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

646 Chapter 21 The Laplace transform
Taking the inverse Laplace transformyields
Hence
{ }
L −1 s −1
= (0.5 + √ 1.5j)e at + (0.5 − √ 1.5j)e bt
(s −a)(s −b)
= (0.5 + √ 1.5j)e (−2+√ 1.5j)t
+ (0.5 − √ 1.5j)e (−2−√ 1.5j)t
= e −2t {(0.5 + √ 1.5j)e √ 1.5jt
+ (0.5 − √ 1.5j)e −√ 1.5jt }
= e −2t {(0.5 + √ 1.5j)(cos √ 1.5t +jsin √ 1.5t)
+ (0.5 − √ 1.5j)(cos √ 1.5t −jsin √ 1.5t)}
= e −2t (cos √ 1.5t −2 √ 1.5sin √ 1.5t)
{ }
L −1 s −1
= e−2t
2s 2 +8s+11 2 (cos √ 1.5t −2 √ 1.5sin √ 1.5t)
EXERCISES21.8
AsseenfromExample21.17,whencomplexnumbersareallowed,allthefactorsinthe
denominatorarelinear.Theunknown constantsareevaluatedusingparticularvaluesof
s orequating coefficients.
1 Expressthe followingexpressions aspartialfractions,
usingcomplex numbers ifnecessary. Hence findtheir
inverse Laplacetransforms.
(a)
3s−2
s 2 +6s+13
(b)
2s+1
s 2 −2s+2
(c)
(e)
s 2
(s 2 /2)−s+5
2s+3
−s 2 +2s−5
(d)
s 2 +s+1
s 2 −2s+3
Solutions
3
1 (a) 2 +11j/4
+
s −a
3
2 −11j/4
s −b
wherea=−3+2j,b=−3−2j,
e −3t[ 3cos2t − 11
2 sin2t ]
(b) 1 −3j/2 + 1 +3j/2
s −a s −b
wherea=1+j,b=1−j,
e t (2cost+3sint)
(c) 2 +8j/3 + 2 −8j/3 +2
s −a s −b
wherea=1+3j,b=1−3j,
(
e t 4cos3t − 16 )
3 sin3t +2δ(t)
3
(d)1+ 2 −j/(2 √ 2)
+
s −a
(e)
3
2 +j/(2√ 2)
s −b
wherea=1+ √ 2j,b=1− √ 2j,
)
δ(t)+e t (3cos √ 2t + 1 √
2
sin √ 2t
−1 +5j/4
s −a
+
−1 −5j/4
s −b
wherea=1+2j,b=1−2j,
e t (−2cos2t − 5 2 sin2t)