notes on state variables for students

8/7/2019 Notes on State Variables for Students

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***************** Begin Lecture Notes on State Variables (1) *****************

Dr. Shoane

General Solution of the State Equations

A solution to the state equation, Eq can be found in any one of the texts on LinearControl Theory listed in the References. The solution is expressed in terms of an n n matrix ( t), called the transition matrix of the system.

+=t

duBtxttx0

)()()0()()(

The transition matrix depends solely on the system matrix A. One method for finding

( t) uses a definition based on an infinite series.

Note that

...!3!2

1!

32

0

++++==

=

xxx

n

xe

n

n

X(6.27a)

We define the state transition matix as

Atet =)(

(6.27b)

Hence,

...!3

)(

!2

)(

!

)()(

32

0

++++===

=

tAtAtAI

k

tAet

k

k

At

(6.27c)or

...)(!3

1)(

!2

1)()( 32 ++++= tAtAtAIt

It can further be shown that the terms of A with power n can be written as a sum ofterms up to power n-1, where n is the linear dimension of the A square matrix. Thus, thehighest term that needs to be written is n-1. Hence, we can write

1

110)(...)()()(

+++==n

n

tAAtAtItet (6.28a)

************************************************************************

Example A - Initial Condition Only (Zero Input)

The solution to this unforced network is given by


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)0(xex At= (6.28b)

Now we would like to find the state transition matrix. Since A is a 2x2 (i.e., n = 2)

matrix, we have

AIAAeAfk

k

k

n

k

k

k

tA

1

1

0 0

1

0

)( +====

=

= (6.28c)

The solution to the characteristic equation 0= IA is

As an illustration of how the transition matrix is used to solve the linear state

equations, suppose the system matrix for an autonomous system ( 0u = ) is

0 1

2 3A

=

0-3-2-

1=

=

IA (6.28d)

0232 =++ (6.28e)

Hence 1 = -1 and 2 = -2.

The Cayley-Hamilton theorem states that every matrix A is a zero of its own

characteristic equation. That is, Eq (6.28c) is also satisfied for the eigenvalues 1 and 2. Therefore, we have

1

110)(...)()()(

+++==n

n

tAAtAtItet (6.28f)

2102

1101 an d

+=+=

te

te (6.28g)

or

10102a n d

2 ==

te

te (6.28h)

Subtracting the two equations in (6.28h) gives


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te

te

2

1

= 6.28i)

Substitute Eq (6.28i) into the first equation in Eq (6.28h) gives

te

te

22

0

=(6.28j)

Substitute Eqs (6.28h) and (6.28i) into Eq (6.28c) gives

0

0

0

233-

222-

2

22

22

10

tetetete

tete

tete

tete

AI +

+

+=+ (6.28k)


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Therefore

22

222

2

2

2

=

+

+

tetetete

tete

tete( t ) (6.28l)

Another way to obtain the state transition matrix is via the transformed domain.

It can be shown that in the transform domain, the general solution is given by

)()()0()()( sBUsxssX += (6.28m)

where )(s is called the resolvent matrix, which is also the Laplace transform of the

state transition matrix. It can be shown that the resolvent matrix is given by

1)()(

= AsIs (6.28n)

So that for our example (Example A),


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1

1

3s2

1-

3-2-

10

s0

0)()(

1

+== =

ssAs Is

(6.28o)


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++++

+++++

=++

+

=++

+

=

21( s

s

2 )1 )( s

2-

)2)1(

1)2) (1(

3

23

s2-13

2)3(

s2-13

2

ssss

s

ss

s

ss

s

(6.28p)

Using partial fraction expansion, we get


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++

+++

+

++

++

+

+=

2s

2

1s

1-

2

2

1

2

2

1-

1s

1

2

1

1

2

ss

sss(6.28q)

The inverse Laplace transform of the resolvent matrix, )(s , is the state transitionmatrix. Hence,

22222

2

22

=

++

tetetete

tetetete( t ) (6.28r)

which is the same as that obtained above in Eq. (6.28k).

Then we can solve for x(t), with u(t)=0 (zero input).

+=t

duBtxttx0

)()()0()()( (6.27 again)

Example B - Input Only (Zero Initial Conditions)

2x 1x

u1


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The equations for the model are given by

21

222

1211

xxy

uaxx

uxaxx

+=

+=

++=

(6.28A)

.11a10

01

0

1H e ,

B,

- aA (6.28B)

The characteristic equation can be obtained from

0--0

1=

=

a

aIA (6.28C)

or

y

u2

-a -a


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0222=++ aa (6.28D)

two.oftymultipliciwith,2

442 22

2,1a

aaa=

=

(6.28E)

1

110)(...)()()(

+++==n

n

tAAtAtItet (6.28f again)

)(10

+==

tef (6.28F)

and because of the multiplicity of two, we take the derivative of f( ) with respect to

)(1

=

=t

ted

df

(6.28G)

Substituting = -a into the above equation give

atte=1

(6.28H)

Substituting (6.28H) back to (6.28F), with = -a, gives

a ta ta ta t

a t eea t ee

+== 00

(6.28I)

Hence, the state transition matrix is given by


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+

+

+=+==

a

aa

a ta t

a ta t

ta

tt ea

t eae

t eaeAIAft

00

0)()(

10

=

a

aa t

e

t ee

0

(6.28J)

The response of a system, with zero initial conditions and no direct connection between

input and output (i.e., D = 0), is given by the following convolution integral

dutHtyt

)()()(0

= (6.28K)

where H(t) is the impulse response matrix is given by

BtCtH )()( = (6.28L)

or


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=

= +

aa

a t

a ta t

ee

e

t eetH )(11001

0

11)(

(6.28M)

For inputs u1 = u(t) and u2 = )(tuet

, the response is given by

[ ]

[ ]

)()1(

112

)1(1

)()1)(()(

)(

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

22

0

))1()(

0

)()(

0

tueaaa

teaa

a

dueetue

due

uetedxtty

att

taatta

ttata

t

+

+

+=

++=

+==

(6.28N)

-----------------------------------------------------------------------------------------------------------

-

Now consider the transformed domain. The solution with zero initial condition isgiven by (see Eq. 6.28m)

)()()( sBUssX = (6.28O)

where the resolvent matrix given by

1)()(

= AsIs (6.28P)

Hence, we have


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1

1

as0

1-

a-0

1a

s0

0)()(

1

+

+== = ass

As Is(6.28Q)


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+

++=++

+

+

=++

+

+

=

)(

1

0

(1

(1

(

as01

(

as01

2

a

asas

a )a )s

as

a )a ) (s

as

(6.28R)

Taking the Inverse Laplace transform gives


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=

a

aa t

e

t ee

0

( t )(6.28T)

which is the same as that obtained using the Cayley-Hamilton approach.

But we also want to obtain the output Y(s), which is given by


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[ ] [ ]

+++

+++

+=

++

+++

+

=

+

+

++

=

==

a )) ( s( s

a )( ssa )s ( s

a )) ( s( s

a( ss

a )s ( s

s

s

a )( s

a )( s

a )( s

B U ( s )sCH ( s ) U ( sY ( s )

1

1

)1(

11

1

1

)1(

11

11

1

1

1

10

11

11

)(

2

22

(6.28U)


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+

++

+

+

+

+

+

+=

asa

sa

as

a

as

a

s

a

asa

sa 1

1

11

1)1(

1

)(

)1(1

1

)1(111

Y(s)

obtainweexpansion,fractionpartialApplying

2

2

2

(6.28V)

(6.28X))()1(

1

1

2

)1(

1)(

givestransformLaplaceInverse

(6.28W))(

1)1(

11)1(

11

21

1)1(s

1a1Y(s)

)(

1

)1(

11

)1(

1)1(2

1

1

)1(s

1

a

1Y(s)

)(

1

)1(

11

)1(

1

)1(

11

1

1

)1(

1

)1(

1

s

1

a

1Y(s)

)1(

1)1()1(

ds

dwhere

22

222

222

222

2

21

tueaaa

te

a

a

aty

asaasaaasaa

asaasaa

aa

sa

a

asaasaaasaa

ass

att

asas

++

+=

+ + ++ +=

+

+

++

+

=

+

++

++

+

+

=

=+=+

which is the same as the solution above (Eq. 6.28N) using the Cayley-Hamilton theorem.

-----------------------------------------------------------------------------------------------------------

-

Summary of Equations

Time Domain

+=t

dutHxtCty0

)()()0()()( (6.28Y)

Zero Input Zero State

where BtCtH )()( = Impulse Response Function

and1

110)(...)()()(

+++== nn

tAAtAtItet

State Transition Matrix

Transformed Domain


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MatrixResolvent()(and

MatrixFunctionTransferewher

StateZeroInputZero

)0()(

1sI-A)s

(s)BCH(s)

H(s)U(s)xsCY(s)

-=

=

+=

(6.28Z)

***************** End Lecture Notes on State Variables (1) *****************

notes on state variables for students

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