D.S.G. Pollock 253For matrix representations of these transforms, one may define:U=T−^1 /^2 [exp{−i2πtj/T};t,j=0,...,T− 1 ],U ̄=T−^1 /^2 [exp{i2πtj/T};t,j=0,...,T− 1 ],(6.20)which are unitary complex matrices such thatUU ̄=UU ̄ =IT. Then:
x=T^1 /^2 U ̄ξ ←→ ξ=T−^1 /^2 Ux, (6.21)wherex=[x 0 ,x 1 ,...xT− 1 ]′andξ=[ξ 0 ,ξ 1 ,...ξT− 1 ]′are the vectors of the data
and of their Fourier ordinates respectively.
6.4 Spectral representations of a stationary process
The various equations of the Fourier analysis of a finite data sequence can also be
used to describe the processes that generate the data. Thus, within the equation:
yt=∑nj= 0{
αjcos(ωjt)+βjsin(ωjt)}=ζ 0 +∑nj= 1{
ζjeiωjt+ζj∗e−iωjt}
,(6.22)the quantitiesαj,βjcan be taken to represent independent real-valued random
variables, and the quantities:
ζj=αj−iβj
2and ζj∗=αj+iβj
2(6.23)can be regarded as complex-valued random variables.
The autocovariance of the elementsytandysis given by:
E(ytys)=∑nj= 0∑nk= 0E[
ζjζkei(ωjt+ωks)+ζjζk∗ei(ωjt−ωks)+ζj∗ζkei(ωks−ωjt)+ζj∗ζk∗e−i(ωjt+ωks)]
.(6.24)The condition of stationarity requires that the covariance should be a function
only of the temporal separation|t−s|ofytandys. For this, it is necessary that:
E(ζjζk)=E(ζj∗ζk∗)=E(ζj∗ζk)=E(ζjζk∗)=0, (6.25)wheneverj=k. Also, the conditions:
E(ζj^2 )=0 and E(ζj∗^2 )=0, (6.26)