D.S.G. Pollock 291To demonstrate this result, we must consider the Fourier representation of a real-
valued square-integrable functionx(t)defined over the real line. The following are
the corresponding expressions for the functionx(t)and its Fourier transformξ(ω):
x(t)=
1
2 π∫∞−∞eiωtξ(ω)dω←→ξ(ω)=∫∞−∞e−iωtx(t)dt. (6.156)By samplingx(t)at intervals ofπ/ωd, a sequence:
{xτ=x(τ[π/ωd]);τ=0,±1,±2,...},is generated. The elements of the sequence and their Fourier transformξs(ω)are
given by:
xτ=
1
2 ωd∫ωd−ωdexp{iωτ[π/ωd]}ξS(ω)dω←→ξS(ω)=∑∞τ=−∞xτexp{−iωτ[π/ωd]}.(6.157)Sinceξ(ω)=ξS(ω)is a continuous function defined on the interval[−ωd,ωd],
it may be regarded as a function that is periodic in frequency, with a period of
2 ωd. Putting the right-hand side of (6.157) into the left-hand side of (6.156), and
taking the integral over[−ωd,ωd]in consequence of the band-limited nature of the
functionx(t), gives:
x(t)=
1
2 π∫ωd−ωd{ ∞
∑τ=−∞xτe−iωτ[π/ωd]}
eiωtdω=
1
2 π∑∞τ=−∞xτ∫ωd−ωdeiω(t−[τπ/ωd])dω.(6.158)The integral on the right-hand side is evaluated as:
∫ωd−ωdeiω(t−[τπ/ωd])dω= 2
sin(tωd−τπ)
t−τ[π/ωd]. (6.159)
Putting this into the right-hand side of (6.158) gives:
x(t)=∑∞τ=−∞xτ
sin(tωd−τπ)
π(t−τ[π/ωd])
=∑∞k=−∞xτφd(τ−k), (6.160)where:
φd(t−τ)=
sin(tωd−τπ)
π(t−τ[π/ωd]). (6.161)