Why do i have to take the abs when using fft and ifft?

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

Answer 1

[Ivan van der Kroon]

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.

[Brenden]

Thanks for the in put.

So the graph of the fourier itself is not correct for a continuous equation, only the discrete data points that where taken to approximate it? and to find the proper fourier for the continuous equation you simply take the abs of the fft when you plot?

[Walter Roberson]

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.

Answer 2

[Walter Roberson]

The fourier of a gaussian might be a gaussian, but your data is only the approximation of a gaussian.

[Brenden]

Sorry i don't quite understand, could you explain a bit more.

In what cases must one use the abs function with fft and ifft, as it is easy to get false graphs if one didnt know the answer beforehand.

Thank you, BN

[Walter Roberson]

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]

Excuse my wording, the program is only as good as the the code it is made up of. How then can i fix my code to get the proper fourier of the equations i am entering?

sorry i am new to matlab

thank you for your time,

BN