-
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
)(
)(
$ignal flo+ graph ason 'ule.:
2
M
x
fkx
xfv
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
3/14
( ) [ ]( )[ ] KsFMsMsKsFMs
sTvv ++
=+
=22
2 1
/1
1/1
Transfer function t+o degrees of freedom / t+o linearly independent motion
roblem: Find the transfer function, )(/)(2 sFsX , for the system of Figure 2.1; (a)
Figure 2!"0:a) '!o"degrees of freedom translational
mechanical system b) block diagram$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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
5/14
&'2
244242
2
22141
2
1)()()()()( KsfKKsffsMKKsffsM vvvvv +++++++++=
enyelesaian 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
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
1) %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
42
2242
2
2
2
21221
41
1112
1
2
1
+
+=
+
=
+
2) 'aking the 0aplace transform, assuming ero initial conditions,
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )
( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )sXsXKsXsXsFssXFsXKsXsM
sXsXKsXsXsFssXFsXKsFsXsM
vv
vv
21221422242
2
2
21221411111
2
1
++=
=
4) eparate the input signal( combination of other signal), systemand output
signal
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )sXsXKsFsXsFKsFsMsX
sXsXKsXsXsFssXFsXKsFsXsM
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
++
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
+++++
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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
-
8/12/2019 Notes 3(a) Translational Mechanical System Transfer Function (Terkini)
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