notes 3(a) translational mechanical system transfer function (terkini)

8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)

1/14

Chapter 3 : 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

roblem: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

1


2/14

%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 =++

&' )()()(2

sFsXKsfMsv =++

ol&ing for the transfer function yields

KsfMssF

sXsG

v ++==

2

1

)(

)()(

!hich is represented in Figure 2.15 (b)

$olution (( :

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

33 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!

( ) ( ) ( ) ( )dttdxftkxtf

dttxdm v=

+

2

2

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

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

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

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

signal

( ) ( ) ( ) ( )[ ]sXsFKsFMs

sX v+= 1

2

6) 7ra! the block diagram using the abo&e information

( ) ( ) ( ) ( )[ sXsFKsFMssX v+= 1 2

8utput ystem 9nput ummingsignal signal :unction

)loc* diagram reduction:

( )KsFMssF

sXsT

v ++==

2

1

)(

)(

$ignal flo+ graph ason 'ule.:

2

M

x

fkx

xfv


3/14

( ) [ ]( )[ ] KsFMsMsKsFMs

sTvv ++

=+

=22

2 1

/1

1/1

Transfer function t+o degrees of freedom / t+o linearly independent motion

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

Figure 2!"0: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 be

re+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


4/14

Figure 2!2* a) Forces on 2M due only to motion of 2M b) forces on 2M due only to

motion of 1M c) all forces on 2M

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

sho!n on the mass*

)()()(

)()(

)()(

/

)(

)(

)()(

)(

)(

)(

)(/

2422

412

2

2

2121

4

22

2

4121

1

412

1

2

11

txKKdt

tdxff

dt

txdMtxK

dt

tdxfF

txKdt

tdx

ftxKKdt

tdx

ffdt

txd

MtfF

vvv

vvvM

+++++==+

++++++==+

/)()()(

)()(

)()(

)()()(

)()()(

)()(

2422

412

22

2121

4

222

41211

412

1

2

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

[ ][ ]

)(

)()()()()(

)()()()()(

242422

2124

224121412

1

==

++++++++++ sF

sXKKsffsMsXKsf

sXKsfsXKKsffsM

vvv

vvv

ol&ing for the transfer function using the Cramers rule:

[ ][ ]

=

+++++

+++++

/

)(

)(

)(

)()()(

)()()(

2

1

4242

2

224

242141

2

1 sF

sX

sX

KKsffsMKsf

KsfKKsffsM

vvv

vvv

[ ][ ])()()(

)()()(

/)()()()(

)(

4242

2

224

242141

2

1

24

2141

2

1

2

KKsffsMKsf

KsfKKsffsM

KsfsFKKsffsM

sX

vvv

vvv

v

vv

+++++

+++++

+++++

=

[ ]

[ ][ ])()()(

)()()(

)()()(

4242

2

224

242141

2

1

242

KKsffsMKsf

KsfKKsffsM

KsfsFsX

vvv

vvv

v

++++++++++

+=

[ ][ ] 224424222214121242

)()()()()(

)(

)(

)(

KsfKKsffsMKKsffsM

Ksf

sF

sX

vvvvv

v

++++++++++

=

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

+

==)(

)(

)()( 242

Ksf

sF

sXsG v

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

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

4242

2

224

242141

2

1

KKsffsMKsfKsfKKsffsM

vvv

vvv

++++++++++=

6


5/14

&'2

244242

2

22141

2

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

enyelesaian dalam *elas:

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


( )

( ) ( )

( ) ( ) ( )

( ) ( )( )

( )( )

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

dt

tdx

dt

tdxf

dt

tdxftxk

dt

txdm

txtxkdt

tdx

dt

tdx

fdt

tdx

ftxktfdt

txd

m

vv

vv

21221

42

2242

2

2

2

212

21

4

1

1112

1

2

1

+

+=

+

=

+

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

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

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

sXsXKsXsXsFssXFsXKsFsXsM

vv

vv

21221422242

2

2

21221411111

2

1

++=

+

=

+

eparate the &ariables,

( )( ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) 24224

2

2124

22414121

2

1

=++++++

=+++++

sXsFsFKKsMsXKsF

sFsXKsFsXsFsFKKsM

vvv

vvv

>rrange it in a &ector"matri< form

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

( )

( )

( )

=

+++++

+++++

/2

1

4224

2

224

244121

2

1 sF

sX

sX

sFsFKKsMKsF

KsFsFsFKKsM

vvv

vvv

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

5

1x2x

f

11xK

11xf

V

( )214

xxfV

( )212 xxK

22xf

V

24xK

start3

9nput

Force

7isplacement

8utput 1

7isplacement

8utput 2


6/14

( )

( )( ) ( )( )

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

KsFsFsFKKsM

KsF

sFsFsFKKsM

sX

vvv

vvv

v

vv

4224

2

224

244121

2

1

24

4121

2

1

2

++++++++++

+++++

=

( )( )

( )( )( ) ( )( ) ( ) 2244224

2

24121

2

1

242

KsFsFsFKKsMsFsFKKsM

KsFsFsX

vvvvv

v

++++++++++=

%e could also sol&e the problem using the bloc* diagram / signal flo+ graph


( )( ) ( )

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

( )( )

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

dt

tdx

dt

tdxf

dt

tdxftxk

dt

txdm

txtxkdt

tdx

dt

tdxf

dt

tdxftxktf

dt

txdm

vv

vv

21221

42

2242

2

2

2

21221

41

1112

1

2

1

+

+=

+

=

+

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

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

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


vv

vv

21221422242

2

2

21221411111

2

1

++=

=

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

signal

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

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsFsMsX


vv

vv

21241112

1

1

21221411111

2

1

1

++=

=

ystem 9nput umming :unction

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )sXsXKsXsXsFssXFsXKsXsM vv 212214222422

2 ++=

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

sX vv 212422422

2 1

+++=

8utput ystem umming :unction

6) 7ra! the block diagram using the abo&e information

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsFsMsX vv 21241112111 ++=

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

sX vv 212422422

2 1

+++=

#

umming

:unction


7/14

)loc* 4iagram 'eduction :

( )

( )

( )( )

( )( ) ( ) ( )[ ]sFKsMsFKsMsFKsMsFKsM sFK

sFKsMsFKsM

sFK

sF

sX

VV

VV

V

VV

V

112

1242

2

24

2

211

2

1

42

24

2

211

2

1

42

2

2

1

+++++++++ ++

+++++

=

( )

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

sF

sX

VVVVV

V

11

2

124

2

24224

2

211

2

1

42

2

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

+=

$ignal Flo+ 5raph ason 'ule.:

?

2

2

1

sM

2X

F1

X2

1

1

sM

2XsFK V11+

sFK V24+

sFK V42+

connect

connect

sFKsM v242

2

1

++

2X

F 1X

sFKsM V112

1

1

++

sFK V42+

sFK V42+

2X

F

sFK V42+

sFKsM

sFK

v

V

24

2

2

42

++

+sFK

sFKsM

V

v

42

24

2

2

+++

sFKsM V112

1

1

++sFKsM V11

2

1 ++

2X

F sFKsM

sFK

v

V

24

2

2

42

+++

( ) ( )sFKsMsFKsM Vv 112

124

2

2 +++++

sFKsM V112

1

1

++


8/14

33 $lease !rite all the for!ard path

gains, loop gains, non"touching loop

gains and

( )

( )

( )( )

(( )

( )( )

(( )

( )( )

(( )22

24

2

1

42

2

2

42

2

1

11

2

2

24

2

1

11

2

2

42

2

2

24

2

1

42

2

1

11

2

2

2

1

42

1

11

sM

FK

sM

sFK

sM

FK

sM

sFK

sM

FK

sM

sFK

sM

sFK

sM

sFK

sM

sFK

sM

sFK

sMsMsFK

sT

VV

VV

VV

V

V

V

V

V

++

++

++

+

+

+

+

+

+

=

6ote :

@heck your ans!ers to make sure that they are correct. 'he ans!er !ill be *

( )

2

2444

224241124121

2

21142141

4

14

2

12

4

12

2

14

4

24

2

22

4

21

2

21

6

21

42

sFFsFK

sFKKKsFFsFKsFKKKsFFsFKsFKKK

sMFsMKsMFsMKsMFsMKsMFsMKsMM

sFKsT V

++

+++++++++++

+++++++++

+=

7uestion*

8btain the transfer function model of this system*

;

2

2

1

sM

2X

F1

X2

1

1

sM

sFK V11+

sFK V24+

sFK V42+

connect

connect

F

( )sFK 11+

2

1

1sM

1

( )sFK 42+2

2

1

sM

( )sFK 24+

1

F 21

1sM

1

( )sFK 42+2

2

1

sM

( )sFK 24+1( )sFK 11+

8oopA a path !ithout passingthrough any other node more than

once BB


9/14

>pplying e!ton-s la! of motion, the force

e+uation can be !ritten as

yBKyFyM =

+

8rdt

dyBKyF

dt

ydM =

+

2

2

'aking the 0aplace transform (assuming

ero initial conditions), !e obtain

( ) ( ) ( ) ( ) ( )syBsKsFsyMs +=2

eparate the &ariables,( ) ( ) ( )sFsyBsKMs =++2

'herefore,

( )

( ) BsKMssF

sy

++=

2

1

)loc* diagram :

(b)

>pplying e!ton-s la! of motion, the

force e+uation can be !ritten as

( ) ( ) ( )iOiOO xxBxxKxM =+'aking the 0aplace transform (assuming

ero initial conditions), !e obtain

( ) ( ) ( )( )

( )( )Oi

iOO

xxBsK

xxBsKsxMs

+=

+=

2

eparate the &ariables,

( ) ( ) ( ) ( )sxBsKsxBsKMsiO

+=++2

'herefore,

( )( ) BsKMs

BsK

sx

sx

i

O

+++

=2

)loc* diagram :

( ) ( ) ( ) ( )OiO xxBsKsxMs +=2

8btain the transfer function model ( ) ( )sFsy1 of this system(@.)

=

y

FM

Ky

yB

9 mass-damper-spring

system

F)4

y

M

K

BFForce,

(a)

OxM

( )iO xxK ( )iO xxB

$implified suspension system

F)4

output,ntdisplaceme x

P

M

K B

ixinput,ntdisplaceme

9 mechanical system

1K

1B

2y

M2FForce,

M1

1y2

K

2B

11yK

11yB

2y

M2

M1

1y ( )122 yyK

F( )122 yyB

Free )ody 4iagram


10/14

( )( )

( ) ( )( ) ( )( )

( )( )

( ) ( ) ( )( ) ( )( )tytyk

dt

tdy

dt

tdyB

dt

tdyBtyk

dt

tydm

tytykdt

tdy

dt

tdyBtf

dt

tydm

12212

21

1112

1

2

1

12212

22

2

2

2

+

+=

+

=

+

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

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

sysyksysysBsfsysm

12212211111

2

1

1221222

2

2

++=

+

=

+

)loc* diagram reduction from above diagram.:

( )( ) ( )( ) ( )( )2

22211

2

122

2

2

22

sMsBKsBKsMsBKsM

sBK

sF

sy

+++++++

=

$ignal flo+ graph*

( ) ( )( )( )

( )( )

(( )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

+++

+

+

+

+=

333 $lease !rite all the for!ard path gains, loop gains, non"touching loop gains and

1

2

1

1

sM

1yF2y

2

2

1

sM

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

22sM

sBKsM 222

2

1

++

F

2

1

1

sM

1

( )sBK 22+2

2

1

sM

( )sBK 11 +

1y

1


11/14

@heck your ans!er.*

( )

sBBsKBsBKKK

sMBsMBsMBsMKsMKsMKsMM

sBKsT

21121212

4

21

4

22

4

22

2

21

2

22

2

12

6

21

22

++++

+++++++

+=

(d) 8btain the transfer function model ( ) ( )sUsY of this system

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

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

+=+

=+

dt

tdy

dt

tdxBtytxk

dt

tydm

tutxkdttdy

dttdxk

dttdy

dttdxB

dttxdm

22

2

2

122

2

1

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

( ) ( ) ( ) ( )( ) ( ) ( )( )sYsXBssYsXKsYsMsXsUKsXsYKsXsYBssXsM

sUsXKsYsXKsYsXBssXsM

+=+

++=

=+

2

2

2

12

2

1

12

2

1

8C

)loc* 4iagram 'eduction:

( )

( )

( )

( ) ( )( )12

1221

2

1

2

2

21

KsMBsKBsKKsMsM

BsKK

sU

sY

+++++++

= Cefer to D. 8gata-s ook pg "32.

$ignal flo+ graph:

11

9 motorcycle

suspension

system

2K B

x

M1

M2

y

1Ku ( )uxK 1Free )ody 4iagram

( )yxB

xM1

M2

y

start3( )xuK 1

Free )ody 4iagram

( )yxK 2 ( )yxB

xM1

M2

y&'

start3( )uxK 1

Free )ody 4iagram

( )xyK 2 ( )xyB

xM1

M2

y&'start3

2

2

1

sM

YU X2

1

1

sM 1K BsK +2

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 2

2

1

sM

1

1

2

1

K

KsM +

YU

( )( )BsKKsMBsKK

++++

21

2

1

21

2

2

1

sM

1

1

2

1

K

KsM +

( )yxK 2


12/14

( ) ( )( )

( )

( )

( )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) 8btain the transfer function model ( ) ( )sXsX iO of this system

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

( ( ) ( ) ( )( ) ( ) yKyxB

yxBxxKxxB

O

OioiO

22

211

/

/

+=+

++=+

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

( ) ( )( ) ( )sYKsYsXsB

sYsXsBsXsXKsXsXsB iOi

2/2

/2/11

=

=+

)loc* 4iagram 'eduction:

12

1

1K ( )sBK 22+

2

2/1 sM 1y

1

U2

1/1 sM

1

9 mechanical system

1K 1B

outputntdisplaceme,Ox

inputntdisplaceme,i

x

2K

2B

outputntdisplaceme,y

Free )ody 4iagram

( )iO xxK

1 ( )iO xxB

1

Ox

yK2

( )yxB o 2

y

start3

OX

iX ( )YXsB O2

+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

+++++


13/14

( )( )

sK

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

+

+

+

+

+

= 'efer to ! &gatas )oo* page "33.

$ignal flo+ graph:

Couth table *2s 21BB 21KK1

s 22122 KBKBK ++ //

s 21KK /

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

( ) 211

211

211211

/

/1

//

KsBK

sBsBK

KsBKsBsBKsT

++

+++=

( ) ( )( ) ( ) sBsBKKsBKsBK

sBKsBsBKK

21121122

112112+++++++=

( ) ( )( ) ( )sBKsBKsBK

sBKsBK

221122

2211+++

++=

2

2112212122

2

21122121

sBBsBKsBKKKsBK

sBBsBKsBKKK

+++++++

=

;quation of motion by inspection :

roblem: %rite but do not sol&e, the e+uations of motion for the mechanical net!ork ofFigure 2.21

Figure 2!2" :'hree" degrees of freedom translational mechanical system

$olution* 'he system has 4 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.

+ E um of impedances connected to the motion at 1x ( )sX1 E um of impedancesbet!een 1x and 2x ( )sX2 E um of impedances bet!een 1x and 4x ( )sX4 = E sum of applied forces at 1x

14

sBK 11+ sB2/1 /X

2/1 K1

iX


14/14

E um of impedances bet!een 1x and 2x ( )sX1 + Eum of impedancesconnected to the motion at 2x ( )sX2 E um of impedances bet!een 2x and 4x

( )sX4 = E sum of applied forces at 2x

E um of impedances bet!een 1x and 4x ( )sX1 Eum of impedances bet!een2x and 4x ( )sX2 + E um of impedances connected to the motion at

4x ( )sX4 = E sum of applied forces at 4x

1M has t!o springs, t!o &iscous dampers and mass associated !ith its motion. 'here is

one spring bet!een 1M and 2M and one &iscous damper bet!een 1M and 4M . 'hus,

/)()()()()( 4422121412

1 =++++ ssXfsXKsXKKsffsM vvv

imilarly, for 2M

)()()()()( 4622622

212 sFssXfsXKsffsMsXK vvv =++++

>nd for 4M

/)()()()( 4642

42614 =+++ sXsffsMssXfssXf vvvv

16

notes 3(a) translational mechanical system transfer function (terkini)

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