• Why are Fourier analysis and transform only applicable for LTI systems?

  • What if the system is not LTI, won't Fourier analysis or transform be possible?

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    $\begingroup$ @Gilles: excellent edit; should have thought of making this as clear as you did. $\endgroup$ May 24, 2016 at 21:08
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    $\begingroup$ @Gilles I use this opportunity to thank you for your efficient edits in general $\endgroup$ May 24, 2016 at 21:09
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    $\begingroup$ @Gilles and yes seconding mr. Duval, you can't imagine how many times a day I wish you were around to deal with the mess of Latex I'm strugling in ! :) $\endgroup$
    – Fat32
    May 24, 2016 at 21:13
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    $\begingroup$ Thanks guys ! I'm just doing my best to make sure people in the years to come find the answers to their DSP questions presented neatly and clearly on this awesome site. :) $\endgroup$
    – Gilles
    May 24, 2016 at 21:19

2 Answers 2


The family of Fourier Transforms are specificaly developed for analysing frequency contents of the signals for which there is no definition of linearity or time invariance. Hence we can define the Fourier transform of any signal, as long as it's integrable (i.e. stable).

On the other hand we can also define a Frequency Response $H(\omega)$ for a particular class of systems (those being LTI) when they are characterised by a unique signal denoted by $h(t)$ which is called as the impulse response of the LTI system.

The Fourier transform $H(\omega)$ of the impulse response $h(t)$ is important because the complex exponentials $e^{j\omega t}$ are eigenfunctions of such LTI systems and when the input is decomposed into a sum of such complex exponentials, via a Fourier transform, so will be the output, multiplied with a complex gain given by the Frequency Response $H(\omega)$ applied for each particular frequency. This gives a considerable ease for the development of the mathematical theory of signals and systems.

On the other hand consider the following nonlinear system defined as: $$y(t) = x(t)^2$$
For which you can apply CTFT to both sides to derive the result in frequency: $$Y(\omega) = \frac{1}{2\pi} X(\omega) \star X(\omega)$$

Also consider the following linear but Time-Varying system defined by: $$y(t) = g(t)x(t)$$ For which you can also express the relation in Frequency domain via CTFT as: $$Y(\omega) = \frac{1}{2\pi} G(\omega) \star X(\omega)$$

Which I hope convinces you about the scope of Fourier Transforms and their applications to systems other than LTI.


Why are Fourier Analysis & Transform only applicable for LTI systems?

That's simply not true.

Won't Fourier analysis or Transform be possible?

They are. The question is just whether they are meaningful.

The point is that the continuous Fourier Transform (FT) is defined to integrate over all times from $t=-\infty$ to $t=\infty$; that works excellently if the thing you're looking at doesn't change over time (ie. a signal is periodic, or a system being LTI).

Now, the reason why we do FT on LTI systems is that we usually represent their effect on a time signal $s(t)$ by the convolution with that system, e.g. the receive signal $r$ after a channel $h$ is

$$r(\tau) = s(\tau)*h(\tau)$$ $*$ being the convolution here; thanks to the properties of the FT, we can directly tell that $R(f)$, the FT of $r(t)$, is

$$\mathcal F\left\{r(f)\right\} =R(f)= \mathcal F\left\{s(f)\right\}\,\mathcal F\left\{h(f)\right\}$$

That falls apart if $h$ doesn't only depend on the relative time $\tau$, but also on an absolute time $t$. So that simply makes analysis harder, and you'd typically have to fix either of the parameters of $h(t,\tau)$ to make statements.

  • $\begingroup$ I would have the same. Perhaps a little mention of ways to stationarize signals using (overlapping) windows would be a plus?? $\endgroup$ May 24, 2016 at 20:42
  • $\begingroup$ @LaurentDuval agreed; but I really don't know where to start for OP. I'd be required to explain Gibb's phenomenon – which is very linked to the problem we're looking at. That would require understanding of the FT that directly incorporates what I've explained here. So: Yes, I do fully agree with you, this answer could use a lot of explanation on stationarity, windowing, linearity of the FT, but that's just too much for one answer. $\endgroup$ May 24, 2016 at 21:07
  • $\begingroup$ You are so d*** right on this $\endgroup$ May 24, 2016 at 21:08
  • $\begingroup$ @LaurentDuval :D $\endgroup$ May 24, 2016 at 21:09

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