0.1 Fourier series in complex form

The Fourier series expansion of a Riemann integrable real function f on the interval [−p, p] is

f(t) = + ∑ n=1∞,

(1)

where the coefficients are

an = ∫ −ppf(x) cos dt, bn = ∫ −ppf(x) sin dt.

(2)

If one expresses the cosines and sines via Euler formulas with exponential function, the series (1) attains the form

f(t) = ∑ n=−∞∞c ne .

(3)

The coefficients c_n could be obtained of a_n and b_n, but they are comfortably derived directly by multiplying the equation (3) by e⁻ and integrating it from −p to p. One obtains

cn = ∫ −ppf(t)e dt (n = 0, ±1, ±2, …).

(4)

We may say that in (3), f(t) has been dissolved to sum of harmonics (elementary waves) c_ne with amplitudes c_n corresponding the frequencies n.

0.2 Derivation of Fourier integral

For seeing how the expansion (3) changes when p →∞, we put first the expressions (4) of c_n to the series (3):

By denoting ω_n := and Δ_nω := ω_n+1−ω_n = , the last equation takes the form

It can be shown that when p →∞ and thus Δ_nω → 0, the limiting form of this equation is

f(t) = ∫ −∞∞eiωtdω ∫ −∞∞f(t)e−iωtdt.

(5)

Here, f(t) has been represented as a Fourier integral. It can be proved that for validity of the expansion (4) it suffices that the function f is piecewise continuous on every finite interval having at most a finite amount of extremum points and that the integral

converges.

For better to compare to the Fourier series (3) and the coefficients (4), we can write (5) as

f(t) = ∫ −∞∞c(ω)eiωtdω,

(6)

where

c(ω) = ∫ −∞∞f(t)e−iωtdt.

(7)

0.3 Fourier transform

If we denote 2πc(ω) as

F(ω) = ∫ −∞∞e−iωtf(t) dt,

(8)

then by (5),

f(t) = ∫ −∞∞eiωtF(ω) dω.

(9)

F(ω) is called the Fourier transform of f(t). It is an integral transform and (9) represents its inverse transform.

N.B. that often one sees both the formula (8) and the formula (9) equipped with the same constant factor in front of the integral sign.

References