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

764 Chapter 24 The Fourier transform
Therefore, fromthe first shift theorem witha = 2 wehave
F{e 2jt u(t)e −3t } =
1
3+j(ω−2)
Consequently the function whose Fourier transform is
1
3+j(ω−2) isu(t)e−(3−2j)t .
Example24.6 Findthe Fourier transformof
{ e
−3t
t0
f(t)=
e 3t t<0
Deducethe functionwhoseFourier transformisG(ω) =
∫ ∞
Solution F(ω) = f (t)e −jωt dt
−∞
Now
=
=
∫ 0
−∞
∫ 0
−∞
e 3t e −jωt dt +
e (3−jω)t dt +
∫ ∞
0
∫ ∞
0
e −3t e −jωt dt
e −(3+jω)t dt
[ e
(3−jω)t 0 [ e
−(3+jω)t ∞
= +
3−jω]
−∞
−(3+jω)]
0
= 1
3−jω + 1
3+jω
= 6
9+ω 2
G(ω) =
6
10+2ω+ω = 6
2 (ω+1) 2 +9 =F(ω+1)
6
10+2ω+ω 2.
Then, using the first shift theorem F(ω + 1) will be F{e −jt f (t)}; that is, the required
function is
{ e
(−3−j)t
t 0
g(t) =
e (3−j)t t < 0
24.3.3 Secondshifttheorem
IfF(ω)isthe Fourier transform of f (t)then
F{f(t −α)} =e −jαω F(ω)