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

21.8 Finding the inverse Laplace transform using complex numbers 645
(b)
2s+3
s 2 +6s+13 = 2s+3
(s −a)(s −b) = A
s −a + B
s −b
wherea = −3 +2jandb= −3 −2j. Hence,
2s+3=A(s−b)+B(s−a)
Puts=a=−3+2j
−3 +4j =A(4j)
A=1+0.75j
Equate the coefficients ofs
Hence,
2=A+B
B=1−0.75j
2s+3
s 2 +6s+13 = 1 +0.75j + 1 −0.75j
s −a s −b
(c)
Taking the inverse Laplace transform yields
{ }
L −1 2s+3
= (1 +0.75j)e at + (1 −0.75j)e bt
s 2 +6s+13
= (1 +0.75j)e (−3+2j)t + (1 −0.75j)e (−3−2j)t
s −1
2s 2 +8s+11 = 1 (
2
= (1 +0.75j)e −3t (cos2t +jsin2t)
+ (1 −0.75j)e −3t (cos2t −jsin2t)
= e −3t (2cos2t −1.5sin2t)
)
= 1 (
2
s −1
s 2 +4s+5.5
s −1
(s −a)(s −b)
)
wherea = −2 + √ 1.5j,b = −2 − √ 1.5j.Applyingthemethodofpartialfractions
produces
Hence,
s −1
(s −a)(s −b) = A
s −a + B
s −b
s−1=A(s−b)+B(s−a)
By lettings =a, thens =binturn, gives
A=0.5+ √ 1.5j B = 0.5 − √ 1.5j
Hence we may write
s −1
(s −a)(s −b) = 0.5 + √ 1.5j
+ 0.5 − √ 1.5j
s −a s −b