Top Banner
ME 330 Control Systems SP 2011 Lecture 5 ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10 ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10
12
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Me330 lecture5

ME 330 Control Systems

SP 2011

Lecture 5

ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10,zscan=0):ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10,zscan=0):

Page 2: Me330 lecture5

Mechanical Systems

)()()()( tkxtxctxmtf f(t)

)()( 2 sXkcsmssF

From the Laplace transform

kcsms 21

F(s) X(s)Often seen in block-diagram representation

Newton’s 2nd Law governs mechanical systems, resulting equation of motion describes dynamical system.

Page 3: Me330 lecture5

Mechanical Systems Modeling Determining the equation of motion

)()()()( tkxtxctxmtf

f(t)

)()()()( 2 skXscsXsXmssF

Free-Body Diagram

)(tkx

m)(txc

)(txm

)(tf

)(skX

m)(scsX

)(2 sXms

)(sF

Summation of forces

Impedance

Summation of impedances

Page 4: Me330 lecture5

Mechanical Components

Table 2.4Force-velocity, force-displacement, and impedancefor springs, viscous dampers, and mass

Page 5: Me330 lecture5

Mechanical Components Table 2.5

Torque-angular velocity, torque-angular displacement, and rotational impedance for springs, viscous dampers, and inertia

Note: rotational mechanics are analogous to translational mechanics force => torque damper => damper mass => inertia

Page 6: Me330 lecture5

Multiple Degrees of Freedom Number of equations of motion required =

number of linear independent motions Linear independence: point of motion is allowed to

move even if all other points of motion are fixed. Also known as degrees of freedom.

Page 7: Me330 lecture5

Multiple Degrees of Freedom Analyze impedances for each degree of freedom.

1c

)(tf

2c

3c

)(11 sXk

m1)(11 ssXc

)(sF

)(12

1 sXsm

)(12 sXk

)(13 ssXc

)(22 sXk)(23 ssXc

)()()()(0

)()()()()(

232322

2123

223121312

1

sXkksccsmsXksc

sXkscsXkksccsmsF

)(22 sXk

m2

)(22 ssXc

)(22

2 sXsm

)(12 sXk)(13 ssXc

)(23 ssXc)(23 sXk

Page 8: Me330 lecture5

Transfer Function Laplace transform of equations of motion

Can solve for any transfer function

)()()(

)()()(

)()(

)(

)(

32322

223

2331312

1

32322

21

kksccsmksc

ksckksccsm

kksccsm

sF

sX

)()()(

)()()()(

)(

32322

223

2331312

1

232

kksccsmksc

ksckksccsm

ksc

sF

sX

Page 9: Me330 lecture5

Equations by Inspection For two degree of freedom system, the general

form is given bySum of

impedances connected to

the motion at x1

Sum of impedances

between x1 and x2

X1(s) X2(s)– = Sum of

applied forces at x1

Sum of impedances connected to

the motion at x2

Sum of impedances

between x1 and x2

X2(s)X1(s)– = Sum of

applied forces at x2

+

)()()()()()()()()(

)()()()()()()()()(

)()()()()()()()()(

332211

2233222121

1133222111

sFsXssXssXssXs

sFsXssXssXssXs

sFsXssXssXssXs

NNNNNN

NN

NN

Higher degree of freedom systems

Page 10: Me330 lecture5

Torsion Mechanical Systems Same rules apply as in translational mechanical systems

Sum of impedances connected to

the motion at 1

Sum of impedances

between 1 and 2

1(s) 2(s)– = Sum of

applied forces at 1

Sum of impedances

between 1 and 3

3(s)–

Sum of impedances connected to

the motion at 2

Sum of impedances

between 1 and 2

2(s)1(s)– = Sum of

applied forces at 2

Sum of impedances

between 2 and 3

3(s)– +

Sum of impedances connected to

the motion at 3

Sum of impedances

between 1 and 3

3(s)1(s)– = Sum of

applied forces at 3

Sum of impedances

between 2 and 3

2(s)– +

Page 11: Me330 lecture5

Torsion Mechanical Systems Summation of impedances

)()()(00

)()()()(0

0)()()()(

3232

322

33222

21

2112

1

ssDsDsJssD

ssDsKsDsJsK

sKsKsDsJsT

Page 12: Me330 lecture5

Next Lectures

Derivation of more mechanical system models Derivation of electrical system models