Transcript Superposition Principle & the Method of Undetermined

Inverse Laplace Transform
• Consider a given function F(s), is it possible to
find a function f(t) defined on [0,  ), such that
L{ f (t )}  F (s) ?
• If this is possible, we say f(t) is the inverse
Laplace transform of F(s), and we write
f (t )  L1{F (s)}.
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Some Examples.
• First we note that the Inverse Laplace
Transform is a “ Linear Operator”.
s-a
1. F(s) 
,
2
2
(s-a)  b
b
2. F(s) 
,
2
2
(s-a)  b
3s  2
3. F(s) 
,
2
s  2 s  10
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Applications
• Consider the Initial Value Problem:
y"2 y'5 y  8e t ;
y(0)  2, y' (0)  12.
• We shall use Laplace Transform and Inverse
Laplace Transform to solve this I.V.P.
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Next let us consider a D.E. with
variable coefficents
• Given the following I.V.P: (#36, P. 406)
ty"ty ' y  2
y (0)  2
y ' (0)  1 .
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Laplace Transform of Discontinuous
and Periodic Functions
• Discontinuous functions play a very
important role in Engineering, for example:
u (t ) : 0,
t 0,
1,
0 t.
• This is known as the unit step function.
1
t
0
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Shifts and scalar multiples of the
Unit Step Functions
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Unit step functions can be used to represent
any piecewise continuous function.
• For example: …
• What is the Laplace Transform of u(t- a), a > 0?
L{u (t  a )}(s ) 


0
e  st u (t  a ) dt
 1  st N
e |a
a
N  s
1  sa 1  sN 
 lim  e
 e

N 
s
s


1  sa

e . It follows t hat
s
1  sa
T heinverseLaplaceT ransformof
e
s
1
is
L1{ e  sa }  u (t  a ).
s


e -st dt  lim
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Next, what is: L{ f (t  a)u(t  a)}(s)  ?

Since L{ f (t  a)u (t  a)}(s)   e  st f (t  a)u (t  a)dt
0



a
e  st f (t  a)dt , let v  t  a, dv  dt,

  e  s ( v  a ) f (v)dv  e  sa L{ f }( s ) , moreover,
0
we haveInverseLaplaceT ransform:
1
 sa
L {e L{ f }( s)}  f (t  a)u (t  a).
In particular, if we let g(t)  f(t - a), then wehave
L{g (t )u (t  a)}(s)  e
 sa
L{g (t  a)}(s)
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Some Examples
• Let us consider the following :
1. L{t u (t  1)}(s )  ?
2. L{(cost )u (t   )}  ?
2
3 s
e
3. L { 2 }(t )  ?
s
1
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Laplace Transform of Periodic
Functions
• 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?
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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 .
• It is not difficult to see that f (t) can be written
as the sum of translates of f T (t). Namely,

f (t )   fT (t  nT )u (t  nT ) .
n 0
Hence we have
T heorem: If f is a piecewise cont inuousperiodicfunct ion
wit h period T , t hen
L{ fT }( s )
L{ f }( s ) 
.
 sT
1 e
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Example: The square wave with
period 2
• Note that in this case f T (t) is given by :
1
1
t
2
• f T (t) = 1 - u(t - 1) . Hence
1 es 1
L{ fT }( s )  
 (1  e  s ) .
s
s
s
T herefore:
L{ fT }( s )
1
L{ f }( s ) 

.
2 s
s
1 e
s (1  e )
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Some problems in the exercise
• #10, 15, 31,
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Convolution Operator “ * “.
• Definition: Given two functions f (t) and g(t)
piecewise continuous on [0,). The convolution
of f and g , denoted by f  g , is defined by :
t
( f  g )(t )   f (t  s ) g ( s )ds .
0
4
t
For example,11  t , 1 f  f, t  t 2 
.
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• Covolution is 1.commutative, 2. distributive, 3.
associative and with 4. existence of zero.
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Laplace Transform of convolution
• An important property of convolution is the
• Theorem:
For any t wofunct ion f(t) and g(t) piecewise
cont inuouson [0, ) , and of exponent ia
l order
 . If L{f}(s)  F(s)and L{g}(s)  G(s), t hen
L{ ( f  g )(t ) }(s)  F(s)  G(s), consequently,
L {F(s) G(s)}(t)  (f  g)(t).
-1
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Proof of Convolution theorem
can be done by
• 1. Writing

( f  g )(t )   u(t  s) f (t  s) g (s)ds
0
• 2. Then apply the Fubini’s theorem on
interchanging the order of integration.
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Applications
• Solve the initial value problem
y" y  g (t ) ; y (0)  0 , and y ' (0)  0.
Whereg(t) is pieceewise continuouson [0,) and
of exponentia
l order . We found afterusing
LaplaceT ransformthat
L{ y}( s )  L{sin t}( s )  L{g (t )}(s ) .
T hus by theconvolution theorem,we have
t
y (t )  sin t  g (t )   sin(t  s ) g ( s )ds.
0
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Example 2: Integral-Differential Equation
• Consider the following equation:
t
y ' (t )  1   y (t  s )e  2 s ds , y( 0 )  1 .
0
(An Equat ionintroducedby V. Volt erra on thest udy
of populationgrowt h).T hisequation can be writtenas;
y ' (t )  1  y (t )  e  2t , which can be solved using
LaplaceT ransformand Convolut ion theorem.(solving
for Y(s) using partialfraction).
2
1
We find L{ y}( s )  
, this y (t )  2  e t .
s s 1
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Transfer function and Impulse
response function
• Consider the linear system governed by the I.V.P:
ay"by'cy  g (t ) , for t  0 .
y (0)  0 and y ' (0)  0 .
• 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,
L{ y}( s)
 H ( s) is called t he t ransferfunct ionof
L{g}( s)
Y ( s)
t helinear syst em. i.e. H ( s) 
G( s)
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For our example, take the Laplace
transform of the I.V.P
• we get
a{s 2Y ( s )  sy (0)  y ' (0)}  b{sY ( s )  y (0)}  cY ( s )  G ( s ).
Since we have t rivialinit ialcondit ions, t heeq. becomes
as2Y(s)  bY(s) cY(s)  G(s) , t hus t he t ransferfunct ionis
Y (s)
1
H (s) 
 2
.
G ( s ) as  bs  c
• 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!!
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This function h(t) is the unique solution to
the homogeneous problem
• Namely:
ah"bh'ch  0 ; with h(0)  0 and h' (0)  1 / a .
• This can be checked easily (using Laplace
transform). Now to solve a general I.V.P. such as
ay"by'cy  g (t ) , with
y( 0 )  y0 ,
y'( 0 )  y1
• This is a non-homogeneous eq with non-trivial
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initial values.
We shall split the given I.V.P into two
problems
• They are the equivalent to the original I.V.P.
Namely:
(*) ay"  by'  cy  g(t) , y( 0 )  0 and y'(0 )  0 ,
(**) ay" by'  cy  0 ,
y(0 )  y0 and y'(0 )  y1 .
T hesolut ionof (*) is found t o be y(t)  (h  g)(t) and
t hesolut ionof (**)is given by , say, yk (t). T hen t he
solut ionof t heoriginalI.V.P .is of t heform
y(t)  (h  g)(t)  yk (t). T hiscan be checkedeasily.
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Theorem on solution using
Impulse Response Function
• Theorem: Let I be an interval containing the
origin. The unique solution to the initial value
problem
(P ):
ay"  by'  cy  g ; y(0 )  y0 , y'( 0 )  y1 ,
where a, b,and c are const ant sand g is cont inuous
on I, is given by
(S) :
y(t)  (h  g)(t) yk (t) ,
where h is t heimpulse responsefunct ionfor
t hesyst emand yk (t ) is t heunique solut ion
t o : ay"  by'  cy  0 , y( 0 )  y0 , y'( 0 )  y1 .
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Example
• #24, P.428
• Let a linear system be governed by the given
initial value problem.
y' '9 y  g (t ); y(0)  2, y' (0)  0
• 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)
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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: (1)  (t )  0 , for all t  0 , and
(2)


-
 (t ) dt  1,
• 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



f (t ) (t  a)dt  f (a) .
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Remarks on Theory of
Distribution.
• Symbolic function, generalized function,
and distribution function.
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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
t1
Impulse  F(t)dt . By Newton's 2nd law of motion,
t0
dv
t0 F(t) dt  t0 m dt dt  m v(t1 )-m v(t0 ) .
T hismeansthat" Impulse Change in momentum".
t1
t1
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This heuristic leads to conditions
1 and 2.
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What is the Laplace Transform of
-function?
• By definition, we have


L{ (t  a)}(s)   e  (t  a)dt   e  st  (t  a)dt  e  as .
 st
0

Note that when
t  a,
t  a,
(*)
t



t
-
 ( x  a)dx  0 , and
 ( x  a)dx  1. It follows that

t
-
δ (x a)dx  u(t  a).
Differentiate both sides of (*), we get (by the F.T .C.):
δ (t  a)  u ' (t  a).
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Application:
• Consider the symbolic Initial Value Problem:
x"9 x  3 (t   ) ; x( 0 )  1 , x'(0 )  0 .
T hisrepresentsa mass at tachedto a springis releasedfrom rest
1 meterbelow theequilibrium positionfor themass - spring
system.It begins to vibrate,but  seconds lat er,themass is struck
by a hammerexertingan impulse on t hemass., where x(t)is the
displacement from theequilibrium positionat timet.
We shall solveit by themethodof LaplaceT ransform.
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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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