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

∫ 2
[ ] e
Solution (a) F(ω)=3 e −jωt −jωt 2
dt = 3
−2 −jω
−2
( e −2jω −e 2jω )
=3
−jω
( e 2jω −e −2jω )
=6
2jω
= 6 ω sin2ω
24.3 Some properties of the Fourier transform 763
(b) Wehave F{f(t)} =F(ω) = 6sin2ω . Using the first shift theorem witha = −1
ω
we have
(c) e −jt f (t) =
F{e −jt f(t)} =F(ω+1) = 6 sin2(ω +1)
ω +1
{
3e
−jt
−2t2
0 otherwise
So toevaluate its Fourier transformdirectly wemustfind
F{e −jt f(t)} =3
as required.
∫ 2
−2
e −jt e −jωt dt
∫ 2
= 3 e −(1+ω)jt dt
−2
[ e
−(1+ω)jt 2
= 3
−j(1+ω)]
−2
= 6
1 + ω
( ) e 2(1+ω)j −e −2(1+ω)j
2j
= 6
1 + ω sin2(1+ω)
Example24.5 UsethefirstshifttheoremtofindthefunctionwhoseFouriertransformis
given that F{u(t)e −mt } =
Solution Fromthe given resultwe have
Now
1
m+jω ,m>0.
F{u(t)e −3t } = 1
3+jω =F(ω)
1
3+j(ω−2) =F(ω−2)
1
3+j(ω−2) ,