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8/12/2019 L2 - Modelling Systems
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Todays Lecture
System modellingdetermining
differential equations
Using the Laplace operator
Determining the transfer function
Standard forms
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System Modelling
Determine differential equation that
represents the system
Mechanical or Electrical
Standard equations for both
Analogies (covered next week)
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Mechanical Systems
Force (F)-displacement (x) relationships
Spring: coefficient k
Damper: coefficient c
Mass (inertia): NSL: ma
Also apply to rotational systems
Force Torque; x ; mass J (MoI)
kxFS
dt
dxcxcFD
2
2
dt
xdmxmFI
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8/12/2019 L2 - Modelling Systems
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Electronic Systems
Voltage (V) - Charge (Q) - Current (i)
relationships
Capacitor
Resistor
Inductor
dtiCC
QVC 1
iRdt
dQ
RVR
dt
diL
dt
QdLVL 2
2
8/12/2019 L2 - Modelling Systems
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Electronic Example
LCR Circuit
Relate Vinand i
Vin
L
C
R
i
dtiCVC
1
iRVR dt
diLVL
dtiC
iRdtdiLV
VVVV
in
CRLin
1
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s-domain
Differential equations can be awkward to
deal with
Using Laplace operator, s, can help in
manipulationusing Laplace transforms,
LDE are converted to simple algebraic
equations
Transient and Steady state components of
the solution can be obtained
8/12/2019 L2 - Modelling Systems
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Laplace Transforms
Definition of Laplace Transform of a
function f(t):
Complex variable, s, is Laplace operator
More details in your Maths notes
dtetfsFtfL st
0 )()()(
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Laplace transforms
Derivatives
)()(
)()(
)()(
3
3
3
2
2
2
sXsdt
txdL
sXsdt
txdL
ssX
dt
tdxL
Multiplication by s in the Laplacedomain corresponds to differentiation in
the time domain
Replace occurrence of d/dt with s:
)(
)(
sXsdt
txd
L n
n
n
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Laplace Transforms
Integrals
Integral in the time domain is converted to
division in the s domain.
Second term almost always zero, hence:
0
00
)(1)(
)( dttfss
sFdttfL
s
sFdttfL )(
)(0
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Mechanical Example
Mass/spring/damper system
When initial conditions are zero
m
k
c
f(t)
x(t)
)()()()(2
2
tftkxdttdxc
dttxdm
)()()()(2 sFskXscsXsXms
0)0()0( xx
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Electronic Example
LCR Circuit
v(t)
L
C
R
i(t)
dttiC
tvC )(1
)(
RtitvR )()(
dt
tdiLtvL
)()(
dtti
C
Rti
dt
tdiLtv )(
1)(
)()(
)(1
)()()( sICs
sRIsLsIsV
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Transfer Function
Single order
Flow in
qin(t)
h(t)
Flow out
kh(t)
sk
AksQsH
AsksQ
sH
kAs
sH
sQ
sHkAssQskHsAsHsQ
in
in
in
in
in
111
)()(
1
)(
)()(
)(
)()()()()(
)(
)(
)( tkhdt
tdh
Atqin
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Transfer Function
Mass/spring damper)()()()(2 sFskXscsXsXms
2
2
2
2
1
1
)(
)(
1
)(
)(
)(
)(
)()(
sk
ms
k
ck
sF
sX
kcsmssF
sX
kcsmssX
sF
sFsXkcsms
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Transfer Function
LCR Circuit )(1)()()( sICs
sRIsLsIsV
CsRLs
sV
sI
CsRLs
sI
sV
sICs
RLssV
1
1
)(
)(
1
)(
)(
)(1
)(
21)(
)(
1
1
)(
)(
byMultiply
LCsRCs
Cs
sV
sI
Cs
Cs
CsRLs
sV
sI
Cs
Cs
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Standard Forms
Second order
2
22
1
1
)(
)(
nn
s
s
sR
sC
gain
[]ratiodamping
[rad/s]frequencynaturalundamped
n
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Standard Forms
Second order
e.g. Mass/Spring/Damper
21
11
)(
)(
sk
ms
k
cksF
sX
k
m
k
k
c
m
kn
1
2
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Todays lecture
Standard equations for Mechanical and
Electrical systemsthese will be supplied
Use the Laplace operator, s, to simplify dealing
with LDEs Transfer functions represent input-output
relationship (output/input)
Standard forms help predict performance Tutorial Sheet 2: On Blackboard. Determining
transfer functions for systems.
