Question

这是我的傅立叶级数图的m文件：

clear
clc
syms n
a0=input('Enter coefficient a0: ');
an=input('Enter coefficient an: ');
bn=input('Enter coefficient bn: ');
a=input('Enter lower boundary: ');
b=input('Enter upper boundary: ');
t=linspace(a,b,10000);
suma=0;
for n=1:10 %% n could be any number, bigger n - better approximation
suma=suma+(subs(an,'n',n).*cos(2.*n.*pi.*t./(b-a))+subs(bn,'n',n).*sin(2.*n.*pi.*t./(b-a)));
end
series=a0+suma;
plot(t,series)
grid

问题是，它太慢了！为了提高速度，我应该在代码中更改什么？

编辑：参考macduff的评论：像这样的东西？

clear
clc
a0=input('Enter coefficient a0: ');
an=input('Enter coefficient an: ','s');
bn=input('Enter coefficient bn: ','s');
a=input('Enter lower boundary: ');
b=input('Enter upper boundary: ');
t=linspace(a,b,10000);
suma=0;
for n=1:10000 %% n could be any number, bigger n - better approximation
ebn = evalin('caller',bn);
ean = evalin('caller',an);
suma = suma + (ean.*cos(2.*n.*pi.*t./(b-a)) + ebn.*sin(2.*n.*pi.*t./(b-a)));
end
series=a0+suma;
plot(t,series)
grid

Answer 1

我会尝试摆脱符号工具箱。尝试将an和bn表达式输入的字符串输入为可由Matlab解释的字符串（我猜它现在甚至是现在）。然后每个循环，评估调用者工作区中的字符串。

for n=1:10 %% n could be any number, bigger n - better approximation
  ebn = evalin('caller',bn);
  ean = evalin('caller',an);
  suma = suma + (ean.*cos(2.*n.*pi.*t./(b-a)) + ebn.*sin(2.*n.*pi.*t./(b-a)));
end

Answer 2

这样的东西？

clear
clc
a0=input('Enter coefficient a0: ');
an=input('Enter coefficient an: ','s');
bn=input('Enter coefficient bn: ','s');
a=input('Enter lower boundary: ');
b=input('Enter upper boundary: ');
t=linspace(a,b,10000);
suma=0;
for n=1:10000 %% n could be any number, bigger n - better approximation
ebn = evalin('caller',bn);
ean = evalin('caller',an);
suma = suma + (ean.*cos(2.*n.*pi.*t./(b-a)) + ebn.*sin(2.*n.*pi.*t./(b-a)));
end
series=a0+suma;
plot(t,series)
grid

傅里叶级数 - 用Matlab绘制

2 个答案: