1
• Consider a given function F(s), is it possible to find a function f(t) defined on [0, ), such that
• If this is possible, we say f(t) is the inverse Laplace transform of F(s), and we write
Inverse Laplace Transform
? )()}({ sFtfL
)}. ( { ) (1
s F L t f
2
• First we note that the Inverse Laplace Transform is a “ Linear Operator”.
Some Examples.
,ss
s . F(s)
,b(s-a)
b . F(s)
,b(s-a)
s-a . F(s)
102
233
2
1
2
22
22
3
• Consider the Initial Value Problem:
• We shall use Laplace Transform and Inverse Laplace Transform to solve this I.V.P.
Applications
.12)0(' ,2)0(
;85'2"
yy
eyyy t
4
• Given the following I.V.P: (#36, P. 406)
Next let us consider a D.E. with variable coefficents
. 1)0('
2)0(
2'"
y
y
ytyty
5
• Discontinuous functions play a very important role in Engineering, for example:
• This is known as the unit step function.
Laplace Transform of Discontinuous and Periodic Functions
. t 0 1,
0, t 0, :)(
tu
t0
1
6
Shifts and scalar multiples of the Unit Step Functions
7
• For example: …
• What is the Laplace Transform of u(t- a), a > 0?
Unit step functions can be used to represent any piecewise continuous function.
).(}s
1{ is
s
1 of Transform Laplace inverse The
thatfollowsIt .s
1
1
s
1 lim
|1
lim
)())}(({
1
0
atueL
e
e
es
e
es
dte
dtatuesatuL
sa
sa
sa
sNsa
N
Na
st
Na
-st
st
8
Next, what is: ?))}(()({ satuatfL
))}(({))}(()({
have then we,let weif ,particularIn
).()()}}({{
:Transform Laplace Inverse have we
moreover, , )}({)(
, ,let , )(
)()())}(()({ Since
1
0
)(
0
satgLesatutgL
f(t - a)g(t)
atuatfsfLeL
sfLedvvfe
dtdvatvdtatfe
dtatuatfesatuatfL
sa
sa
saavs
a
st
st
9
• Let us consider the following :
Some Examples
?)}({ .3
?)} (){(cos .2
?))}(1({ .1
2
31
2
t
s
eL
tutL
stutL
s
10
• Periodic functions play a very important role in the study of dynamical systems
• Definition: A function f (t) is said to be periodic of period T, if for all t D(f) , we have f (t
+ T) = f (t).• For examples, sine waves, cosine waves and
square waves are periodic functions.• What can we say about the transforms of periodic
functions?
Laplace Transform of Periodic Functions
11
• It is not difficult to see that f (t) can be written as the sum of translates of f T (t). Namely,
Let f T (t) be the part of f over the basic period [0, T]. This is known as the Windowed
version of the periodic function f .
. 1
)}({)}({
thenT, periodwith
function periodic continuous piecewise a is f If :Theorem
have weHence
. )()()( 0
sTT
nT
e
sfLsfL
nTtunTtftf
12
• Note that in this case f T (t) is given by :
• f T (t) = 1 - u(t - 1) . Hence
Example: The square wave with period 2
t1 2
1
. )1(
1
1
)}({)}({
:Therefore
. )1(11
)}({
2 ssT
ss
T
ese
sfLsfL
ess
e
ssfL
13
• #10, 15, 31,
Some problems in the exercise
14
• Definition: Given two functions f (t) and g(t) piecewise continuous on [0,). The convolution of f and g , denoted by
• Covolution is 1.commutative, 2. distributive, 3. associative and with 4. existence of zero.
Convolution Operator “ * “.
. 12
t tf,f1 ,11 example,For
. )()())((
:by defined is ,
42
0
tt
dssgstftgf
gf
t
15
• An important property of convolution is the
• Theorem:
Laplace Transform of convolution
.
ly,consequent , ))((
then, and If .
order lexponentia of and , ) [0,on continuous
piecewise and function any twoFor
1 g)(t) (fG(s)}(t){F(s)L
G(s) F(s) }(s) t gfL{
G(s)L{g}(s) F(s)L{f}(s)
g(t)f(t)
-
16
• 1. Writing
• 2. Then apply the Fubini’s theorem on interchanging the order of integration.
Proof of Convolution theorem can be done by
0
)()()())(( dssgstfstutgf
17
• Solve the initial value problem
Applications
tdssgsttgtty
stgLstLsyL
yytgyy
0.)()sin()(sin)(
have wen theorem,convolutio by the Thus
. ))}(({)}({sin)}({
that Transform Laplace
usingafter found We. order lexponentia of
and )[0,on continuous pieceewise is g(t) Where
.0)0(' and , 0)0( ; )("
18
• Consider the following equation:
Example 2: Integral-Differential Equation
.2)( this, 1
12)}({ find We
fraction). partial using Y(s)for
(solving n theorem.Convolutio and Transform Laplace
using solved becan which ,)(1)('
as; written becan equation This growth). population of
study on the Volterra V.by introducedEquation An (
10 , )(1)('
2
0
2
t
t
t s
etyss
syL
etyty
. )y(dsestyty
19
• Consider the linear system governed by the I.V.P:
• Thus given g(t) we wish to find the solution y(t). g(t) is called the input function and y(t) the output. The ratio of their Laplace Transforms,
Transfer function and Impulse response function
. 0)0(' and 0)0(
. 0for , )('"
yy
ttgcybyay
)(
)()( i.e. system.linear the
offunction transfer thecalled is )()}({
)}({
sG
sYsH
sHsgL
syL
20
• we get
• The inverse Laplace Transform of H(s), written h(t) = L-1{H(s)}(t) is called the Impulse response function for the system. Graph!!
For our example, take the Laplace transform of the I.V.P
. 1
)(
)()(
isfunction transfer the thus,
becomes eq. the,conditions initial trivialhave weSince
).()()}0()({)}0(')0()({
2
2
2
cbsassG
sYsH
G(s)cY(s)bY(s)Y(s)as
sGscYyssYbysysYsa
21
• Namely:
• This can be checked easily (using Laplace transform). Now to solve a general I.V.P. such as
• This is a non-homogeneous eq with non-trivial initial values.
This function h(t) is the unique solution to the homogeneous problem
. /1)0(' and 0)0( with ; 0'" ahhchbhah
1
0
0
0
with , )('"
y)y'(
, y) y(
tgcybyay
22
• They are the equivalent to the original I.V.P. Namely:
We shall split the given I.V.P into two problems
easily. checked becan This .
form theof is I.V.P. original theofsolution
Then the . say, ,by given is (**) ofsolution the
and be tofound is (*) ofsolution The
.000
0000 (*)
10
(t) y g)(t) (hy(t)
(t)y
g)(t) (hy(t)
y) and y'(y) , y( cyby'(**) ay"
,) and y'()y( g(t) , cyby' ay"
k
k
23
• Theorem: Let I be an interval containing the origin. The unique solution to the initial value problem
Theorem on solution using Impulse Response Function
. 00 ,0 :to
solution unique theis )( and system the
for function response impulse theis where
, : (S)
by given is I,on
continuous is and constants are and where
00 :(P)
10
10
k
k
y), y'( y)y( cyby'ay"
ty
h
(t)yg)(t)(hy(t)
gca, b,
,y), y'( y) g ; y(cy by'ay"
24
Example
• #24, P.428
• Let a linear system be governed by the given initial value problem.
• Find the transfer function H(s), the impulse response function h(t) and solve the I.V.P.
• Recall: y(t) = (h*g)(t) + yk(t)
0)0(' ,2)0( );(9'' yytgyy
25
Dirac Delta Function • Paul A. M. Dirac, one of the great physicists
from England invented the following function:
• Definition: A function (t) having the following properties:
• is called the Dirac delta function. It follows from (2) that for any function f(t) continuous in an open interval containing t = 0, we have
- 1, dt (t) (2)
and , 0 tallfor , 0(t) )1(
. )()( )( afdtattf
26
Remarks on Theory of Distribution.
• Symbolic function, generalized function, and distribution function.
27
Heuristic argument on the existence of -function.
• When a hammer strikes an object, it transfer momentum to the object. If the striking force is F(t) over a short time interval [t0, t1], then the total impulse due to F is the integral
momentum".in Change Impulse " that means This
motion, of law 2nd sNewton'By .dt F(t)Impulse
1
0
1
0
1
0
01
t
t
t
t
t
t) . )-mv(t mv(t dt
dt
dvm F(t) dt
28
This heuristic leads to conditions 1 and 2.
29
What is the Laplace Transform of -function?
• By definition, we have
).(')
:F.T.C.) (by theget we(*), of sidesboth ateDifferenti
(*)
thatfollowsIt .1)( ,
and , 0)( ,
that whenNote
. )( )( ))}(( {
t
0
atuaδ (t
a). u(ta)dxδ (x
dxaxat
dxax a t
edtatedtatesatL
t
-
-
t
asstst
30
Application:• Consider the symbolic Initial Value Problem:
x +9x=3δ \( t - π \) ; ital x \( 0 \) = 1 ital , x ' \( 0 \) = 0 . } {} # {} # size 12{ Thisrepresents a mass attached ¿a spring is released¿
rest } {} # size 12{1meter below the equilibrium position for the mass−spring } {} # size 12{ system . begins ¿vibrate , but π secondslater , the mass is struck } {} # size 12{ by a hammer exerting an impulse on the mass . , where x \( t \) is the} {} # size 12{ displacement ¿ the equilibrium position at time t . } {} # size 12{We shall solveby the method of ℒ Transform . } {} } } {
31
Linear Systems can be solved by Laplace Transform.(7.9)
• For two equations in two unknowns, steps are:• 1. Take the Laplace Transform of both equations
in x(t) and y(t),• 2. Solve for X(s) and Y(s), then• 3. Take the inverse Laplace Transform of X(s)
and Y(s), respectively.• 4. Work out some examples.
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