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

634 Chapter 21 The Laplace transform
sothatd =3andF(s) = 1 s2. If we let
L{f(t)} =F(s) = 1 s 2
then f (t) =t and so f (t −3) =t −3.Now, usingthe second shift theorem
L{u(t −3)f(t −3)} = L{u(t −3)(t −3)} = e−3s
s 2
Hence the required function isu(t −3)(t −3) as shown inFigure 21.2.
f(t)
3 t
Figure21.2
Thefunction: f(t) =u(t −3)(t −3).
21.4.4 Finalvaluetheorem
This theorem applies only to real values of s and for functions, f (t), which possess a
limitast → ∞.
Thefinal valuetheorem states:
limsF(s) = lim f(t)
s→0 t→∞
Some care is needed when applying the theorem. The Laplace transform of some functions
exists only for Re(s) > 0 and for these functions taking the limit ass → 0 is not
sensible.
Example21.9 Verifythe final value theoremfor f (t) = e −2t .
Solution ∫ ∞
0
e −st e −2t dt exists provided Re(s) > −2 and so
F(s) = 1
s +2
Re(s) > −2
Since weonly require Re(s) > −2, itispermissibletolets → 0.
s
limsF(s) =lim
s→0 s→0 s +2 = 0
Furthermore
lim
t→∞ e−2t = 0
So lim s→0
sF(s) = lim t→∞
f (t)and the theorem isverified.