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

(b) Usethe first shift theorem witha = −3.
L{e 3t f(t)} =F(s−3)
=
=
2(s−3)+1
(s−3)(s−3+1)
2s−5
(s −3)(s −2)
21.4 Properties of the Laplace transform 633
21.4.3 Secondshifttheorem
If L{f(t)} =F(s)then
L{u(t −d)f(t −d)} =e −sd F(s) d >0
The function, u(t − d)f (t − d), is obtained by moving u(t)f (t) to the right by an
amount d. This is illustrated in Figure 21.1. Note that because f (t) is defined to be
0fort < 0,then f (t −d) = 0fort <d.Itmayappearthatu(t −d)isredundant.However,
it is necessary for inversion of the Laplace transform, which will be covered in
Section21.6.
u(t) f(t)
u(t – d) f(t – d)
(a)
t
Figure21.1
Shifting the functionu(t)f (t)to the rightbyan amountd yields the function
u(t−d)f(t−d).
(b)
d
t
Example21.7 Given L{f (t)} =
2s ,find L{u(t −2)f(t −2)}.
s +9
Solution Usethe second shift theorem withd = 2.
L{u(t −2)f(t −2)} = 2se−2s
s +9
Example21.8 TheLaplace transform ofafunctionis e−3s
. Find the function.
s2 Solution Theexponentialterminthetransformsuggeststhatthesecondshifttheoremisused.Let
e −sd F(s) = e−3s
s 2