the laplace transform
DESCRIPTION
THE LAPLACE TRANSFORM. LEARNING GOALS. Definition The transform maps a function of time into a function of a complex variable. Two important singularity functions The unit step and the unit impulse. Transform pairs Basic table with commonly used transforms. Properties of the transform - PowerPoint PPT PresentationTRANSCRIPT
THE LAPLACE TRANSFORM
LEARNING GOALS
DefinitionThe transform maps a function of time into a function of a complexvariable
Two important singularity functionsThe unit step and the unit impulse
Transform pairsBasic table with commonly used transforms
Properties of the transformTheorem describing properties. Many of them are useful as computational tools
Performing the inverse transformationBy restricting attention to rational functions one can simplifythe inversion process
Convolution integralBasic results in system analysis
Initial and Final value theoremsUseful result relating time and s-domain behavior
RoCs
s
defined well is integral the
ONE-SIDED LAPLACE TRANSFORM
limitlower the as consider tonecessary be willIt 0t
A SUFFICIENT CONDITION FOR EXISTENCE OF LAPLACE TRANSFORM
js
00)( ttf for assumes one transform the of uniqueness insure To
THE INVERSE TRANSFORM
Contour integralin the complex plane
Evaluating the integrals can be quite time-consuming. For this reason wedevelop better procedures that apply only to certain useful classes of function
Transform exists for
Re{ }s
TWO SINGULARITY FUNCTIONS
Using the unit step to build functions
Unit step (Important “test” functionin system analysis)
)()()( tutftf functions time positiveFor
This function has derivative that is zeroeverywhere except at the origin.We will “define” a derivative for it
Using square pulses to approximatean arbitrary function
The narrower the pulse the betterthe approximation
Computing the transform of the unit step
T
sxT
sx dxedxesU0
0lim1)(
Tsx
T es
sU0
1lim)(
)(lim1
)( jss
e
ssU
sT
T
0}Re{;1
)( sss
sURoC
j
ee
ssU
TjT
T
lim1
)(
An example of Region of Convergence (RoC)
)Re(s
)Im(s
0}Re{ s
Complex Plane
THE IMPULSE FUNCTION
Sifting or samplingproperty of the impulse
Representation of the impulse
Height is proportionalto area
Laplace transform
0ste
defined NOT is integral the or For 1020 tttt
0
)(
t
t
assumed islimit lower
the for transform valida have toorder In
Approximationsto the impulse
These two conditions are notfeasible for “normal” functions
(Good model for impact, lightning, and other well known phenomena)
LEARNING BY DOING
20 0 t
cos
2100 t
0
We will develop properties that willpermit the determination of a largenumber of transforms from a smalltable of transform pairs
Linearity
Multiplication by exponential
Multiplication by time
Time shifting
Time truncation
Some properties willbe proved and used asefficient tools in thecomputation of Laplacetransforms
LEARNING EXAMPLE
atetf )(for transform the Find
dtedteesF tasstat
0
)(
0
)(
( )
0
1 1( ) s a tF s e
s a s a
Basic Table of Laplace Transforms
We develop properties that expand thetable and allow computation of transformswithout using the definition
LINEARITY PROPERTY
Homogeneity
Additivity
Follow immediately from the linearityproperties of the integral
APPLICATION
0 21 1
[ ] [ ]2 2
j t j tst
j t j t
e ee dt
L e L e
ja ja
jsjssF
1
2
11
2
1)(
))((
)()(
2
1
jsjs
jsjs
22 s
s
With a similar use of linearity one shows
22][sin
s
tL
Additional entries for the table
LEARNING EXAMPLE
)3/3cos()( ttx
for transform Laplace the Find
tttx 3sin3/sin3cos3/cos)(
9
33/sin
93/cos)( 22
ss
ssX
4( ) 3 ( ) 3 tx t t e LEARNING EXAMPLE
1 1( ) 3 1 3
4X s
s s
Application of Linearity
Notice that the unit step is not shown explicitly. Hence
equivalent are
and )(33 tu
MULTIPLICATION BY EXPONENTIAL
( )
0 0
[ ( )] ( ) ( )at at st s a tL e f t e f t e dt f t e dt
( )F s a
LEARNING EXAMPLE
3( ) cos(10 )ty t e t
2( ) cos10 ( ) (From table)
100
sf t t F s
s
3
2
3( ) ( ) ( ) ( 3)
( 3) 100t s
y t e f t Y s F ss
LEARNING EXAMPLE
ttetx t 4sin3/sin4cos3/cos)( 2
16)2(
43/sin
16)2(
23/cos)( 22
ss
ssX
4,2 ba
New entries for the table of transform pairs
MULTIPLICATION BY TIME
Differentiation under an integral1
( ) ( )
1( )
u t U ss
dtu t
ds s
2
1
s
LEARNING EXAMPLE stepunit the be Let )(tu
)()( ttutr
function ramp the of transform the Find
1
!))(( n
n
s
ntut
shows oneproperty the of
napplicatio succesiveBy
2
2 1)(
sds
dtut 3
2
s
This result, plus linearity, allowscomputation of the transform of anypolynomial
3621)( tttx
LEARNING BY DOING
2 4
1 1 3!( ) 2 6X s
s s s )()()( tutxtx
Hence 0.tfor zero be to
functions theconsider wethat Remember
TIME SHIFTING PROPERTY
)()()()()()( 000 sFettuttfsFtutf st
LEARNING EXAMPLE
elsewhere0
311)(
ttf
)(tf
t1 3
1 )1( tu
)3( tu
)3()1()( tututf
ssss eess
es
esF 33 111)(
LEARNING EXTENSION
)1()1()( )1()1( tuetutetf tt
FOR TRANSFORM THE FIND
One can apply the time shifting propertyif the time variable always appears as itappears in the argument of the step. Inthis case as t-1
)1()1()11()( )1()1( tuetuettf tt
)1()1(
)1(
)1()1()1()(
)1(
)1(
)1()1(
tuet
tue
tuetuettf
t
t
tt
2
2
)1(
1)(
1)(
stute
sttu
t
( 1)2
( 1) ( 1)( 1)
st e
t e u ts
)1()1()( tuetetf t
write also could One
property
truncation time theapply And
)(tg
)]1([)()1()()( tgesFtutgtf sL
2
)1(
)1(
1)]1([
)1(
stg
teetetg tt
L
The two properties are only differentrepresentations of the same result
PERFORMING THE INVERSE TRANSFORM
FACT: Most of the Laplace transforms that we encounter are proper rational functions of the form
Zeros = roots of numeratorPoles = roots of denominator
nm
KNOWN: PARTIAL FRACTION EXPANSION
ii
iii
nPsQ
sP
sQ
sPKsF
nnnQ
sQsQsQ
)deg(;
)(
)(
)(
)()(
,)()deg(
)()()(
2
2
1
10
21
then
withr denominato the of
ionfactorizat COPRIME a is If
If m<n and the poles are simple
Simple, complex conjugate poles
Pole with multiplicity r
THE INVERSE TRANSFORM OF EACHPARTIAL FRACTION IS IMMEDIATE. WE ONLY NEED TO COMPUTE THE VARIOUS CONSTANTS
...)()(
)(22
222
1
s
C
s
sC
SIMPLE POLES
)/( ips
LEARNING EXAMPLE
)5)(4)(2(
)3)(1(12)(
ssss
sssF
Write the partial fraction expansion
542)( 4321
s
K
s
K
s
K
s
KsF
Determine the coefficients (residues)
10
9
542
3112)(
01
sssFK
1)3)(2)(2(
)1)(1(12)()2(
22
ssFsK
8
36
)1)(2)(4(
)1)(3(12)()4(
43
ssFsK
5
32
)1)(3)(5(
)2)(4(12)()5(
54 ssFsK
Get the inverse of each term and writethe final answer
)(5
32
8
36
10
9)( 542 tueeetf ttt
The step function is necessary to make the function zero for t<0
“FORM” of the inverse transform
)()( 54
43
221 tueKeKeKKtf ttt
COMPLEX CONJUGATE POLES
1 1| | | | jK K e
2cos
jj ee
Identity sEuler'
...)cos(||2)( 1 teKtf t
USING QUADRATIC FACTORS
...)()(
)(
)()(
)(22
222
122
1
1
s
C
s
sC
ssQ
sP
...sincos)( 21 teCteCtf tt Avoids using complex algebra. Must determine the coefficients in different way
The two forms are equivalent !
LEARNING EXAMPLE)54(
)2(10)( 2
sss
ssY
1)2(
)12)(12(542
2
s
jsjsss
1212)12)(12(
)2(10)(
*110
js
K
js
K
s
K
jsjss
ssY
45
20
)12)(12(
)2(10)(
00
jjssYKs
43.1535
5
)2)(12(
)1(10)()12(
121 jj
jsYjsK
js 43.153236.2 678.2236.2 je
2( ) 4 2 2.236 cos( 2.678) ( )ty t e t u t Using quadratic factors
1)2(1)2(
)2(
)54(
)2(10)( 2
22
102
s
C
s
sC
s
C
sss
ssY
)54(
)2()1)2((2
212
0
sss
sCssCsC
sCssCsCs 212
0 )2()1)2(()2(10
0520:0 Cs For
00
210
102
520:
2410:
0:
Cs
CCCs
CCs
40 C
401 CC
22 C 20 20:2 CCs For
210210:1 CCCs For
Alternative way to determine coefficients
2 20 1 2( ) ( cos sin ) ( )t ty t C C e t C e t u t
...)cos(||2)( 1 teKtf t
MUST use radians inexponent
MULTIPLE POLES ptnn et
nps
1
)!1(
1
)(
11-L
The method of identification of coefficients, or even the method of selectingvalues of s, may provide a convenient alternative for the determination of theresidues
rps )/( 1
LEARNING EXAMPLE
10)1(
)1(10)()2( 322
ssFsK
20)1(
)2(10)()1(
1
313
ssFsK
1
1
312 2
)3(10)()1(
s
s s
s
ds
dsFs
ds
dK
2 21 1
10( 2) 10( 3) 1010
( 2) ( 2)s s
s s
s s
21
32
2
11)2(
10
!2
1)()1(
!2
1
sds
dsFs
ds
dK
s
10
)2(
10
)2(
)2(210
2
1
13
1411
ss ss
sK
)(2
1)( 2
22
131211 tueKetKteKeKtf tttt
2113 0: KKs
212112 340: KKKs
21312111 33510: KKKKs
21312110 22230: KKKKs
)2(
)1()2()2)(1()2()1(
)2(
)3(10)()1(
321312
211
3
s
sKsKssKssK
s
ssFs
Using identification of coefficients
CONVOLUTION INTEGRAL
1
1 0 01
0
CLAIM: Given an ODE
... ...
there exists a function, ( ), 0, such that
( ) ( ) ( ) ( ) ( )
is a particular solution of the equation for 0
(Actually, the
n n m
n mn n m
t
d y d y d xa a y b b x
dt dt dth t t
y t h t x d h t x t
t
zero state response)
PROOF
Shifting
0;)()(
)(
0
)( ttdxxyety
sYt
xt
FINDEXAMPLE
ttyety t )()(2
1)(
2
1)(
ssY
ssY
2
1)(
2
11
ssY
s
)3(
2)( 2
ss
ssY
functions time positive are , ,If :RESULT 21 ff
)()()( 21 sFsFsF
LEARNING EXAMPLEUsing convolution to determine a network response
5
10
)(
)()( 0
ssV
sVsH
S
function Network
ssVS
1)(
Input
)(sH)(sVS )(0 sV
)()()(0 sVsHsV Sss
sV1
5
10)(0
)(1
)(105
10 5
tus
tues
t
)()()( 50 tutuetv t functions time positive are , ,If :RESULT 21 ff
)()()( 21 sFsFsF
t
t detv )(50 10)(
tt dee
0
5510 5 5
0
110
5
tte e
5 50 ( ) 2 1t tv t e e 0;12 5 te t
0tFor
In general convolution is not an efficient approach to determine theoutput of a system. But it can be a very useful tool in special cases
INITIAL AND FINAL VALUE THEOREMS
These results relate behavior of a function in the time domain with the behaviorof the Laplace transform in the s-domain
INITIAL VALUE THEOREM
)(lim)(lim
,),(
0 ssFtf
dtdftf
st Then transform.
Laplace have boththat Assume
0][lim
)0()(][
dt
df
fssFdt
df
s L
L
then bletransforma is derivative the if And
FINAL VALUE THEOREM
)(lim)(lim
)(lim
,),(
0 ssFtf
tfdt
dftf
st
t
Then exists. that and transform
Laplace have boththat Assume
00
0
)0()(lim)(
0
)0()()(
fssFdttdt
df
s
fssFdtetdt
df
s
st
as limits Taking
0
)()(lim
s
sFtft
at pole
single amost at andpart real negative with
poles has ifexist will :NOTE
LEARNING EXAMPLE
)(
.)22(
)1(10)( 2
tf
sss
ssF
for valuesfinal and initial the Determine
Given
)(lim)0( ssFf s
22
)1(10lim)0( 2
ss
sf s 0
)(lim)(lim 0 ssFtf st
F(s) has one pole at s=0 and the others have negative real part. The final valuetheorem can be applied.
22
)1(10lim)(lim 20
ss
stf st 5
)4
3cos(255)(
tetf t
gets one inverse the Computing :NOTE
Clearly, f(t) has Laplace transform. And sF(s) -f(0) is also defined.
LEARNING EXTENSION
)(
.)22)(2(
)1()( 2
2
tf
ssss
ssF
for valuesfinal and initial the Determine
Given
)22)(2(
)1(lim)0( 2
2
sss
sf s 0
)22)(2(
)1(lim)(lim 2
2
0
sss
stf st
4
1
Laplace