Fourier series in complex form and Fourier integral

Fourier series in complex form

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

$\displaystyle f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos{\frac{n\pi t}{p}}+b_n\sin{\frac{n\pi t}{p}}\right),$ (1)

where the coefficients are

$\displaystyle a_n = \frac{1}{p}\int_{-p}^{\,p}f(x)\cos{\frac{n\pi t}{p}}\,dt, \quad b_n = \frac{1}{p}\int_{-p}^{\,p}f(x)\sin{\frac{n\pi t}{p}}\,dt.$ (2)

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

$\displaystyle f(t) = \sum_{n=-\infty}^\infty c_ne^{\frac{in\pi t}{p}}.$ (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^{-\frac{im\pi t}{p}}$ and integrating it from $ -p$ to $ p$ . One obtains

$\displaystyle c_n = \frac{1}{2p}\int_{-p}^{\,p}f(t)e^{\frac{-in\pi t}{p}}\,dt \qquad (n = 0,\,\pm1,\,\pm2,\,\ldots).$ (4)

We may say that in (3), $ f(t)$ has been dissolved to sum of harmonics (elementary waves) $ c_ne^{\frac{in\pi t}{p}}$ with amplitudes $ c_n$ corresponding the frequencies $ n$ .

Derivation of Fourier integral

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

$\displaystyle f(t) = \sum_{n=-\infty}^\infty e^{\frac{in\pi t}{p}}\frac{1}{2p}\int_{-p}^{\,p}f(t)e^{\frac{-in\pi t}{p}}\,dt$

By denoting $ \omega_n := \frac{n\pi}{p}$ and $ \Delta_n\omega := \omega_{n+1}\!-\!\omega_n = \frac{\pi}{p}$ , the last equation takes the form

$\displaystyle f(t) = \frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{i\omega_nt}\Delta_n\omega \int_{-p}^{\,p}f(t)e^{-i\omega_nt}\,dt.$

It can be shown that when $ p \to \infty$ and thus $ \Delta_n\omega \to 0$ , the limiting form of this equation is

$\displaystyle f(t) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty} e^{i\omega t}d\omega\int_{-\infty}^{\,\infty} f(t)e^{-i\omega t}dt.$ (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

$\displaystyle \int_{-\infty}^{\,\infty}\vert f(t)\vert\,dt$

converges.

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

$\displaystyle f(t) \,=\, \int_{-\infty}^{\,\infty}c(\omega)e^{i\omega t}d\omega,$ (6)

where

$\displaystyle c(\omega) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty}f(t)e^{-i\omega t}dt.$ (7)

Fourier transform

If we denote $ 2\pi c(\omega)$ as

$\displaystyle F(\omega) \,=\, \int_{-\infty}^{\,\infty} e^{-i\omega t}f(t)\,dt,$ (8)

then by (5),

$\displaystyle f(t) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty}e^{i\omega t}F(\omega)\,d\omega.$ (9)

$ F(\omega)$ 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 $ \displaystyle\frac{1}{\sqrt{2\pi}}$ in front of the integral sign.

Bibliography

1
K. V¨AISÄLÄ: Laplace-muunnos. Handout Nr. 163.    Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).



Contributors to this entry (in most recent order):

As of this snapshot date, this entry was owned by pahio.