state space many methods

8/22/2019 State Space Many Methods

1/20

Lecture: 6 Part-2 Many methods of

state space representations

By Mr S M Wijewardana

PhD student QMUL

04-05-2013

Learning Objectives:

1. Introduction to Phase-Variable Method

2. Controller Canonical form3. Observer Canonical form


2/20

The Method of selecting state variables:

1

.

1

2

.

2

3

.

3

xxdt

d

xxdt

d

xxdt

d

nn xx

xx

xx

1

.

32

.

21

.

This is the phase variable form of representation. In this method each subsequent state

variable is defined to be the derivative of the previous state variable. This known as thePhase-variable canonical form(PVCF) or controller canonical form(CCF)


3/20

n

nn

n

nn

x

x

x

y

u

bx

x

x

x

aaaax

x

x

x

2

1

1

2

1

1210

.

3

.2

.1

.

001

.

0

.

10.0.0

.....

0100

0010

Or we can write:

DuCxyBuAxx

.


4/20

Find the state-space representation in phase-variable form for the transfer function:

964

3

)(

)(22

ssssR

sC

Solution:

Cross multiplication yields:

)(3)()964( 23 sRsCsss

Taking the Inverse Laplace transform of, both sides of the equation and assuming zero

initial conditions:

rcccc 3964......

Selecting state variables as successive derivatives we can get:

...

2

..

3

...

2

.

3

.

1

.

2

.

2

111

cxxcxx

cxxcx

xxorcx


5/20

Now we can write:

rxxxx

xx

xx

3964 1233

.

32

.

21

.

Output equation:

1xcy

3

2

1

3

2

1

3

.2

.1

.

001

3

0

0

469

100

010

x

x

x

y

r

x

x

x

x

x

x


6/20

R(s) sX3(s) X3(s) sX2(s) X2(s) sX1(s) X1(s) Y(s)

equationOutputxcy

equationsInput

rxxxx

xx

xx

1

3213

.

32

.

21

.

3469

-43

-9

1/s

11/s 1/s

1

-6


7/20

Cascade Form: We have introduced the cascade form in the previous lecture.

Consider the equation:

1)3)(2)(1(

3

)(

)(

eq

ssssR

sC

The above transfer function can be separated into individual blocks as shown below:

3

R(s)

)1(

1

s )2(

1

s )3(

1

s

C(s)

3)1(

1

s )2(

1

s )3(

1

s

C(s)R(s)

Cascade method

The output of each first order block has been labelled as a state variable from end

block. (Note here this is not the phase variable form of representation)

X3(s) X2(s)X1(s)


8/20

The signal flow graph for each first order block can be drawn by transforming each

block in to an equivalent differential equation.

Let us take the block of , and assume:

)3(

1

s )3(

1

)(

)(

1

1

ssR

sC

From cross multiplication:

(s+3) C1(s) = R1(s)

Taking Inverse Laplace transform, we have:

sC1(s) = R1(s) -3C1(s)

2)()(3)( 111 eqtrtctcdt

d

R1(s) 1 sC1(s) 1/s C(s)

-3State diagram


9/20

R(s) 3 1/s X3(s) 1 1/s X2(s) 1 1/s X1(s) 1 C(s)

-1 -2 -3

Now we can derive the state equation from state diagram:

1)(

:

:

3

2

3

33

.

322

.

211

.

xtcy

equationOutput

inspectionBy

rxx

xxx

xxx

Hence the state space representation:

xy

rxx

001

3

0

0

100

120

013.


10/20

Parallel Form:

In this method we take the partial fractions first and expand the transfer function.

)4(

4

)3(

6

)1(

2

)4)(2)(1(

12

)(

)()4)(2)(1(

12

)(

)(

sssssssR

sCssssR

sC

We can write the same equation as shown below:

3.)4(

4).(

)3(

6).(

)1(

2).()(

eq

s

sR

s

sR

s

sRsC

Eq-3 can be considered as the sum of individual first-order subsystems.Recognizing C(s) as the sum of three terms, each term can be consideredas a subsystem with R(s) as the input. According to this idea the SFG canbe drawn as shown in the next slide.


11/20

X2(s)1/s

-1

-3R(s) -6 1/s X2(s) 1 C(s)

1/sX1(s)

-4

4

1

1321

33

.

22

.

11

.

)(

44

63

2

xxxtcy

rxx

rxx

rxx

xy

rxx

111

4

6

2

400

030

001.

The advantage of this representation is that each equation is a first-order differential

equation. Therefore we can solve these equations independently. Hence this type of

equations are said to be decoupled


12/20

Controller Canonical Form(CCF):

CCF is another way of representing phase variables in state-space method. This method is

nothing else other than representing the phase-variable form in reverse order.

Consider a transfer function:

22187

23

)(

)()(

23

2

sss

ss

sR

sCsG

3

2

1

3

2

1

3

.2

.1

.

132

1

0

0

71822

100

010

x

x

x

y

r

x

x

x

x

x

x

State space representation in

phase-variable form is:

b d


13/20

1

2

3

2

3

1

. 2

.3

.

132

10

0

171822100

010

x

x

x

y

rxx

x

xx

x

If we re-number the phase variables in reverse order:

Now we can arrange equations and so as the matrices to give the ascending order:

3

2

1

3

2

1

3.

2

.1

.

231

0

0

1

010

001

22187

x

x

x

y

r

x

x

x

x

x

x

Next slide for

more details:


14/20

RUsUUsUsand

UsUUscylet

sssR

Uand

ssUC

sss

ss

R

U

U

C

sss

ss

sR

sC

22187

23

22187

1

23

22187

23.

22187

23

)(

)(

23

2

23

2

23

2

23

2

Now we have to define the state variables in reverse order:

...1

.

1

..

2

.

2

.

3

.

33

ux

xux

xux

uxx

321

3211

.

...

1

.

1

..

2

.

2

.

3

.

33

23

22187

xxxcy

rxxxx

ux

xux

xux

uxx

3

2

1

3

2

1

3

.2

.1

.

231

0

0

1

010

001

22187

x

x

x

y

r

x

x

x

x

x

x


15/20

1 1/s 1/s 1/s 2

Now to construct the state diagram is also very easy:

321

3211

.

...1

.

1

..

2

.

2

.

3

.

33

23

22187

xxxcy

rxxxx

ux

xux

xux

uxx

R(s) X1(s) X2(s) X3(s)

C(s)

-22

-18

-7

1

3

State diagram in s-domain for Controller Canonical Form


16/20

Observer Canonical Form:

Consider the same example:22187

23

)(

)()(

23

2

sss

ss

sR

sCsG

By dividing both Numerator and denominator by the highest power of s :

)(22187

1)(231

221871

231

)(

)(

3232

32

32

sCsss

sRsss

sss

sss

sR

sC

Rearranging the equation:

)](22)(2[1)](18)(3[1)](7)([1)(32

sCsRs

sCsRs

sCsRs

sC

Consider only: C(s)=1/s[R(s)-7C(s)]

SFG for this term is:

----eq4


17/20

C(s)=1/s[R(s)-7C(s)] :SFG for this term is:

X1(s)

C(s)

R(s)1

1/s1

-7

)])(22)(2[1

)](18)(3[1

)](7)(([1

)(2

sCsRs

sCsRs

sCsRs

sC

According to the eq4 we can see that first terms under 1/s is integrated

once and the second terms 1/s2[3R(s) -18C(s)] is integrated twice so on.

Hence we have to add the rest of terms to the State diagram as shown in

the next slide:

R(s)


18/20

1/s X3(s) 1 1/s X2(s) 1 1/s X1(s) 1

R(s)

23

1

C(s)

-22

-18-7

By looking at the state diagram

now we can deduce the state

equations:

By observing left to the right

1

211

.

312

.

13

.

7

318

222

xy

rxxx

rxxx

rxx

1


19/20

3

2

1

3

2

1

3

.2

.1

.

001

2

3

1

0022

1018

017

x

x

x

y

r

x

x

x

x

x

xThe state- space

representation: Observer canonical Form

3

2

1

3

2

1

3

.2

.1

.

231

0

0

1

010

001

22187

x

x

x

y

r

x

x

x

x

x

x

Controller canonical form

1

2

3

3

2

1

3

2

1

3

.2

.1

.

132

1

0

0

71822

100

010

x

x

x

y

r

x

x

x

x

x

x

Controller canonical form

Phase variable form


20/20

Stability of a StateSpace Model

State space model:

DuCxyBuAxx

.

State equation

Output equation

The above state-space model is stable if and only if the eigenvalues of the system

matrix A that is ,the roots of the systems characteristic equation,

0AsI

All lie in the left half of the s-plane.

21

12ALet

e.g:

The associated eigenvalues are -3, -1

Hence the system is stable.

state space many methods

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