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

21.13.2 Periodicfunctions
21.13 Laplace transforms of some special functions 677
Recallthedefinitionofaperiodicfunction, f (t).GivenT > 0, f (t)isperiodicif f (t) =
f (t +T ) for all t in the domain. If f (t) is periodic and we know the values of f (t)
over a period, then we know the values of f (t) over its entire domain. Hence, it seems
reasonable that the Laplace transform of f (t) can be found by studying an appropriate
integral over an interval whose length is just one period. This is indeed the case and
formsthe basis of the following development.
Let f (t)be periodic, with periodT. The Laplace transformof f (t)is
L{f(t)} =F(s) =
=
∫ T
0
∫ ∞
0
e −st f (t)dt
∫ 2T ∫ 3T
e −st f(t)dt + e −st f(t)dt + e −st f(t)dt+···
T
2T
Lett =xinthefirstintegral,t =x +T inthesecond,t =x +2T inthethirdandsoon.
F(s) =
∫ T
0
∫ T
+
0
∫ T
e −sx f(x)dx +
0
e −s(x+T ) f(x+T)dx
e −s(x+2T ) f(x+2T)dx+···
Since f isperiodic with periodT then
So,
f(x)=f(x+T)=f(x+2T)=···
F(s) =
∫ T
0
∫ T
∫ T
e −sx f(x)dx + e −sT e −sx f(x)dx + e −2sT e −sx f(x)dx + ···
0
0
=(1+e −sT +e −2sT +···)
∫ T
0
e −sx f(x)dx
We recognize the terms in brackets as a geometric series whose sum to infinity
1
is . Hence,
1 −e−sT F(s) =
∫ T
0 e−st f (t)dt
1 −e −sT
Example21.33 Awaveform, f (t), isdefinedasfollows:
f(t)=
{ 2 0<t1.25
0 1.25<t1.5
and f (t)isperiodicwith periodof1.5.Findthe Laplace transformofthe waveform.