complex output of the inverse fourier transform

Hello, I have written a simple algorithm that applies a fourier transform on an image X to get F as a result. Then I have inversed F and I got a complex output Y. I did not use the predefined functions fft and ifft. Is it logical to get complex output? could you help me to fix the problem in this code

 X=imread('test.png');
 [N,N]=size(I); 
 F=zeros(N,N);

 for k=1:N,
     for n=1:N

         F(:,k)=F(:,k)+X(:,n)*exp(2*pi*(1i)*(n-1)*(k-1)/N);
     end

 end

 y=zeros(N,N);

for n=1:N,
   for k=1:N,
       y(:,n)=y(:,n)+F(:,k)*exp(-2*pi*(1i)*(n-1)*(k-1)/N);
   end
end
y=y/N;

[Pragya singh]

if x is real then you should get back real output.

Maybe the problem lies in calling positive and negative frequency.

x=\summation X(k+) +\summation X(k-)

so if you look into this then all you need for real value is that the complex part of positive and negative frequencies cancel each other (which might not be happening in your case)

Answer 1

[Matt J]

If the imaginary components are small, it's probably just floating point noise. Use real() to discard it.

[Matt J]

ines Commented:

Thank you for your answer.

I try to use the predefined functions fft and ifft but the problem is that I must inverse the sign of their exponential term.

Have I the possibility to change the sign of the exponential term when using the fft ad the ifft functions. Thank you for your help.

[Matt J]

fft(X) will produce linear combinations of X(i) with negatively signed exponentials

length(X)*ifft(X) will do the same but with positively signed exponentials.

[ines]

In fact, I want to inverse the signs. So,I want the fft to produce combinations of x(i) with positively signed exponentials and ifft(X) do the same but with negatively signed exponential. Could I use fft and ifft and change the exponetial signes. Thank you.

[Matt J]

No, you can't, but I don't see the rationale for wanting this. As long as you have a routine that produces the linear combination you're after, what difference does it make if the routine is named fft(), ifft(), bob(), or steve()?

[ines]

I want to use fft because it is faster and the inversion is more precise. The reconstructed image is approximative to the original image.

[Matt J]

If X is real, then conj(fft(X)) is also equivalent to flipping the exponential sign. If X is not real, then conj(fft(conj(X))) is needed.