D.1 Computing Fourier integrals using FFT (not finished, to be deleted)

Consider the calculation of the following integral:

      ∫ b
I(ω ) =   eiωth(t)dt.
       a
(141)

Divide the interval [a,b] into M uniform sub-intervals and define

    b− a
Δ = ----,tj = a + Δj,hj = h(tj),j = 0,1,2,...,M
     M
(142)

Then the integration in Eq. (141) can be approximated as

         M∑−1
I(ω) ≈ Δ    hj exp (iωtj).
         j=0
(143)

Define ωm = 2πm∕(MΔ) with integer m and M∕2 < m < M∕2. Consider the calculation of I(ωm). Using Eq. (143), we obtain

            M∑ −1
I(ωm)  =  Δ     hjexp[iωm (a+ Δj)]
             j=0
                 M−1
       =  Δeiωma ∑   hjexp(iωm Δj)
                 j=0
                 M−1      (      )
       =  Δeiωma ∑   hjexp i2πm-j
                 j=0         M
             iωma
       =  Δe iω aHm                               (144)
       =  Δe  m [DFT (h0,h1,h2,...,hM −1)]m.        (145)
Equation (145) indicates the value of the integration I(ωm) can be obtained by calculating the discrete Fourier transformation of hj. However, as discussed in Ref. [2], equation (145) is not recommended for practical use because the oscillatory nature of the integral will make Eq. (145) become systematically inaccurate as ω increases. Next, consider a new method, in which h(t) is expanded as
      M∑     ( t− tj)     ∑         ( t− tj)
h(t) ≈   hjψ  --Δ--  +         hjφj  --Δ--
      j=0              j=endpoints
(146)

Apply the integral operator abdtexp(iωt) to both sides of Eq. (146), we obtain

∫ b          ∑M    ∫ b (      )         ∑       ∫ b   (     )
   h(t)eiωtdt ≈   hj    ψ  t−-tj eiωtdt+         hj   φj  t-− tj eiωtdt.
 a           j=0    a     Δ           j=endpoints    a      Δ
(147)

Make the change of variables s = (ttj)Δ in the first integral and s = (ta)Δ in the second integral, the above equation is written as

∫ b            ∑M    ∫ b                   ∑       ∫ b
   h(t)eiωtdt ≈ Δ   hj    ψ(s)eiω(Δs+tj)ds+ Δ         hj   φj (s− j)eiω(Δs+a)ds
 a             j=0    a                  j=endpoints    a
(148)

Define 𝜃 = ωΔ and make use of tj = a + jΔ, the above equation is written as

∫                          ∫                            ∫
  b    iωt       iωa M∑    i𝜃j  b    i𝜃s     iωa   ∑         b        i𝜃s
 a h(t)e  dt ≈ Δe      hje   a ψ(s)e  ds+ Δe            hj a φj(s− j)e ds
                   j=0                        j=endpoints
(149)

Define

       ∫ b
W (𝜃) =   ψ (s)ei𝜃sds
        a
(150)

       ∫ b         i𝜃s
αj(𝜃) = a φj(s− j)e  ds
(151)

Then Eq. (149) is written as

                   ⌊                               ⌋
∫ b                      ∑M           ∑
   h (t)eiωtdt ≈ Δeiωa⌈W (𝜃)   hjei𝜃j +        hjαj(𝜃)⌉.
  a                      j=0        j=endpoints
(152)

 

 

 

 

 


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Figure 8: Older version of Fig. ??, created by Metapost, the new version is created by the vector graphic editor in TeXmacs.