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SYSTEM DYNAMICS SYSTEM DYNAMICS (KJM 497) (KJM 497) System Dynamics – KJM 497 Fakulti Kejuruteraan Mekanikal State-Space Representation of Two DOF System

Exercise State Space Representation (TDOF)

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SYSTEM DYNAMICSSYSTEM DYNAMICS(KJM 497)(KJM 497)

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

State-Space Representationof Two DOF System

OBJECTIVESOBJECTIVES• TO DETERMINE THE STATE SPACE

REPRESENTATIOIN OF A MDOF MECHANICAL SYSTEM.

• TO UNDERSTAND THE STEPS INVOLVE IN OBTAINING THE STATE SPACE REPRESENTATION.

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

State-Space Equations:STATE SPACE REPRESENTATION:STATE SPACE REPRESENTATION:

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

BuAxx +=&

DuCxy +=

where,

⎥⎥⎥⎥

⎢⎢⎢⎢

=

nx

xx

x 2

1

⎥⎥⎥⎥

⎢⎢⎢⎢

=

nnnn

n

n

aaa

aaaaaa

A

21

22221

11211

⎥⎥⎥⎥

⎢⎢⎢⎢

=

nrnn

r

r

bbb

bbbbbb

B

21

22221

11211

⎥⎥⎥⎥

⎢⎢⎢⎢

=

ru

uu

u 2

1

State EquationOutput Equation

STATE SPACE REPRESENTATION:STATE SPACE REPRESENTATION:

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

where,

⎥⎥⎥⎥

⎢⎢⎢⎢

=

my

yy

y 2

1

⎥⎥⎥⎥

⎢⎢⎢⎢

=

mnmm

n

n

ccc

cccccc

C

21

22221

11211

⎥⎥⎥⎥

⎢⎢⎢⎢

=

mrmm

r

r

ddd

dddddd

D

21

22221

11211

⎥⎥⎥⎥

⎢⎢⎢⎢

=

ru

uu

u 2

1

EXAMPLE:EXAMPLE:

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

Consider a two DOF system as shown in figure below. Obtain the Consider a two DOF system as shown in figure below. Obtain the state space representation.state space representation.

m2

x2

k2

c2

m1

x1

f(t)

k1

c1

Solution:Solution:

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

1. Draw the Free Body Diagram (FBD).1. Draw the Free Body Diagram (FBD).

m2

x2

k2

c2

m1

x1

f(t)

k1

c1

m2m1

x2x1

f(t)( )211 xxc && −

( )211 xxk − 22xk

22xc &

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

2.Apply Newton2.Apply Newton’’s Second Law.s Second Law.

m2m1

x2x1

f(t)( )211 xxc && −

( )211 xxk − 22 xk

22 xc &

Mass 1, mMass 1, m1:1: xxmaF =+→ ∑

1211211 )()()( xmxxkxxctf &&&& =−−−−

)(2111211111 tfxkxkxcxcxm =−+−+ &&&& ( )1Mass 2, mMass 2, m2:2: 222222211211 )()( xmxkxcxxkxxc &&&&& =−−−+−

( )2022222111211122 =++−++− xkxcxkxkxcxcxm &&&&&

( ) 0)( 221112211122 =−−+++− xkkxkxccxcxm &&&&

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

2.Define State Variables.2.Define State Variables.

)(2111211111 tfxkxkxcxcxm =−+−+ &&&& ( )1( )2( ) 0)( 221112211122 =−−+++− xkkxkxccxcxm &&&&

[ ]tfxkxkxcxcm

x (121112111

11 ++−+−= &&&&

( )[ ]22111221112

2 )(1 xkkxkxccxcm

x −+−+−= &&&&

24

23

12

11

xxxxxxxx

&

&

====

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

3.Define State Equation (from state variables).3.Define State Equation (from state variables).

24

23

12

11

xxxxxxxx

&

&

====

24

43

12

21

xxxxxxxx

&&&

&

&&&

&

====

[ ]tfxkxkxcxcm

x (121112111

11 ++−+−= &&&&

( )[ ]22111221112

2 )(1 xkkxkxccxcm

x −+−+−= &&&&

Need to be substitute Need to be substitute using state variablesusing state variables

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

3. State Equation.3. State Equation.

[ ]

( )[ ]32111421212

24

43

311141211

12

21

)(1

(1

xkkxkxccxcm

xx

xx

tfxkxkxcxcm

xx

xx

−+−+−==

=

++−+−==

=

&&&

&

&&&

&

4. Output Equation.4. Output Equation.

1xy =

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

3. State3. State--Space Equation in Matrix Form.Space Equation in Matrix Form.

( ) ( )⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

+

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

+−−

−−

=

⎥⎥⎥⎥

⎢⎢⎢⎢

00

)(0

1000

0010

1

4

3

2

1

2

21

2

21

2

1

2

1

1

1

1

1

1

1

1

1

4

3

2

1

mtf

xxxx

mcc

mkk

mc

mk

mc

mk

mc

mk

xxxx

&

&

&

&

[ ]⎥⎥⎥⎥

⎢⎢⎢⎢

=

4

3

2

1

0001

xxxx

y

CONCLUSION:CONCLUSION:

• Know how to determine the state space representation of MDOF system.

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

EXAMPLE:EXAMPLE:

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

Consider massConsider mass--springspring--damper system as shown in figure below. The damper system as shown in figure below. The displacement x of the mass is the output of the system and the displacement x of the mass is the output of the system and the external force external force f(tf(t) is the input to the system. Derive the mathematical ) is the input to the system. Derive the mathematical model of the system.model of the system.

m

x1f(t)

k1

c1030

x2

c2

k2

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

m

x1 x2

30cos)(tf

( )211 xxk −

( )211 xxc && −22xc &

22xk

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

m

x1 x2

30cos)(tf

( )211 xxk −

( )211 xxc && −

−22xc &

22xk

−30cos)(tf ( )−− 211 xxc && ( )211 xxk − 1xm &&=

( )−− 211 xxk( )+− 211 xxc &&

22xc &

022 =xk

( )1

( )2

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal

−22xc &

−30cos)(tf ( )−− 211 xxc && ( )211 xxk − 1xm &&=

( )−− 211 xxk( )+− 211 xxc && 022 =xk

( )1

( )2

24

23

12

11

xxxxxxxx

&

&

====

2.Define State Variables.2.Define State Variables.

24

43

12

21

xxxxxxxx

&&&

&

&&&

&

====

System Dynamics – KJM 497Fakulti Kejuruteraan Mekanikal