Matlab中的Fourier级数图

这是我的傅里叶级数图的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

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

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

---

我会试着离开象征性的工具箱。尝试使用

an

和

bn

表达式作为字符串输入的东西，该字符串可以被Matlab解释（我猜现在也是这样）。然后在每个循环中，计算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

像这样的

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

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

---

是的，这就是我的意思，希望它有用！现在太快了：）非常感谢！是否可以检查代码是否产生与代码相同的结果？使用输入未定义函数或变量“n”时出错。