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

21.8 Finding the inverse Laplace transform using complex numbers 643
(h)
(i)
(j)
8s−5
(s +1)(s +2)(s +3)
3s+5
(s +1)(s 2 +3s +2)
2s−8
(s +2)(s 2 +7s +6)
2 Expressthe following fractions aspartialfractions
andhence findtheir inverse Laplacetransforms:
(a)
(b)
3s+3
(s −1)(s +2)
5s
(s +1)(2s −3)
(c) 2s+5
s +2
s 2 +4s+4
(d)
s 3 +2s 2 +5s
1 −s
(e)
(s +1)(s 2 +2s +2)
s +4
(f)
s 2 +4s+4
(g)
(h)
2(s 3 −3s 2 +s−1)
(s 2 +4s +1)(s 2 +1)
3s 2 −s+8
(s 2 −2s +3)(s +2)
Solutions
1 (a) 8e −2t −3e −t
(b) 5e −3t −2e −2t
(c) 15e −4t −11e −3t
35
(d)
2 e−5t − 23
2 e−3t
1
(e)
6 + 7 2 e−2t − 11
3 e−3t
1
(f)
4 +6e−3t − 25
4 e−4t
7
(g)
8 + 5 4 e−2t − 17
8 e−4t
(h) 21e −2t − 29
2 e−3t − 13
2 e−t
(i) e −t +2te −t −e −2t
(j) 3e −2t −e −6t −2e −t
2 (a) 2e t +e −2t
(b) e −t + 3e3t/2
2
(c) 2δ(t) +e −2t
(d) 1 5 (4 +e−t cos2t + 11 2 e−t sin2t)
(e) e −t (2−2cost −sint)
(f) e −2t (1 +2t)
(
(g) e −2t 3cosh √ 3t − √ 8 sinh √ )
3t −cost
3
(h) e t (cos √ 2t + √ 2sin √ 2t) +2e −2t
21.8 FINDINGTHEINVERSELAPLACETRANSFORMUSING
COMPLEXNUMBERS
InSections21.6and21.7wefoundtheinverseLaplacetransformusingstandardforms
and partial fractions. We now look at a method of finding inverse Laplace transforms
usingcomplexnumbers.Essentiallythemethodisoneusingpartialfractions,butwhere
all the factors in the denominator are linear -- that is, there are no quadratic factors. We
illustrate the methodusingExample21.13.
Example21.17 Findthe inverse Laplace transformsofthe following functions:
(a)
s +3
s 2 +6s+13
(b)
2s+3
s 2 +6s+13
(c)
s −1
2s 2 +8s+11