How to use multi-part function to find Fourier series coefficients with this program?

2 次查看(过去 30 天)
Hi I am writing a script to find Fourier series coefficients. Code is given below. In the code, I have function f=x*x defined on interval -L to L. However if I have some function defined, say
f(x) = -1 -L<=x<0
= 1 0<=x<=L
then, how I can use such function to find Fourier series coefficients? Any ideas?
close all;
clc;
syms x k L n
evalin(symengine,'assume(k,Type::Integer)'); %k is treated as integer
f=x*x; %f is the function to find Fourier series
a = int(f*cos(k*pi*x/L)/L,x,-L,L); %Fourier series coefficient a_n
b = int(f*sin(k*pi*x/L)/L,x,-L,L); %Fourier series coefficient b_n
afun=@(f,x,k,L) int(f*cos(k*pi*x/L)/L,x,-L,L);
bfun = @(f,x,k,L) int(f*sin(k*pi*x/L)/L,x,-L,L);
fs = vpa(afun(f,x,0,L)/2 + symsum(afun(f,x,k,L)*cos(k*pi*x/L) + ... bfun(f,x,k,L)*sin(k*pi*x/L),k,1,3),5); %Partial sum of Fourier series coefficients
fs=subs(fs,L,1); %making L=1
fs1=symfun(fs,x); %defining symbolic function fs1 for plotting
% disp(a);
% disp(b);
% disp(fs);
figure();
ezplot(fs1,[-5*pi,5*pi]);

回答(1 个)

Star Strider
Star Strider 2014-7-13
编辑:Star Strider 2014-7-13
You are not defining the integral limits correctly.
For example:
syms a b k L w0 x
krnl = symfun(a*cos(-j*k*w0*x) + j*b*sin(-j*k*w0*x), x) % Transform kernel
F1 = int(-1*krnl, x, -L, 0); % -L -> 0
F2 = int(+1*krnl, x, 0, L); % 0 -> +L
FT = F1 + F2; % Add integrals
FT = subs(FT, L, 1) % L = 1
FT = simplify(collect(FT))
produces (in R2014a):
FT =
(4*b*sinh((k*w0)/2)^2)/(k*w0)
MATLAB (correctly) substituted sinh for the sin of an imaginary argument, so this is equivalent to:
FT =
(4*b*sin((j*k*w0)/2)^2)/(k*w0)
The individual ‘b’ coefficients are functions of k. The ‘a’ coefficients are identically zero.
The frequency argument, w0, is the fundamental frequency of the sampled signal. I refer you to the Wikipedia article on Fourier series for the details.

类别

Help CenterFile Exchange 中查找有关 Assumptions 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by