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

770 Chapter 24 The Fourier transform
1
Solution We know F(ω) =
1+jω is the Fourier transform of f (t) = u(t)e−t . Therefore
2π(u(−ω)e ω 1
)is the Fourier transform of
1+jt .
24.6 FOURIERTRANSFORMSOFSOMESPECIALFUNCTIONS
WesawinSection24.4thattheFouriertransformtellsusthefrequencycontentofasignal.IfweweretofindtheFouriertransformofasignalcomposedofonlyonefrequency
component, for example f (t) = sint, we would hope that the exercise of finding the
Fourier transform would resultinaspectrum containing thatsinglefrequency.
Unfortunately if we try to find the Fourier transform of, say, f (t) = sint problems
arisesince the integral
∫ ∞
−∞
sinte −jωt dt
cannot be evaluated in the usual sense because sint oscillates indefinitely as |t| → ∞.
In particular, Condition (2) of Section 24.2 fails since ∫ ∞
|sint|dt diverges. There are
−∞
many other functions which give rise to similar difficulties, for instance the unit step
function, polynomials and so on. All these functions fail to have a Fourier transform
in its usual sense. However, by making use of the delta function it is possible to make
progress even with functions like these.
24.6.1 TheFouriertransformof δ(t −a)
Example24.10 Usethe propertiesofthe deltafunction todeduceits Fourier transform.
Solution Bydefinition
F{δ(t −a)} =
∫ ∞
−∞
δ(t −a)e −jωt dt
Next, recall the following propertyofthe delta function
∫ ∞
−∞
f(t)δ(t −a)dt = f(a) (24.9)
foranyreasonablywell-behavedfunction f (t).UsingEquation(24.9)with f (t) = e −jωt
wehave
F{δ(t −a)} =
∫ ∞
−∞
e −jωt δ(t −a)dt = e −jωa
In particular, ifa = 0 wehave F{δ(t)} = 1.This resultisdepictedinFigure 24.9.