MEM1833 Topic 3 State Space Realizations HANDOUT

8/6/2019 MEM1833 Topic 3 State Space Realizations HANDOUT

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Chapter 3hapter 3hapter 3hapter 3StateState--Space RealizationsSpace Realizations

Assoc. Prof. Dr. Mohamad Noh Ahmad.

Faculty of Electrical Engineering

Universiti Teknologi Malaysia

u a

[email protected] / [email protected] 012-7379299


2/27

.

:

)()()( ttt BuAxx +=& )()()( tutt BxAx +=&

uy =)()( tt Pxx =


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&v - :

)()()(

ttt DuCxy += . .

xPxPxx-1

==&

PBuPAxxPx +== & PBuxPAP +=

DuxCPDuCxy +=+= 1

DDCPCPBBPAPA ==== ,,, --

(**))()()(;)()()( uDxCyuBxAx tttttt +=+=&

Let

(*) and (**) are said to be equivalent to each other

and the procedure from (*) to (**) is called an

equivalent transformation.


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The feedforward matrix D between the input and output has nothing

to do with the state space and is not affected by the equivalenttransformation.

The characteristic equation for (*) is:

For (**), we have

AI = )( = APPPP =

APPPIP11

)(

= PAIP )(1

=

AIPP = 1 AI =

and hence the same set of eigenvalues.


5/27

Recall: othereachsimilar toareand- APAPA =

They have same eigenvalues, same stability perf.

Similar transfer functions!!

1 DBAICG += ss DBAICG += 1ssand

)()( ss GG =To verify,

DBAICG += 1)()( ss

DPBPAPPPCP += 111-1

)(s1111

)( = XYZXYZDPBPAIPCP += -1-1-1 ))-(( s

DPBPAIPCP += 111 )(s

)()1

s(s GDBAIC =+=


6/27

.

x. . :

From the circuit (via observation):

11 xx =

=

2

1

2

1

11

01

x

x

x

x

1)( 212 = xxx)()( tt xPx =Or

=

1

1

1 01 xx = 101 x

22

11 xx

2

11 x


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-

Two state equations are said to be zero-state equivalent

if they have the same transfer matrix or

BAICDBAICD 11 )()( +=+ ss

Note that:

L+++= 23211

)( AAIAI ssss,

LL ++++=++++ 3221321 ssssss BACBACBCDBCACABCBD 2


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-

Theorem 3.1

are zero-state equivalent or have the same transfermatrix iff and

, ,, ,,,

DD =

,...2,1,0; == mmm BACBCA

- In order for two state equations to be equivalent, they

.- This is, however, not the case for zero-state equivalence.


9/27

.

Consider:

)(5.0)( = tuty )()( = txtx&

0=== B

5.0;5.0;0;1

0.5u(t))(5.0)(

====

+=

DCBA

txty

5.0== DD

Note that:

0== BACBCA mm

The two s stems are zero-state e uivalent.~ Theorem 3.1


10/27

0123

Consider: )()()(

1

0)(

134

012)( tuttutt bAxxx +=

+

=&

It can be shown that:

2

YES!,,


11/27

en:

1700

Thus the representation of A w.r.t. the basis is:}{2bAAb,B,

=

510

1501A


12/27

12)3(

1700 ny spec a

)1(34

0)1(2)(

23 ==

==

I

=

510

1501A

000 43214 Companion-form matrices:

(-1)

0100

0010;

100

010

1

2

3

(-1) -

Transpose

100

010;

1000

0100

3

2

1

All have similar

characteristic

00041234

43

2

2

3

1

4

)( ++++=


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Q. What is "realization"?

For a iven , find a corres ondin state-)()( ss NG =space equation

&

s

)()()( ttt DuCxy +=

Implicitly implies LTI systems

Shall start with multi-variable systems and willsometimes specialize to single-variable systems


14/27

.

Q. If (s) is realizable, how many possible

realizations? Infinite ~ in view of equivalent transformations and the

possibility of adding un-controllable or un-observable

componentsQ. Which one is the "good" realization?

- A good realization is the one with the minimal order

~ Irreducible realization Will be discussed in Topic 6


15/27

.

Q. Under what condition is (s) realizable by an

LTI system?-Recall that the transfer function of the dynamic equation is

DBAICDBAICG +=+= 11 sss

Theorem 3.2(s) is realizable by a dynamic equation iff it is a

proper rational function (order of numerator order ofdenominator)

In fact, the part contributed by C(sI - A)-1B iss r c y proper


16/27

.

To determine the realization of , decompose it

as:

)( sG

)()()( ss spGGG +=

[ ]BAICAIBAICG ).()(:)(1

==

sAdjssssp

rr

rr ssssd ++++=

1

1

1)( LLet

ssp.e. eas common enom na or o a en r es o .


17/27

.

Then the realization of is given by

IIIII L pprprpp 121

)(sG

u0

0

x00I0

000I

x

+

= L

L

& p

p

~Block companion form

00I00 L p

r r [ ] uGxNNNNy )(121 += rrL

(q rp)where Ip is thep p unit matrix and every 0 is ap p zero matrix.


18/27

.

Proof:

Define, Z

Z

12

1

Z

r

M i .

Then,Simplifying, BZAI = )(s (*)BAZZ += s

0

I

Z

Z

000I

IIII

Z

Z

L

Lp

p

prprpp

s

s

2

1121

2

1

+

=

0I00

MM

L

MMMM

L

M

ps 33

rpr


19/27

.

Using the shifting property of the companion form of A, we obtain

12312 ,,, === rrsss ZZZZZZ Lwhich implies

1,,

1,

1ZZZZZZ === L

Substituting these into the 1st block of equation (*) yields

s IZZZZ += L

sss

pr

r

ss

IZ +

+++=

112

1

L

or

r sds IZ ==

++++ 1

21

)( L

sss


20/27

.

Thus we have,

rr

ss IZIZIZ 121

===

L

Then,

prppsdsdsd )()()(

)()(1

+

GBAIC s

[ ] )()(

2

2

1

1 ++++=

GNNN rrr

sssd L

)()( += GG ssp

sG=


21/27

.3104 s

,

)2(

1

)2)(12(

1212

)(

2

+

+

++

++=

s

s

ss

sssGConsider a proper rational matrix

Determine a realization of this transfer matrix.Solution:

312

3104sDecompose into a strictly proper rational matrix)(sG

+

+

++

+

=

2)2(

1

)2)(12(

100

s

s

ss

ss)()( sspGG +=

+

+

++

=

2)2(

1

)2)(12(

1)(

s

s

ss

sssG

265.4)2)(5.0()(232 +++=++= ssssssd

The monic least common denominator of is)(sspG


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.

+++

+++

+++=

)5.0)(1()2(5.0

)5.0)(2(3)2(6

265.4

12

23

sss

sss

sss

spGThus

+++

++++=

)5.05.1()2(5.0

.

)(

12 sss

ssss

sd

22

+++=

5.05.115.0

.

)( 2 ssssd

3245.72436

+

+

=5.015.15.010)(

sssd


23/27

.

Given,

,11

212)(

+

++

=s

ss

s

sG )(

sG)2()2)(12(

2 +++ sss

Realizationthe realization is,

10

01

20605.40

020605.4

+

=

2

1

00

00

000100

000010 uxx&

DC

=S

023245.72436 u

00001000 { }DC,B,A,=

+

=2005.015.15.010 u

xy


24/27

.

Consider again the proper rational matrix in Ex. 3.3.

3104s

+

+

++

++=

2)2(

1

)2)(12(

1212

)(

s

s

ss

sssG

The 1st column is

+

=1

12)( 1

sscG

++

=1

)2)(12( ss

++

=

1

2522 ss

ss

++ )2)(12( ss ++ )2)(12( ss ++ 2522 ss


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. . .

yields the following realization for the 1st

column of :

s ng : n1= 4 2 20;0 0 1 ;d1= 2 5 2 ; a,b,c,d =tf2ss n1,d1

)(sG

11111110

1

01

15.2uu

+

=+= xbxAx&

111111105.00 uuc +

=+= xdxCy

Similarly, the function tf2ss can generate the following realization for the2nd column of :)( sG

2222222

063

001uu

+

=+= xbxAx&

2222222

011

uuc

+

=+= xdxCy


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. . .

ese two rea zat ons can e com ne as

0bx0Ax11111

+

= u

&

&

[ ] [ ]uddxCCyyy

xx

212121

22222

+=+= cc

u

or

00

01

0001

0015.2

10

01

20605.40

020605.4

uxx

+

=

00

10

0100

4400&

+

=2

1

00

00

000010

000001

u

uxx&

(4x4) (4x2) (6x6) (4x2)uxy

+

=

00

02

115.00

63126

00001000

1023245.72436 uC D 2005.015.15.010 ux x

(2x6) (2x2)


27/27

.

Special Case: p = 1 (For simplicity, let r= 4 and q =2.)

Consider a 21 proper rational matrix:

+++

+++

+++++

=

2423

2

22

3

21

1413

2

12

3

11

43

2

2

3

1

4

2

1 1)(

sss

sss

ssssd

dsG

Then, the realization is 14321 Note: The

u

+

=

0

0

0010

0001xx&

controllable-canonical-form

ud

+

= 114131211 xy

rea za on can eread out from the

224232221

.

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MEM1833 Topic 3 State Space Realizations HANDOUT