group 2 bhadouria , arjun singh glave , theodore dean han, zhe
DESCRIPTION
Group 2 Bhadouria , Arjun Singh Glave , Theodore Dean Han, Zhe. Chapter 5. Laplace Transform Chapter 19. Wave Equation. Laplace Transform. Chapter 5. 5. Laplace Transform. 5.1. Introduction & Definition 5.2. Calculation of the Transform 5.3. Properties of the Transform - PowerPoint PPT PresentationTRANSCRIPT
Group 2
Bhadouria, Arjun SinghGlave, Theodore Dean
Han, Zhe
Chapter 5. Laplace TransformChapter 19. Wave Equation
Laplace Transform
Chapter 5
5. Laplace Transform
• 5.1. Introduction & Definition• 5.2. Calculation of the Transform• 5.3. Properties of the Transform• 5.4. Application to the Solution of Differential
Equations• 5.5. Discontinuous Forcing Functions; Heavisid
e Step Function• 5.6. Impulsive Forcing Functions; Dirac Impuls
e Function• 5.7. Additional Properties
5.1. Introduction & Definition
• The Laplace transform is a widely used integral transform. Denoted , it is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms it to a function F(s) with a complex argument s. The Laplace transform has the useful property that many relationships and operations over the originals f(t) correspond to simpler relationships and operations over the images F(s). The Laplace transform has many important applications throughout the sciences.
• The Laplace transform is used for solving differential and integral equations. In physics and engineering, it is used for analysis of linear time-invariant systems. In this analysis, the Laplace transform is often interpreted as a transformation from the time-domain, in which inputs and outputs are functions of time, to the frequency-domain, where the same inputs and outputs are functions of complex angular frequency, in radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.
http://en.wikipedia.org/wiki/Laplace_transform
Basic idea
Tim e d o m a inu n kn o w n f(t), d /d t, D iff E q s
F req u en c y d o m a inu n kn o w n F (s ), A lg E q s
L ap laceTra n s fo rm a tio n
S o lv eA lgeb ra icE q u a tio n s
F req u en c y d o m a inkn o w n F (s )
T im e d o m a inkn o w n f(t)
S o lveD iffe ren tia lE q u a tio n s
In ve rs eL ap laceTran s fo rm
http://faculty.mercer.edu/olivier_pd/documents/Ch2LaplaceTransforms.ppt
• Given a known function K(t,s), an integral transform of a function f is a relation of the form
• The Laplace Transform of f is defined as
where the kernel function is K(s,t) = e-st , a=0, b= . F (s) is the symbol for the Laplace transform, L is the Laplace transform operator, and f(t) is some function of time, t.
b
adttfstKsF )(),()(
0)()()( dttfesFtfL st
How to solve problems
tttt xyyy 12
tt eet 2
21
21y
s
t
sss
1X
231)H( 2
2311
2 sss
1x t
Time Domain Frequency Domain
Solve algebraic equation
Laplace transform
Inverse Laplace transform
Douglas Wilhelm Harder, University of Waterloo.
5.2. Calculation of the Transform
Since the Laplace Transform is defined by an improper integral, thus it must be checked whether the transform F(s) of a given function f(t) exists, that is
whether the integral converges.
0)()( dttfesF st
EXAMPLE 2.1. Consider the following improper integral.
We can evaluate this integral as follows:
Note that if s = 0, then est = 1. Thus the following two cases hold:
0dtest
1lim1limlim0
00
sb
b
bst
b
b st
b
st ess
edtedte
0. if,diverges and0; if,100
sdtess
dte stst
EXAMPLE 2.2.Consider the following improper integral.
We can evaluate this integral using integration by parts:
Since this limit diverges, so does the original integral.
0cos tdtst
1cossinlim
cossinlim
sinsinlim
coslimcos
00
00
00
bsbsb
tstst
tdtstst
tdtsttdtst
b
bb
b
bb
b
b
b
Exponential orderSuppose that f is a function for which the following hold:
(1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | Kect when t T, for constants c, K, M, with K, M > 0.
A function f that satisfies the conditions specified above is said to to have exponential order as t
EXAMPLE 2.3. Is f(t)=exp(4t)cost of exponential order?Solution: Yes.
ttt eetetf 444 10cos)(
Therefore, K=10, c=4, and T=10 for instance
EXAMPLE 2.4. Is f(t)=ln(t) of exponential order?Solution: Yes.
011)ln( limlim
ctt
ctt cete
t
Therefore, K=100, c=1, and T=10 for instance
Piecewise ContinuousIf an interval [a, b] can be partitioned by a finite number of
points a = t0 < t1 < … < tn = b such that
Then the function f is piecewise continuousOr we can say f is piecewise continuous on [a, b] if it is
continuous there except for a finite number of jump discontinuities.
nktf
nktf
k
k
tt
tt
,,1,)(lim)3(
1,,0,)(lim)2(
) t,(teach on continuous is f (1)
1
1kk
Picture from Paul's Online Math Notes
EXAMPLE 2.5. Piecewise continuous functionConsider the following piecewise-defined function f.
• (a)
From this definition of f, and from the graph of f below, we see that f is piecewise continuous on [0, 3].
32121,310,
)(
2
tttttt
tf
• (b)
From this definition of f, and from the graph of f below, we see that f is NOT piecewise continuous on [0, 3].
32,421,210,1
)( 1
2
ttttt
tf
http://ebookbrowse.com/ch06-laplace-transform-ppt-d116265990
THEOREM 2.1 Existence of the Laplace Transform
Let f(t) satisfy these conditions:(i) f(t) is piecewise continuous on for every A>0, (ii) f(t) is of exponential order as
That is to say(1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | Kect when t T, for constants c, K, M, with K, M > 0.
At0
t
Then the Laplace Transform of f exists for s > c. finite )()()(
0 dttfesFtfL st
Inverse Laplace transform operatorBy definition, the inverse Laplace transform operator, L-1, converts an s-domain function back to the
corresponding time domain function:
Importantly, both L and L-1 are linear operators.Thus,
1 1{ } ( )2
i sti
f t L F s F s e dsi
{ } { } { }L au t bv t aL u t bL v t aU s bV s
1{ }L aU s bV s au t bv t
Examples of Calculation2.6.
2.7
2.8
0 0
)( 1)(;)(as
dtedteesFetf tasstatat
attf sin)(
0,)()(1
sin/)sin(lim1
cos/)cos(lim
sinlimsin)sin()(
222
2
00
00
00
sas
asFsFas
a
ateasaate
as
a
ateasaate
atdteatdteatLsF
b stbst
b
b stbst
b
b st
b
st
;tan tconsaf(t) 0
0
0st sta a aF s ae dt es s s
Laplace transform table
5.3. Properties of the Transform
THEOREM 3.0The transform of an expression that is multiplied by a constant is the constant multiplied by the transform. That is:
1 1( ) ( ) and ( ) ( )L kf t kL f t L kF s kL F s
• A various types of problems that can be treated with the Laplace transform include ordinary and partial differential equations as well as integral equations.
• Suppose u and v are functions whose Laplace transforms exist for s > a1 and s > a2, respectively.
• Then, for s greater than the maximum of a1 and a2, the Laplace transform of au(t) + bv(t) exists. That is,
With
Therefore
finite is )()()()(0 dttbvtauetbvtauL st
)( )()( )()()(00
tvbLtuaLdttvebdttueatbvtauL stst
THEOREM 3.1 Linearity of the Transform
)( )()()( tvbLtuaLtbvtauL
EXAMPLE 3.1.f (t) = 5e-2t - 3sin(4t) for t 0.
by linearity of the Laplace transform, and using results of Laplace transform table, the Laplace transform F(s) of f is:
0,
1612
25)4sin(35
)4sin(35)}({)(
22
2
sss
tLeL
teLtfLsF
t
t
For any U(s) and V(s) such that the inverse transforms
For any constants a,b.
THEOREM 3.2 Linearity of the Inverse Transform
1{ }L aU s bV s au t bv t
1 1{ } { } .L U s u t and L V s v t exist
Basic idea : Consider a general expression,
Expand a complex expression for F(s) into simpler terms, each of which appears in the Laplace Transform table. Then you can take the L-1 of both sides of the equation to obtain f(t).
1
n
ii
N s N sF s
D s s b
5
1 4sF s
s s
Perform a partial fraction expansion (PFE)
1 25( )
1 4 1 4sF s
s s s s
where coefficients and have to be determined.1 2
EXAMPLE 3.2.
To find : Multiply both sides by s + 1 and let s = -11
1 21 4
5 4 5 14 3 1 3s s
s ss s
To find : Multiply both sides by s + 4 and let s = -42
1 1
1 1 4
4 1 1 1( ) { ( )} { }3 1 3 4
4 1 1 1 4 1{ } { }3 1 3 4 3 3
t t
f t L F s Ls s
L L e es s
Therefore,
Let f(t) be continuous and f’(t) be piecewise continuous on 0≤t≤t。For every finite t。, and let f(t) be of exponential order as t so that there are constants K, c, T such that | f(t) | Kect when t T. Then L{f’(t)} exists for all s>c.
• This is a very important transform because derivatives appear in the ODEs we wish to solve.
• Similarly, for higher order derivatives:
where:
{ '( )} { ( )} 0L f t sL f t f
1 2 1( ) 1 2{ ( )} 0 0 ... 0 0n nn n n nL f t s F s s f s f sf f
0
0n
nn
t
d ffdt
THEOREM 3.3 Transform of the Derivative
Deriving the Laplace transform of f (t ) often requires integration by parts. However, this process can sometimes be avoided if the transform of the derivative is known:
For example, if f (t ) = t then f ‘ (t ) = 1 and f (0) = 0 so that, since:
That is:
( ) { ( )} (0) then {1} { } 0L f t sL f t f L sL t
2
1 1{ } therefore { }sL t L ts s
Proof
It has already been established that if:( ) { ( )} and ( ) { ( )}F s L f t G s L g t
then: { ( )} ( ) (0) and { ( )} ( ) (0)L f t F s f L g t G s g
Now let ( ) ( ) so (0) (0) and ( ) ( ) (0)g t f t g f G s sF s f
so that: { ( )} { ( )} ( ) (0) ( ) (0) (0)L g t L f t sG s g s sF s f f
Therefore: 2{ ( )} ( ) (0) (0)L f t s F s sf f
For example, if:
then:
and
Similarly:
And so the pattern continues.
( ) { ( )}F s L f t
{ ( )} ( ) (0)L f t F s f
2{ ( )} ( ) (0) (0)L f t s F s sf f
3 2{ ( )} ( ) (0) (0) (0)L f t s F s s f sf f
EXAMPLE 3.3
Then substituting in:
yields
So:
2( ) sin so that ( ) cos and ( ) sinf t kt f t k kt f t k kt
2{ ( )} ( ) (0) (0)L f t s F s sf f
2 2 2{ sin } {sin } {sin } .0L k kt k L kt s L kt s k
2 2{sin } kL kts k
)}sin({1
11
)1(11
)0(1
)}cos({ 22
22
2
2
2
2
tLss
sss
sfs
stdtdL
EXAMPLE 3.4
•COMMENT: Difference in )0(&)0(),0( fff
The values are only different if f(t) is not continuous at t=0
Example of discontinuous function: u(t)
1)0()0(
1)(lim)0(
0)(lim)0(
0
0
uf
tuf
tuf
t
t
*Additional Section
EXAMPLE 3.5Try to solve the differential equation:
( ) ( ) 1 where (0) 0f t f t f take the Laplace transform of both sides of the differential equation to yield:
( ) ( ) 1 so that ( )} { ( ) 1L f t f t L L f t L f t L
1 1( ) (0) ( ) which means that ( 1) ( )sF s f F s s F ss s
1( )( 1)
F ss s
Resulting in:
The right-hand side can be separated into its partial fractions to give:1 1( )
1F s
s s
From the table of transforms it is then seen that:
1 1 11 1 1 1( ) 11 1
tf t L L L es s s s
( ) 1 tf t e Thus,
THEOREM 3.4 Laplace Convolution Theorem
If both exist for s>c, then
{ } { }L f t F s and L g t G s
10
{ } ( ) ( )t
L F s G s f g t d
0{ ( ) ( ) } .
tor L f g t d F s G s for s c
0( )( ) ( ) ( )
tdefine f g t f g t d As Laplace convolution of f and g.
Proof: { }L f t F s
0)()( duufesF su
0
)()( dvvgesG svWe have,
})()({
])()([
)()(
)()(
])([])([)()(
0
0 0
0 0
0 0
)(
00
duutgufL
dtduutgufe
dudtutgufe
dudvvgufe
dvvgeduufesGsF
t
t
t
u
st
t
t
u
st
vus
svsu
10
{ } ( ) ( )t
L F s G s f g t d therefore
EXAMPLE 3.6 If f(t)=exp(t), g(t)=t, then
10
)(0
)()(*0
tet
eet
etdtetgf tt
EXAMPLE 3.7 ))((
1}*{bsas
eeL btat
)(*}))((
1{ )(
0
1 babaeedeeee
bsasL
btattbt abtat
Schiff. Joel L. The Laplace transform: theory and applications. P92
5.4. Application to the Solution of Differential Equations
• Laplace transforms play a key role in important engineering concepts and techniques.
Examples: • Transfer functions • Frequency response• Control system design• Stability analysis• …
Solution of ODEs by Laplace Transforms
Procedure:
1. Take the L of both sides of the ODE.
2. Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., F(s).
3. Perform a partial fraction expansion.
4. Use the L-1 to find f(t) from the expression for F(s).
Equation with initial conditions
Laplace transform is linear
Apply derivative formula
Rearrange
Take the inverse
EXAMPLE 4.1.
'' 1, (0) '(0) 0f f f f
{ ''} { } {1}L f L f L
2 1( ) (0) '(0) ( )s F s sf f F ss
2 21 1( )
( 1) 1sF s
ss s s
( ) 1 cosf t t
'' 1, (0) '(0) 0f f f f Solve
resulted in the expression 3 21
6 11 6F s
s s s s
31 2 41
1 2 3 1 2 3
1/ 6 1/ 2 1/ 2 1/ 61 2 3
F ss s s s s s s s
s s s s
''' 6 '' 11 ' 6 1, (0) 0f f f f f
EXAMPLE 4.2.
Take L-1 of both sides:1 1 1 1 11/ 6 1/ 2 1/ 2 1/ 6{ ( )} { } { } { } { }
1 2 3L F s L L L L
s s s s
2 31 1 1 16 2 2 6
t t tf t e e e
EXAMPLE 4.3.3 ' 2 4 2, (0) 0xf f e f
Taking Laplace transforms of both sides of this equation gives:
4 2 6 23[ ( ) (0)] 2 ( )1 ( 1)
ssF s f F ss s s s
6 2 27 1 1 4 1( ) ( ) ( )( 1)(3 2) 5 3 2 5 1
27 1 1 4 1( ) ( )15 2 / 3 5 1
sF ss s s s s s
s s s
2 /39 4( ) 15 5
x xf t e e
K.A. Stroud. Engineering Mathematics. P1107
EXAMPLE 4.4.A mass m is suspended from the end of a vertical spring of constant k (force required to produce unit stretch). An external force f(t) acts on the mass as well as a resistive force proportional to the instantaneous velocity. Assuming that x is the displacement of the mass at time t and that the mass starts from rest at x=0,(a) Set up a differential equation for the motion(b) Find x at any time t
Solution:(a)The resistive force is given by –βdx/dt. The restoring force is given by –kx. Then by Newton’s law, 2
2 ( ) (1)d x dxm kx F tdtdt
2
2 ( ) (1)d x dxor m kx F tdtdt
0, '(0) 0 (2)where x x
(b) Taking the Laplace transform of (1), using we obtain
So that on using (2)
Where
There are three cases to be considered.
EXAMPLE 4.4.
{ } , ( )L f t F s L x X
2[ (0) '(0)] [ (0)] ( )m s X sx x sX x kx F s
2 2( ) ( ) (3)
[( / 2 ) ]F s F sX
ms s k m s m R
2
24kRm m
• Case 1, R>0. In this case letWe have
Then we find from (3)
• Case 2, R=0. In this case
thus
EXAMPLE 4.4.2R
1 /22 2
1 sin{ }( / 2 )
t m tL es m
( )/20
1 ( ) sin ( )t t u mx F u e t u du
m
1 /22
1{ }( / 2 )
t mL tes m
( )/0
1 ( )( )t t u mx F u t u e du
m
• Case 3. R<0, In this case letWe have
EXAMPLE 4.4.
2R
1 /22 2
1 sinh{ }( / 2 )
t m tL es m
( )/20
1 ( ) sin ( )t t u mx F u e h t u du
m
Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists. P115, Problem 4.46
5.5. Discontinuous Forcing Functions; Heaviside Step Function
0, 0( )
1, 0t
H tt
The unit step function is widely used. It is defined as:
0 0
1 1{ ( )} ( ) [ ]0
t tst st st
t t
tL H t H t e dt e dt e
ts s
Because the step function is a special case of a “constant”, it follows
More generally,0,
( )1,
t aH t a
t a
{ ( ) ( )} ( )asL H t a f t a e F s
It is possible to express various discontinuous functions in terms of the unit step function.
0{ ( )} (0) (1) 0 0
st asa st st
a
e eL H t a e dt e dt if sas s
{ ( )} 0aseL H t a s
s
1{ } ( ) 0aseL H t a s
s
Therefore, or
EXAMPLE 5.10
1 1( ) ( ( ) ( ))
0
a b
t a
H t H t a H t b a t bb a b a
t b
{ ( )}( )
as bse eL H ts b a
Schiff. Joel L. The Laplace transform: theory and applications. P92
EXAMPLE 5.2Determine L{g(t)} for
2
0, 0 1( )
( 1) , 1t
g tt t
2 23 3
2 2{ ( )} { ( 1)( 1) } { }s
s s eL g t L H t t e L t es s
Schiff. Joel L. The Laplace transform: theory and applications. P92
5.6. Impulsive Forcing Functions; Dirac Impulse Function
Pictorially, the unit impulse appears as follows:
Mathematically:
01)(0
0
0
dtttt
t
tt
tt0
0/1)( 0
tt
tt0
0);( 0
0lim
Picture from web.utk.edu
2
1 2010
20100 ,0
)()()(
t
t ttttttttg
dttttg
)0()()(0
gdtttg
We can prove that
Where 010 ttt
)()()(1)()( 0100
10
00
limlimlim tgtgtgdttttg
thus
)(t is known as Dirac delta function, or unit impulse function
further
The Laplace transform of a unit impulse:if we let f(t) = (t) and take the Laplace
1)()}({ 0
0
sst edtettL
* Rectangular Pulse Function
0 for 0
for 00 for
w
w
tf t h t t
t t
1 wt shF s es
*Additional Section
EXAMPLE 6.1Some useful properties of Dirac Impulse Function
1{ } ( ) 2 ats aL t aes a
1( ) ( ) duax dx u
a a
( )( ) xaxa
( ) ( ) ( )f t t T dt f T
5.7. Additional Properties
THEOREM 7.1 S-plane (frequency) shiftIf exists for s>s0, then for any real constant a,
for s+a>s0, or equivalently,)()}({ asFtfeL at
{ }L f t F s
)()}({1 tfeasFL at
Proof : )()()()}({0
)(
0
asFdtetfdtetfetfeL tasstatat
EXAMPLE 7.1.
22)()}sin({
asteL at
EXAMPLE 7.2.
22sin2cos2}
4)1(1{}
4)1()1({2
}4)1(1)1(2{}
4)1(12{}
5212{
21
21
21
21
21
tetes
Ls
sL
ssL
ssL
sssL
tt
THEOREM 7.2 Time ShiftIf exists for s>s0, then for any real constant
a>0,
for s>s0, or, equivalently,
{ }L f t F s
)()}()({ sFeatfatHL as
)()()}({1 atfatHsFeL as
)()()(
,
)()()()}()({
00
)(
0
sFedueufedueuf
autatuset
dteatfdteatfatHatfatHL
assuasa
aus
a
stst
Proof :
EXAMPLE 7.3.
aseetHL
sta
10)10( })10({
THEOREM 7.3 Multiplication by 1/s (Integrals)If exists for s>s0, then
for s>max{0,s0}, or, equivalently,
{ }L f t F s
ssFtfDLdfL
t )()}({})({ 100
dfssFL
t
0
1 )(})({
Proof : ststte
svdtedvdttfdudftfDtguset
1,,)(,)()()(0
10
ssFdtetf
setg
s
tgdes
etgs
des
tgdtetgtgLdfL
stst
ststststt
)()(1])(1[
))((1])(1[)1)(()()}({})({
0
000
0
0,)()(,)()(for 0,g(t) 0, tIf00
stst ethanslowertgdttfsodtetft
Notice:
EXAMPLE 7.4. )}{sin(
11)
1)(1(})cos({})({ 2200
tLss
ss
dLdfLtt
COMMENT:Another way to prove Time Integration Property
ssF
dτeτfs
tdτdeτf
tdτdeτf
dtedττfdττfL
st
τ
st
τ
st
sttt
0
0
0
0 00
1
From Douglas Wilhelm Harder, University of Waterloo.
THEOREM 7.4 Differentiation with Respect to s (Multiplication by tn)If exists for s>s0, then
for s>s0, or, equivalently,
{ }L f t F s
)()1(})({1 tftds
sFdL nnn
n
Proof :
EXAMPLE 7.5.
)()1()}({ sFdsdtftL n
nnn
)()1()()1(
)()1()()()}({
0
000
sFs
dtetfs
dtes
tfdtettfdtetfttftL
n
nnst
n
nn
stn
nnstnstnn
1
!)1()1(}1{)}({ nn
nnnn
sn
sdsdtLtutL
00
000
cossin1
cossin1sin1sin
dtettdtets
dtettts
etts
dtett
stst
ststst
00
000
sincos1
sincos1cos1cos
dtettdtets
dtettts
etts
dtett
stst
ststst
EXAMPLE 7.6.
tttf sin)(
0000
sincos1sin1sin dtettdtets
dtets
dtett stststst
Firstly
where
substitute
Douglas Wilhelm Harder, [email protected]
EXAMPLE 7.6.
tttf sin)(
0000
sincos1sin1sin dtettdtets
dtets
dtett stststst
222
22
sin1
11
1
sin1
11
11sin
sttL
ssss
ttLs
ssss
ttL
substitute
1
11
2sin
121sin
222
22
2
sdsd
ssttL
ssssttL
therefore
THEOREM 7.5 Integration with Respect to sIf there is a real number s0 such that exists for s>s0, and exists, then
for s>s0, or, equivalently,
{ }L f t F s
EXAMPLE 7.7. To evaluate
')'(})({ dssFttfL
s
0lim /t f t t
ttfdssFL
s
)(}')'({1
}{ln1
sasL
sassas
dsd 11ln
')'
1'1(lnln
11
1 dssass
ass
as s
s
')'
1'1(ln ds
sassas
s
1}11{)( 1
ate
sasLtf
te
sasL
at
1}{ln1
THEOREM 7.6 Large s Behavior of F(s)Let f(t) be piecewise continuous on for each finite and of exponential order as , then(i) (ii)
00 t t
Proof : Since f(t) is of exponential order as then there exist real constants K and c, with K 0 such that | f(t) | Kect for all t T. Since f(t) is piecewise continuous onThere must be a finite constant M such that | f(t) | M, onFor all s>c:
0tt
( ) 0F s as s ( )sF s is bounded as s
t
00 t t
00 t t
csK
sM
tcseK
seMdtKedtMe
dtetfdtetfdtetfdtetfsF
tcsst
t
tcst st
t
stt ststst
0
)()(
0
000
)(1
)()()()()(
0
0
0
0
0
THEOREM 7.7 Initial-Value TheoremLet f be continuous and f’ be piecewise continuous on for each finite , and let f and f’ be of exponential order as then
00 t t
Proof : with the stated assumptions on f and f’, it follows that
0t t [ ( )] (0)lim
ssF s f
NOTE: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.
{ '( )} ( ( )) (0) ( ) (0)L f t sL f t f sF s f
Since f’ satisfies the conditions of THEOREM 4.7, it follows that
{ '( )} 0L f t as s
[ ( )] (0)lims
sF s f
thus
Given22 51
2
)(s)(sF(s)
Find f(0)
1262
2
25122]
512[][0
2222
222
2
2
22
lim
limlimlim
)s(ssssssss
ssss
)(s)(sssF(s))f(
s
sss
EXAMPLE 7.8.
THEOREM 7.8 Final Value TheoremLet f be continuous and f’ be piecewise continuous on, and let f and f’ be of exponential order as then
0 t
EXAMPLE 7.9.
Suppose
then
t
0[ ( )] ( )lim lim
s tsF s f t
NOTE:It can be used to find the steady-state value of a closed loop system (providing that a steady-state value exists.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.
5 25 4sF s
s s
0
5 2lim 0.5lim5 4t s
sf f ts
*Additional Section
Given:tte(s)Fnote
)(s
)(sF(s) t 3cos12322
2322 2
)(ffind
.0]
232)2(
232)2([)]([)( limlim00
s
ssssFfss
EXAMPLE 7.10.
THEOREM 7.9 Transform of Periodic FunctionIf f is periodic with period T on and piecewise continuous one period, then
0 t
EXAMPLE 7.11.
0for s 0
1{ ( )} ( )1
T stsTL f t f t e dt
e
(a)cos(t) is repeated with period 2 (b)cos(t) is repeated with period tcos
tL cos
tf
tfL
πs
πs
sπ
πst
es
sse
e
dtettfL
11
1
cos2
0From Douglas Wilhelm Harder, University of Waterloo.
Consider f(t) below:(a) (b)
s
s
s
s
s
st
ese
es
e
e
dtetfL 222
1
0
11
1
1
1
tf tu
s
tuL 1 tfL
EXAMPLE 7.12.
ss
dtetuL st 110
From Douglas Wilhelm Harder, University of Waterloo.
THEOREM 7.10 Scaling in TimeIf exists for s>s0, then for any real constant a>0,
for s>s0
Proof :
EXAMPLE 7.13.
)(1)}({asF
aatfL
{ }L f t F s
)(1)(1)()}({
1,,
0
)(
0 asF
adueuf
adteatfatfL
dua
dtautatulet
a uas
st
2222
2
2)(1)1
)(1(1)}{sin(
ssstL
222 1)/(/1cos
ωss
ss
ωωtL
*Additional Section
THEOREM 7.11 Time delayTime delays occur due to fluid flow, time required to do an analysis (e.g., gas chromatograph). The delayed signal can be represented asthen
EXAMPLE 7.14.
θ θ time delayf t
θ{ θ } sL f t e F s
θ2 2{ θ } {cos ( θ)} s sL f t L t e
s
*Additional Section
Common Transform Properties Tablef(t) F(s)
)()(
)(1)(
)(1
1)(
')'()(
)0(10...)0('2)0(1)()(
)()1()(
)(
0)(
)(0),()(
)()(
0
sFetfasF
aatf
dtetfe
tf
dssFttf
fnfsfnsfnssFnsndt
tfnd
dssFdtft
ssFt
df
sFaseaatfatH
asFtfate
s
T stsT
s
n
nnn
EXAMPLE 7.15Solve the ODE,
5 4 2 0 1df f fdt
First, take L of both sides of the equation,
25 1 4sF s F ss
Rearrange,
5 25 4sF s
s s
Take L-1, 1 5 2{ }
5 4sf t L
s s
From Table, 0.80.5 0.5 tf t e
References• Schiff. Joel L. The Laplace transform: theory and applications.• Murray. R. SPIEGEL. Schaum’s Outline of Theory and
Problems of Advanced Mathematics for Engineers and Scientists.
• K.A. Stroud. Engineering Mathematics. • http://en.wikipedia.org/wiki/Laplace_transform• Douglas Wilhelm Harder, Math. University of Waterloo• http://faculty.mercer.edu/olivier_pd/documents/Ch2Laplace
Transforms.ppt• http://ebookbrowse.com/ch06-laplace-transform-ppt-d1162
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