Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
determination of Fourier coefficients (Algorithm)

Suppose that the real function $f$ may be presented as sum of the Fourier series:

$\displaystyle f(x) \;=\; \frac{a_0}{2}+\sum_{m=0}^\infty(a_m\cos{mx}+b_m\sin{mx})$ (1)
Therefore, $f$ is periodic with period $2\pi$.  For expressing the Fourier coefficients $a_m$ and $b_m$ with the function itself, we first multiply the series (1) by $\cos{nx}$ ( $n \in \mathbb{Z}$) and integrate from $-\pi$ to $\pi$.  Supposing that we can integrate termwise, we may write
$\displaystyle \int_{-\pi}^\pi\!f(x)\cos{nx}\,dx \,=\, \frac{a_0}{2}\!\int_{-\pi... ...^\pi\!\cos{mx}\cos{nx}\,dx+b_m\!\int_{-\pi}^\pi\!\sin{mx}\cos{nx}\,dx\right)\!.$ (2)
When  $n = 0$,  the equation (2) reads
$\displaystyle \int_{-\pi}^\pi f(x)\,dx = \frac{a_0}{2}\cdot2\pi = \pi a_0,$ (3)
since in the sum of the right hand side, only the first addend is distinct from zero.

When $n$ is a positive integer, we use the product formulas of the trigonometric identities, getting

$\displaystyle \int_{-\pi}^\pi\cos{mx}\cos{nx}\,dx = \frac{1}{2}\int_{-\pi}^\pi[\cos(m-n)x+\cos(m+n)x]\,dx,$

$\displaystyle \int_{-\pi}^\pi\sin{mx}\cos{nx}\,dx = \frac{1}{2}\int_{-\pi}^\pi[\sin(m-n)x+\sin(m+n)x]\,dx.$
The latter expression vanishes always, since the sine is an odd function.  If  $m \neq n$,  the former equals zero because the antiderivative consists of sine terms which vanish at multiples of $\pi$; only in the case  $m = n$  we obtain from it a non-zero result $\pi$.  Then (2) reads
$\displaystyle \int_{-\pi}^\pi f(x)\cos{nx}\,dx = \pi a_n$ (4)
to which we can include as a special case the equation (3).

By multiplying (1) by $\sin{nx}$ and integrating termwise, one obtains similarly

$\displaystyle \int_{-\pi}^\pi f(x)\sin{nx}\,dx = \pi b_n.$ (5)
The equations (4) and (5) imply the formulas

$\displaystyle a_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos{nx}\,dx \quad (n = 0,\,1,\,2,\,\ldots)$
and

$\displaystyle b_n \;=\; \frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin{nx}\,dx \quad (n = 1,\,2,\,3,\,\ldots)$
for finding the values of the Fourier coefficients of $f$.



Anyone with an account can edit this entry. Please help improve it!

"determination of Fourier coefficients" is owned by pahio. [ full author list (2) ]

View style:

See Also: generalized Fourier transform, generalized Fourier and measured groupoid transforms, 2D-FT MR- Imaging and related Nobel awards

Keywords:  Fourier series coefficients, discrete Fourier transform

Cross-references: identities, formulas, function

This is version 3 of determination of Fourier coefficients, born on 2009-04-18, modified 2009-04-18.
Object id is 660, canonical name is DeterminationOfFourierCoefficients.
Accessed 296 times total.

Classification:
Physics Classification: 02. (Mathematical methods in physics)
 02.30.Nw (Fourier analysis)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "