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

632 Chapter 21 The Laplace transform
{
(b) L −e −t + sint+cost }
= −1
2 s +1 + 1
2(s 2 +1) + s
2(s 2 +1)
= −1
s +1 + 1 +s
2(s 2 +1)
21.4.2 Firstshifttheorem
If L{f(t)} =F(s)then
L{e −at f (t)} =F(s +a)
aconstant
We obtainF(s +a) by replacing every ‘s’ inF(s) by ‘s +a’. The variableshas been
shifted by anamounta.
Example21.5 (a) UseTable 21.1 tofind the Laplace transform of
f(t)=tsin5t
(b) Usethe first shift theoremtowritedown
L{e −3t tsin5t}
Solution (a) FromTable 21.1 wehave
L{f(t)} = L{tsin5t} =
10s
(s 2 +25) 2 =F(s)
(b) Fromthe first shift theorem witha = 3 we have
L{e −3t f(t)} =F(s +3)
10(s +3)
=
((s +3) 2 +25) 2
=
10(s +3)
(s 2 +6s +34) 2
Example21.6 TheLaplace transformofafunction, f (t), isgiven by
F(s) = 2s+1
s(s +1)
State the Laplace transformof
(a) e −2t f (t)
(b) e 3t f (t)
Solution (a) Usethe first shift theorem witha = 2.
L{e −2t f(t)} =F(s +2)
=
=
2(s+2)+1
(s+2)(s+2+1)
2s+5
(s +2)(s +3)