ee 529 circuits and systems analysis mustafa kemal uyguroğlu lecture 9

EE 529 Circuits and Systems Analysis

Mustafa Kemal Uyguroğlu

Lecture 9

EEastern astern MMediterranean editerranean UUniversityniversity

State vectorState vector

a listing of state variables in vector form

(t)x

(t)x

(t)x

(t)x

(t)x

n

1n

2

1


State equationsState equations

(t)uB(t)xA

(t)x

(t)x

(t)x

(t)x

n

2

1

(t)uD(t)xC(t)y

System dynamics

Measurement

Read-out mapOutput vector

Inpu

t vec

tor

Stat

e ve

ctor


x:n-vector (state vector)

u:p-vector (input vector)

y:m-vector (output vector)

A:nxn

B:nxp

C:mxn

D:mxp

nn

nn

mm

mm

nn

pp

nn

pp

System matrix

Input (distribution) matrix

Output matrix

Direct-transmission matrix


Solution of state eq’nsSolution of state eq’ns

Consists of:

Free response Forced sol’n&

(Homogenous sol’n) (particular sol’n)


Homogenous solutionHomogenous solution

Homogenous equation

xAx has the solution

0xΦ(t)(t)x

State transition matrix X(0)


State transition matrixState transition matrix

An nxn matrix (t), satisfying

( ) A ( ), (0)

where is identity matrix.

0

t t

n n

I

I

x 0 = x 0


Determination of Determination of (t):(t): transform methodtransform method

Laplace transform of the differential equation:


Determination of Determination of (t):(t): transform methodtransform method

11 tt s e

AI AL


Determination of Determination of (t):(t): time-domain solutiontime-domain solution

( ) . ( )t a t

Scalar case

( ) att e

where

0

12211

k

kkk

at taatate !! .........)(

0

x ax

x t t x


Determination of (t): time-domain solution

( ) . ( )t t AFor vector case, by analogy

( ) tt e A

where

21 12! !

0

1 ( ) .........t k kk

k

e At At A t

A

Can be verified by substitution.


Φ(t2-t0)

Properties of TMProperties of TM

(0)=I

-1(t)= (-t)

Ф(t2-t1)Φ(t1-t0)= Φ(t2-t0)

[Φ(t)]k= Φ(kt)

Φ(t)Φ(-t)

Φ(t1-t0) Φ(t2-t1)

t0 t1 t2

Φ(t) Φ(t) Φ(t) Φ(t) Φ(t) Φ(t)Φ(kt)


General solutionGeneral solutionScalar case


General solutionGeneral solutionVector case


General solutionGeneral solution: transform method: transform method

uBxAx L{ }

(s)uB(s)xA(0)x(s)xs ˆˆˆ

1 1ˆ ˆx(s) (sI A) x(0) (sI A) Bu(s)


Inverse Laplace transform yields:

(t)uB*Φ(t)(0)xΦ(t)(t)x

t

0

)d(uBe(0)xe(t)x )-A(tAt


t

t0

) )d(uBe)(txe(t)x )-A(t0

t-A(t 0

For initial time at t=t0


The outputThe output

y(t)=Cx(t)+Du(t)

(t)uD(t)duBeC)(txCe(t)yt

t

)-A(t

0

)t-A(t 0 0

Zero-inputresponse Zero-state response


ExampleExample

Obtain the state transition matrix (t) of the following system. Obtain also the inverse of the state transition matrix -1(t) .

1

2

0 1

2 3

x x

x x

1

2

For this system

0 1

2 3

A

the state transition matrix (t) is given by

1 1( ) [( ) ]tt e s LA I A

since0 0 1 1

0 2 3 2 3

s ss

s s

I A =


ExampleExample

2 21

2 2

2( )

2 2 2

t t t tt

t t t t

e e e et e

e e e e

A

1

2 1 1 11 2 1 21

2 2 1 21 2 1 2

3 11( )

2( 1)( 2)

( ) s s s s

s s s s

ss

ss s

s

I A

I A

The inverse (sI-A) is given by

Hence

2 2

2 2

2( )

2 2 2

t t t t

t t t t

e e e et

e e e e

Noting that -1(t)= (-t), we obtain the inverse of transition matrix as:


Exercise 1Exercise 1

Find x1(t) , x2(t)

The initial condition


Exercise 1 (Solution)Exercise 1 (Solution)

x = (t)x(0)

11

2 21

2

2 2

( )

4 11 4 11 4 5 4 5

5 4 5 54 5

4 5 4 5

t sI A

ss s s s s ssI A sI As s ss s

s s s s

L


Example 2Example 2


Exercise 2Exercise 2

Find x1(t) , x2(t)

The initial condition

Input is Unit Step


Exercise 2 (Solution)Exercise 2 (Solution)


Matrix Exponential eMatrix Exponential eAtAt


P=

1 1 1

1 2

12

22 2

11

21 1

n

n

n nnn

The transformation where

1,2,…,n are distinct eigenvalues of A. This transformation will transform P-1AP into the diagonal matrix

n

2

1

0

0

=APP 1

ˆx = Px


Example 3Example 3


Method 2:

1 1

1

2

1 1

2

[( ) ]

0 0 1 1

0 0 2 0 2

1 1 1 1 1 12 1 21 2 2

( )02 11

0022

11 1

2[( ) ]1

02

t

t

t

t

e s

s ss

s s

s s s s s s ss

ss s

ss

ee s

e

L

L

A

A

I A

I A =

I A

I A


Matrix Exponential eMatrix Exponential eAtAt


Example 4Example 4


Laplace TransformLaplace Transform

ee 529 circuits and systems analysis mustafa kemal uyguroğlu lecture 9

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