Using ifft to get the Fourier Coefficient

3 次查看（过去 30 天）
Raunak Raj
Raunak Raj 2016-6-26
编辑： Jan Orwat 2016-6-27
What exactly does the ifft() gives me?
I have a real data in 'x' where,
f=summation over -N to N-1 [C(n)exp(2*pi*1i*x/L)]
So, here f is known at every point (2N points in total). fftshift(ifft(f)) also gives an array of 2N size. So, does it gives me the coefficients C(n). If so, then can you please check the following code.
N=256;
X=2*N;
L=2*pi;
x=linspace(-pi,pi,X);
c=0;
for n=1:2*N
k(n)=2*pi*(n-N-1)/L;
end
y=x;
z=fftshift(ifft(y));
for i=1:2*N
c=c+z(i)*exp(1i*k(i)*x);
end
plot(x,y);hold on;plot(x,c);
Here, if ifft() gave the coefficients, then shouldn't the plots have matched?
  3 个评论
显示 1更早的评论
Raunak Raj
Raunak Raj 2016-6-27
编辑：Raunak Raj 2016-6-27
Hi, I am sorry, that wasn't the intended code. I have edited the code correctly. fftshift is just to shift the values from 0 to 2N to -N to N-1 frequencies (actually wave numbers). Moreover, the code works as intended for y=sin(x) but for other functions there appears a shift in the graphs.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Jan Orwat
Jan Orwat 2016-6-27
N=256;
X=2*N;
L=2*pi;
x=linspace(-pi,pi,X);
c=0;
k = 2*pi*((1:2*N)-N-1)/L; % vectorised
y = sin(x); % don't understand why it is here, why not defined earlier
z = ifftshift(ifft(y)); % would be more logical to use fft here
for i=1:2*N
c=c+z(i)*exp(1i*k(i)*(pi-x));
end
plot(x,y);hold on;plot(x,real(c));
  1 个评论
Jan Orwat
Jan Orwat 2016-6-27
编辑：Jan Orwat 2016-6-27
I'm still not sure why you calculate ifft of signal, then dft of ifft and compare with original signal. From mathematical point of view it makes no difference, because y, ifft(fft(y)) and fft(ifft(y)) are equal (within numerical precision), but it's logically weak.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by