How can i get the fourier series of any given function

Question

Murloc50 2021-11-25

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1596104-how-can-i-get-the-fourier-series-of-any-given-function

评论： Friedel Hartmann 2021-11-27

采纳的回答： Friedel Hartmann

Since there is no direct way of finding/computing the fourier seris of a function, e.g. cos(x), how would I do this?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Friedel Hartmann 2021-11-26

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1596104-how-can-i-get-the-fourier-series-of-any-given-function#answer_841309

在 MATLAB Online 中打开

clear

syms x f(x) k fh(x) p(x) ph(x) % a0 = a0 Formel /2

iTerme = 8;

sco = zeros(1,iTerme);

cco = zeros(1,iTerme);

% T = 2*pi, omega = 1

f(x) = piecewise(x <= pi/4, -1/(2*pi) * x, pi/4 < x,-1/(2*pi) * x + 1); % Function

for k=1:iTerme

sco(k) = 1/pi * int(sin(k * x) * (f(x)),[0,2*pi]); % coefficients

cco(k) = 1/pi * int(cos(k * x) * (f(x)),[0,2*pi]);

end

cnum = 1:iTerme; % Table with coefficients

Tabelle.num = cnum';

Tabelle.scof = sco';

Tabelle.ccof = cco';

T = struct2table(Tabelle)

T = 8×3 table

num scof ccof ___ __________ _________ 1 0.22508 -0.22508 2 9.8393e-18 -0.15915 3 -0.075026 -0.075026 4 -0.079577 0 5 -0.045016 0.045016 6 3.2798e-18 0.053052 7 0.032154 0.032154 8 0.039789 0

fplot(f(x),[0,2*pi]), title('Original'), yline(0);

figure

a0G = vpa(1/2 * 2/(2*pi) * int(f(x),[0,2*pi]),3)

a0G =

0.375

fh(x) = 1/(2*pi) * int(f(x),[0,2*pi]); % 'Mittelwert' 1. Term ist a0

for k = 1:iTerme

fh(x) = fh(x) + sco(k) * sin(k * x) + cco(k) * cos(k * x);

end

fplot(fh(x),[0,2*pi]), title('Fourier series'), yline(0);

vpa(fh(0),3)

ans =

0.046

p(x) = x/(2*pi); % 2nd Function

for k=1:iTerme

sco(k) = 1/pi * int(sin(k * x) * (p(x)),[0,2*pi]);

cco(k) = 1/pi * int(cos(k * x) * (p(x)),[0,2*pi]);

end

cnum = 1:iTerme;

Tabelle.num = cnum';

Tabelle.scof = sco';

Tabelle.ccof = cco';

T = struct2table(Tabelle)

T = 8×3 table

num scof ccof ___ _________ ____ 1 -0.31831 0 2 -0.15915 0 3 -0.1061 0 4 -0.079577 0 5 -0.063662 0 6 -0.053052 0 7 -0.045473 0 8 -0.039789 0

fplot(p(x),[0,2*pi]), title('Original'), yline(0);

figure

a0L = vpa(1/2 * 2/(2*pi) * int(p(x),[0,2*pi]),3)

a0L =

0.5

ph(x) = 1/pi * int(p(x),[0,2*pi]) * 1/(2); % 'Mittelwert' 1. Term ist a0/2

for k = 1:iTerme

ph(x) = ph(x) + sco(k) * sin(k * x) + cco(k) * cos(k * x);

end

fplot(ph(x),[0,2*pi]), title('Fourier series'), yline(0);

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Friedel Hartmann 2021-11-27

The series starts with a constant term, the average value, called 'Mittelwert' The routine first calculates the average value, assigns this number to ph(x) and then begins the loop for k = 1:iTerme etc. and adds the following sin and cos functions.

请先登录，再进行评论。