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Laplace transform: Review
Use
Definition of Laplace transform
Existence
Laplace transforms of some functions of time
Exponential
Impulse Ramp
Sine, Cosine
.
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time domain Laplace or sdomain
DifferentialEquation
Solution
AlgebraicEquation
LaplaceTransform of
Solution
L-1
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Properties of Laplace Transforms
Translation:
L f t e F ss( ( )) ( ) =
Multiplication by exponential:
L e f t F st( ( )) ( ) = +
Differentiation:
Ldf t
dts F s f(
( )) ( ) ( )= 0
Ld f t
dts F s s f
df t
dt t(
( )) ( ) ( )
( )2
22
00=
=
Integration
L f dF s
s
t( ( )
( ) =
0
2
t
f(t)
f(t-)
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Laplace transforms (Section 2.3):
Final value theorem
Use: Can findf() even if you do not knowf(t).
lim ( ) lim ( )t sf t sF s = 0
Conditions:f, df/dtLaplace transformable and
poles ofsF(s) have negative real parts.
Initial value theorem:
lim ( ) lim ( )t sf t sF s + =0
Conditions:f, df/dtLaplace transformable
Example: Mass suspended by a spring and a
damper it hit by a hammer. Find initialdisplacement, velocity
and final value ofdisplacement.
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Inverse Laplace Transform
Idea for finding inverse Laplace transforms.
Break downF(s) into components of which youknow the inverse
Laplace transforms. Add theinverse Laplace transforms of
thesecomponents.
F sk s z s z s z
s z s z s zm
n( )
( )( )...( )
( )( )...( )=
+ + +
+ + +
1 2
1 2
Cases: a) Distinct real polesb) Complex polesc) Multiple
poles
a) F sa
s p
a
s p
n
n
( ) ...=+
+ +
+
1
1
f t L F s a e a ep t
np tn( ) ( ( )) ...= = + +
11
1
b) Complex poles
Use: L e t s aat( sin )
( )
=
+ +
2 2
L e t s a
s a
at( cos )( )
=
+
+ +
2 2
c) Multiple poles
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F ss
s s s s s( )
( )( ) ( )=
+
+ +
=
+
+
+
+
+
5 2
1 2
3
1
3
2
8
22 2
f t e e tet t t( ) = + + 3 3 82 2
Why not use:
F ss
s s
a
s
b
s( )
( )( ) ( )=
+
+ +
=
+
+
+
5 2
1 2 1 22 2
Try to find a and b:
5 2 4 2 12
s a s s b s+ = + + + +( ) ( )
3 equations, 2 unknowns. Need to introduceanother fraction so
that you have 3 unknowns.
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Solving differential equations using Laplacetransforms
Example: mass-spring system
Importance of ultra simple models in real life
Problem:Designed aircraft wing for low vibration.
Large safety margin.
The boss wants to redesign the wing to reduceweight. Recommends
reducing plate thickness
by 10%. Need to find if vibration level will stillbe acceptable.
Need a quick answer.
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.
Simplified model 2
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Simplification
Beam, sameIandA asaverageIandA of wingsection
F(t)=sint
F(t)=sint
m c x(t)
Actual structure
Simplifiedmodel 1
x(t)
x(t)
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Findx(t) using simplified model 2
Check if it still acceptable.
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Equivalentstatic
stiffness, k
F(t)=sint
mass of wing/2
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Example: vibration of aircraft wing if engine
detaches from wing.
000
0
0 ==
=++
)(x,)(x
)t(kx)t(xc)t(xm
kms
ms)s(X
+
=2
0
)t
m
kcos()t(x 0=
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static equilibrium
position (no engine)m
k
x(t)0
x(t)
t
period= mk
/2
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Linearize non linear systems
Non linear systems very difficult to analyze
Linearize using Taylor series expansion aboutstatic equilibrium
position
Example:
0=+ sin
l
g
Nonlinear because of the sin.
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sinl
g
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Taylor expansion:
)(d
)(df)(f)(f
.T.O.H)(d
)(df)(f)(f
00
00
0
0
+
++=
=
=
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linear approximation
sin
