Usually, when we have a stereo signal, we process each channel separately in order to extract the frequencies using Fourier transform.

However, Fourier transform can also be applied to complex numbers as well. So what if I represent the original stereo signal with complex numbers, do a Fourier transform, and back?

I know that in this case, for DTF, I would need to keep all N frequency numbers as opposed to N/2 for DFT on the real data. Should I then be able to recover my original signal from the frequencies? (My guess would be "yes".)

Now, what if I want to detect if a specific frequency is present in the signal? I could use the formula for DFT:

enter image description here

So, what if instead of real $x_n$, I pass a complex number containing values of the stereo channels as real and imaginary parts? Would that make any sense? What would that detect?


1 Answer 1


For your first claim. Yes it's true. You should keep the full N-DFT complex frequency samples as opposed to half, N/2, for the real input case. And also yes, you will retain back the original stereo signal perfectly by an inverse DFT.

Whether this artificially produced complex signal adds any new information that cannot be obtained from individual channels is not very clear, however. At least from the linearity of DFT we can guess that the resultant DFT of the complexed-stereo pair will be exactly given by those individual left/right channel DFTs, and this suggests to me that, you wont be obtaining some interesting benefit here.

Keep in mind that, the (geometric) spatial relationship between a pair of stereo microphones can be investigated by a joint processing of those channels. However, a DFT alone does not care for any spatial relationship between real and imaginary part of its time domain complex input. Hence I suspect any good use of that for spatial processing.

Yet, in some cases, there may be a computational benefit due to performing a single complex input DFT instead of two real input DFTS. In such cases you are favored to do so.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.