[bisby]

Which is the point. A discrete fourier transform can never recreate a square wave unless you have infinite samples. It can only recreate the finite sampled signal (which is only an approximation of the square wave, and not a real square wave).

This means that sure, the Fourier transform itself isn't lossy (garbage in, garbage out) but Fourier transforms would be used in contexts where loss are introduced. If I have a real perfect square wave, and I want to a take a fourier transform of it, the sampling is going to introduce loss, so to associate sampling losses with the transform itself is fair. Real square wave ran through a DFT program on my computer is going to spit out an approximation of a square wave -- loss.

[davidgay]

The good news of course is that if you were sampling a real signal, then that signal was not actually a perfect square wave. So the fact that you can't (re)construct a perfact square wave is somewhat moot...

[bisby]

Generally yes, but it's a perfectly reasonable assumption that a natural source could generate a signal that is beyond the bounds of what we can record. Any real signal generated by a computer is going to fit within the constraints of what we can generate, but inevitably something like a whale, or a quasar or something will generate a wave that will be lossy.

But also, the question this is all responding to was effectively "why would engineers associate Fourier transforms with loss" and the answer is simply "because the techniques used in calculating most Fourier transforms are going to inherently put a frequency limit and anything beyond that will be lost or show up as an artifact". Engineers work with real world constraints and tend to be hyper aware of those constraints even if they often don't matter.

[kybernetikos]

I think it's the discontinuous jump that causes the problem, and natural signals do sometimes have sudden jumps.