082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

644 Chapter 21 The Laplace transform
Solution (a) We first factorize the denominator. To do this we solves 2 +6s +13 = 0 using the
formula
s = −6 ± √ 36 −4(13)
2
= −6 ± √ −16
2
= −6±4j
2
=−3±2j
It then follows that the denominator can be factorized as (s − a)(s − b) where
a = −3 +2jandb=−3 −2j. Then, usingthe partialfractions method
s +3
s 2 +6s+13 = s +3
(s −a)(s −b) = A
s −a + B
s −b
The unknown constantsAandBcan now be found:
s+3=A(s−b)+B(s−a)
Puts=a=−3+2j
2j =A(−3 +2j −b) =A(4j)
A = 1 2
Equate the coefficients ofs
1=A+B
B = 1 2
So,
s +3
s 2 +6s+13 = 1 ( 1
2 s −a + 1 )
s −b
The inverse Laplace transform can now be found:
{ }
L −1 s +3
= 1 { 1
s 2 +6s+13 2 L−1 s −a + 1 }
s −b
= 1 2 (eat +e bt ) = 1 2 (e(−3+2j)t +e (−3−2j)t )
= 1 2 e−3t (e 2jt +e −2jt )
= 1 2 e−3t (cos2t +jsin2t +cos2t −jsin2t)
= e −3t cos2t