3. Why use Laplace Transforms?
Find solution to differential equation
using algebra
Relationship to Fourier Transform
allows easy way to characterize
systems
No need for convolution of input and
differential equation solution
Useful with multiple processes in
system
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4. How to use Laplace
Find differential equations that describe
system
Obtain Laplace transform
Perform algebra to solve for output or
variable of interest
Apply inverse transform to find solution
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5. What are Laplace transforms?
∫
∫
∞+σ
∞−σ
−
∞
−
π
==
==
j
j
st1
0
st
dse)s(F
j2
1
)}s(F{L)t(f
dte)t(f)}t(f{L)s(F
t is real, s is complex!
Inverse requires complex analysis to solve
Note “transform”: f(t) → F(s), where t is integrated
and s is variable
Conversely F(s) → f(t), t is variable and s is
integrated
Assumes f(t) = 0 for all t < 0school.edhole.com
6. Evaluating F(s) = L{f(t)}
Hard Way – do the integral
∫
∫ ∫
∫
∞
−
∞ ∞
+−−−
−
∞
−
=
=
+
===
=
=−−==
=
0
st
0 0
t)as(stat
at
0
st
dt)tsin(e)s(F
tsin)t(f
as
1
dtedtee)s(F
e)t(f
s
1
)10(
s
1
dte)s(F
1)t(flet
let
let
Integrate by parts
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7. Evaluating F(s)=L{f(t)}- Hard Way
remember ∫ ∫−= vduuvudv
)tcos(v,dt)tsin(dv
dtsedu,eu stst
−==
−== −−
∫
∫ ∫
∞
−−
∞ ∞
−∞−−
−−
=−−=∴
0
stst
0 0
st
0
stst
dt)tcos(es)1(e
dt)tcos(es)tcos(e[dt)tsin(e ]
)tsin(v,dt)tcos(dv
dtsedu,eu stst
==
−== −−
∫∫
∫
∞
−−
∞
−∞−
∞
−
+−=+−
=∴
0
stst
0
st
0
st
0
st
dt)tsin(es)0(edt)tsin(es)tsin(e[
dt)tcos(e
]
2
0
st
0
st2
0 0
st2st
s1
1
dt)tsin(e
1dt)tsin(e)s1(
dt)tsin(es1dt)tsin(se
+
=
=+
=−=
∫
∫
∫ ∫
∞
−
∞
−
∞ ∞
−−
let
let
Substituting, we get:
It only gets worse…
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8. Evaluating F(s) = L{f(t)}
This is the easy way ...
Recognize a few different transforms
See table 2.3 on page 42 in textbook
Or see handout ....
Learn a few different properties
Do a little math
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9. Table of selected Laplace
Transforms
1s
1
)s(F)t(u)tsin()t(f
1s
s
)s(F)t(u)tcos()t(f
as
1
)s(F)t(ue)t(f
s
1
)s(F)t(u)t(f
2
2
at
+
=⇔=
+
=⇔=
+
=⇔=
=⇔=
−
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11. Note on step functions in Laplace
∫
∞
−
=
0
st
dte)t(f)}t(f{L
0t,0)t(u
0t,1)t(u
<=
≥=
Unit step function definition:
Used in conjunction with f(t) → f(t)u(t)
because of Laplace integral limits:
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12. Properties of Laplace Transforms
Linearity
Scaling in time
Time shift
“frequency” or s-plane shift
Multiplication by tn
Integration
Differentiation
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17. Properties: Multiplication by tn
)s(F
ds
d
)1()}t(ft{L n
n
nn
−=
Example :
1n
n
n
n
n
s
!n
)
s
1
(
ds
d
)1(
)}t(ut{L
+
=−
=
Proof :
)s(F
s
)1(dte)t(f
s
)1(
dte
s
)t(f)1(
dtet)t(f
dte)t(ft)}t(ft{L
n
n
n
0
st
n
n
n
0
st
n
n
n
0
stn
0
stnn
∂
∂
−=
∂
∂
−
=
∂
∂
−
=
==
∫
∫
∫
∫
∞
−
∞
−
∞
−
∞
−
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18. The “D” Operator
1. Differentiation shorthand
2. Integration shorthand
)t(f
dt
d
)t(fD
dt
)t(df
)t(Df
2
2
2
=
=
)t(f)t(Dg
dt)t(f)t(g
t
=
= ∫∞−
)t(fD)t(g
dt)t(f)t(g
1
a
t
a
−
=
= ∫if
then then
if
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19. Properties: Integrals
s
)s(F
)}t(fD{L 1
0 =−
Example :
)}t{sin(L
1s
1
)
1s
s
)(
s
1
(
)}tcos(D{L
22
1
0
+
=
+
=−
Proof :
let
stst
0
st
1
0
e
s
1
v,dtedv
dt)t(fdu),t(gu
dte)t(g)}t{sin(L
)t(fD)t(g
−−
∞
−
−
−==
==
=
=
∫
∫
∫
=
=+−= −∞−
t
0
st
0
st
dt)t(f)t(g
s
)s(F
dte)t(f
s
1
]e)t(g
s
1
[
∫
∞
−
∞<⇒∞=
0
)()( dtetft st
If t=0, g(t)=0
for so
slower than∫
∞
∞→=
0
)()( tgdttf 0→−st
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20. Properties: Derivatives
(this is the big one)
)0(f)s(sF)}t(Df{L +
−=
Example :
)}tsin({L
1s
1
1s
)1s(s
1
1s
s
)0(f
1s
s
)}tcos(D{L
2
2
22
2
2
2
2
−=
+
−
+
+−
=−
+
=−
+
=
+
Proof :
)s(sF)0(f
dte)t(fs)]t(fe[
)t(fv,dt)t(f
dt
d
dv
sedu,eu
dte)t(f
dt
d
)}t(Df{L
0
st
0
st
stst
0
st
+−
=+
==
−==
=
+
∞
−∞−
−−
∞
−
∫
∫
let
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21. Difference in
The values are only different if f(t) is not
continuous @ t=0
Example of discontinuous function: u(t)
)0(f&)0(f),0(f −+
1)0(u)0(f
1)t(ulim)0(f
0)t(ulim)0(f
0t
0t
==
==
==
+
−
→
+
→
−
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