IMTS Mechanical Eng. (Applied mathematics)

APPLIED MATHEMATICS

= 3e Example 6

t

130

2t 2t  e2t  e2t   t e  e 3t t  5 e      4e  e 2 2    

14s  10 49s 2  28s  13  5 14  s   14s  10  7 = 2 4 13  49s  28s  13  49  s 2  s   7 49   5 2 3 s (s  )  2 2 7 7 7 = = 2 2 2 7 2 9 7 2 3 s   s    7  49 7 7  

Find the inverse Laplace transform of

 14s  10  2  L1  = e 2  49s  28s  13  7

=

2 72t e 7

 3   s   1  7 L  2  s2   3       7 

     3      s  1   7  L1  L  2 2   2 3   s 2   3    s          7   7   =

5.3.3

2 t 7

2 72t e 7

3t 3t   cos 7  sin 7 

Inverse Laplace Transform by second shifting theorem

If

 f (t  a), t  a L1 F (s)  f (t ), thenL1 e as F (s)  G(t ) where G (t )    0, t  a

Or G(t) = f(t-a) H(t-a) Or Equivalently if

L1 F (s)  f (t ) , then L1 e as F (s) = f(t-a) H(t-a) Where H(t-a) is Heaviside ‘s unit step function

Example 7

Find

 e3s  L1  .

4  ( s  2) 

Solution Let

F(s) =

1 ( s  2) 4

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