傅里叶级数展开matlab 实现给个例子说明下:将函数
y=x*(x-pi)*(x-2*pi),在(0,2*pi)的范围内傅里叶级数展开syms x fx=x*(x-pi)*(x-2*pi);
[an,bn,f]=fseries(fx,x,12,0,2*pi)%前12 项展开latex(f)%将f 转换成latex 代码an = [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] bn = [ -12, 3/2, -4/9, 3/16, -12/125, 1/18, -12/343, 3/128, -4/ 243, 3/250, -12/1331, 1/144] f =
12*sin(x)+3/2*sin(2*x)+4/9*sin(3*x)+3/16*sin(4*x)+12/ 125*sin(5*x)+1/18*sin(6
*x)+12/343*sin(7*x)+3/128*sin(8*x)+4/243*sin(9*x)+3/ 250*sin(10*x)+12/1331* sin(11*x)+1/144*sin(12*x) ans = 12\,\sin \left( x \right) +3/2\,\sin \left( 2\,x \right)
+4/9\,\sin \left( 3\,x \right) +3/16\,\sin \left( 4\,x \right) +{\frac {12}{125}}\,\sin \left( 5\,x \right) +1/18\,\sin
\left( 6\,x \right) +{\frac {12}{343}}\,\sin \left( 7\,x \right) +{\frac {3}{128}}\,\sin \left( 8\,x \right) +{\frac
{4}{243}}\,\sin \left( 9\,x \right) +{\frac {3}{250}}\,\sin \left( 10\,x \right) +{\frac {12}{1331}}\,\sin \left( 11\,x \right) +{\frac {1}{144}}\,\sin \left( 12\,x \right) function [an,bn,f]=fseries(fx,x,n,a,b) %傅里叶级数展开% %an 为fourier 余弦项系数%bn 为fourier 正弦项系数%f 为展开表达式%f 为给定函数%x 为自变量%n 为展开系