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I'm currently trying to understand Fourier transform and I've got curious about why Fourier transform exists.

Let's suppose that we have a 10 seconds of non-periodic wave. For example:

a nonperiodic wave

As far as I understand, the concept of Fourier transform is to think the wave's period is infinite. If I describe what is in my mind with a picture, it is as same as the following:

a nonperiodic wave with an infinite period

And if we apply Fourier transform to that nonperiodic wave, we will get a continuous spectrum which shows components of the nonperiodic wave.

Meanwhile, let's just treat the nonperiodic wave's period as 10 seconds. Then we become to be able to calculate Fourier coefficients. And if we make a discrete spectrum with the coefficients we just got, we can still describe the nonperiodic wave.

And, I guess the shape of both spectrum will be same. (Please correct me if this is false)

So... I wonder why we need Fourier transform...

Of course, unlike Fourier coefficients, which give us nth multiples of fundamental frequency, Fourier transform can give us any frequency... but is it important? I think Fourier coefficients are enough information!

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There's a few misconceptions here:

the concept of Fourier transform is to think the wave's period is infinite

No, that's not the case. Also, from an argument by limits, it would follow that any signal with infinite period must be a constant.

Also, Fourier theory inherits that any signal that's periodic would have a line spectrum, i.e. be a countable sum of weigthed Dirac deltas in Fourier domain. That's not the case for your signal.

picture, it is as same as the following:

This illustrates a mathematical fallacy when dealing with inifinities. What you draw is non-sensical. You can't have something at "$\infty + 1$ seconds". Infinity is not just a large number. And this doesn't work out if you replace infinity with "just an arbitrarily large number".

Meanwhile, let's just treat the nonperiodic wave's period as 10 seconds. Then we become to be able to calculate Fourier coefficients. And if we make a discrete spectrum with the coefficients we just got, we can still describe the nonperiodic wave.

Right, repeating your signal periodically makes it periodic, and thus, it can be represented by the countable (but potentially infinite) set of Fourier coefficients.

And, I guess the shape of both spectrum will be same. (Please correct me if this is false)

That's false, no matter how you'd turn it. The Fourier domain and time domain contain (up to countably many discontinuities) exactly the same info, so one Spectrum == one time signal.

So... I wonder why we need Fourier transform...

I don't think this question, based on these misconceptions, makes a lot of sense, I'm afraid. The Fourier transform doesn't exist because we need it, it just exists; it's math. The fact that it has a name is just because Joseph Fourier nerded hard with function theory and discovered it.

I think Fourier coefficients are enough information!

No. You can only describe periodic signals with a discrete set of Fourier coefficients, and most signals are not periodic. You cannot equate a non-periodic signal with a periodic repetition of the same, they are simply two different things, not to speak of the fact that a signal can be both infinite in time and non-periodic, so that this mechanism in general won't work.

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  • $\begingroup$ "> the concept of Fourier transform is to think the wave's period is infinite - - - No, that's not the case. Also, from an argument by limits, it would follow that any signal with infinite period must be a constant." $$ $$ I would not say so flatly "that's not the case." The continuous Fourier Transform really is set up as a Fourier Series where a non-periodic function is truncated at $[-\frac{T}2, \ +\frac{T}2]$, periodically extended, Fourier Series calculated and the spectral lines placed at multiples of $\frac1T$. Then we let $T \to \infty$ and see what happens. $\endgroup$ Commented Nov 12, 2023 at 20:09
  • $\begingroup$ @robertbristow-johnson fair! but: I was looking at it in the context of the visualization posted below; and as you say, we let the integral's bounds approach infinity, but at that point we certainly aren't speaking of a signal with infinite period. There isn't an "infinitely long" streak of 0, then the signal repeats, as integrand. $\endgroup$ Commented Nov 12, 2023 at 20:49
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Fourier Transform in wavetable synthesizers like Serum allows you to make very fine tunings to the waves harmonics where you can create a filter-like sound or do additive synthesis adding in frequencies here and there to make new textures. Here's some info by ADSR

"FFT can be used for a variety of tasks from convolution reverb, to vocoding and more"

If I'm not mistaken Fourier Transform is part of Aphex Twin's secret ingredient There's a pretty sweet free VST Plugin called DtBlkFx that has some unique effects. Seems like FFT is an under rated tool in audio production.

"FFT's are also used for fault analysis, quality control, and condition monitoring of machines or systems"

Hopefully this is relevant to your question

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The Taylor series attempts to model a limited piece of a function or a data series using a weighted sum of polynomials. This only works for relatively small values of x or t.

The Fourier transformation attempts to model a data series using a weighted sum of sin and cos functions with given frequencies. Beforehand, the data series is automatically supplemented with copies of this data series on both sides an infinite number of times. This often creates annoying cracks at the connections, which can be dampened with windowing. FFT is usually done to see which frequencies are particularly important for imitation. Then it is said that these frequencies are included in the data series.

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