# L17 The Inverse Laplace Transform (Section 7.4) by zuu19905

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```									L17 The Inverse Laplace Transform (Section 7.4)                             Inverse Laplace Transform
Definition: Given a function F ( s ) , if there is a function f ( t ) that
The Laplace transform is defined by                                         is continuous on [0, +∞) and satisfies
∞
L { f }( s ) = ∫ e − st f ( t ) dt ,                                       L {f}= F,
0                                                   Then we say that f ( t ) is the inverse Laplace transform of F ( s )
that is, L : f ( t ) → F ( s ) ,                                            and use the notation
then the inverse Laplace transform, L −1 : F ( s ) → f ( t ) .                    f = L −1{ F } .

Example: y′′ + y = t , y ( 0 ) = 0,           y ′ ( 0 ) = 1.                To determine the inverse Laplace transform of a given function
F ( s ) , we refer to the tables.

Example: Determine L −1{ F } , where
6
(a) F ( s ) = 4
s
s
(b) F ( s ) = 2
s + 25
2
(c) F ( s ) = 2
s − 2s + 5

The linearity of the inverse Laplace transform L −1 is inherited
from the linearity of the operator L.
Note: y ( t ) = t is not the only function whose Laplace transform is
1                                ⎧t , t ≠ 3                1               Theorem: Assume that L −1{ F } , L −1{ F1} , and L −1{ F2 } exist and
, for instance, if g ( t ) = ⎨          , then L{ g} = 2 . Thus, the
⎩0, t = 3
2
s                                                         s                are continuous and let c be any constant. Then,
inverse Laplace transform is not defined uniquely, but among all                 (1) L −1{ F1 + F2 } = L −1{ F1} + L −1{ F2 }
those functions there is only one which is continuous on [0, +∞ ) .              (2) L −1{cF } = c L −1{ F }

132                                                               133
⎧ 3        2s      4      ⎫   The Inverse Laplace Transform of Rational Functions
Example: Determine L −1 ⎨      + 2    + 2         ⎬
⎩ s − 5 s + 16 s + 2s + 5 ⎭          P(s)
R(s) =      , where P and Q are polynomials and deg P < deg Q.
Q(s)
Method of Partial Fractions
1. Nonrepeated linear factors
2. Repeated linear factors

Case 1. Nonrepeated linear factors:
⎧ 7 ⎫
⎪          ⎪                       Q ( s ) = ( s − r1 )( s − r2 ) ...( s − rn ) , ri ≠ rj    (i ≠ j ) .
Example: Determine L −1 ⎨         4⎬
⎪ ( s − 3) ⎪
⎩          ⎭
The partial fraction expansion has the form
P(s)         A          A                A
= 1 + 2 + ... + n
Q ( s ) s − r1 s − r2                  s − rn
2s − 1
Example: Determine L −1{ F } , where F ( s ) =                                          .
( s − 1)( s + 2 )( s + 3)
⎧ 2s − 3 ⎫
Example: Determine L −1 ⎨ 2            ⎬
⎩ s − 4 s + 20 ⎭

134                                                              135
Case 2. Repeated linear factors:                                             Case 3. Quadratic factors:
P(s)         P(s)                                                              P(s)                  P(s)
=                                                                                =                              ,
Q ( s ) Q1 ( s )( s − r )m                                                      Q ( s ) Q ( s ) ⎡( s − α ) 2 + β 2 ⎤ m
1
⎣               ⎦
The portion of partial fraction expansion that corresponds to
where ( s − α ) + β cannot be reduced to a product of linear
2      2
( s − r ) is
m

factors.
A1     A2                Am
+         + ... +                                               The portion of the partial fraction expansion that corresponds to
s − r (s − r) 2
(s − r)
m
[( s − α ) + β 2 ]m is
2

3s 2 + 4s + 7                         C1s + D1             C2 s + D2                    Cm s + Dm
Example: Determine L −1{ F } , where F ( s ) =                           .                             +                     + ... +                     .
( s − 2 )( s + 1)
2
( s − α ) + β [( s − α ) + β ]                   [( s − α ) + β 2 ]m
2      2             2    2 2                      2

It is more convenient to express it in the equivalent form
A1 ( s − α ) + β B1 A2 ( s − α ) + β B2              A ( s − α ) + β Bm
+                      + ... + m                    .
(s −α ) + β          [( s − α ) + β ]              [( s − α ) + β 2 ]m
2      2              2     2 2                     2

7 s 2 − 41s + 84
Example: Determine L −1{ F } , where F ( s ) =                                     .
( s − 1) ( s 2 − 4s + 13)

136                                                                                 137

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