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

The t−ω duality principle 769
f(t)
1
F(v)
2 sint
——— t
2pf(v)
2p
–1 1
t
v
t
–1 1
v
Figure24.7
Illustrating thet--ω duality principle.
Figure24.8
Illustrating thet--ω duality principle.
where
F(ω) =
∫ ∞
−∞
f (t)e −jωt dt (24.7)
is the Fourier transform of f (t). In Equation (24.6), ω is a dummy variable so, for example,
Equation (24.6) could be equivalently written as
f(t)= 1
2π
∫ ∞
−∞
F(z)e jzt dz (24.8)
Then, from Equation (24.8),replacingt by −ω wefind
f(−ω) = 1
2π
∫ ∞
−∞
F(z)e −jωz dz = 1
2π
∫ ∞
−∞
F(t)e −jωt dt
which werecognize as 1 timesthe Fourier transformofF(t).
2π
We have the following result:
IfF(ω)isthe Fourier transformof f (t)then
f(−ω)is 1 ×(theFourier transform ofF(t))
2π
which isknown as thet--ω duality principle.
We have seen inExample 24.2 that if
{ 1 |t|1
f(t)=
0 |t|>1
thenF(ω) = 2sinω
ω
.ThisisdepictedinFigure24.7.Fromthedualityprinciplewecan
immediately deduce that
{ } 2sint
F =2πf(−ω)=2πf(ω)
t
since f is an even function (Figure 24.8). Unfortunately it is very difficult to verify this
result in most cases because while one of the integrals is relatively straightforward to
evaluate, the other is usually very difficult. However, we can use the result to derive a
number of new Fourier transforms.
Example24.9 GiventhattheFouriertransformofu(t)e −t is
the transform of
1
1+jt .
1
1+jω usethedualityprincipletodeduce