ECE 550 Lecture Notes 1

8/6/2019 ECE 550 Lecture Notes 1

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ECE 550

LINEAR SYSTEM THEORYLECTURE NOTES

Prof. Mario Edgardo Magaa

EECSOregon State University


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System Description:

A system Nis a device that maps a set of admissible inputs Uto a set ofadmissible output responses Y.

Mathematically, N: U Yory() = N[u()].

Alternatively, a system can be described either by differential or differenceequations in the time domain, or by algebraic equations in the complex

frequency domain.

Example: Let us describe the relationship between the input and the output ofthe following active filter system:

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


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Consider the first amplifier. Using nodal analysis,

The second amplifier yields

Assuming infinite input impedance, we get . Hence,

Furthermore, the two currents are described by

021

6

01

2

1

1

1 =++

++

cci ii

R

vv

R

v

R

vv

03

21

1=+

R

veic

05

02

4

22 =

+

R

ve

R

ve

1 2 0e e= =

1

52 3 0 2

4

= andcR

v R i v vR

=


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Thus,

( )1

1

1

1 1 1 1 1

c

c

dv dvd

i c c v e cdt dt dt = = =

( )22

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= = dvcRv 2131

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 Rdt R dt

=

1 4 4

0 1 03 5 1 3 5 1

dv R R

v v v d dt R R c R R c = =


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In terms of the input and output voltages,

0 1 1 2

1 1 21 2 6 1 6

1 1 1

0

iv v d v d v d v

v c c R R R R R d t d t d t

+ + + + =

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 1 0iv v dv R R R c Rv 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 2

0 23 5 1 1 2 6 1 1 3 5 6 5 4

1 1 1 1 1

1 0i

dv dv d v R R c R R c

v R 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 0i

R dv R dv d vcv c

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

+ + + + + =

or2

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 dvvdt c c R R R c dt R c c R R R R R c dt

+ + + + + =


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

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

relationship between input and output 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.

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

R sR R c

V s V sR

s sc c R R R c R c c R R R

=

+ + + + +


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Input-Output Linearity:

A system Nis 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

[ ] [ ] [ ]22112211

uNcuNcucucN +=+

)(tfKxxBxM a=++ &&&

)(tfa

( )kf xx

( )af t


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Letfa(t) =fa1(t) +fa2(t) .

Letx1(t) be the solution whenfa(t) is replaced byfa1(t) andx2(t) be the solutionwhenfa(t) is replaced befa2(t) .

Then,

x(t) =x1(t) +x2(t) .

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

This time, however,x(t) x1(t) +x2(t) , i.e., the linearity property does not holdbecause the system is now nonlinear.

Time Invariance and Causality

Let Nrepresent a system and y() be the response of such system to the inputstimulus u(), i.e., y() = N[u()]. If for any real T, y( - T) = N[u( - T)], then thesystem is said to be time invariant . In other words, a system is time invariant if

delaying the input by Tseconds merely delays the response by Tseconds.

)(2 tfKxxBxM a=++ &&&


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

If the same system is also causal, then fort 0,

Example: Let a system be described by the ordinary, constant coefficientsdifferential 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 u h t d h u t d = =

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

1

)( tutyatyatyaty nnnn =++++


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Example: Consider the dynamic system described by

According to the previous definition, this is a distributed-parameter system.

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

i(t): Current flowing through circuity(t): Output variable (current flowing through inductor)

)1()()1()( 10 ++=+ tubtubtayty&


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Fort

t0, i(t0) = i0 , y(t) = i(t),

And the solution is given by

Hence, regardless of what vi(t) is fort < t0, all we need to know to predict thefuture of the output y(t) is the initial state i(t0) and the input voltage vi(t), 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

10i i

di di Rv Ri L i v

dt dt L L + + = = +

0

0

( )( )

0 0

1( ) ( ) ( ) ,

t Rt

t tLi

t

i t e v d e i t t t L

= +


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Example:

By KVL, fort t0,

After taking the time derivative of the last equation, we get

0 R L C + + =

0

0

1( ) ( ) 0

t

C

t

di Ri L i d t

dt C

+ + + =

01

2

2

=++ i

LCdt

di

L

R

dt

id


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To solve this homogeneous differential equation, we may proceed as follows:

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

then, fort t0,

C1 and C2 can be uniquely obtained as follows:

From the knowledge ofi(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=


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Using a state variable approach, letx1(t) = vc(t) andx2(t) = i(t), then fort 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

=

+== 444 3444 21

,)(

)(

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 t =& x x x


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Specifically, fort 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 (state) clear.

Linear State Models for Lumped-Parameter Systems

Consider the system described by the following block diagram

Mathematically, for t 0,

0( )0( ) ( )

A t t t 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)


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wherex(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 piecewise continuous

functions of time.

Definitions:

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

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

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

x(t0)=0.

The zero-state system response is the response y() to an input u() wheneverx(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().


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Example:

Now,

or

or

Let and , then fort t0, the state model is

0)(5.0)(3)( =++ tvtvtu &

)(2)(6)( tutvtv +=&

.)(,)(,)(2)(6)( 00 tytytutyty &&&& +=

)()(1 tytx = )()()(2 tvtytx == &

=

+

=

=

)()()(,)(

20

)()(

6010

)()()(

02

01

0

2

1

2

1

txtxtxtu

txtx

txtxtx

&&&

[ ]

=

)(

)(01)(

2

1

tx

txty

0.5m Kg=

( )input force u t

( )3riction force v t

( )output position y t

( )velocity v t


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Now, the state solution is given by

where fort t0= 0, the matrix exponential eAt is described by

1. Zero-input state response: u(t) = 0, t 0andx(0) 0.

+=t

t

tAttA dBuetxetx

0

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

=

t

tAt

e

ee

6

6

0

6

1

6

11

+=

==

20

6

206

10

20

10

6

6

61

61

061

611

)0()(

xe

xex

x

x

e

exetx

t

t

t

tAt


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2. Zero-input system response: u(t) = 0, t 0andx(0) 0.

3. Zero-state state response: x(0) = 0.

[ ]6

10 20 610 20

6

20

1 11 1

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

t

At

t

x e x y t Cx t Ce x x e x

e x

+

= = = = +

d

e

eBdedBuetx

t

t

tt

tA

t

s

tA

===

2

0

0 6

1

6

11

)()(0

)(6

)(6

0

)(

0

)(

( )

( )

=

=

=

t

t

t

t

t

t

t

t

t

e

et

de

de

d

e

e6

6

0

)(6

0

)(6

0)(6

)(6

13

1

118

1

3

1

2

3

1

3

1

23

1

3

1


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

Fort 0, the complete state response is then given by

and the complete system response by

[ ]( )

( )( )t

t

t

et

e

ettCxty 6

6

6

118

1

3

1

13

1

118

1

3

1

01)()(

=

==

( )

( )

+

+=+=

t

t

t

tt

tAAt

e

et

xe

xex

dBuexetx6

6

20

6

20

6

10

0

)(

13

1

1

18

1

3

1

6

1

6

1

)()0()(

[ ] ( )tt etxextxtx

txtCxty 620

6

101

2

11

18

1

3

1

6

1

6

1)(

)(

)(01)()( +

+==

==


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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.

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

Let us implicitly solve this equation, namely,

)(

111

)1(

1

)(

......

m

mmnn

nn

ubububyayayay +++=++++ +

&&

2

2

123322

2

13

3

)()()()()()()(dt

tudbdt

tdubtubtyadt

tdyadt

tydadt

tyd ++=+++

2

2

123322

2

13

3 )()()()(

)()()(

dt

tudb

dt

tdubtubtya

dt

tdya

dt

tyda

dt

tyd+++=

( ) ( ) ( ))()()()()()( 33221122

tyatubtyatubdtdtyatub

dtd ++=


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Integrating this last equation on 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 )()()()()()()( 33221122

( ) ( ) ( )

++= t ddyaubyaubtyatubdt

tdy

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

)(

332211

( ) ( ) ( )

dddyaubyaubyaubtyt

++= )()()()()()()( 332211


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Block diagram representation:

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

3

13123

23212

3331

xy

ubxaxx

ubxaxx

ubxax

=

+=

+=

+=

&

&

&

1x&

3x&

2x&


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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, assuming zero initial conditions, i.e.,

or

uBxAu

b

b

b

x

a

a

a

x 00

1

2

3

1

2

3

10

01

00

+=

+

=&

[ ] xCxy 0100 ==

++=+++

2

2

123322

2

13

3 )()()()(

)()()(

dt

tudb

dt

tdubtubtya

dt

tdya

dt

tyda

dt

tydL

( ) ( ) )()(2

12332

2

1

3

sUsbsbbsYasasas ++=+++


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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.

( )( ) 332211

3

3

2

2

1

1

32

2

1

3

2

123

1)()(

+++++=

+++++=

sasasasbsbsb

asasassbsbb

sUsY

( )1 2 31 2 3 1 2 31 2 3

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

Y s Y s Y s b s b s b sU s U s a s a s a sY s

= = + + + + +

3

3

2

2

1

11

1

)(

)(

+++= sasasasUsY

3

3

2

2

1

1)(

)( ++= sbsbsbsY

sY


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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 anintegrator in the time domain.

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

1s

1s

1s

1a

2a

3a

+

( )Y s

1 ( )s Y s 2 ( )s Y s 3 ( )s Y s

1

s

1

s

1

s

1b 2b 3b

+

+ +

( )Y s


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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 3&

3x 2x 1x1s

1

s

1

s

1a

2a

3a

+

1b 2b 3b

+

+ +

( )y t

2x&

1x&


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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 dynamicsystem, i.e.,

3 2 1

0 1 0 00 0 1 0

1

c c x x u A x B u

a a a

= + = +

&

[ ]3 2 1 cy b b b x C x= =

cc

cccc

xCy

uBxAx

=

+=&


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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 the 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

T

c

T

c

T

cT

c

Tc

Tc BCandCBAA

xBy

uCxAx ===

=+=

000

0

00 ,&

)(

1

)1(

211

)1(

1

)( ...... nnnnnnnn ububububyayayay ++++=++++ +

&&

)()()(

)()()(

tDutCxty

tButAxtx

+=

+=&


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wherex(t) Rn, u(t), y(t) R, A,B,Cand D are constant matrices of appropriate

dimensions.

Objective: Determine the initial state vectorx(0)=[x1(0) xn(0)]T from the initial

conditions and the input initial values

In the derivation of both observable and controllable canonical forms from an

ordinary linear differential equation with scalar constant coefficients we found that

D = 0, hence,

)0(),...,0(),0( )1( nyyy & )0(),...,0(),0( )1( nuuu &

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

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

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

)0()0(

)2()3(321)1(

2

+++++=

++=+==

+==

=

nnnnnn CBuCABuuBCABuCAxCAy

uCBCABuxCAuCBxCAxCy

CBuCAxxCy

Cxy

L&

M

&&&&&&&

&&


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In matrix form,

where Rnxn, TRnxn.To get a unique solutionx(0) for the last algebraic equation, it will be necessary

that the matrix be non-singular, i.e.,

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 statex(0) from input and output

measurements, the system must be observable, i.e., -1 must exist. In fact, iscalled the observability matrix.

)0()0(

)0(

)0(

)0()0(

0

00000

)0(

)0(

)0(

)0()0(

)0(

)1(321

2

)1(

TUx

u

u

uu

CBBCABCA

CBCAB

CB

x

CA

CA

CAC

y

y

yy

Y

nnnnn

+=

+

=

=

M

&&

&

LMOOMM

MOO

LLLL

MM

&&

&


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Method 2: 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

withx(0-) =x0, A Rnxn, B Rn, describe an nth order scalar differentialequation and

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

We know that for t 0--

( ) ( ) ( ), 0x t Ax t Bu t t = + &

)()()()( )1(110 ttttun

n

+++= L&

+=

t

tAAt dBuexetx0

)( )()0()(

( ) ( 1)

0 1 1

0

(0 ) ( ) ( ) ( )

t

At A t n

ne x e B d

= + + + + & L


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But, the ith term in the integral can be rewritten as

Therefore,x(t) is given by

where

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

00 0

( ) ( ) ( )

t t

t A t i A t i At A ie B d e B e e AB d

= + ( ) ( 2) 2 ( 2)

0

0

( ) 1

0 0

( ) ( )

( ) ( )

tt

A t i At A i

tt

A t i At A i At i

e AB e e A B d

e A B e e A B d e A B

= +

= + =

M

1

1

10)0()( ++++= n

nAtAtAtAt BAeABeBexetx L

{ }1(0 ) At ne x B AB A B = + %L

[ ]Tn 110~

= L


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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 ifand only if the inverse ofQ exists, namely,

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 x x B AB A B x Q + = + = + % %L

[ ])0()0(~ 1 + = xxQ

))(),(,()(

)()),(),(,()( 00

tutxtgty

xtxtutxtftx

=

==&


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

interval t [t0, tn] for a given initial statex(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.

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

Case 1 (forward Euler formula):

+=

t

t

duxftxtx

0

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


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In this case,

Over one time interval,

At time tk,

Case 2 (backward Euler formula):

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

10

0

fttfttfttduxf

t

t+++ L

11 )())(),(,(1

kkk

t

tfttduxf

k

k

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


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For the kth time interval,

Therefore,

and the approximate solution is given by

Case 3 (trapezoidal rule):

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

0

fttfttfttduxf

t

t

+++ L

kkk

t

t

fttduxf

k

k

)())(),(,( 11

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


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In this case the integral is approximately equal to

Therefore, the solution at time tk is approximately equal to

Example: Obtain an approximate solution of the following linearized pendulumstate model at equally spaced time instants, tk tk-1 = 0.5.

with initial conditions

2)()(

2)()(

2)()())(),(,( 109910

2112

1001

10

0

ffttffttffttduxf

t

t

++++++ L

[ ]))(),(,())(),(,()(2

1)()( 11111 kkkkkkkkkk tutxtftutxtftttxtx ++

=

)t(x

)t(x

)t(x

)t(x

1

2

2

1

4&

&

1

2

x (0)40

x (0)0

=


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Forward Euler Method:

Backward Euler Method:

Trapezoidal Rule Method:

+=

+

)t(x)t(x)t(x.)t(x

)t(x)t(x.

)t(x)t(x

)t(x)t(x

kk

kk

k

k

k

k

k

k

1112

1211

11

12

12

11

2

1

250

450

+=

+

)t(x)t(x

)t(x.)t(x

)t(x

)t(x.

)t(x

)t(x

)t(x

)t(x

kk

kk

k

k

k

k

k

k

112

211

1

2

12

11

2

1

2

50

450

( )( )

+

++=

++

)t(x)t(x)t(x

)t(x)t(x.)t(x)t(x)t(x

)t(x)t(x)tt(.

)t(x

)t(x

)t(x

)t(x

kkk

kkk

kk

kk

kk

k

k

k

k

11112

12211

111

122

1

12

11

2

1

25044

50


40/181

40

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, the system is discretized at time t = tk. This is illustrated in the

figure below (a sampled-data system).

t=tk t=tk

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

h(t)


41/181

41

Consider a linear, discrete-time system described 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)


42/181

42

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

In addition, ifu(kT) = 0 fork < 0, then

State representation of discrete-time dynamic systems:

Consider a linear discrete-time dynamic system described by the differenceequation

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

=

=k

n

nTunTkThkTy )()()(

= =k

n

nTunTkThkTy0

)()()(

)()1()2()1()( 121 kyakyankyankyanky nn +++++++++ L

)()1()( 11 mkubkubkub mm +++++= + L


43/181

43

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 one at a time:

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( 331

2211

1 kyakubqkyakubkyakubkykyq ++++=+=+

[ ] [ ]{ })()()()()()()1()2( 331221111 kyakubqkyakubqkyakubkykyq ++=+=+

[ ] [ ] [ ])()()()()()()()1( 331

22

1

11

11 kyakubqkyakubqkyakubqkykyq ++==+


44/181

44

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 =


45/181

45

In matrix form,

In general, a discrete-time system can be represented by (assuming T = 1)

wherex(k) Rn, u(k) Rm, y(k) Rr,A, B, Cand D are constants matrices of

appropriate dimensions.

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

measurements.

)()(

10

01

00

)1(

1

2

3

1

2

3

ku

b

b

b

kx

a

a

a

kx

+

=+

[ ] )(100)( kxky =

)()()(

)()()1(

kDukCxky

kBukAxkx

+=+=+


46/181

46

Iteratively,

In matrix form,

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

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

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

321

2

++++++++=+

+++++=+++=+

+++=+++=++=

nkDunkCBukBuCAkBuCAkxCAnky

kDukCBukCABukxCAkDukCxky

kDukCBukCAxkDukCxkykDukCxky

nnn L

M

)()(

)1(

)2(

)1(

)(

0

00

000

)(

)1(

)2(

)1(

)(

)(

321

2

kTukx

nku

ku

ku

ku

DCBBCABCA

DCBCAB

DCB

D

kx

CA

CA

CA

C

nky

ky

ky

ky

ky

nnn

+=

+

+

+

+

=

+

+

+

=

M

L

OOMM

M

L

L

MM


47/181

47

Ifhas full rank, thenx(k) = -#[y(k)-Tu(k)], i.e., if the system is observable then

we can reconstruct the state at time kusing input, output measurements up to timek+n-1, where -# is the pseudoinverse of.

Solution of the discrete-time state equation

Iteratively,

and

=

+=

++++=+=+++=+=

++=+=

+=

1

0

1

234

23

2

)()0()(

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

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

)0()0()1(

k

l

lkk lBuAxAkx

BuABuBuABuAxABuAxxBuABuBuAxABuAxx

BuABuxABuAxx

BuAxx

M

)()()0()(1

0

1 kDulBuACxCAkyk

l

lkk +

+=

=


48/181

48

From the last equation,

where Q is the controllability matrix

and

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 sate at time k, the following

relationship must be satisfied

where is the pseudo inverse ofQ

In other words, the system must be controllable.

)()()( kQUkxAnkx n =+

][ 1BAABBQ n= L

+

+

=)(

)2(

)1(

)(

ku

nku

nku

kU M

#( ) ( ) ( )nU k Q x k n A x k = + #Q


49/181

49

Linearization of Nonlinear Systems

Consider the following scalar nonlinear 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 analyticon N, i.e., it is infinitely differentiable on N, then forx such thatx0+xN, we get the following Taylor series expansion

For small x,

g()x(t) y(t)

0x

0 0

22

0 02

1( ) ( )

2! x x x x

dg d g g x x g x x x

dx dx= =+ = + + +L

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

dg dg y g x x g x x y x

dx dx= == + + = +


50/181

50

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 xdg y y xdx

= =

0 y y m x =

0x x

dgm

dx==

0 ln 1 ( )s

iv v g i

i

= + =

0 0

0

0

1 1

1i i o i d

s s

s

vdgm v r

idi i i i

i

=

= = = =

+ +

0v


51/181

51

The linearized model is then given by

Using the parameters = 2, 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

Volts


52/181

52

If, on the other hand, y=g(x1,x2,,xn), then if g() is analytic on the setN={xn:a


53/181

53

In other words,

Assume that we know xn(t). Perturb the state and the input by taking x(t0)

= x0 + x0 and u(t) = un(t0) + u(t).

We want to find the solution to

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

of xn(t) and un(t), we get

where

0 0 0( ) ( ( ), ( ), ) , ( ) , for n n n nx t f x t u t t x t x t t = = &

0 0 0 0( ) ( ( ) ( ), ( ) ( ), ) , ( ) ( ) , for n n nx t f x t x t u t u t t x t x t x t t = + + = + &

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

( )

( , , ) higher order terms( )nn

n n n n

x xu u

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

x t

f x u t u t==

+ + =

+ + ( , , )

f ff x u t

x u

=


54/181

54

Let and

Then

Therefore,

Example: Let

Clearly,

Now,

( ( ), ( ), )( )

n

n

x xu u

f x t u t t a t

x==

( ( ), ( ), )

( )n

n

x xu u

f x t u t t b t

u==

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

dx tdx t d x t f 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

+ =

2 0 0( ) ( ) ( ) , ( )x t x t u t x t x= + =&

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

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

n

n

x x nu u

f x t u t t a t x t

x==

=


55/181

55

( ( ), ( ), )( ) 1

n

n

x xu u

f x t u t t b t

u==

=

and

So,

Let

Finally,

0 0

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

d x tx t x t u t x t x

dt

+ =

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

2

1 2 1 22

( ) ( ) , (1) 1

( ) 1 1and ( ) , 1

( )( )

n n n

nn

nn

dx t x t xdt

dx tdt c t c c c x t t

x t t x t

= =

= + = + = =

0

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

d x tx t u t x x t

dt t

+ =


56/181

56

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( ) ( ( ), ( ), ) , ( ) , ( ) , ( ) for n mt t t t t t t t t = = &x f x u x x x u

0 0 0( ) ( ( ), ( ), ) , ( ) , ( ) , ( ) for n m

i it t t t t t t t t = = &x f x u x x x u

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

( )

( , , ) higher order terms( )nn

i n n i n n

i

f t t t t t f t t t

t

f t t==

+ + =

+ + x xu u

x x u u x u

x

x u u

( , , ) ( , , ) ( , , )i i i f t f t f t = x u x u x u x u


57/181

57

and

or

1 2

( , , ) i i iin

f f f f t

x x x

=

Lx x u

1 2

( , , ) i i iim

f f f f t

u u u

=

Lu 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 t d

dt

x x

f t f t

= == =

= == =

= == =

= +

M M

M M

x x x u x x u u u u

x x x u x x u u u u

x x x u x x u u u u

x u x u

x u x u

x u x u

1

2

m

uu

u

M


58/181

58

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 nn n

n n

n n

n n

n n

p n

p p

g t g t

y x

y g t x g t t

y x

g t g t

= == =

= == =

= =

= =

= = +

M MM M

x x x u x x u u u u

x x x u x x u u u u

x x x u x x

u u u u

x u x u

x u x uy

x u x u

1

2

m

u

u

u

M


59/181

59

Example: Suppose we have 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 axisu1(t) is the thrust in the radial direction

u2

(t) is the thrust in the tangential direction

m is the mass of the orbiting body

m

u1(t)u2(t)

(t)r(t)

orbit


60/181

60

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 K r t u t

dtdt r t

= +

2

22

( ) 2 ( ) ( ) 1( )

( ) ( )

d t d t dr t u t

r t dt dt r t dt

= +

1 2 3 4

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

dr t d t x 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

=

= +

&

&


61/181

61

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 t x t u t

x t x t

=

= +

&

&

2

21 4 12

1

4

2 4 2

1 1

( ( ), ( ), ) .

2

x

K x x u

xt t t

x

x x u

x x

+

= +

f x u

0 0( ) 0 , 0T

n t R t t = x

( )d


62/181

62

where rn(t) =R and

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

0 3

( ).n

d t K

dt R

= =

[ ]1( , , ) 0 1 0 0nn

f t =

=

= x x x u u

x u

2 22 4 1 4 0 03

1

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

n

n

n

n

K f t x x x R

x =

=

= = + = =

x x x u u

x xx u

u u

[ ]3 ( , , ) 0 0 0 1nn

f t ==

= x x x u u

x u

02 4 2 4 24 2

1 11

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

n

n

n

x x u x xf t x x Rx

==

= = = = x x x

u u

x xx uu u


63/181

63

Likewise,

In state form,

[ ]1( , , ) 0 0nn

f t ==

=u x xu u

x u

[ ]2 ( , , ) 1 0nn

f t ==

=u x xu u

x u

[ ]3 ( , , ) 0 0nn

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 ux u

20 0

0

0 1 0 0 0 0

1 03 0 0 2

0 00 0 0 1

100 2 0 0

R

RR

= +

x x u


64/181

64

Existence of solution of differential equations

Consider the following unforced, possibly nonlinear dynamic system described by

where x(t0) = x0, x(t) Rn and f(,):RxRn Rn.

Then the state trajectory ( ; t0,x0) is a solution to over the time interval [a,b] ifand only if(t0; t0, x0) = x0and for all t [a,b].

Def. Let D RxRn be a connected, closed, bounded set. Then the function f(t,x)

satisfies a local Lipschitz condition at t0on D with respect to (t0, x) D if there

exists a finite constant k (t0,x1), (t0,x2) D,

where kis the Lipschitz constant.

( ) ( , ( ))t t t=& x f x

0 0 0 0( ; , ) ( , ( ; , ))t t x t t t =& x

0 1 0 2 1 2 22( , ) ( , )t t k f x f x x x

Gl b l E i t d U i


65/181

65

Global Existence and Uniqueness:

Assumptions:1. SR+ [0, ) contains at most a finite number of points per unit interval.

2. For each xRn, f(t, x) is continuous at tS.

3. For each ti S, f(t, x) has finite left and right hand limits at t = ti.

4. f(,) : R+xRn Rn satisfies the global Lipschitz condition, i.e., there exists a

piecewise continuous function k():R+R+ such that

for all tR+ and all x1, x2Rn.

Theorem: Suppose that assumptions (1) (4) hold. Then for each t0R+ and x0

Rn there exists a unique continuous function (; t0, x0) : R+ Rn such that(a) and (b) ( t0; t0,x0) =x0, tR+ and tS.By uniqueness we mean that if1 and 2 satisfy conditions (a) and (b) then1(t; t0, x0) = 2(t; t0, x0) tR+.

1 2 1 22 2

( , ) ( , ) ( )t t k t f x f x x x

0 0 0 0( ; , ) ( , ( ; , ))t t t t t =& x f x


66/181

66

Consider now the unforced, linear, time-varying system

,x(0) =x0

whereA(t) Rnxn and its components aij(t) are piecewise continuous.

Theorem: IfA() is piecewise continuous, then for each initial conditionx(0), asolution (; 0, x0) to the equation exists and is unique.

Proof: Define the sets Dj= [j 1, j) forj = 1, 2, .

Then, ,

SinceA() is piecewise continuous on R+, it must be piecewise continuous for

each t Dj, j = 1, 2, . Therefore, for arbitraryx1, x2 Rn ,

U

=

+=1j

j RD

221,221221)())(()()( xxtAxxtAxtAxtA

jD=

( ) ( ) ( )t A t t =&x x


67/181

67

for all Dj, j = 1, 2,

Let , tR+, then , where k(t) is a

piecewise continuous function fortR+. Therefore, for eachx(0) Rn, a unique

solution (; 0, x0) to exists.

Example: Verify that the differential equation

with initial conditionx(0) = [1 0]T has a unique solution.

First of all, all the entries ofA(t) are continuous functions of time.

But,

which implies that there exists a unique solution since k(t) = 1 + 2tis continuous t

R+.

Consider now the linear time-varying unforced dynamic system described by

jD

tAtk,

)()(

= 1 2 1 22 2

( ) ( ) ( )t x A t x k t x x

)(1

12)( tx

t

ttx

=&

221221)21()()()(21)( xxtxtAxtAtkttA +=+=

Th L t A(t) Rnxn b i i ti Th th t f ll l ti f


68/181

68

Theorem: LetA(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 ifi = 0, i = 1, 2, , n; and i() be thesolutions of with initial conditions i(t0) = i, i = 1, 2, , n.

Suppose that the is, i = 1, 2, , n are linearly dependent, then i R, i = 1, 2, ,

n, such that tR+.

At t = t0R+,

the is are linearly dependent, which is an outright contradiction of the

hypothesis that the is are linearly independent. Therefore, the is are linearly

independent for all tR+.

0

~2211 =+++ nn L

0~

)()()( 2211 =+++ ttt nn L


69/181

69

Let be any solution of and (t0) = . Since the is 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 .

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.

=

=n

i

ii

1

==

n

i

ii t1

0 )(

====

===

n

i

ii

n

i

ii

n

i

ii

n

i

ii ttAttAttdt

d

1111

)()()()()()( &

E l C id h d i l d ib d b


70/181

70

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

00)( tx

ttx =

&

=

0

11

=

1

02

= 21

2

11

)(t

t

=

1

0)(2 t


71/181

71

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

where the is, i= 1, 2, , n are the basis solutions, 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,

[ ]0 0( ; , ) 0 0 1 0 0T

i i it t = = L L

)()()( tMtAtM =&

[ ]),;(),;(),;(),( 02021010 nn ttttttttdt

d &L&&=

),;()(),;( 00 iiii tttAtt =&

Therefore


72/181

72

Therefore,

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

3. (t, t0) is uniquely defined.

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

then (t, t0) is also uniquely determined byA(t).

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

Proof: At t= t0, (t0, t0)x0 =Ix0 = x0.

We already know that .

Therefore,

In other words, , satisfies the differential equation.

[ ]),;()(),;()(),;()(),( 02021010 nn tttAtttAtttAtt L& =

[ ] ),()(),;(),;(),;()( 00202101 tttAtttttttA nn == L

),()(),( 00 tttAtt =&

0000 ),()(),( xtttAxtt =&

00 ),( xtt

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


73/181

73

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

Then,

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

We can show that

)()()()(00

tAdAdAtA

t

t

t

t

=

0

( )

0( , )

t

t

d

t t e

=

)(10

1)(

2

txe

tx

t

=&

( )

+

== 0

2

0

22

0

0

)(2

1)()()()(

0

00 tt

etteetttAdAdAtA

tttt

t

t

t

H


74/181

74

Hence,

and

0

0 0 0

2 3( )

0

1 1( , ) ( ) ( ) ( )2! 3!

t

t

A d t t t

t t tt t e I A d A d A d

= = + + + + L

( ) ( )( )

( )( )( )

( ) +

+

+=

!20

!2

1

!20

2

1

10

012

0

22

0

2

0

0

22

0

0

0

tt

eetttt

tteett

tttt

( ) ( ) ( )( )

L+

!30

!323

!33

0

22

2

0

3

0 0

tt

eetttt tt

)(3

0

2

000110)(

!3

1)(

!2

1)(1),(

ttetttttttt

=++= L

( ) ( ) ( ) 2222222 1311


75/181

75

Finally,

Hence, the state transition matrix is given by

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

infinity, therefore, the system is unstable.

( ) ( ) ( ) L+= 202202222012 )(!3

1

2

3)(

!2

1

2

1),( 000 tteetteeeett

tttttt

( )0),(),( 011022tt

etttt==

( ) ( )( )

=

++

0

000

02

1

),(3

0

tt

tttttt

e

eeett

Theorem:A(t) and commute ift

dA )(


76/181

76

( )

1. A() is constant

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

3. , i() :R R and the Mis are constant matrices such that

MiMj = MjMi i, j.

Proof:

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

(2) IfA(t) = (t)M, then

t0

)(

=

=k

i

ii MttA1

)()(

2

00

22 )()(

00

AttttAIdAAdAt

t

t

t

===

2

0

2

)(00

AttAIdAAd

t

t

t

t ==

But


77/181

77

But,

(3) If then

But,

2

0000

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

t

t

t

t

t

t

t

t ===

=

=k

i

ii MttA1

)()(

= == = ==k

i

t

tjij

k

ji

k

i

t

tjj

k

jii MMdtMdMt 1 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 = = = =

= =


78/181

or, tttt


79/181

79

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 tR+

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 andx(t) = M(t)vbe 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) everywhereA(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

),(

L

)()()( tztAtz =&

,)()()( tMtAtM =&

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


80/181

80

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

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

Theorem (Semigroup Property): For all t1, t0and t, we have (t,t0) = (t,t1) (t1,t0).

Proof: We know 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 conditionsx(t0

)

andx(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).

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


81/181

81

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

for all t, t0R+.

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

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

Fort0= 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

withx(t0) =x0 .

Theorem: The solution to the state equation is given by

( )=

t

t

dAtr

ett 0)(

0 )],(det[

)()()()()( tutBtxtAtx +=&

)()()()()( tutDtxtCty +=

0

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

t

t

x t t t x t B u d = +

where


82/181

82

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

t

t

duBt

0

)()(),(

000000

0

0

)()(),(),()( xduBtxtttx

t

t

=+=

+=

+

t

t

t

t

duBtt

xttduBtxttdt

d

00

)()(),(),()()(),(),( 0000 &

++=

t

t

duBtt

tutBttxtttA

0

)()(),()()(),(),()(00

+=

=

t

t

t

t

t

dtft

tfdtft

00

),(),(),(

Hence


83/181

83

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) = Cand

D(t) = D, whereA, B, Cand D are constant matrices.

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

[ ] ++=

t

t

duBttAtutBxtttAdt

d

0

)()(),()()()(),()(00

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

0

00 tutBtxtAtutBduBtxtttAt

t

+=+

+=

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

0

00 tutDduBttCxtttCtyt

t

++=

)(

000)0,(),(

ttAetttt

==

Proof: Since A and commute, we have thatt

Ad


84/181

84

Proof: SinceA 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

Ad

0

( )[ ] )()0,(exp),( 0000 0 ttttttAett

t

t

Ad

===

=

+=+=t

t

AAtttA

t

t

tAttA dBueexedBuexetx

0

0

0

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

0

)(

)()()(0

0

0

)(

tDudBueCexCety

t

t

AAtttA

++=

)(1

0

1 000 )()()(ttAAtAtAtAt eeeeetMtM

===

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


85/181

85

Def. IfA is an nxn matrix, C, eC and the equation

Ae= e, e0

is satisfied, then is called an eigenvalue ofA and e is called an eigenvector ofA

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

polynomial, i.e.,

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

the non negative real number

The right eigenvectoreiofA associated with the eigenvalue isatisfies the

equationAei

= i

ei

, whereas the left eigenvectorwi

Cn ofA 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 eand e*, respectively.

)())(()det()( 2111

1 nnnnn

A aaaAI =++++== LL

{ }AA ii = :max)(

Example: Find the right eigenvectors of the matrix


86/181

86

p g g

The characteristic polynomial of A is given by

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

Now,

or

For1

= -1, we get

=

010

011

552

A

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

0~~)(~~ == iiiii eAIeeA

0

~~

10011

552

=

+

+

i

i

i

i

e

551


87/181

87

or

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

For2 = -1 j2,

0~~

110

001

551

1 =

e

13121312

1111

131211

000

055

eeeeee

eee

====

=++

[ ]Te 110~1 =

0~~

2110

021

5521

2

=

e

j

j

j

or055)21( eeej =++


88/181

88

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

and

Finally,

Theorem: LetA be an nxn constant matrix. ThenA is diagonalizable if and only if

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

A.

Proof: IfA 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 +=

[ ]Tjjee 12124)*~(~23

+==

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


89/181

89

Now, TAT T [Ae1Ae2 Aen] T [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 ofA.

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

AT = TD.

Let T= [t1 t2 tn], thenAT= [At1At2 Atn] = [t1d11 t2d22 tndnn] = TD Ati= diiti, which implies

that the ith column ofTis an eigenvector ofA associated with the eigenvalue dii.

Since Tis nonsingular, there are n linearly independent eigenvectors.

Now, ifA is diagonalizable, then eAt= TeDtT-1 because

( ) ( ) L++++== 3312211!3

1

!2

11tTDTtTDTtTDTIee tTDTAt

( )( ) ( )( )( )1 1 1 2 1 1 1 31 12! 3! I TDT t TDT TDT t TDT TDT TDT t = + + + +L

L++++= 3132121!3

1

!2

1tTTDtTTDtTDTI

2 2 3 3 1 11 12! 3!

DtT I Dt D t D t T Te T = + + + + = L

552


90/181

90

Example: For the given matrix, compute eAt.

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

The inverse ofTis

=

010

011

552

A

++

=

111

21211

24240

jj

jj

T

++=

2121121211

1022

8

11

jjjjT

Therefore,


91/181

91

Therefore,

which implies that

Finally, eAt= TeDtT-1.

+

=2100

0210

001

j

jD

=

+

tj

tj

t

Dt

e

e

e

e

)21(

)21(

00

00

00

Proposition: SupposeD is a block diagonal matrix with square blocksDi, i = 1, 2, , n, i.e.,


92/181

92

Then,

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

Let

=

nD

D

D

D

L

MOMM

L

00

0

00

2

1

=

tD

tD

tD

Dt

ne

e

e

e

L

MOM

M

L

00

0

00

2

1

[ ]

==

011

211

240

}Im{}Re{ 221 eeeT

then, 511


93/181

93

and

which implies that

But,

Finally,

=

220

111

511

4

11T

=

==

2

11

0

0

120

210

001

D

DATTD

=

tD

tDDt

e

ee

2

1

0

0

==

tttteeee ttDttD

2cos2sin2sin2cosand 21

( )11

2cos2sin02sin2cos0

001

==T

ttttTeTeTe

tDtAt

Def. The impulse response matrix of a linear, lumped, time-varying system is a matrix map


94/181

94

H(,) :RxR Rrxm given byH(t, ) = [h1(t, ), , hm(t, )], where each column hi(t, )

represents the response of the system to the impulsive input u(t) = (t - ) i, i Rm, i = [0 0 1 0 0]T (1 occurs as the ith component), is the time of application of theinput and tis the observation time.

Recall that ifH(t, ) 0 for any given t, and t < , then the system is noncausal. On the

other hand, ifH(t, ) = 0 fort < , then the system is causal.

In block diagram form,

Hence,

Consider the linear, time-varying system with state model

= dtHt )(),()( uy

0)(,)()()()()( =+= xuxx ttBttAt&

)()()()()( ttDttCt uxy +=

Theorem: The impulse response matrix for the above system is given by


95/181

95

Proof: The response of the above time-varying system to an input u is given by

But,x(-) = 0, implies that

Let u(t) = (t - ) i, i=[0 0 1 0 0], where the 1 appears at the ith location, then fort

Now,

t. Thus, fortt


96/181

96

andy(t) = 0, fort < .

Ifu(t) = (t - ) i, then

Hence,H(t, ) = C(t)(t, )B() +D(t)(t - ), t

= 0 t <

IfA(),B(), C() andD() are constant matrices, then

Consider again the time-invariant system state model

=t

dqqqtHt )(),()( uy

deg(A

()), we can solve forh() directly from , i.e.,

Let h() 0 + 1+ + n-1n-1, then if the eigenvalues ofA are distinct, the

js can be computed from the n linear equations

f(i) = q(i) A(i) + h(i) = h(i), i = 1, , n.

Example: Calculatef(A) =A10 + 3A with

=21

10A

The characteristic polynomial of A is A() = 2 + 2+ 1 = (+1)2 1 = 2 = -1.


102/181

102

Let h() = 0 + 1. Then withf() = 10 + 3, we get

f(1) =f(-1) = (-1)10 + 3(-1) = -2 = 0 - 1.

But, f(2) =f(1) ! For the repeated eigenvalue case, the solution procedure is modified as

follows:

Let

Since deg(h()) n 1, the coefficientsj, j = 0, 1, , n 1, can now be obtained from

the following set of n equations:

f(l)(i) = h(l)(i), i = 0,1, , ni - 1; i = 1, 2, , m.

Going back to the previous example, n1 = 2 and

==

==m

i

i

m

i

n

iA nni

11

)()(

==+=+===

73)1(10310)( 91

9)1(

1 f 1

)1(

1

)(

==

h

We must solve the equations 0 - 1 = -2 and 1 = -7. This implies that 0 = -9.

M


103/181

103

Moreover,

f(A) =A10

+3A = h(A) = 0I + 1A = -9I 7A =

Suppose again the eigenvalues ofA are distinct, then

where

and

Consequently, fort0

57

79

n

n

A s

R

s

R

s

R

s

sRAsI +++==

L2

2

1

11

)(

)()(

=

=++++=

n

i

inn

nn

A sasasass

1

1

1

1 )()( L

=

)(

)()(lim s

sRsR

A

i

si

i

{ } Atn

i

t

i

n

i i

i eeRs

RLAsIt i ==

==

==

11

111 )()(

L

Example: Consider the system . Obtain (t) using the partial fractionexpansion method with the system matrix A given by

)()( tAt xx =&


104/181

104

expansion method with the system matrixA given by

Now,

and the characteristic polynomial ofA is

The residue matrices are found as follow:

=

3210A

+=

+

=

s

ssR

s

sAsI

2

13)(

32

1

))(()2)(1(23)det()( 212

=++=++== ssssssAsIsA

=

+

+=

+=

12

12

2

13

2

1

)(

)()1(

limlim 111 s

s

ss

sRsR

sAs

=

+

+=

+=

22

11

2

13

1

1

)(

)()2( limlim

222

s

s

ss

sRsR

sAs

The inverse ofsI-A is equal to


105/181

105

and the state transition matrix of the system is

or

Suppose now that the matrixA contains repeated eigenvalues, then the adjoint

matrixR(s) and A(s) have common factors. Consider, for example, the matrix

2

22

11

1

12

12

)( 1

+

++

= ss

AsI

tt eet 2

22

11

12

12)(

+

=

++

=

tttt

tttt

eeee

eeeet

22

22

222

2)(

=1

1

1

0000

01

A

Then,11

2

1 010)(

sss


106/181

106

We know from the Cayley-Hamilton theorem that ifA() is the characteristic

polynomial of the nxn matrixA, then A(A) = 0.

Def. A polynomialp() such thatp(A) = 0 is called an annihilating polynomial of

the matrixA.

Def. The monic polynomial of least degree which annihilates the matrixA is called

the minimal polynomial ofA and is denoted by A().

Supposeg() is a polynomial of arbitrary degree, thenp() =g()A() is also anannihilating polynomial ofA.

Theorem: For every nxn matrixA, the minimal polynomial A() divides the

characteristic polynomial A(). Moreover, A() = 0 if and only ifis aneigenvalue ofA, so that every root ofA() = 0 is a root ofA() = 0.

2

1

1

1

3

1

21

2

1

1

)(

00

00

)(

)(00

0)(0

)(

)()(

=

== s

s

s

s

s

s

s

sRAsI

A

Proof: IfA() annihilatesA and ifA() is a monic polynomial of minimum

degree that annihilatesA, then deg (A()) deg(A()).


107/181

107

degree that annihilates , then deg (A()) deg(A()).

By the Euclidean algorithm there exists polynomials h() and r() such thatA() = A()h()+ r() and deg(r()) < deg(A())

But, 0 = A(A) = A(A)h(A) + r(A) = 0h(A) + r(A) r(A) = 0.

However, deg(r()) < deg(A()), and by definition A() is the polynomial of

minimum degree such that A(A) = 0 r() 0 A() divides A().

This result implies that every root ofA() = 0 is a root ofA() = 0 and hence

every root ofA() = 0 is an eigenvalue ofA.

If (A) and ifx0 is its corresponding eigenvector, thenAx= x and

0 = A(A)x= A()x A() = 0.

IfA has repeated eigenvalues 1, 2, , , i j, ij, i, j = 1, 2, , , then for

m1+m2++mn the minimal polynomial A() has the structure

mmm

A )()()()(21

21 = L

Theorem: LetA Rnxn, then


108/181

108

where

In this particular case, fort0 the state transition matrix is given by

This is true because

= =

===

1 1

1

)()(

)(

)(

)()(

i

m

jj

i

j

i

AA

i

s

R

s

sR

s

sRAsI

=

1)()()!(

1lim AsIsds

d

jmR i

i

i

i

m

ijm

jm

si

j

i

= =

=

1 1

1

)!1(

)(i

m

j

tj

j

i

i

ie

j

tRt

= = = =

=

=

=

1 1 1 1

11

11

)!1()(

1

}){()(

i

m

j i

m

j

tj

j

ij

i

j

i

i i

iej

tR

sR

AsIt

L

L

Example: Consider a dynamic system with theA matrix


109/181

109

Then

Using the Faddeev-Leverrier algorithm, we get

N1 =I

N2 =N1A + a1I=

N3 = N2A + a2I=

=

452

100

010

A

2,5,4

254)2()1()det()(

321

232

===

+++=++==

aaa

sssssAsIsA

052

140

014

020

002

145

So, R(s) =N1s2 + N2s + N3 and

)(R


110/181

110

In this case, the minimal polynomial is the same as the characteristic polynomial,

i.e., A(s) = A(s) =s3 + 4s2 +5s + 2.

The inverse of sI A is

where the residue matrices are

)2()1(

)()(

2

1

++=

ss

sRAsI

2)1(1)(

1

2

2

2

1

1

11

++

++

+=

s

R

s

R

s

RAsI

3232

2

1

12

1

1

1 232

)()1(!1

1limlim NNIs

NsNIs

ds

dAsIs

ds

dR

ss

+=

+++

=

+=

=

384

252

120

++132

2 NsNIs


111/181

111

Therefore, fort0, the state transition matrix is given by

Consider now the following block diagonalA matrix

Then, its characteristic polynomial is the same as that of the last example, i.e.,

=+=

+

++=

132

132

2

3232

1

2

1 lim NNIs

NsNIsR

s

=+=

+

++=

484

242

121

24

)1(322

32

2

2

1

2 lim NNIs

NsNIsR

s

tttAt

eRteReRte

21

2

2

1

1

1)(

++==

=200

010

011

A

)2()1()()( 2 ++== ssss AA

In this case,

tt tee 0


112/181

112

But,

provided that a nonsingularTcan be constructed

Suppose that

where

=

t

ttA

e

ee2

00

00

11 == TTeeTATA tAAt

=

PJ

J

J

J

LL

MOM

M

0

0

000

2

1

iixnn

i

i

i

i

i JJ

= ,

01

10001

LLLOM

MOOM

ML

then

tJe 0001


113/181

113

where

In this case, matrix is said to be in the Jordan canonical form.

LetA have eigenvalues 1, 2, , p each with multiplicity ni, i = 1, ,p,

n1 + n2++ np = n. SupposeA hasp independent eigenvectors

associated with the eigenvalues 1, 2, , p.

=tJ

tJ

Jt

Pe

e

e

LL

MOM

M

0

0 2

=

t

t

t

i

ntt

tJ

i

i

i

i

ii

i

e

e

en

ttee

e

00

0

)!1(

1

LMOM

M

L

11

2

1

1 ...,,, peee

Then the set of eigenvectors generated by{ }11 11 1,..., ,..., ,..., pnn

p pe e e e


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114

i = 1, 2, ,p forms a basis.

Theorem: The generalized eigenvectors ofA associated

with the eigenvalues {1,, p} each with multiplicity ni, i = 1, 2, ,p, are

linearly independent.

Example: Let . We already know that (A) = {-1, -1, -2}.

{ 11 11 1,..., ,..., ,..., pnn

p pe e e e

1

23

12

1

)(

)(

)(

0)(

=

=

==

ii n

i

n

ii

iii

iii

ii

IA

IA

IA

IA

ee

ee

ee

e

M

=

452

100

010

A

Clearly, n1= 2 and n2 = 1. Moreover,


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115

Clearly, are two linearly independent eigenvectors.

Furthermore,

and implies that the three eigenvectors are linearly independent

Let us construct the similarity transformation Tas follows:

==

=

1

1

1

0

352

110

011

)( 111

13

1

12

1

11

1

11 ee

e

e

e

IA

==

=

4

21

0

252

120012

)( 121

23

1

22

1

211

22 ee

e

ee

IA

1 11 2ande e

==

1

0

1

)( 21

1

1

2

12

eeeIA

[ ] 1det 122111 =eee

[ ]2

1

1

1

1

2 eee=T

In other words,

012

111

T


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116

The new system matrix, in Jordan canonical form is given by

The matrix exponential of the equivalent system is

=

114

012T

JATTA =

==

100

110002 1

==

t

tt

t

JTtA

e

tee

e

ee

00

0

002

The matrix exponential of the original system is therefore given by


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117

Discrete-Time Systems:

Consider the linear time-invariant discrete-time system

Taking the one-sided Z transform, yields

The state transition matrix is given by (k) =Ak=Z-1{z(zI-A)-1}.

The transfer function matrix of a discrete-time linear dynamic system is defined by

2

1

1 1 1 0 0 1 2 1

2 1 0 0 2 5 2

4 1 1 0 0 2 3 1

t

At J t t t

t

e

e Te T e te

e

= =

)0(,)()()1( 0 xxuxx =+=+ kBkAk

)()()( kDkCk uxy +=

1 1

0

1 10

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

z z zI A zI A B z

z zC zI A C zI A B D z

= +

= + +

X x U

y x U

DBAzICzH += 1)()(

On the other hand, the unit sample response matrix is then given by

h(k) Z 1{H( )}


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h(k) =Z-1{H(z)}.

Def. MRnxn is symmetric ifMT= M.

Theorem: The eigenvalues ofM = MTRnxn are real.

Proof: Let be an eigenvalue ofMand v its eigenvector. Then Mv = v. Now, if

v*is the complex conjugate transpose ofv,

v*Mv = v*(v) = (v*v)

But, v*(Mv) and v*v are real must be real, all eigenvalues ofMmust be

real. This can be verified from the fact that

(v*Mv)* = (Mv)*v = v*M*v = v*MTv = v*Mv.

Theorem: Let M = MTRnxn. Then there exists an orthogonal matrix

Q = [q1 q2 ...qn ] such that M = QDQTorD = QTMQ, whereD is a diagonal matrix

constructed from the eigenvalues ofM, qi , i = 1, 2 , n is the normalized version

of eigenvectorvi associated with the eigenvalue i ofM, i = 1, 2 , n.

Theorem: A matrix M= MTRnxn is positive definite (positive semi-definite) if and

onl if


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119

only if

a) Every eigenvalue ofMis positive (zero or positive).

b) All leading principal minors ofMare positive (all principal minors ofMare zero

or positive).

c) There exists an nxn nonsingular matrixN(an nxn singular matrixNor an mxn,

m < n matrixN) such that M = NTN.

We denote positive definiteness (semi-definiteness) by M> 0 ifxTMx> 0 (M 0 if

xTMx0 ) x 0.

Example: Consider the following 2x2 matrix M:

then MT= M, {1, 2} = {4, 2}. The corresponding eigenvectors are:

= 31

13M

= 1

1,

1

1},{ 21 vv

The normalized eigenvectors are given by

11


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120

Therefore, the Q andD matrices are given by

Singular value decomposition (SVD)

LetHRmxn

and define MHT

HRnxn

. Then M = MT

0. Let rbe the totalnumber of positive eigenvalues ofM, then we may arrange them such that

Letp = min{m, n}, then the set {1

2

r> 0 = r+1 = = p} is calledthe singular values ofH, where and r = rank(H).

Example: Let a rectangular matrixHbe given by

=

21

21

21

21

21,},{ qq

=

21

21

2

1

2

1

Q

= 20

04D

1 2 10r r n + > = = =L L

= 4221

12

H

i i =

Then

==89

HHM T


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121

Now, det(I M) = 2 - 30+ 125 eigenvalues (M) = {25, 5} and the singular

values ofHare the square root of the eigenvalues ofM, i.e., {5, 5}.

Example: LetHnow be described by

Then

and

218

HHM

=

120

004H

==

120

240

0016

HHM T

)5)(16(

120

240

0016

det)det( =

= MI

which implies that the set of eigenvalues ofMis {16, 5, 0} singular values ofH

are {4 5} since min{m n} = min{2 3} = 2 Also rank(H) = 2


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122

are {4,5}, since min{m, n} = min{2, 3} = 2. Also, rank(H) = 2.

Theorem: LetH Rmxn, thenH = RSQTwithRTR = RRT=Im, QTQ = QQT= In,

and SRmxn with the singular values ofHon its main diagonal and such that

QTHTHQ = D = STSwithD a diagonal matrix with the squared singular values of

Hon its main diagonal.

Example: LetHbe given by

then the eigenvalues ofM = HTH {16, 5, 0} give rise to the normalized

eigenvectors

=

120

004H

[ ]T

q 001~

1 =[ ]Tq

51

52

2 0~ =

[ ]Tq5

25

13 0

~ =

Thus,

[ ]

==5

15

2321 0

001~~~ qqqQ


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123

Stability of Dynamic SystemsDynamic system stability is a very important property. It enables the tracking of

desired signals or the suppression of undesired signals. System stability is

described either in terms of input-output stability or in terms of internal stability.

[ ]

5251

55321

0

0qqqQ

=

=

==

00

00

050

004

0

0

001

120

004

2

1

52

51

51

52

HQS

2

1

2

22

3

4 0 16 0 0 0 04 0 0

0 5 0 5 0 0 00 5 00 0 0 0 0 0 0

T

S S

= = =

Input-output Stability:


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124

A single input, single output (SISO) linear, time-invariant (LTI), continuous time

dynamic system is bounded-input, bounded-output (BIBO) stable if and only if its

impulse response h(t) is absolutely integrable, i.e.,

where Mis a real constant.

Proof: Let the input u(t) be bounded, i.e., |u(t)| k1 < , t 0. Then

y(t) is bounded.

Suppose h(t) is not absolutely integrable. Then for a causal, linear time-invariant

system, with u(t) = k1 > 0 and h(t) > 0, t 0, with nondecreasing envelope.


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Fort 0 , this implies that

as t ,y(t) is not bounded even when u(t) is bounded. Thus, h(t) must be absolutely

integrable.

Theorem: A SISO LTI, continuous-time dynamic system is BIBO stable if and

only if every pole of its transfer functionH(s) lies on the left-half of thes-plane.

Proof: LetH(s) be a proper rational function ofs, then if every pole located at

s = - pi, pi > 0, has multiplicity ni, such that

1

0

( ) ( )t

y t k h d =

0

( )

t

h d

nnm

i

i ==1

we get

+=m n

j

iji

ps

ksH

)()(


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126

and the impulse response is given by,

t 0, which is absolutely integrable.

Consider a multiple-input, multiple-output (MIMO), LTI, continuous-time dynamic

system described by the impulse response matrixH(t) = [hij(t)]. Such a system isBIBO stable if and only if i, j,

Alternatively, a MIMO, LTI, continuous-time dynamic system described by the

proper rational transfer function matrixH(s) =L{H(t)} = [Hij(s)] is BIBO stable if

and only if every pole ofHij(s) is located on the left half of thes-plane.

= = +i j

j

ips1 1 )(

{ } = =

= =

=

+==

m

i

n

j

tpjijm

i

n

jj

i

ij

i

i

i

etj

k

psLksHLth

1 1

1

1 1

11

)!1()(

1)()(

0

( )ij ijh t dt K

= <

As we already know, the solution of the state equation is given byt


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127

Moreover, in the s-domain we get

Suppose that the input is identically zero and the initial state is nonzero, i.e.,

and

then

and

LetHin(s) (sI - A)-1be the internal transfer function matrix of some continuous

time LTI dynamic system. Then, with u(t) = 0 and for some

+= tAAt duBexetx0

)( )(~)0(~)(~

)(~

)()0(~)()(~ 11

sUBAsIxAsIsX

+=

0~

)(~

0~

)(~ == sUtu or

0~)0(~ x

)0(~)()(~ 1xAsIsX =

)0(~)(~ xetx tA=


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and

1. The unforced system is marginally stable if and only if the poles ofHin(s) (or the

eigenvalues ofA) have either zero or negative real parts, and those with zero real

parts are simple roots of the minimal polynomial ofA

2. The unforced system is asymptotically stable if and only if all the poles ofHin(s)

(all eigenvalues ofA) lie strictly on the left half of thes-plane.

Asymptotic stability also implies that

Observation: These two concepts deal with internal stability only.

Example: Let an unforced dynamic system be described by

0~

)(~lim =

txt

)(~

2

2)(~

21

21

txtx

=&

then

+++

)()()( 4

5

21

45

21

ssss

s

H


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129

the poles ofHin(s) are located ats = 0 ands = -5/4 system is marginallystable.

Suppose now that the input stimulus is nonzero and that the initial condition is

zero, i.e., and

Then

and

Furthermore,

+++

=

)(2

)(2

)()()(

45

45

44

sss

ss

sHin

0~

)0(~ =x .0~

)(~ tu

)(~

)()(~ 1 sUBAsIsX =

=t

tA duBetx0

)( )(~)(~

[ ] )(~)()(~)()(~ 1 sUsHsUDBAsICsY =+=

Clearly, every pole ofH(s) is an eigenvalue ofA if every eigenvalue ofA has a

negative real part then all poles of H(s) lie on the left half of the s plane the


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negative real part, then all poles ofH(s) lie on the left-half of t

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ECE 550 Lecture Notes 1