FourierSeriesInComplexFormAndFourierIntegral 2009-04-18 05:55:09 2009-04-18 05:55:09 Derivation Fourier series Fourier integral % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \PMlinkescapeword{expansion} \PMlinkescapeword{harmonics} \subsection{Fourier series in complex form} The Fourier series expansion of a Riemann integrable real function $f$ on the interval \,$[-p,\,p]$\, is \begin{align} 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), \end{align} where the coefficients are \begin{align} 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. \end{align} If one expresses the cosines and sines via \PMlinkname{Euler formulas}{ComplexSineAndCosine} with \PMlinkname{exponential function}{ComplexExponentialFunction}, the series (1) attains the form \begin{align} f(t) = \sum_{n=-\infty}^\infty c_ne^{\frac{in\pi t}{p}}. \end{align} 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 \begin{align} c_n = \frac{1}{2p}\int_{-p}^{\,p}f(t)e^{\frac{-in\pi t}{p}}\,dt \qquad (n = 0,\,\pm1,\,\pm2,\,\ldots). \end{align} We may say that in (3), $f(t)$ has been dissolved to sum of {\em harmonics} (elementary waves) $c_ne^{\frac{in\pi t}{p}}$ with amplitudes $c_n$ corresponding the frequencies $n$. \subsection{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): $$f(t) = \sum_{n=-\infty}^\infty e^{\frac{in\pi t}{p}}\frac{1}{2p}\int_{-p}^{\,p}f(t)e^{\frac{-in\pi t}{p}}\,dt$$ 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 $$f(t) = \frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{i\omega_nt}\Delta_n\omega \int_{-p}^{\,p}f(t)e^{-i\omega_nt}\,dt.$$ It can be shown that when\, $p \to \infty$\, and thus\, $\Delta_n\omega \to 0$,\, the limiting form of this equation is \begin{align} 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. \end{align} Here, $f(t)$ has been represented as a {\em 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 $$\int_{-\infty}^{\,\infty}|f(t)|\,dt$$ converges. For better to compare to the Fourier series (3) and the coefficients (4), we can write (5) as \begin{align} f(t) \,=\, \int_{-\infty}^{\,\infty}c(\omega)e^{i\omega t}d\omega, \end{align} where \begin{align} c(\omega) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty}f(t)e^{-i\omega t}dt. \end{align} \subsection{Fourier transform} If we denote $2\pi c(\omega)$ as \begin{align} F(\omega) \,=\, \int_{-\infty}^{\,\infty} e^{-i\omega t}f(t)\,dt, \end{align} then by (5), \begin{align} f(t) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty}e^{i\omega t}F(\omega)\,d\omega. \end{align} $F(\omega)$ is called the {\em Fourier transform} of $f(t)$.\, It is an integral transform and (9) \PMlinkescapetext{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.