

( ) ( )

d t d t dr tu t

r t dt dt r tdt

1 2 3 4

( ) ( )( ) ( ), ( ) , ( ) ( ), ( )

dr t d tx t r t x t x t t and x t

dt dt

1 2

22 1 4 12

1

( ) ( )

( ) ( ) ( ) ( )( )

x t x t

Kx t x t x t u t

x t

67

which implies that

For a circular orbit and u1n(t) = u2n(t) = 0 and t0 = 0, we have

3 4

2 44 2

1 1

( ) ( )

( ) ( ) 1( ) 2 ( )

( ) ( )

x t x t

x t x tx t u t

x t x t

2

21 4 12

1

4

2 4 2

1 1

( )( ( ), ( ), ) .

2

x

Kx x u

xd tt t t

dt x

x x u

x x

x= f x u

0 0( ) 0 , 0Tn t R t t x

68

where rn(t) = R and

Linearizing about xn(t) and un(t), yields

0 3

( ).nd t K

dt R

1( , , ) 0 1 0 0n

n

f t

x x xu u

x u

2 22 4 1 4 0 03

1

2( , , ) 0 0 2 3 0 0 2

n

n

n

n

Kf t x x x R

x

x x xu u

x xx u

u u

3( , , ) 0 0 0 1n

n

f t

x x xu u

x u

02 4 2 4 24 2

1 11

2( , , ) 2 0 2 0 2 0 0

n

n

n

n

x x u x xf t

x x Rx

x x xu u

x xx u

u u

69

Likewise,

In state form,

1( , , ) 0 0n

n

f t

u x xu u

x u

2 ( , , ) 1 0n

n

f t

u x xu u

x u

3( , , ) 0 0n

n

f t

u x xu u

x u

41

1 1( , , ) 0 0

n n

n n

f tx R

u x x x x

u u u u

x u

20 0

0

0 1 0 0 0 0

1 03 0 0 2

0 00 0 0 1

100 2 0 0

Rd

dt

RR

x x u

70

Theorem: Let A(t) Rnxn be piecewise continuous. Then the set of all solutions of

forms an n-dimensional vector space over the field of the real numbers.

Proof: Let { 1, 2, , n} be a set of linearly independent vectors in Rn, i.e.,

, if and only if i = 0, i = 1, 2, , n; and the i()’s be the

solutions of with initial conditions i(t0) = i, i = 1, 2, , n.

Suppose that t R+ the i’s, i = 1, 2, , n are linearly dependent, then i R,

i = 1, 2, , n, different than zero such that

But, at t = t0 R+,

the i’s are linearly dependent, which is an outright contradiction of the

hypothesis that the i’s are linearly independent. Therefore, the i’s are linearly

independent for all t R+.

1 1 2 2 0n n

1 1 2 2( ) ( ) ( ) 0n nt t t

0( ) ( ) ( ),t A t t t t x x

1 1 0 2 2 0 0( ) ( ) ( )n nt t t 1 1 2 2 0n n

71

Let (t) be any solution of and (t0) = . Since the i’s are linearly independent

vectors in Rn, can be uniquely represented by

But, is a solution of with initial condition .

This is because

In other words, the linear combination

satisfies the differential equation and the initial condition .

Therefore, () = implies that every solution of is a linear combination

of the basis of solutions i(), i = 1, 2, , n, i.e., the set of all solutions of forms

an n-dimensional vector space.

1

n

i i

i

0

1 1

( )n n

i i i i

i i

t

1 1 1 1

( ) ( ) ( ) ( ) ( ) ( )n n n n

i i i i i i i i

i i i i

dt t A t t A t t

dt

1

( )n

i i

i

1

( )n

i i

i

1

( )n

i i

i

1

n

i i

i

72

Example: Consider the dynamical system described by

Let the vectors 1 and 2 be described by and .

Then, and are two independent solutions to the system

with initial conditions 1(0) = 1 and 2(0) = 2.

Therefore, any solution (t) will be given by

i R, i = 1, 2.

0 0( ) ( )

0t t

t

x x

1

1

0

2

0

1

1 2

1

( ) 1

2

tt

2

0( )

1t

1

0

2

11

)()()( 2212211 t

ttt

73

Def. The state transition matrix of the differential eq. is given by

where the i’s, i = 1, 2, , n are the basis solutions and i = [0 … 0 1 0 … 0]T.

Properties of the state transition matrix:

1. (t0, t0) = I

Proof: Recall that

Thus (t0, t0) = I.

2. (t, t0) satisfies the differential eq. , M(t0) = I, M(t) Rnxn.

Proof: The time derivative of the state transition matrix is given by

However,

),;(),;(),;(),( 02021010 nn tttttttt

0 0( ; , ) 0 0 1 0 0T

i i it t

)()()( tMtAtM

0 1 0 1 2 0 2 0( , ) ( ; , ) ( ; , ) ( ; , )n n

dt t t t t t t t

dt

0 0( ; , ) ( ) ( ; , )i i i it t A t t t

( ) ( ) ( )t A t tx x

ith location

74

Therefore,

Also, from part (1), (t0, t0) = I.

3. (t, t0) is uniquely defined.

Proof: Since each i(t) is uniquely determined by A(t) for each initial condition i

then (t, t0) is also uniquely determined by A(t).

Proposition: The solution to , x(t0) = x0 is x(t) = (t, t0)x0 t.

Proof: At t = t0, (t0, t0)x0 = Ix0 = x0, it satisfies the initial condition.

We already know that .

Therefore, .

In other words, , satisfies the differential equation.

0 1 0 1 2 0 2 0( , ) ( ) ( ; , ) ( ) ( ; , ) ( ) ( ; , )n nt t A t t t A t t t A t t t

1 0 1 2 0 2 0 0( ) ( ; , ) ( ; , ) ( ; , ) ( ) ( , )n nA t t t t t t t A t t t

),()(),( 00 tttAtt

0 0 0 0( ) ( , ) ( ) ( , ) ( ) ( )t t t A t t t A t t x x x x

0 0( , )t t x

( ) ( ) ( )t A t tx x

75

If t and t0, and A(t) has the following commutative property:

Then, the state transition matrix is given by

Example: Compute the state transition matrix (t, t0) for the differential equation,

We can show that and are commutative, i.e.

)()()()(

00

tAdAdAtA

t

t

t

t

0

( )

0( , )

t

t

A d

t t e

21( ) ( )

0 1

tet t

x x

0

0 0

22 2

0 0

0

1( )

( ) ( ) ( ) ( ) 2

0

tt tt t

t t

t t e e t t eA t A d A d A t

t t

0

( ) ( )

t

t

A t A d 0

( ) ( )

t

t

A d A t

76

Hence,

and

0

0 0 0

2 3( )

0

1 1( , ) ( ) ( ) ( )

2! 3!

t

t

A d t t t

t t t

t t e I A d A d A d

!20

!2

1

!2

02

1

10

012

0

22

0

2

0

0

22

0

0

0

tt

eetttt

tt

eetttt

tt

!30

!32

3

!33

0

22

2

0

3

0 0

tt

eetttt tt

)(3

0

2

000110)(

!3

1)(

!2

1)(1),(

ttetttttttt

77

Finally,

Hence, the state transition matrix is given by

and .

We can show that the norm of the state transition matrix blows up as time t goes to

infinity, therefore, the system is unstable.

2

0

22

0

2222

012 )(!3

1

2

3)(

!2

1

2

1),( 000 tteetteeeett

tttttt

00000 3)(223

0

2

00

22

2

1

2

1)(

!3

1)(

!2

1)(1

2

1 tttttttttt eeeeettttttee

0),( 021 tt

0),(),( 011022

ttetttt

0 0 0

0

3

0

1

2( , )

0

t t t t t t

t t

e e et t

e

0 0

1 0( , )

0 1t t

78

Theorem: A(t) and commute if

1. A() is constant

2. A(t) = (t)M, () : R R and M is a constant square matrix

3. , i() : R R and the Mi’s are constant square matrices

such that MiMj = MjMi i, j.

Proof:

(1) If A() is a constant matrix, i.e., A() = A, then

(2) If A(t) = (t)M, then

t

t

dA

0

)(

k

i

ii MttA1

)()(

2

00

22 )()(

00

AttttAIdAAdA

t

t

t

t

2

0

2 )(

00

AttAIdAAd

t

t

t

t

2

0000

)()()()()()()()( MdtMIdMtMdMtdAtA

t

t

t

t

t

t

t

t

79

But,

(3) If then

But,

2

0000

)()()()()()()()( MdtMtMdIMtMdtAdA

t

t

t

t

t

t

t

t

k

i

ii MttA1

)()(

k

j

t

t

jj

k

i

ii

t

t

k

j

jj

k

i

ii

t

t

MdMtdMMtdAtA1111

000

)()()()()()(

k

i

t

t

jij

k

j

i

k

i

t

t

jj

k

j

ii MMdtMdMt1 11 1

00

)()()()(

k

i

ii

k

j

j

t

t

j

k

i

ii

t

t

k

j

jj

t

t

MMdMdMtAdA1111

)()()()()()(

000

0 01 1 1 1

( ) ( ) ( ) ( )

t tk k k k

j j i i i j j i

j i j it t

d M t M t d M M

80

and, if MiMj = MjMi i, j (i ≠ j), then

Corollary: If A(t) satisfies condition (3) of the previous theorem, then

Proof: Since when A(t) satisfies condition (3), then

But,

)()()()(

00

tAdAdAtA

t

t

t

t

0

( )

0

1

( , )

t

i i

t

M dk

i

t t e

)()()()(

00

tAdAdAtA

t

t

t

t

t

t

dA

ett 0

)(

0),(

k

i

t

t

ii

t

t

k

i

ii MddMk

i

ii eettMttA1

001

)()(

0

1

),()()(

81

or,

Def. Any nxn matrix M(t) satisfying the matrix differential equation

M(t0) = M0, where det(M0) ≠0 is a fundamental matrix of solutions.

Theorem: If det(M0) ≠0 then det(M(t)) ≠ 0 t R+

Proof: (By contradiction) Suppose there exists t1 R+ such that det(M(t1)) = 0.

Let v = [v1 v2 … vn]T ≠ 0 such that M(t1)v = 0 and x(t) = M(t)v be the solution to the

vector differential equation , x(t1) = 0. Notice also that z() 0 is a

solution to , z(t1) = 0. By the uniqueness theorem we conclude that

x(t) = z(t) everywhere A(t) is piecewise continuous.

But, z(t0) = x(t0) = M(t0)v = 0 det(M0) = 0, which is a contradiction. Hence,

det(M(t)) ≠ 0 t, i.e., M(t) is nonsingular t.

k

i

dMMdMdMd

t

t

ii

t

t

kk

t

t

t

t eeeett1

)()()()(

0000

22

0

11

),(

( ) ( ) ( )t A t tz z

,)()()( tMtAtM

( ) ( ) ( )t A t tx x

82

Def. Let M(t) be any fundamental matrix of .. Then, t R+, the

state transition matrix (t,t0) of is given by, (t,t0) = M(t)M-1(t0).

Theorem (Semigroup Property): For all t1, t0 and t, we have (t,t0) = (t,t1) (t1,t0).

Proof: It is known from the existence and uniqueness theorem that

x(t) = (t,t0)x(t0) for any t,t0 (a)

x(t1) = (t1,t0)x(t0) for any t1,t0 (b)

and x(t) = (t,t1)x(t1) for any t,t1 (c)

are solutions to the differential equation with initial conditions x(t0)

and x(t1). But, from (c) and (b)

x(t) = (t,t1)x(t1) = (t,t1) (t1,t0)x(t0) (d)

Comparing (a) and (d) leads us to conclude that

(t,t0) = (t,t1) (t1,t0) for any t, t1 and t0.

Theorem (The Inverse Property): (t,t0) is nonsingular t, t0 R+ and -1(t,t0) =

(t0,t).

( ) ( ) ( )t A t tx x

( ) ( ) ( )t A t tx x

( ) ( ) ( )t A t tx x

83

Proof: Since (t,t0) is a fundamental matrix of , then it is nonsingular

for all t, t0 R+.

Now, from the semigroup property we know that for arbitrary t0, t1, t R+ ,

(t,t0) = (t,t1) (t1,t0).

For t0 = t, we get,

(t,t) = I = (t,t1) (t1,t) -1(t,t1) = (t1,t) and since t1 is arbitrary we have that

-1(t,t0) = (t0,t).

Theorem (Liouville formula):

Consider now the linear, time-varying dynamic system modeled by

with x(t0) = x0 .

Theorem: The solution to the state equation is given by

t

t

dAtr

ett 0

)(

0 )],(det[

( ) ( ) ( ) ( ) ( )t A t t B t t x x u

( ) ( ) ( ) ( ) ( )t C t t D t t y x u

0

0 0( ) ( , ) ( , ) ( ) ( )

t

t

t t t t B d x x u

( ) ( ) ( )t A t tx x

84

where

1. (t,t0)x0 is the zero-input state response, and

2. is the zero-state state response.

Proof: At t = t0, the solution to the differential equation is given by

Now,

since

0

( , ) ( ) ( )

t

t

t B d u

0

0

1

0 0 0 0 0 0

0

( ) ( , ) ( , ) ( ) ( )

n n

n

t

tI

t t t t B d

x x u x

0 0

0 0 0 0( , ) ( , ) ( ) ( ) ( , ) ( , ) ( ) ( )

t t

t t

dt t t B d t t t B d

dt t

x u x u

0

0 0( ) ( , ) ( , ) ( ) ( ) ( , ) ( ) ( )

t

t

A t t t t t B t t t B dt

x u u

t

t

t

t

t

dtft

tfdtft

00

),(),(),(

85

Hence,

The complete response should be given by

Let us now consider the time-invariant case, i.e., A(t) = A, B(t) = B, C(t) = C and

D(t) = D, where A, B, C and D are constant matrices.

Theorem: The state transition matrix of the time-invariant state model is

0

0 0( ) ( , ) ( ) ( ) ( ) ( , ) ( ) ( )

t

t

dA t t t B t t A t t B d

dt x u u

0

0 0( ) ( , ) ( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

t

t

A t t t t B d B t t A t t B t t

x u u x u

0

0 0( ) ( ) ( , ) ( ) ( , ) ( ) ( ) ( ) ( )

t

t

t C t t t C t t B d D t t y x u u

)(

000)0,(),(

ttAetttt

86

Proof: Since A and commute, we have that

The complete state and system responses are now given by

This follows from the fact that (t,t0) = M(t)M-1(t0) since we can always let

M(t) = eAt, i.e.,

t

t

Ad

0

)()0,(exp),( 00000 ttttttAett

t

t

Ad

0 0

0 0

( ) ( )( )

0 0( ) ( ) ( )

t t

A t t A t tA t At A

t t

t e e B d e e e B d x x u x u

0

0

( )

0( ) ( ) ( )

t

A t t At A

t

t Ce Ce e B d D t y x u u

)(1

0

1 000 )()()(ttAAtAtAtAt eeeeetMtM

87

Def. If A is an nxn constant matrix, C, e Cn and the equation

Ae = e, e ≠0

is satisfied, then is called an eigenvalue of A and e is called an eigenvector of A

associated with . Also, the eigenvalues of A are the roots of its characteristics

polynomial, i.e.,

The set (A) = {1, 2,…, n} is called the spectrum of A. The spectral radius of A is

the non negative real number

The right eigenvector ei Cn of A associated with the eigenvalue i satisfies the

equation Aei = iei, whereas the left eigenvector wi Cn of A associated with i

satisfies the equation wi*A = iwi*, where ()* designates the complex conjugate

transpose of a vector. If (A) and is complex then * (A). The eigenvectors

associated with and * will be e and e*, respectively.

)())(()det()( 211

1

1 nnn

nn

A aaaAI

( ) max :i ii

A A

88

Example: Find the right eigenvectors of the matrix

The characteristic polynomial of A is given by

Therefore, its spectrum is described by (A) = {-1, -1 - j2, -1 + j2}

Now,

or

For 1 = -1, we get

010

011

552

A

)52)(1(573)det()( 223 AIA

( ) 0i i i i iAe e I A e

0~~

10

011

552

i

i

i

i

e

89

or

let e13 = 1, then e12 = -1, then

For 2 = -1 – j2,

1 1 1

1 5 5

( ) 1 0 0 0

0 1 1

I A e e

13121312

1111

131211

0

00

055

eeee

ee

eee

Te 110~1

2 2 2

1 2 5 5

( ) 1 2 0 0

0 1 1 2

j

I A e j e

j

90

or

let e23 = 1, then e22 = -1 – j2, e21 = -j2(-1 – j2) = -4 + j2

and

Finally, , because .

Theorem: Let A be an nxn constant matrix. Then A is diagonalizable if and only if

there is a set of n linearly independent vectors, each of which is an eigenvector of

A.

Proof: If A has n linearly independent eigenvectors e1, e2, … , en, form the

nonsingular matrix T = [e1 e2 … en].

23222322

22212221

232221

)21(0)21(

202

055)21(

ejeeje

ejeeje

eeej

Tjje 12124~2

3 2( ) 4 2 1 2 1T

e e j j 3 2

91

Now, T-1AT = T-1A [e1 e2 … en] = T-1[Ae1 Ae2 … Aen] = T-1[1e1 2e2 … nen]

= T-1[e1 e2 … en]D = T-1TD = D

where D = diag [1, 2, … , n], and i, i = 1, 2, … , n are the eigenvalues of A.

Conversely, suppose there exists a matrix T such that T-1AT = D is diagonal. Then

AT = TD.

Let T = [t1 t2 … tn], then

AT = [At1 At2 … Atn] = [t1d11 t2d22 … tndnn] = TD Ati = diiti, which implies

that the ith column of T is an eigenvector of A associated with the eigenvalue dii.

Since T is nonsingular, there are n linearly independent eigenvectors.

Now, if A is diagonalizable, then eAt = TeDtT-1 because

1 2 3

1 1 1 2 1 31 1

2! 3!

At TDT te e TT TDT t TDT t TDT t

1 1 1 2 1 1 1 31 1

2! 3!I TDT t TDT TDT t TDT TDT TDT t

3132121

!3

1

!2

1tTTDtTTDtTDTI

2 2 3 3 1 11 1

2! 3!

DtT I Dt D t D t T Te T

92

Example: For the given matrix, compute eAt.

We already know that (A) = {-1, -1 - j2, -1 + j2}. Now,

The inverse of T is

010

011

552

A

111

21211

24240

jj

jj

T

1

2 2 101

1 1 2 1 28

1 1 2 1 2

T j j

j j

93

Therefore,

which implies that

Finally,

2100

0210

001

j

jD

tj

tj

t

Dt

e

e

e

e)21(

)21(

00

00

00

1

4cos(2 ) 2sin(2 ) 10sin(2 ) 10sin(2 )

1 cos(2 ) 2sin(2 ) 1 5cos(2 ) 5 5cos(2 ) ( )4

1 cos(2 ) 1 cos(2 ) 2sin(2 ) 5 cos(2 ) 2sin( )

2

At Dtt

e Te T

t t t te

t t t t t

t t t t t

94

Proposition: Suppose D is a block diagonal matrix with square blocks Di , i = 1, 2, …, n, i.e.,

Then,

Example: For the same A matrix of the previous example, compute eAt.

Let the similarity transformation T be given by

nD

D

D

D

00

0

00

2

1

tD

tD

tD

Dt

ne

e

e

e

00

0

00

2

1

1 2 2

0 4 2

Re{ } Im{ } 1 1 2

1 1 0

T

e e e

95

then,

and

which implies that

But,

Finally,

220

111

511

4

11T

11

1 2

2

1 0 00 1 2

0 1 2 , 1,0 2 1

0 2 1

DD T AT D D

D

tD

tD

Dt

e

ee

2

1

0

0

tt

tteeee ttDttD

2cos2sin

2sin2cosand 21

1 1

1 0 0

0 cos 2 sin 2

0 sin 2 cos 2

At Dt te T e T e T t t T

t t

Explicitly,

4cos(2 ) 2sin(2 ) 10sin(2 ) 10sin(2 )

1 cos(2 ) 2sin(2 ) 1 5cos(2 ) 5 5cos(2 ) ( )4

1 cos(2 ) 1 cos(2 ) 2sin(2 ) 5 cos(2 ) 2sin(2 )

tAt

t t t te

e t t t t t

t t t t t

.

At 0t , 3 3(0) I , hence, it is a valid state transition matrix.

96