MTH 352
The Laplace Transform (CH. 21, 22, 24) - Rev 2
Prof. Townsend
Fall 2009
Definition:
F ( s ) = L { f ( t )} =
"
# f (t ) e
! st
dt
0
There will be conditions on s such that the integral converges.
Recall
! cf ( x ) dx = c ! f ( x ) dx hence L {cf (t )} = cL { f (t )}
Note that since the Laplace Transform is an integral, we know that
L { f ( t ) + g ( t )} = L { f ( t )} + L { g ( t )}
Also note that since the Laplace Transform is an integral
L { f ( t )ig ( t )} ! L { f ( t )}iL { g ( t )}
The integral of a product is NOT the product of the integrals.
Let's start building the table.
1) f ( t ) = c where c is a constant and we noted above that L {cf ( t )} = cL { f ( t )} .
Therefore
F (s) =
#
"
0
ce
! st
"
dt = c # e
! st
0
$ e! st '
dt = c &
% !s )(
"
=!
t =0
{
c
lim e! st ! e0
s t *"
}
we know that e0 = 1 but we need to be careful at the upper limit. If s >0 then lim e# st = 0 .
t !"
However, if s <0 then lim e
# st
= ".
t !"
c
s
L {c} =
Hence
s>0
2) f ( t ) = eat where a is a constant.
F (s) =
#
"
0
at ! st
e e
dt =
#
"
0
e
! ( s ! a )t
$ e! ( s ! a )t '
dt = &
% ! ( s ! a ) )(
"
=!
t =0
{
1
lim e! ( s ! a )t ! e0
( s ! a ) t *"
}
This form is very similar to the one for a constant. If s > a then lim e# ( s # a )t = 0 . However, if
s < a then lim e
# ( s # a )t
t !"
L {eat } =
t !"
= ".
1
s!a
Hence
s>a
(Note the minus sign in the denominator)
3) f ( t ) = cos ( bt ) where b is a constant.
Let's recall Euler's theorem
equivalently
ei! = cos! + i sin !
e!i" = cos" ! i sin "
If we solve for cos! and sin ! , we get
ei! + e"i!
ei! " e"i!
cos! =
sin ! =
and
2
2i
We can use these relationships to find the Laplace Transforms of cos! and sin ! .
ibt
"e
"
+ e!ibt ! st
1 " ibt ! st
L {cosbt } = #
e dt =
e
e
dt
+
e!ibt e! st dt
#
#
0
0
0
2
2
{
}
{ }
1
. We can therefore write down that
s!a
ibt
"e
+ e!ibt ! st
1$ 1
1 '
L {cosbt } = #
e dt = %
+
(
0
2
2 & s ! ib s ! ib )
But we just found L eat =
We combine the terms using the LCD of LCD = ( s + ib ) ( s ! ib ) = s 2 + b 2 .
Therefore
L {cosbt } =
1 " s + ib
s ! ib % 1 " 2s %
s
+ 2
= # 2
= 2
# 2
2
2 &
2 &
2 $s + b
s + b ' 2 $ s + b ' s + b2
Using the same technique, you prove that
b
L {sin bt } = 2
s + b2
4) The hyperbolic sine and cosine behave similarly.
e x + e! x
e x ! e! x
cosh x =
sinh x =
and
2
2
s
b
and
L {cosh bt } = 2
L {sinh bt } = 2
2
s !b
s ! b2
5) One more that you can prove using integration by parts is
n!
where n! = n ! (n " 1) ! (n " 2)...3 ! 2 !1
L t n = n +1
s
{ }
Find the Laplace Transforms of the following functions of time.
f ( t ) = 5t + 2
f ( t ) = 10t 2 ! 5
f ( t ) = e3t
f (t) = cos ( 3t )
f (t) = cosh ( 3t )
f ( t ) = e5t ! 7
f ( t ) = sin ( 5t + 2 )
f ( t ) = sinh ( 5t + 2 )
Note: sin ( A + B ) = sin A cos B + cos Asin B
sinh ( A + B ) = sinh A cosh B + cosh Asinh B
cos ( A + B ) = cos A cos B ! sin Asin B
cosh ( A + B ) = cosh A cosh B + sinh Asinh B
Check out Trig Expand in the F2 menu of the TI-89.