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INC 341 – Feedback Control Systems
Modelling of
Mechanical Systems
S Wongsa
sarawan.won@kmutt.ac.th
Transfer Functions of Physical Systems
Today’s goal
Mechanical system models
Review of the Laplace transform & transfer function
Translation Rotation
Transfer Functions of Physical Systems
Linear time-invariant systems
Input OutputSystem
X(s) Y(s)
x(t)y(t)
)()()(
)()()(*)()(0
sHsXsY
dthxthtxty
t
=
−== ∫ τττ
The system is called time-invariant if system parameters do not change in time.
time
domain
frequency
domain
Transfer Functions of Physical Systems
Laplace transform
dtetftfsF st∫∞
−
−==0
)()()( L
F(s) is the frequency domain representation of f(t)
s is a complex number
ωσ js +=
where σ and ω are real numbers with units of frequency, i.e. Hz.
Inverting the Laplace transform
[ ] dsesYj
sYtyj
j
st∫∞+
∞−
− ==σ
σπ)(
2
1)()( 1
L
Using tables is much easier!
Transfer Functions of Physical Systems
Why the Laplace transform?
We can transform an ordinary differential equation (ODE) into an algebraic equation
(AE) and easily find the rather complicated solution of the ODE.
ODE AE
Partial fraction
expansion
Solution to ODE
t - domain s - domain
L
-1L
1
2
3
Transfer Functions of Physical Systems
Laplace transform of some commonly used functions
From Table 2.1, Nise, Norman S., Control Systems Engineering. 5th Ed. John Wiley, 2008.
Impulse function / Dirac function
Properties
Unit energy
Sifting
∫+∞
∞−=1)(tδ
∫+∞
∞−= )0()()( ftftδ
Transfer Functions of Physical Systems
Laplace transform of some commonly used functions
From Table 2.1, Nise, Norman S., Control Systems Engineering. 5th Ed. John Wiley, 2008.
2
3
4
5
6.
7
Transfer Functions of Physical Systems
Properties of Laplace Transform
From Table 2.2, Nise, Norman S., Control Systems Engineering. 5th Ed. John Wiley, 2008.
Transfer Functions of Physical Systems
The Transfer Function
Given a nth-order, LTI differential equation,
)(...)()(
)(...)()(
01
1
101
1
1 trbdt
trdb
dt
trdbtca
dt
tcda
dt
tcda
m
m
mm
m
mn
n
nn
n
n +++=+++ −
−
−−
−
−
If all initial conditions are zero, taking the Laplace transform of
both sides gives
0
1
1
0
1
1
...
...)(
)(
)(
asasa
bsbsbsG
sR
sCn
n
n
n
m
m
m
m
++++++
== −−
−−
G(s) is known as the transfer function.
Transfer Functions of Physical Systems
The Transfer Function
Transfer functions permit cascaded interconnection of several
subsystems.
Transfer Functions of Physical Systems
Mechanical system components : translation
Transfer Functions of Physical Systems
Example: One degree of freedom
[sum of impedances] X(s)=[sum of applied forces]
)()()()( tftKxtxftxM v =++ &&&
L
)()()()(2 sFsKXssXfsXMs v =++
)(sG
Transfer Functions of Physical Systems
Example : Two degrees of freedom
(a) Forces on M1 due only to motion of M1
(b) Forces on M1 due only to motion of M2
(c) All forces on M1
Forces on M1
)()(][)(])([ 22121
2
1 331sFsXKsfsXKKsffsM vvv =+−++++
Transfer Functions of Physical Systems
(a) Forces on M2 due only to motion of M2
(b) Forces on M2 due only to motion of M1
(c) All forces on M2
Forces on M2
0)(])([)(][ 232
2
212 323=++++++− sXKKsffsMsXKsf vvv
Example : Two degrees of freedom
Transfer Functions of Physical Systems
Example : Two degrees of freedom
Equations of motion
0)(])([)(][
)()(][)(])([
232
2
212
22121
2
1
323
331
=++++++−
=+−++++
sXKKsffsMsXKsf
sFsXKsfsXKKsffsM
vvv
vvv
Transfer Functions of Physical Systems
Equations of motion can also be formulated by inspection
]at x forces applied of sum[
)(] xand between x imp. of sum[)(]at xmotion the toconnected imp. of [sum
1
22111 =− sXsX
)()(][)(])([ 22121
2
1 331sFsXKsfsXKKsffsM vvv =+−++++
Forces on M1
]at x forces applied of sum[
)(] xand between x imp. of sum[)(]at xmotion the toconnected imp. of [sum
2
12122 =− sXsX
Forces on M2
0)(])([)(][ 232
2
212 323=++++++− sXKKsffsMsXKsf vvv
Transfer Functions of Physical Systems
Example : Two degrees of freedom
Equations of motion
0)(])([)(][
)()(][)(])([
232
2
212
22121
2
1
323
331
=++++++−
=+−++++
sXKKsffsMsXKsf
sFsXKsfsXKKsffsM
vvv
vvv
Transfer function
=
0
)(
)(
)(
2
1 sF
sX
sX
dc
ba
=
−
0
)(
)(
)(1
2
1 sF
dc
ba
sX
sX
∆
−
−
=
0
)(
)(
)(
2
1
sF
ac
bd
sX
sX
∆+
==)(
)()(
)( 232 KsfsG
sF
sX
∆−
==)(
)()(
)(2 scFsG
sF
sX
dc
ba=∆
where
See Example 2.18 and try Skill-assessment Exercise 2.8
Transfer Functions of Physical Systems
K-Spring constant, D – coefficient of viscous friction, J – moment of inertia
Mechanical system components : Rotation
Transfer Functions of Physical Systems
Example : Two equations of rotational motion
(a) Torques on J1 due only to motion of J1
(b) Torques on J1 due only to motion of J2
(c) All torques on J1
Torques on J1
)()(][)(][ 211
2
1 sTsKsKsDsJ =−++ θθ
Transfer Functions of Physical Systems
Example : Two equations of rotational motion
(a) Torques on J2 due only to motion of J2
(b) Torques on J2 due only to motion of J1
(c) All torques on J2
Torques on J2
0)(][)(][ 22221 =+++− sKsDsJsK θθ
Transfer Functions of Physical Systems
Example : Two equations of rotational motion
)()(][)(][ 211
2
1 sTsKsKsDsJ =−++ θθ
0)(][)(][ 22221 =+++− sKsDsJsK θθ
Equations of motion
Transfer Functions of Physical Systems
Let’s get this done by inspection
]at torquesapplied of sum[
)(] and between imp. of sum[)(]at motion the toconnected imp. of [sum
1
22111
θ
θθθθθ =− ss
)()(][)(][ 211
2
1 sTsKsKsDsJ =−++ θθ
See Example 2.20 and try Skill-assessment Exercise 2.9
]at torquesapplied of sum[
)(] and between imp. of sum[)(]at motion the toconnected imp. of [sum
1
22111
θ
θθθθθ =− ss
0)(][)(][ 22221 =+++− sKsDsJsK θθ
Torques on J1
Torques on J2
Transfer Functions of Physical Systems
Mechanical system components: rotation: gears
2
1
2
1
2
1
1
2
T
T
N
N
r
r===
θθ
Transfer Functions of Physical Systems
Gear transformations
)()()( 22
2 sTsKDsJs =++ θ1
212
2 )()()(N
NsTsKDsJs =++ θ
1
211
2
12 )()()(N
NsTs
N
NKDsJs =++ θ
(1) (2)
(3)
)()( 11
2
2
1
2
2
12
2
2
1 sTsN
NKs
N
NDs
N
NJ =
+
+
θ
Rotational mechanical impedances can be
reflected through gear trains by multiplying
the mechanical impedance by the ratio
(Number of destination teeth/Number of source teeth)2
Transfer Functions of Physical Systems
Example: Reflected impedances
)()( 1
1
22221
2
1
22
21
2
1
2 sTN
NsKsDD
N
NsJJ
N
N
=
+
+
+
+
θ
See Example 2.22 and try Skill-assessment Exercise 2.10
Transfer Functions of Physical Systems
Using the Laplace transform to solve ODEs
From 2.004 Dynamics & Control II, MIT OCW, Fall 2007.
The motor applies torque Ts(t) as the following step function:
)(0,
0,0)( 0
0
tuTtT
ttTs ≡
≥
<=
J = The shaft inertia.
b = Coefficient of viscous friction applied by the bearings.
ω = The shaft rotational speed.
)()()()( 0 tuTtTtbtJ s ==+ ωω&
)(
)()(
)()(
21
0
0
bJs
K
s
K
bJss
Ts
s
TsbJs
++=
+=Ω
=Ω+
+−=Ω
J
bs
sb
Ts
11)( 0
( )τω /0 1)( teb
Tt −−=
where τ = J/b
1/0 =bT
-1L
L
Partial
fraction
expansions
force response
natural response
Summary
Laplace transform
Transfer functions & impedances of mechanical systems
dtetftfsF st∫∞
−
−==0
)()()( L
KDsJs ++2
1T(s) ΩΩΩΩ(s)
ODE AE
Partial fraction
expansion
Solution to ODE
t - domain s - domainL
-1L
1
2
3
Transfer Functions of Physical Systems
Next class
Transfer functions of electrical systems (2.1-2.4 of Ch 2).
Transfer functions of electro-mechanical (DC motor) systems (2.8 of Ch2).
Nonlinearities & linearisation (2.10 & 2.11 of Ch2 and 4.9 of Ch4).
You are highly recommended to read these topics before coming to the next class!
Transfer Functions of Physical Systems
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