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Fourier series

已有 3262 次阅读 2015-11-23 03:23 |系统分类:科研笔记

1) square wave

FourierSeriesSquareWave

Consider a square wave f(x) of length 2L. Over the range [0,2L], this can be written as

 f(x)=2[H(x/L)-H(x/L-1)]-1,
(1)

where H(x) is the Heaviside step function. Since f(x)=f(2L-x), the function is odd, so a_0=a_n=0, and

 b_n=1/Lint_0^(2L)f(x)sin((npix)/L)dx
(2)

reduces to

b_n=2/Lint_0^Lf(x)sin((npix)/L)dx
(3)
=4/(npi)sin^2(1/2npi)
(4)
=2/(npi)[1-(-1)^n]
(5)
=4/(npi){0 n even; 1 n odd.
(6)

The Fourier series is therefore

 f(x)=4/pisum_(n=1,3,5,...)^infty1/nsin((npix)/L).


2) sawtooth

FourierSeriesSawtoothWave

Consider a string of length 2L plucked at the right end and fixed at the left. The functional form of this configuration is

 f(x)=x/(2L).
(1)

The components of the Fourier series are therefore given by

a_0=1/Lint_0^(2L)x/(2L)dx
(2)
=1
(3)
a_n=1/Lint_0^(2L)x/(2L)cos((npix)/L)dx
(4)
=([2npicos(npi)-sin(npi)]sin(npi))/(n^2pi^2)
(5)
=0
(6)
b_n=1/Lint_0^(2L)x/(2L)sin((npix)/L)dx
(7)
=(-2npicos(2npi)+sin(2npi))/(2n^2pi^2)
(8)
=-1/(npi).
(9)

The Fourier series is therefore given by

f(x)=1/2-1/pisum_(n=1)^(infty)1/nsin((npix)/L)
(10)
=1/2+i/(2pi)ln(-e^(-ipix/L))
(11)
=1/2-1/(2pi)arg(-e^(-ipix/L)).


3) triangle

FourierSeriesTriangleWave

Consider a symmetric triangle waveT(x) of period 2L. Since the function is odd,

a_0=0
(1)
a_n=0,
(2)

and

b_n=2/L{int_0^(L/2)x/(L/2)sin((npix)/L)dx+int_(L/2)^L[1-2/L(x-1/2L)]sin((npix)/L)dx}
(3)
=(32)/(pi^2n^2)cos(1/4npi)sin^3(1/4npi)
(4)
=(32)/(pi^2n^2){0 n=0, 4, ...; 1/4 n=1, 5, ...; 0 n=2, 6, ...; -1/4 n=3, 7, ...
(5)
=8/(pi^2n^2){(-1)^((n-1)/2) for n odd; 0 for n even.
(6)

The Fourier series for the triangle wave is therefore

 f(x)=8/(pi^2)sum_(n=1,3,5,...)^infty((-1)^((n-1)/2))/(n^2)sin((npix)/L).
(7)

FourierSeriesTriangleWaves

Now consider the asymmetric triangle wave pinned an x-distance which is (1/m)th of the distance L. The displacement as a function of x is then

 f_m(x)={(mx)/L   for 0<=x<=L/m; 1-m/((m-1)L)(x-L/m)   for L/m<=x<=2L-L/m; m/L(x-2L)   for 2L-L/m<=x<=2L.
(8)

The coefficients are therefore

a_0=0
(9)
a_n=0
(10)
b_n=-(2(-1)^nm^2)/(n^2(m-1)pi^2)sin[(n(m-1)pi)/m].
(11)

Taking m=2 gives the same Fourier series as before.


4) Semi-circle

FourierSeriesSemicircle

Given a semicircular hump

f(x)=sqrt(L^2-(x-L)^2)
(1)
=sqrt((2L-x)x),
(2)

the Fourier coefficients are

a_0=1/2piL
(3)
a_n=((-1)^nLJ_1(npi))/n
(4)
b_n=0,
(5)

where J_1(z) is a Bessel function of the first kind, so the Fourier series is therefore

 f(x)=L[1/4pi+sum_(n=1)^infty((-1)^nJ_1(npi))/ncos((npix)/L)].




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