傅里叶级数 - 情节在Matlab

Question

这里是傅里叶级数的情节我的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，它是如此之慢！为了提高速度，我应该在代码中更改哪些内容？

编辑：在参考macduff的评论： 这样的事情？

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。尝试使用输入为an和bn的表达式作为可由Matlab解释的字符串（我猜测它现在还是）。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

是这样的吗？

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