INTEGRAL TRANSFORMS
c(ω)expiωt
− 1 0
0
1 2 r
−^2 Tπ^2 Tπ^4 Tπ ωr
Figure 13.1 The relationship between the Fourier terms for a function of
periodTand the Fourier integral (the area below the solid line) of the
function.
∆ω=2π/Tbecomes vanishingly small and the spectrum of allowed frequencies
ωrbecomes a continuum. Thus, the infinite sum of terms in the Fourier series
becomes an integral, and the coefficientscrbecome functions of thecontinuous
variableω, as follows.
We recall, cf. (12.10), that the coefficientscrin (13.1) are given by
cr=
1
T
∫T/ 2
−T/ 2
f(t)e−^2 πirt/Tdt=
∆ω
2 π
∫T/ 2
−T/ 2
f(t)e−iωrtdt, (13.2)
where we have written the integral in two alternative forms and, for convenience,
made one period run from−T/2to+T/2 rather than from 0 toT. Substituting
from (13.2) into (13.1) gives
f(t)=
∑∞
r=−∞
∆ω
2 π
∫T/ 2
−T/ 2
f(u)e−iωrudu eiωrt. (13.3)
At this stageωris still a discrete function ofrequal to 2πr/T.
The solid points in figure 13.1 are a plot of (say, the real part of)creiωrtas
a function ofr(or equivalently ofωr) and it is clear that (2π/T)creiωrtgives
the area of therth broken-line rectangle. IfTtends to∞then ∆ω(= 2π/T)
becomes infinitesimal, the width of the rectangles tends to zero and, from the
mathematical definition of an integral,
∑∞
r=−∞
∆ω
2 π
g(ωr)eiωrt →
1
2 π
∫∞
−∞
g(ω)eiωtdω.
In this particular case
g(ωr)=
∫T/ 2
−T/ 2
f(u)e−iωrudu,