Why do i have to take the abs when using fft and ifft?
显示 更早的评论
I have a simple code to test how matlab processes the fft and ifft command:
clear all
close all
N=1e4;
dx=0.01;
Axis=(-ceil(N/2):ceil(N/2));
x=dx*Axis;
y=exp(-(x.^2)/(2*(1^2)));
Fy=fftshift(abs(fft(y)));
IFy=ifftshift(abs(ifft(Fy)));
y1=exp(-((x-3).^2)/(2*(1^2)))+ exp(-((x+3).^2)/(2*1^2)));
Fy1=fftshift((fft(y1)));
IFy1=abs(ifft(Fy1));
figure(1)
subplot(311);
plot(x,y)
subplot(312)
plot(x,Fy)
subplot(313)
plot(x,IFy)
figure(2)
subplot(311);
plot(x,y1)
subplot(312)
plot(x,Fy1)
subplot(313)
plot(x,IFy1)
Firstly, why in the first case, figure 1, do i have to take the abs in the fft and ifft. the fourier of a gaussian is a gaussian and the abs should not have to be taken.
Second, in the second case why can i not use the same code. ie why does the abs have to be left out in the fft and there is still a warning once run:
Warning: Imaginary parts of complex X and/or Y arguments ignored
> In Untitled at 38
采纳的回答
Ivan van der Kroon
2011-5-31
Walter means that discrete Fourier transforms are not exactly as what one would expect from their continuous counterparts. You check the web why and how this works and how to minimize the undesired artifacts in your transformations.
As for the plot of IFY; first you have to have the Fourier transform as input for the ifft and not the absolute values. Secondly, you make a mistake in the order of the operations:
Fy=fftshift(fft(y));
IFy=ifft(ifftshift(Fy));
plot(abs(Fy)), etc
You see, an ifft undoes an fft and an ifftshift undoes an fftshift, hence
y=ifft(ifftshift(fftshift(fft(y))));
for any y, exept for round-off errors. Maybe the 'symmetric' input for the ifft could be useful to you as well.
2 个评论
Brenden
2011-5-31
Walter Roberson
2011-5-31
To find the proper fourier of a continuous equation, use the Symbolic Toolbox. Taking abs() is not sufficient (especially if the original equation involves complex numbers.)
But yes, the plot your are seeing is of the discrete fourier of the approximating data points. You very likely have round-off errors in calculating those approximations that are leading to phase shifts.
更多回答（1 个）
Walter Roberson
2011-5-31
0 个投票
The fourier of a gaussian might be a gaussian, but your data is only the approximation of a gaussian.
3 个评论
Brenden
2011-5-31
Walter Roberson
2011-5-31
One does not get false graphs: one gets correct graphs for the data that was input. fft() cannot (and *should not*) do anything to "correct" the data input to it.
Brenden
2011-5-31
类别
在 帮助中心 和 File Exchange 中查找有关 Discrete Fourier and Cosine Transforms 的更多信息
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
