Notes 3 Translational Mechanical System Transfer Function

8/10/2019 Notes 3 Translational Mechanical System Transfer Function

1/13

Translational mechanical system transfer function

Table 2 Force-velocity, force-displacement, and impedance translational relationships

for springs, viscous dampers, and mass

Transfer function one equation of motion

Problem:Find the transfer function, )(/)( sFsX , for the system of Figure 2.15 (a)

Figure 2.!:a) Mass, spring and damper system b) block diagram

"olution:

egin the solution by dra!ing the free"body diagram sho!n in Figure 2.1#(a). $lace on the

mass all forces felt by the mass. %e assume the mass is tra&eling to!ard the right. 'hus,only the applied force points to the right all other forces impede the motion and act to

oppose it. ence, the spring, &iscous damper and the force due to acceleration point to the

left.

Figure 2.#* a) Free"body diagram of mass, spring and damper system b) transformed free"

body diagram


2/13

%e no! !rite the differential e+uation of motion using e!ton-s la! to sum to ero all of

the forces sho!n on the mass in Figure 2.1# (a)*

)()()()(

)()()(

)(

2

2

2

2

tftKx

dt

tdxf

dt

txdM

tKxdt

tdxf

dt

txdMtfF

v

v

=++

=

+++=+

'aking the 0aplace transform, assuming ero initial conditions,

)()()()(2 sFsKXssXfsXMs v =++

r )()()(2 sFsXKsfMs v =++

ol&ing for the transfer function yields

KsfMssF

sXsG

v ++

==2

1

)(

)()(

!hich is represented in Figure 2.15 (b)

333333333333333333333333333333333333333333333333333333333333333333333

"olution $$:

%e could also sol&e the problem using the block diagram / signal flo! graph.4444 egin the

solution by

dra!ing the

free"body

diagram sho!n

in Figure 2.Figure 2* Free"body diagram of mass, spring and damper system

1) %rite the differential e+uation of motion using e!ton-s econd 0a!

( )( ) ( )

( )

dt

tdxftkxtf

dt

txdm v=

+

2

2

2) 'aking the 0aplace transform, assuming ero initial conditions,

( ) ( ) ( ) ( )ssXFsKXsFsXMs v=2

or ( ) ( ) ( ) ( )sXsFKsFsXMs v+=2

) eparate the input signal (6 combination of other signal), system and output

signal

( ) ( ) ( ) ( )[ ]sXsFKsFMs

sX v+= 1

2

7) 8ra! the block diagram using the abo&e information

( ) ( ) ( ) ( )[ ]sXsFKsFMs

sX v+= 1

2

utput ystem 9nput umming :unction

ignal ignal

%loc& diagram reduction:

( )

KsFMssF

sXsT

v ++

==2

1

)(

)(

"ignal flo' graph ()ason *ule+:

( ) [ ][ ]

( ) KsFMsMsKsF

MssT

vv ++

=

+

=22

2 1

/1

1/1

m

x

fkx

xfv


3/13

Transfer function t'o degrees of freedom t'o linearly independent motion

Problem: Find the transfer function, )(/)(2 sFsX , for the system of Figure 2.1; (a)

Figure 2.:a) '!o"degrees of freedom translational

mechanical system b) block diagram

"olution:

'he system has t!o degrees of freedom, since each mass can be mo&ed in the horiontal

direction !hile the other is held still. 'hus, t!o simultaneous e+uations of motion !ill bere+uired to describe the system. 'he t!o e+uations come form free body diagrams o each

mass. uperposition is used to dra! the free"body diagrams. For e


4/13

'he differential e+uation of motion using e!ton-s la! to sum to ero all of the forces

sho!n on the mass*

)()()(

)()(

)()(

)()(

)()()(

)()(

)(

2522

512

2

2

2121

5

222

51211

512

1

2

11

txKK

dt

tdxff

dt

txdMtxK

dt

tdxfF

txKdt

tdxftxKK

dt

tdxff

dt

txdMtfF

vvv

vvvM

+++++==+

++++++==+

)()()(

)()(

)()(

)()()(

)()()(

)()(

2522

512

22

2121

5

222

51211

512

12

1

=+++++

=++++

txKKdt

tdxff

dt

txdMtxK

dt

tdxf

tftxKdt

tdxftxKK

dt

tdxff

dt

txdM

vvv

vvv

'he 0aplace transform of the e+uations of motion can no! be !ritten from Figures 2.1= (c)

and 2.2 ( c) as

[ ]

)(

)()()()()(

)()()()()(

25252

2

2125

225121512

1

=

=

+++++

+++++ sF

sXKKsffsMsXKsf

sXKsfsXKKsffsM

vvv

vvv

ol&ing for the transfer function using the 0ramer1s rule:

[ ][ ]

=

+++++

+++++

/

)(

)(

)(

)()()(

)()()(

2

1

5252

2

225

252151

2

1 sF

sX

sX

KKsffsMKsf

KsfKKsffsM

vvv

vvv

[ ] [ ])()()( )()()(

)(

)()()(

)(

5252

2

225

252151

2

1

25

2151

2

1

2

KKsffsMKsfKsfKKsffsM

Ksf

sFKKsffsM

sX

vvv

vvv

v

vv

+++++ +++++

+++++

=

[ ]

[ ][ ])()()(

)()()(

)()()(

5252

2

225

252151

2

1

25

2

KKsffsMKsf

KsfKKsffsM

KsfsFsX

vvv

vvv

v

+++++

+++++

+=

[ ][ ] 225525222215121252

)()()()()(

)(

)(

)(

KsfKKsffsMKKsffsM

Ksf

sF

sX

vvvvv

v

++++++++++

=

From this, the transfer function, )(/)(2 sFsX is

+

==)(

)(

)()( 252

Ksf

sF

sXsG v

>s sho!n in Figure 2.1; (b) !here

[ ])()()()()()(

5252

2

225

252151

2

1

KKsffsMKsf

KsfKKsffsM

vvv

vvv

+++++

+++++=

r2

255252

2

22151

2

1 )()()()()( KsfKKsffsMKKsffsM vvvvv +++++++++=


5/13

Penyelesaian dalam &elas:

1) egin the solution by dra!ing the free"body diagram

2) %rite the differential e+uation of motion using e!ton-s econd 0a!

( )( ) ( )

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

( )( )

( ) ( ) ( )( ) ( )( )txtxk

dt

tdx

dt

tdxf

dt

tdxftxk

dt

txdm

txtxkdt

tdx

dt

tdxf

dt

tdxftxktf

dt

txdm

vv

vv

21221

52

2252

2

2

2

21221

51

1112

1

2

1

+

+=

+

=

+

'aking the 0aplace transform, assuming ero initial conditions,( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )sXsXKsXsXsFssXFsXKsXsM

sXsXKsXsXsFssXFsXKsFsXsM

vv

vv

21221522252

2

2

21221511111

2

1

++=

+

=

+

eparate the &ariables,

( )( ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) 25225

2

2125

22515121

2

1

=++++++

=+++++

sXsFsFKKsMsXKsF

sFsXKsFsXsFsFKKsM

vvv

vvv

>rrange it in a &ector"matri< form

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

( )

( )

( )

=

+++++

+++++

/2

1

5225

2

225

255121

2

1 sF

sX

sX

sFsFKKsMKsF

KsFsFsFKKsM

vvv

vvv

ol&ing for the transfer function, )(/)(2 sFsX yields

( )

( )( ) ( )( )

( )( ) ( )( ) ( )( )sFsFKKsMKsF

KsFsFsFKKsM

KsF

sFsFsFKKsM

sX

vvv

vvv

v

vv

5225

2

225

255121

2

1

25

5121

2

1

2

+++++

+++++

+

++++

=

( )

( )

( )

( )( ) ( )( ) ( )2

2552252

251212

1

25

2

2

KsFsFsFKKsMsFsFKKsM

KsF

sF

sX

vvvvv

v

+++++++++

+

=

1x2x

f

11xK

11xf

V

( )215 xxfV

( )212 xxK

22xfV

25xK

9nput

Force

8isplacement

utput 1

8isplacement

utput 2


6/13

%e could also sol&e the problem using the bloc& diagram signal flo' graph

+ %rite the differential e+uation of motion using e!ton-s econd 0a!

( )( ) ( )

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

( )( )

( ) ( ) ( )( ) ( )( )txtxk

dt

tdx

dt

tdxf

dt

tdxftxk

dt

txdm

txtxk

dt

tdx

dt

tdxf

dt

tdxftxktf

dt

txdm

vv

vv

21221

52

2252

2

2

2

21221

51

1112

1

2

1

+

+=

+

=

+

2+ 'aking the 0aplace transform, assuming ero initial conditions,

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

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )sXsXKsXsXsFssXFsXKsXsM


vv

vv

21221522252

2

2

21221511111

2

1

++=

=

+ eparate the input signal(6 combination of other signal), systemand output

signal

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

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsFsM

sX


vv

vv

21251112

1

1

21221511111

2

1

1

++=

=

ystem 9nput umming ?unction

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )sXsXKsXsXsFssXFsXKsXsM vv 212215222522

2 ++=

( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsM

sX vv 212522522

2 1

+++=

utput ystem umming ?unction

3+ 8ra! the block diagram using the abo&e information

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsFsMsXvv 21251112

1

1

1

++=

( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsM

sX vv 212522522

2 1

+++=

2

2

1

sM

2X

F1X

2

1

1

sM

2XsFK V11 +

sFK V52+

sFKV2

+

sFK V52+connect

connect

umming :unction


7/13

%loc& 4iagram *eduction:

( )

( )

( )( )

( )( )( ) ( )[ ]sFKsMsFKsM

sFKsMsFKsM

sFK

sFKsMsFKsM

sFK

sF

sX

VV

VV

V

VV

V

11

2

125

2

2

25

2

211

2

1

52

25

2

211

2

1

52

2

2

1

+++++++++

++

++++

+

=

( )( ) ( )( ) ( )( ) ( )[ ]sFKsMsFKsMsFKsFKsMsFKsM

sFK

sF

sX

VVVVV

V

11

2

125

2

25225

2

211

2

1

52

2

2+++++++++++

+=

"ignal Flo' 5raph ()ason *ule+:

44 $lease !rite all the for!ard path

gains, loop gains, non"touching loopgains and

( )

( )

( )

( )(

(( )

( )(

(( )

( )(

( 225

2

1

52

2

2

2

2

1

11

2

2

5

2

1

11

2

2

52

2

2

25

2

1

52

2

1

11

2

2

2

1

52

1

11

sM

FK

sM

sFK

sM

FK

sM

sFK

sM

FK

sM

sFK

sM

sFK

sM

sFK

sM

sFK

sM

sFKsMsM

sFK

sT

V

VV

VV

V

V

V

V

V

++

++

++

+

+

+

+

+

+

=

sFKsM v252

2

1

++

2X

F 1X

sFKsM V112

1

1

++

sFK V52+

sFK V52+

2X

F

sFK V52+

sFKsM

sFK

v

V

25

2

2

52

+++

sFK

sFKsM

V

v

52

25

2

2

+++

sFKsM V112

1

1

++sFKsM V11

2

1 ++

2X

F

sFKsM

sFK

v

V

252

2

52

++

+

( ) ( )sFKsMsFKsM Vv 112

125

2

2 +++++

sFKsM V1121

1

++

2

2

1

sM

2X

F1X

2

1

1

sM

2XsFK V11 +

sFK V52+

sFK V2 +

sFK V52+connect

connect

F

( )sFK 11 +

2

1

1sM

( )sFK 2 +

( )sFK 2 +2

2

1

sM

( )sFK 2 +

1


8/13


9/13

btain the transfer function model ( ) ( )sFsy1 of this system(@.)

( )( )

( ) ( )( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )tytykdt

tdy

dt

tdyB

dt

tdyBtyk

dt

tydm

tytykdt

tdy

dt

tdyBtf

dt

tydm

12212

21

1112

12

1

12212

22

2

2

2

+

+=

+

=

+

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

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )sysyksysysBtsyBtyksysm

sysyksysysBsfsysm

12212211111

2

1

1221222

2

2

++=

+

=

+

%loc& diagram reduction (from above diagram+:

( )( ) ( )( ) ( )( )222211

2

122

2

2

22

sMsBKsBKsMsBKsM

sBK

sF

sy

+++++++=

"ignal flo' graph*

6 mechanical system

1K

1B

2y

M2FForce,

M1

1y2K

2B

11yK

11yB

2y

M2

M1

1y ( )122 yyK

F( )122 yyB

Free %ody 4iagram

2

1

1

sM

1yF2y

2

2

1

sM

sBK 22+

sBK 11 +

sBK 22+

sBKsM11

2

1

1

++1yF

sBK 22+

sBK 22+2

2sM

2

2

1

sM

sBKsM

sBK

11

2

1

22

+++ 1yF

2

2sMsBKsM 22

2

2

1

++


10/13

( ) ( )( )( )

( )

( )(

( )22

11

2

1

22

2

2

11

2

2

22

2

1

22

2

2

2

122

1

/1/1

sM

sBK

sM

sBK

sM

sBK

sM

sBK

sM

sBK

sMsMsBKsT

+++

+

+

+

+=

444 $lease !rite all the for!ard path gains, loop gains, non"touching loop gains and

'o check the e+u.*

( )

sBBsKBsBKKK

sMBsMBsMBsMKsMKsMKsMM

sBKsT

21121212

5

21

5

22

5

22

2

21

2

22

2

12

7

21

22

++++

+++++++

+=

(d) btain the transfer function model ( ) ( )sUsY of this system

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

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

+=+

=+

dt

tdy

dt

tdxBtytxk

dt

tydm

tutxkdt

tdy

dt

tdxk

dt

tdy

dt

tdxB

dt

txdm

22

2

2

122

2

1

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

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( ) ( )( )sYsXBssYsXKsYsM

sXsUKsXsYKsXsYBssXsM

sUsXKsYsXKsYsXBssXsM

+=+++=

=+

2

2

2

122

1

12

2

1

or


F

2

1

1

sM

( )sBK 22 +

( )sBK 22 + 22

1

sM

( )sBK 11 +

1y

1

6 motorcycle suspension system

2K B

x

M1

M2

y

1Ku ( )uxK 1Free %ody 4iagram

( )yxK 2 ( )yxB

xM1

M2

y

2

2

1

sM

YU X2

1

1

sM

1K BsK +2


11/13

( )( )

( )( ) ( ) ( )12

1221

2

1

2

2

21

KsMBsKBsKKsMsMBsKK

sUsY

+++++++= Aefer to B. gata-s ook (pg 12)

"ignal flo' graph:

( ) ( )( )[ ]

( )

( )

( )2

2

2

2

1

1

2

22

2

12

2

11

2

22

2

11

/

/

/

1

//

sM

BsK

sM

K

sMBsK

sMBsK

sMK

sMBsKsMKsT

++

+

+

+=

(e) btain the transfer function model ( ) ( )sXsX iO of this system

'he differential e+uation of motion for the system*

( ) ( ) ( ) ( )

( ) ( ) yKyxB

xxBxxKxxB

O

iOioiO

22

211

++

++

'aking the 0aplace 'ransform (assuming ero initial condition), !e obtain

2

2

1

sM

YU X

1

2

1

1

KsM

K

+

BsK +2

1/1 K

2

2

1

sM

YU X

2

1

1

sM

K

BsK +2

1/1 K

YU

1/1 K

1

2

1

1

KsM

K

+

BsK +2 22

1

sM

1

1

2

1

K

KsM +

YU

( )

( )BsKKsMBsKK

+++

+

21

2

1

21

2

2

1

sM

1

1

2

1

K

KsM +

1

1K ( )sBK 22 +2

2/1 sM 1y

1

U2

1/1 sM1

6 mechanical system

1K 1B

Ox

ix

2K

2B

y

Free %ody 4iagram

( )iO xxK 1 ( )iO xxB 1

Ox

yK2

( )yxB o 2

y


12/13

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

( ) ( )( ) ( )sYKsYsXB

sYsXBsXsXKsXsXsB iOi

22

211

=

=+


( )

( )s

K

Bs

K

Bs

K

B

sK

Bs

K

B

sX

sX

i

O

1

2

2

2

1

1

2

2

1

1

11

11

+

+

+

+

+

= Aefer to B. gata-s ook (page 1)

"ignal flo' graph:

( ) ( )[ ] ( )[ ]( )

( )211

211

211211

/

/1

//

KsBK

sBsBKKsBKsBsBKsT

+

+

+++=

( ) ( )

( ) ( ) sBsBKKsBKsBK

sBKsBsBKK

21121122

112112+++++++

=

( )( )

( )( )sBKsBKsBK

sBKsBK

221122

2211+++

++=

7quation of motion by inspection

Problem: %rite but do not sol&e, the e+uations of motion for the mechanical net!ork of

Figure 2.21

OXiX ( )YXsB O 2

+

sBK 11+

Y

sB2

1 ( )YXO

2

1

K

OXiX

sBK 11+22

11

KsB+

OXiX

( )( )

sBK

sBKsBK

22

2211 ++

OXiX ( )( )( )( )sBKsBKsBK

sBKsBK

221122

2211

+++

++

sBK 11 + sB2/1 X

2/1 K

1

iX


13/13

Figure 2.2 :'hree" degrees of freedom translational mechanical system

olution* 'he system has degrees of freedom, since each of the three masses can be

mo&ed independently !hile the others are held still. 'he form of the e+uations !ill be

similar to electrical mesh e+uations.

6C um of impedances connected to the motion at

Notes 3 Translational Mechanical System Transfer Function

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Transcript of Notes 3 Translational Mechanical System Transfer Function