浙大845自动控制-15原理王慧chapter transfer function to state

From

transfer

function

to

state

space

modelChapter

4-42007-11-12Outline

of

this

chapterIntroductionDetermination

of

the

overall

transferfunctionSimulation

diagramsSignal

flow

graphFrom

transfer

function

to

state

spacemodelFrom

transfer

function

to

stateChapter

4-42007-11-13space

model

(See

P158,5.9)Parallel

statediagramsfromtransfer

functionParallel

state

diagramDiagonalizing

the

A

matrixUse

of

state

transformation

for

the

state

equationsolutionTransformingintocompanion

formControllable

standard

formObservable

standard

formFrom

block

diagram

to

state

space

modelFrom

transfer

function

to

state

space

model

A

SISO

system

represented

by

the

differential

equation(5.62)

maybe

represented

by

an

overall

transfer

function

of

the

formParallel

state

diagrams(

D

n

aD

n

1n

11w0w

1

(

c

w

D

c

w

1

D

·

c1

D

c

0

)u(5.62)

·

a D

a

)

y w

nw

nsn

sn

1

a sn2n

1

n

2

1

00

a

·

a

s

aY

(s)

G(s)

U

(s)cw

sw

cw1

sw1

·

c

s1

c(5.67)

To

factor

the

denominator

and

express

G(s)

in

partial

fractions,When

there

are

no

repeated

roots

and

w=n:Chapter

4-42007-11-14

i

i

i

inU

(s)f

Z

(s)

fs

iG

(s);

G

(s)

c

G

(s)

n

ii

1

n n

1

1

0

(s

1

)(

s

2

)·(s

n

)c

sn

c

s

n

1

·

c

s

cFrom

transfer

function

to

state

space

modelParallel

state

diagrams0

i

i

i

inU

(s)f

Z

(s)

fs

iG

(s);

G

(s)

c

G

(s)

n

ii

1n n

1

1(s

1

)(

s

2

)·(s

n

)c

sn

c

sn

1

·

c

s

c

The

symbols

zi

and

their

Laplace

transforms

Zi(s)

are

used

for

the

statevariables

in

order

to

distinguish

the

form

of

the

associated

diagonalmatrix.

1

2

·f

U

(

s

)

f

U

(

s

)Y

(

s

)

cnU

(

s

)

s

1

s

2(5.68)z·iiu1+zifiy

i

cnU

(

s

)

f1

Z

1

(

s

)

f

2

Z

2

(

s

)

·

The

state

variables

Zi(s)

are

selected

to

satisfy

this

equation.iiZ

(s)

U

(s)

;sZ

i

(s)

-

i

Zi

(s)

U

(s)s

z·i

i

z

i

uy

i

f

i

z

iChapter

4-42007-11-15From

transfer

function

to

state

space

modelParallel

state

diagrams

The

complete

simulation

diagram

is

drawn

in

Fig.5.31,

P159.

The

initialconditions

of

the

states

zi(t0)

are

omitted

in

the

Fig.,

you

can

add

themyourself.Z1(s)Z2(s)12n:fnf2f1U(s)Y(s)cnFig.5.31

Simulation

of

Eq.(5.67)

byparallelpositionforw=nnZ

(s)FeedforwardpathnG

(s)

cn

Gi

(s)i

1iChapter

4-42007-11-16i

iif

Z

(s)fiU

(s)

s

G

(s)

From

transfer

function

to

state

space

modelParallel

state

diagrams

The

state

space

model

therefore

of

the

formny

f

n

z

u

z

b

uf

·

f

z

c

u

c z

d

u

:z·

1

2

n

n

n

n21:

11

00

:0

·

0

..

.·

01

0(5.70)(5.71)

The

system

matrix

A

appears gonal

or

normal

or

canonical

form(正则型,it

indicated

by

(or

A*),

corresponding

state

variables

are

oftencalled

canonical

variables.Allelements

are

unityw=n,

dn0,

else

dn=0Coefficients

ofthepartial

fractions

For

the

MIMO

system:----b

B,

c

C,

and

d

D.Not

same!Chapter

4-42007-11-17From

transfer

function

to

state

space

modelParallel

state

diagrams

The

diagonal

matrix

A=

means

that

each

state

equation

isuncoupled;

i.e.

each

state

zi

can

be

solved

independently

of

theother

states.

This

fact

simplifies

the

procedure

for

finding

the

state

transitionmatrix

(t).

This

diagonal

form

is

very

useful

for

studying

the

system.,

Ex.,observability

and

controllability.

Transfer

functions

with

repeated

roots

are

not

considered

in

thistextbook.这里讲的系统特征根为各不相同的单根时的正则规范型状态空间描述。是一种最简单的情况。对于存在复根的情况(较少碰到)，A阵为约当阵。如P169,5.12所示。此略。有

的同学可

。Chapter

4-42007-11-18From

transfer

function

to

state

space

modelZ1(s)Z2(s)-1-2421Y(s)Fig.5.32

Simulation

diagramfor

exampleofSec.5.9U(s)Parallel

state

diagrams

3

s

2s

24

s

2

15

s

13G

(

s

)

【Ex-1】For

the

givenG(s),draw

the

parallel

statediagram

and

determine

the

state

equation

(See

P160).Solution：

3

s

2s

24

s

2

15

s

13G

(

s

)

1

2s

2

s

1G

(

s

)

4

2z

4uy

11

0

z

1u;

0

1z·

2w=n,

dn

0,

else

dn=0Chapter

4-42007-11-19From

transfer

function

to

state

space

modelParallel

state

diagrams【Ex-2】设控制系统的传递函数为试求系统的正则规范型状态空间描述。s(s2

3s

2

7s

12)s2U

(s)Y

(s)

1

1

2

3

16

s

3

s

3

2

s

4Y

(

s

)

1U

(

s

)解：将系统传递函数化为分母为因式相乘的形式，系统的特征根为0，－3，－4；相应的留数可求得为1

,

2

,

36

3

2即：000

10

3 0

x

1

u0

4

1x·

03

x232

6y

1

因此可得系统的正则规范型状态空间描述：nw<n,

d

=0Chapter

4-42007-11-110From

transfer

function

to

state

space

modelChapter

4-42007-11-111Diagonalizing

the

A

matrix

(see

P160-168.

5.10)In

the

method

of

partial-fraction

expansion

above,

the

desirablenormal

form

of

the

state

equation

is

obtained,

in

which

the

A

matrixis

diagonal.When

the

system

is

MIMO

or

the

system

equations

are

alreadygiven

in

state

form,

this

partial-fraction

expansion

method

is

notconvenient.Section

5.10

presents

a

more

general

method

for

converting

thestate

equation

by

means

of

a

linear

similarity

transformation.Four

methods

are

presented

obtained

the

modal

matrix

T

for

thecase

where

the

matrix

A

has

distinct

eigenvalues(特征值）.From

transfer

function

to

state

space

modelState

equation

solution

(see

P168,

5.11)It

is

very

useful

to

transform

a

general

system

matrix

A

to

diagonalform

(or

A*)

in

solving

the

state

equation

(we

have

studied

in

Section3.4

slide).The

state

equation:x(t)

Ax

(t)

Bu(t)(5.112)For

a

non-diagonal

matrix

A,

we

can

diagonalize

it ,

that

meansto

evaluate

the

modal

matrix

T

for

diagonalizing

A.

Then:

diag

[

1

,

2

,.

,

n

]e

t

diag

[e

1

,

e

2

,.

,

e

n

]exp(

At

)

T

exp(

t

)T1Its

solution

is:(5.113)tt00

)dBu

(

)d

Φ(t

)x(0)

Φ(t

)Bu

(x(t

)

e

x(0)

eA

(t

)AtA

TT11

2

n

,

,.,

]

T1AT

diag[eΛt

exp(

T1ATt

)

T1

exp(

At)TChapter

4-42007-11-112From

transfer

function

to

state

space

model,

find

exp(At)

by

method

3

1

5

6

Example-3.

Assuming

matrixand

x(t).A

0Solution:

We

can

find

the

eigenvalues

of

A:1

2,

2

3Since

A

is

not

diagonal

matrix,

it

is

required

to

evualute

the

modalmatrix

for

diagonalizing

A.For

this

example,

A

is

not

in

companion

form,

the

modal

matrix

Tshould

be

evaluated.

Here

using

method

2

given

in

5.10

6

1

2

For

1

2

,

adj

[

2

I

A

]

v

3

6

3

,For

21adj

[

3

I

A

]

v

6 6

2

3T

1/

3

11T

0.5 6

e

2

t

e

3

t1e

2

t

2

e

2

t

3e

3

t6

e

2

t

6

e

3

t

3e

2

t

2

e

3

t00e

3

t

Te

A

t

T

eΛ

t

T

1

T

State

equationsolution

(see

P168,

5.11)Chapter

4-42007-11-113From

transfer

function

to

state

space

model,

find

exp(At)

by

method

3and

x(t).Chapter

4-42007-11-114

1

5

6

Example-3.

Assuming

matrixA

0Solution:

e

T

3

t

2

t

e

3

te

3

t

e

2

t

2e

2

t

3e

6e

3

t6

e

2

t

2e

3

t

3e

2

t

00

T

1:e

A

t

T

e

Λ

t

T

1

t

e

ee

2

t

e

3

t21

3

t

2

t

e

3

t0

2

3

t

2

t

e

3

t0

2

t

2

e

3

t

1

3

ex

(

0

)

x

(

0

)

3

e

2

e

2

t6

e

2

t

6

e

3

t

3

e

2

t

2

e

3

t

2

e

3

ex

(

0

)

x1

(

0

)

3

e

2

e

2

t6

e

2

t

6

e

3

t

1d

2

3

t

6

e

2

6

e

3

)

Bu

(

t

)d

Φ

(

t

)

x

(

0

)

Φ

(

2

e

3

t:

x

(

t

)

3

e

2

tState

equation

solution

(see

P168,

5.11)From

transfer

function

to

state

space

model

The

synthsis

of

feedback

control

systems

and

theysis

oftheir

response

characteristics

is

often

considerably

facilitatedwhen

the

plant

and

control

matrices

in

the

matrix

state

equationhave

particular

forms.

Controllable

standard

form

and

observable

standard

form

areobtained

from

transfer

function

directly

are

introduced.(The

transformation

from

a

physical

or

a

general

state

variablexp

to

the

phase

variable

xc

is

discussed

in

the

Section

5.13).Standard

forms

of

state

equation(see

P172,

5.13)Chapter

4-42007-11-115From

transfer

function

to

state

space

modelAC00

:0

a000:11

0

·0

1:0

·

0

a1

·

an

2

an

1

10

0

:0

BCControllable

Standard

form(可控（相伴）

)w

1

(

c

w

D

w

c

w

1

D

·

c1

D

c

0

)u

a

n

1

D

n

1(

D

n

·

a

1

D

a

0

)

y(5.62)w

nw

n10sn

aY

(s)

G(s)

U

(s)n

2sn

1

an

1w

w1

1

0sn

2

·

a

s

ac

sw

c sw1

·

c

s

c(5.67)

A

SISO

system

represented

by

the

differential

equation(5.62)

maybe

represented

by

an

overall

transfer

function

of

the

formx·

Ac

x

Bc

uy

cc

x

Du注意：下标cCompanion

matrix友矩阵，能控标准形(Ac,bc)(只与A、B有关)Phase

variableChapter

4-42007-11-116From

transfer

function

to

state

space

modelControllable

Standard

form(可控（相伴）

)c

nc

0c

n

1n

2c1c2ca1a02

a

an

2

an

1y

(c0

a0

cn

)c

n

0c

n

0(c1

)

.

(cn

1

an

1cn

)x

[cn

]uy

c0c1

.

c

w

0

.

0

x

cxChapter

4-42007-11-117From

transfer

function

to

state

space

modelControllable

Standard

form(可控（相伴）

)Ex-4请写出

系统的传递函数与可控

.s

4

a3

s3

a2

s

2

a1s

a0G(s)

3

2

1

0

Y

(s)

c

s

3

c

s

2

c

s

cU

(s)x3x2x4x1Y(s)U(s)c1c2c3c0Solution:因为该图以相变量为状态变量，可直接由图写出系统的传递函数与可控

。又因w=3<n=4，故由式（5.67）Chapter

4-42007-11-118From

transfer

function

to

state

space

modelControllable

Standard

form(可控（相伴）

)x

Ac

x

Bc

uy

cc

x

Du

1

0

0

0

bC式中

3

21

01

0001010000

a

a

a

0

aAC4c

0c

c

c0

c1

c

2

c

3

x3x2x4x1Y(s)U(s)c1c2c3c0s

4G(s)

3

2

1

0

Ex-4请写出

系统的传递函数与可控

.Y

(s)

c

s

3

c

s

2

c

s

cU

(s)

a3

s3

a2

s

2

a1s

a0Chapter

4-42007-11-1192007-11-1From

transfer

function

to

state

space

modelObservable

Standard

form(可观（相伴）

)w

nU

(s)

an

2sn

sn

1

asn

2n

1

G(s)

w

w11

0Y

(s)

c

sw

csw1·

c

s

c·

a

s1

a

0

(5.67)

Another

important

state

equation

form

is

observable

standard

form.ny

x2x·

a

1

y

c

1

u:x·n

a

n

1

y

c

n

1

u1x·

a

0

y

c

0

ux·

Ao

x

Bo

uy

co

x

Dux·1

a0

y

c0

u

a

0

xn

(c0

a0

cn

)ux·2

a1

y

c1u

a1

xn

(c1

)u:x·n

an

1

y

cn

1u

an

1

xn

(cn

1

an

1cn

)uU(s)x·1x2x3yx·2x·3x·4c2c3c1x1c0c4x4w

n

,

c

n

0

y

x

n

c

n

u21From

transfer

function

to

state

space

modelObservable

Standard

form(可观（相伴）

)w

n

asn

sn

1

an

1

n

20Y

(s)

G(s)

U

(s)cw

sw

cw1

sw1

·

c

s1

c(5.67)x·

Ao

x

Bo

uy

co

x

Dusn

2

·

a

s

a1

0注意：式中的下标Owhere01

an

1

Ao

0

;010

·

0

a

0

0

·

0

a1

·

0

a

2;

;

;0

·

1co

0

0

·

0

1能观标准形(Ao,Co)(只与A、C有关)Chapter

4-4

Another

important

state

equation

form

is

observable

standard

form.nD

c

ac c

B

c

a

c

n1

n1

n

o

2 2

n

c0

c1

a1cn

;w

n

,

c

n

0w

n

,

c

n

0

0

0

;

;

w

c

c0

Bo

D

0From

transfer

function

to

state

space

modelObservable

Standard

form(可观（相伴）

)Ex-5

请写出系统的状态空间表达式2

0

a3

0

0

0

a0

0

0

a

11

0

a

0

0

1A

1oco

0

0

0

1

2

c3

c1

c

c0

BoD

0x1U(s)1x又x2x3x4=yx又23x又x又4c3c2c1c0w

n

,

c

4

01

4

1

1x又

a

x

x

c

u22

2x又3

a2

x4

x

cux又

a3

x4

x3

c3u41

0

4x又

a

x

c0uy

x4Solution:Chapter

4-42007-11-122From

transfer

function

to

state

space

model01sn

an2sn1

an1sn1

cG(s)

n

n1

n2

1

0sn2

…

a

s

ac

sn

c

sn2

…

c

s

c已知传递函数为：x·

Ao

x

Bo

uy

co

x

Dux·

Ac

x

Bc

uy

cc

x

Du可控可观仔细观察两种规范型中的系统矩阵A、控制矩阵B以及输出矩阵C，可以发现可控

与可观

之间存在关系：Ao

(

Ac

)TCo

(Bc

)TBo

(Cc

)T这是普遍规律?可控（相伴）与可观（相伴）Yes!Chapter

4-42007-11-123From

transfer

function

to

state

space

model可控（相伴）与可观（相伴）【例6】设控制系统的传递函数为s2Y

(s)

3s

2进而可得系统的可控

状态空间表达式为

00

x1

0

1

x

0

u2

2

1012

7

x3

1

x1

y

[2 3

1]

x

2

x3

x·1

x·

0x·3

0U

(s)

s(s2

7s

12)试求系统的可控规范型状态空间描述。解：将系统传递函数对照可控规范型公式，可得w

2

,

n

3

,

c

3

0

,

c

2

1,

c1

2a

3

1,

a

2

7

,

a

1

12

,

a

0

0Chapter

4-42007-11-124From

transfer

function

to

state

space

model1

00

x1

0

21

x

0

u2

1012

7

x3

1

x

y

[2 3

1]

x

2

x3

x·1

x·

0

x·3

0可控（相伴）与可观（相伴）【例7】设控制系统的传递函数为s

2U

(s)

s(s

2

7s

12)Y

(s)

3s

2

20

0

0

00

12

x

3

u1

7

1x·

1y

0

0

1x可控试求系统的可观规范型状态空间描述。解：将系统传递函数对照可观规范型公式，因为w

2

,

n

3

,

c

3

0

,

c

2

1,

c

1

2a

3

1,

a

2

7

,

a

1

12

,

a

0

0进而可得系统的可观

状态空间表达式为Chapter

4-42007-11-125Block

diagramState

equation

model由一般的方块图直接得到状态方程Chapter

4-42007-11-126From

transfer

function

to

state

space

model方法一：环节规范化设一阶环节：1U

(s)

Ts

1G(s)

Y

(s)

?ks

aG(s)

ux·因为一个控制系统的方块图主要由比例环节。积分环节、一阶滞后环节（惯性环节）、二阶振荡环节等基本环节所组成，要将其改画成状态变量图是很方便的，只要把其中的一阶惯性环节和二阶振荡环节，按图示的方式改画成局部状态变量图就可以了。U(s)x·

1

x

1

uT

T

xChapter

4-42007-11-127From

transfer

function

to

state

space

model方法一：环节规范化设二阶环节：G

(

s

)

Y

(

s

)

U

(

s

)1T

2

s

2

2

Ts

1UU(s)x又1x又2=x121

1T

2Tx又

2

x

1xx又2

x1

uChapter

4-42007-11-128From

transfer

function

to

state

space

model方法二：直接从方块图计算从已知的控制系统方块图中，选择一阶环节的输出变量作为状态变量，经直接计算将方块图化为状态变量图，从而可得系统的状态空间表达式。【例8】已知如下系统，状态变量如图示，请写出状态空间表达式并画出状态变量图。R(s)Y(s)x1x2x322s

144s

110.5sChapter

4-4-2911-1From

transfer

function

to

state

space

model方法二：直接从方块图计算R(s)Y(s)x1x2x322s

144s

110.5s

1解：由图中的状态变量，可写出：232(R

X

(s))2s

1X

(s)

421X

(s)4s

1X

(s)

113X

(s)0.5s

1X

(s)

Y

(s)

X1

(s)21

14x·

1

x

x2Chapter

4-42007-11-130322x·

1

x

x

rx·3

2x1

2x3y

x1From

transfer

function

to

state

space

model例8的状态变量图方法二：直接从方块图计算x1yx21/s-1/4rx31/s-1/21/s2-2x3x1R(s)Y(s)x1x2x322s

144s

110.5s

121

14x又

1

x

x3Chapter

4-42007-11-1312

22x又

1

x

x

rx又3

2x1

2x3y

x1From

transfer

function

to

state

space

model方法二：直接从方块图计算问题：例8的状态变量信号流图？例8的状态变量图x1yx21/s-1/4rx31/s-1/21/s2-2x3x1

00

2

2

10

14

1

1

x

1

rx.