Inverse fourier transform Imginary Components

When I take an image and do fft on it in Matlab it produces a complex matrix. If I take the ifft of that image, it produces a double array. When doing the ifft, does matlab discard the imaginary component of the image, if any?

Answer 1

[Walter Roberson]

When you have a set of fourier coefficients that are in complex conjugates pairs, then the result of the inverse fourier is always real-valued; see here for one

It is not a matter of discarding any complex part: it is that the result of the inverse transformation you did had only 0's for the complex part, and MATLAB automatically leaves out an all-0 complex part.

If you had done an inverse fft of coefficients that did not have the complex-conjugate symmetries then the result would have been something that had non-zero complex parts.

[DSP Student]

Will this hold even if I convolve the image with a filter, and then do ifft.

[Image Analyst]

Yes, if you convolve the spatial domain image in the spatial domain with a spatial filter. A "round trip" will still be real.

No, if you convolve the spectrum in the Fourier domain with a filter. That may make the spectrum non-Hermitian and thus introduce imaginary components when transformed back to the spatial domain.

Answer 2

[Greg Heath]

Matlab discards nothing of the ifft.

Answer 3

[Image Analyst]

Ikenna, look at the Fourier Transform properties table about a third of the way down this page: http://www.cv.nrao.edu/course/astr534/FourierTransforms.html

I haven't tried it, but if you do a round trip with a real matrix and have some small imaginary component then it could be just some kind of precision/rounding/truncation issue, since the theory says it should be purely real again. But any imaginary component should be very small if it's even there at all.