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Comment by computerfriend | Hacker News Reader

I wonder how Fourier transformations are taught in engineering courses, because the idea that it could be "lossy" is strange and not obvious to me. It has an inverse after all.

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In my engineering courses, Fourier transforms were taught in the context of discrete fourier transforms. Because sampling is a thing that matters (computer audio is discrete data points, not an actual wave).

The Fourier transform of a discrete signal repeats in the frequency domain. For example, [1, -1, 1] could be a sine wave with the exact half of the sampling frequency going from 1 to -1 back to 1 once .... Or it could be a sine wave with double the sampling frequency that is actually going from 1 to -1 to 1 to -1 all within the gap of the first 2 samples. Or it could be 3x the sampling frequency, 4x the sampling frequency, etc. The solution is to only keep the part of the transform that is below the Nyquist limit, because we don't have a sampling rate accurate enough to measure the higher frequencies, so just assuming they dont exist. This also means that if the source signal WAS in fact 4x the sampling frequency, we will see a spike at 1/2 the sampling frequency in the fourier transform, and when we re-create the signal, it will be completely wrong.

So unless you have analog hardware for measuring the Fourier transform (or are working purely in a non-physical mathematical domain, like "i have a sine wave" which can be perfectly represented), you are naturally going to be taking discrete samples of a signal to measure the Fourier transform, which means you are going to be losing any part of the signal that doesn't adhere to sampling rates.

Because my engineering courses were so heavily focused on digital signal processing, when I hear "fourier transform" i immediately think of "discrete fourier transform" and loss is immediately applicable.

Don't engineers at least learn to solve the heat equation analytically using Fourier transforms? It wasn't all numerical algorithms, was it?

It can be lossy and not just because of truncation. Convergence is somewhat finicky in Fourier analysis and was not well understood mathematically before the 60's. Engineers made great use of it anyway.

Remember that it is an integral transform. Basically, any data on a set of vanishing measure can be lost or corrupted. Unfortunately it can even be the case the deviation around some points is unbounded.

Frequency of a signal and time at which you measure it have an uncertainty principle. https://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_... You either know frequency well, but on a large window or shorten the window and get larger error bars on frequency.

A square wave would take an infinite number of sinusoidal waves to perfectly reconstruct. To approximate it, one truncates the coefficients. This is done in any engineering application where memory isn't infinite. Which is all of them.

Of course, a discrete, finite sampling of a square wave at a set of points in time only requires a finite number of coefficients to perfectly reconstruct.

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.

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...

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.

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

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