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

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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

##### 0 Comments

### Accepted Answer

Ivan van der Kroon
on 31 May 2011

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 Comments

Walter Roberson
on 31 May 2011

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.

### More Answers (1)

Walter Roberson
on 31 May 2011

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