Question

我有以下脚本

clc; clear all; close all;

syms x n
f = x;
L = 1;

subplot(2,1,1)
h = ezplot(f,[-L,L])
set(h, 'Color','r','LineWidth',1)

a0 = (1/L) * int(f * cos(0* pi*x/L),-L,L)
an = (1/L) * int(f * cos(n* pi*x/L),-L,L)
bn = (1/L)* int(f* sin(n* pi*x/L),-L,L)
fx = a0/2 + symsum((an* cos(n*pi*x/L) + bn* sin(n*pi*x/L)),n,1,5)
% for n =5, the answer: fx = (2*sin(pi*x))/pi - sin(2*pi*x)/pi + 
%(2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)

hold on
h = ezplot(fx,[-L,L])
grid on

%Solution with complex Fourier
c0 = (1/(2*L))*int(f*exp(-j*0*pi*x/L),-L,L)
cn = (1/(2*L))*int(f*exp(-j*n*pi*x/L),-L,L)

subplot(2,1,2)
h = ezplot(f,[-L,L])
set(h, 'Color','r','LineWidth',1)

fx_c = c0 + symsum(cn*exp(j*n*pi*x/L),n,-5,-1) + ...
          symsum(cn*exp(j*n*pi*x/L),n,1,5) % n for complex -5,5

hold on
h = ezplot(fx_c,[-L,L])
grid on

我的问题：因为fx的答案应该等于fx_c（复数傅立叶）。我们可以从这两个函数产生的数字中看出。他们是一样的。但

    fx =
(2*sin(pi*x))/pi - sin(2*pi*x)/pi + (2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)

和

fx_c =
exp(-pi*x*i)*((pi*i - 1)/(2*pi^2) + (pi*i + 1)/(2*pi^2)) - exp(pi*x*i)*((pi*i - 1)/(2*pi^2) + (pi*i + 1)/(2*pi^2)) - exp(-pi*x*2*i)*((pi*2*i - 1)/(8*pi^2) + (pi*2*i + 1)/(8*pi^2)) + exp(pi*x*2*i)*((pi*2*i - 1)/(8*pi^2) + (pi*2*i + 1)/(8*pi^2)) + exp(-pi*x*3*i)*((pi*3*i - 1)/(18*pi^2) + (pi*3*i + 1)/(18*pi^2)) - exp(pi*x*3*i)*((pi*3*i - 1)/(18*pi^2) + (pi*3*i + 1)/(18*pi^2)) - exp(-pi*x*4*i)*((pi*4*i - 1)/(32*pi^2) + (pi*4*i + 1)/(32*pi^2)) + exp(pi*x*4*i)*((pi*4*i - 1)/(32*pi^2) + (pi*4*i + 1)/(32*pi^2)) + exp(-pi*x*5*i)*((pi*5*i - 1)/(50*pi^2) + (pi*5*i + 1)/(50*pi^2)) - exp(pi*x*5*i)*((pi*5*i - 1)/(50*pi^2) + (pi*5*i + 1)/(50*pi^2))

如何将fx_c转换为fx？

Answer 1

它们与Euler's formula相关联。您可以使用rewrite command：

进行检查

>> rewrite(exp(1i*x), 'cos')

ans =

cos(x) + sin(x)*1i

将它应用到你的函数并简化一下，你可以得到相同的表达式：

>> expand(rewrite(fx_c, 'cos'), 'ArithmeticOnly', true)

ans =

(2*sin(pi*x))/pi - sin(2*pi*x)/pi + (2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)

>> fx

fx =

(2*sin(pi*x))/pi - sin(2*pi*x)/pi + (2*sin(3*pi*x))/(3*pi) - sin(4*pi*x)/(2*pi) + (2*sin(5*pi*x))/(5*pi)

MATLAB：从欧拉（复数傅立叶）转换为正弦函数（bn系数）

1 个答案: