laplace transform
TRANSCRIPT
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The Laplace Transform
The University of TennesseeElectrical and Computer Engineering Department
Knoxville, Tennessee
wlg
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The Laplace Transform
The Laplace Transform of a function, f(t), is defined as;
0
)()()]([ dtetfsFtfL st
The Inverse Laplace Transform is defined by
j
j
tsdsesFj
tfsFL
)(
2
1)()]([1
*notes
Eq A
Eq B
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The Laplace Transform
We generally do not use Eq B to take the inverse Laplace. However,
this is the formal way that one would take the inverse. To use
Eq B requires a background in the use of complex variables and
the theory of residues. Fortunately, we can accomplish the same
goal (that of taking the inverse Laplace) by using partial fraction
expansion and recognizing transform pairs.
*notes
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The Laplace Transform
Laplace Transform of the unit step.
*notes
|0
0
11)]([
stst es
dtetuL
stuL
1)]([
The Laplace Transform of a unit step is:
s
1
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The Laplace Transform
The Laplace transform of a unit impulse:
Pictorially, the unit impulse appears as follows:
0 t0
f(t) (t – t0)
Mathematically:
(t – t0) = 0 t 0
*note
01)(0
0
0
dtttt
t
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The Laplace Transform
The Laplace transform of a unit impulse:
An important property of the unit impulse is a sifting or sampling property. The following is an important.
2
1 2010
20100 ,0
)()()(
t
ttttt
ttttfdttttf
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The Laplace Transform
The Laplace transform of a unit impulse:
In particular, if we let f(t) = (t) and take the Laplace
1)()]([ 0
0
sst edtettL
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The Laplace Transform
An important point to remember:
)()( sFtf
The above is a statement that f(t) and F(s) are transform pairs. What this means is that for each f(t) there is a unique F(s) and for each F(s)there is a unique f(t). If we can remember thePair relationships between approximately 10 of the Laplace transform pairs we can go a long way.
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The Laplace Transform
Building transform pairs:
eL(
e
tasstatat dtedteetueL0
)(
0
)]([
asas
etueL
stat
1
)()]([ |
0
astue at
1
)( A transform pair
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The Laplace Transform
Building transform pairs:
0
)]([ dttettuL st
0 00| vduuvudv
u = tdv = e-stdt
2
1)(
sttu
A transform pair
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The Laplace Transform
Building transform pairs:
22
0
11
2
1
2
)()][cos(
ws
s
jwsjws
dteee
wtL stjwtjwt
22)()cos(
ws
stuwt
A transform
pair
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The Laplace Transform
Time Shift
0 0
)( )()(
,.,0,
,
)()]()([
dxexfedxexf
SoxtasandxatAs
axtanddtdxthenatxLet
eatfatuatfL
sxasaxs
a
st
)()]()([ sFeatuatfL as
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The Laplace Transform
Frequency Shift
0
)(
0
)()(
)]([)]([
asFdtetf
dtetfetfeL
tas
statat
)()]([ asFtfeL at
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The Laplace Transform
Example: Using Frequency Shift
Find the L[e-atcos(wt)]
In this case, f(t) = cos(wt) so,
22
22
)(
)()(
)(
was
asasFand
ws
ssF
22 )()(
)()]cos([
was
aswteL at
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The Laplace Transform
Time Integration:
The property is:
stst
t
stt
es
vdtedv
and
dttfdudxxfuLet
partsbyIntegrate
dtedxxfdttfL
1,
)(,)(
:
)()(
0
0 00
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The Laplace Transform
Time Integration:
Making these substitutions and carrying outThe integration shows that
)(1
)(1
)(00
sFs
dtetfs
dttfL st
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The Laplace Transform
Time Differentiation:
If the L[f(t)] = F(s), we want to show:
)0()(])(
[ fssFdt
tdfL
Integrate by parts:
)(),()(
,
tfvsotdfdtdt
tdfdv
anddtsedueu stst
*note
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The Laplace Transform
Time Differentiation:
Making the previous substitutions gives,
0
00
)()0(0
)()( |
dtetfsf
dtsetfetfdt
dfL
st
stst
So we have shown:
)0()()(
fssFdt
tdfL
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The Laplace Transform
Time Differentiation:
We can extend the previous to show;
)0(...
)0(')0()()(
)0('')0(')0()()(
)0(')0()()(
)1(
21
233
3
22
2
n
nnnn
n
f
fsfssFsdt
tdfL
casegeneral
fsffssFsdt
tdfL
fsfsFsdt
tdfL
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The Laplace TransformTransform Pairs:
____________________________________
)()( sFtf
f(t) F(s)
1
2
!
1
1
1)(
1)(
nn
st
s
nt
st
ase
stu
t
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The Laplace TransformTransform Pairs:
f(t) F(s)
22
22
1
2
)cos(
)sin(
)(
!
1
ws
swt
ws
wwt
as
net
aste
natn
at
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The Laplace TransformTransform Pairs:
f(t) F(s)
22
22
22
22
sincos)cos(
cossin)sin(
)()cos(
)()sin(
ws
wswt
ws
wswt
was
aswte
was
wwte
at
at
Yes !
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The Laplace TransformCommon Transform Properties:
f(t) F(s)
)(1
)(
)()(
)0(...)0(')0()()(
)()(
)([0),()(
)(0),()(
0
1021
00
000
sFs
df
ds
sdFttf
ffsfsfssFsdt
tfd
asFtfe
ttfLetttutf
sFetttuttf
t
nnnnn
n
at
sot
sot
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The Laplace TransformUsing Matlab with Laplace transform:
Example Use Matlab to find the transform of tte 4
The following is written in italic to indicate Matlab code
syms t,slaplace(t*exp(-4*t),t,s)ans = 1/(s+4)^2
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The Laplace TransformUsing Matlab with Laplace transform:
Example Use Matlab to find the inverse transform of
19.12.)186)(3(
)6()(
2prob
sss
sssF
syms s t
ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))
ans = -exp(-3*t)+2*exp(-3*t)*cos(3*t)
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The Laplace TransformTheorem: Initial Value
If the function f(t) and its first derivative are Laplace transformable and f(t)Has the Laplace transform F(s), and the exists, then)(lim ssF
0
)0()(lim)(lim
ts
ftfssF
The utility of this theorem lies in not having to take the inverse of F(s) in order to find out the initial condition in the time domain. This is particularly useful in circuits and systems.
Theorem:
s
Initial Value Theorem
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The Laplace Transform
Initial ValueTheorem:Example:
Given;
22 5)1(
)2()(
s
ssF
Find f(0)
1)26(2
2lim
2512
2lim
5)1(
)2(lim)(lim)0(
2222
222
2
2
22
sssss
ssss
ss
ss
s
ssssFf
ss s
s
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The Laplace TransformTheorem: Final Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)has the Laplace transform F(s), and the exists, then)(lim ssF
s
)()(lim)(lim ftfssF0s t
Again, the utility of this theorem lies in not having to take the inverseof F(s) in order to find out the final value of f(t) in the time domain. This is particularly useful in circuits and systems.
Final ValueTheorem
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The Laplace Transform
Final Value Theorem:Example:
Given:
ttesFnotes
ssF t 3cos)(
3)2(
3)2()( 21
22
22
Find )(f .
03)2(
3)2(lim)(lim)(
22
22
s
ssssFf
0s0s
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