B.2 Fourier transformation

As discussed above, the Fourier transformation of a function h(t) is given by

       ∫ ∞
H (f ) =    h(t)ei2πftdt.
        −∞
(102)

Once the Fourier transformation H(f) is known, the original function h(t) can be reconstructed via

      ∫ ∞       −i2πft
h (t) =    H (f)e     df.
       −∞
(103)

[Note that the signs in the exponential of Eq. (100) and (101) are opposite. Which one should be minus or positive is actually a matter of convention because a trivial variable substitution f= f can change the sign between minus and positive. Proof. In terms of f, Eq.  (103) is written

        ∫
h(t)  =    −∞ H (− f ′)ei2πtf′d(− f′),
         ∞
        ∫ ∞       ′ i2πtf′ ′
     =   − ∞ H(− f)e    df,                 (104)
Define
H(f′) ≡   H(− f′)
          ∫ ∞          ′
      =       h(t)e−i2πtf dt.               (105)
           −∞
Then Eq. (104) is written
      ∫
        ∞ --  ′ i2πtf′ ′
h (t) = −∞ H (f )e     df
(106)

The signs in the exponential of Eqs. (105) and (106) are opposite to Eqs. (102) and (103), respectively.]

[Some physicists prefer to use the angular frequency ω 2πf rather than the frequency f to represent the Fourier transformation. Using ω, equations. (102) and (103) are written, respectively, as

       ∫
H(ω) =   ∞ h(t)eiωtdt,
        −∞
(107)

         ∫
h (t) =-1-  ∞ H (ω)e−iωtdω,
      2π  −∞
(108)

where we see that the asymmetry between the Fourier transformation and its inverse is more severe in this representation: besides the opposite-sign in the exponents, there is also a 12π factor difference between the Fourier transformation and its inverse. Whether the 12π factor appears at the forward transformation or inverse one is actually a matter of convention. The only requirement is that the product of the two factors in the forward and inverse transformation is equal to 12π. To obtain a more symmetric pair, one can adopt a factor 1√2-π- at both the forward and inverse transformation. The representation in Eqs. (102) and (103) is adopted in this note. But we should know how to change to the ω representation when needed.]

[

∫                                ∫                    ∫                      ∫
  ∞ dh(t)e−i2πtfdt = h(t)e−i2πtf|+ ∞−  ∞ h(t)de−i2πtf dt = −  ∞  h(t)de−i2πtfdt = i2πf   ∞ h(t)e−i2πtfdt = i2πfH (f)
 −∞  dt                     − ∞   −∞       dt          −∞       dt            −∞
(109)

]