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There is a system (consisting of several LTI analogue filters) whose input and output signals are being recorded at $20 \ \mathrm{kHz}$ (the sampling period is in the range of 10 times the system's transient response). Suppose I calculate

$$ G = \frac{\mathrm{FFT(output)}}{\mathrm{FFT(input)}}$$

What does that correspond to? Can you call it the frequency response? If it does not have anything to do with the frequency response, can someone provide an (ideally mathematical) explanation why it doesn't? I think, that it is legit if there were no transients present in the measured output. But since they impact the signal significantly I wonder if one can still get useful information from $G$.

EDIT: I guess I should rephrase the question. I know that the frequency response is defined as $$X(\omega) = \frac{Y(\omega)}{U(\omega)}$$. The issue here is that $y(t)$ and hence $Y(\omega)$ are greatly impacted by the system's transient response. May $X(\omega)$ still be called frequency response ?

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    $\begingroup$ Are the analog filters LTI systems? And what about the sampling frequency? $\endgroup$ – Fat32 Mar 25 '17 at 20:37
  • $\begingroup$ lti and 20 kHz. $\endgroup$ – luis Mar 25 '17 at 20:54
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    $\begingroup$ Try taking the magnitude of each before you divide if you are looking for a magnitude response. $\endgroup$ – Dan Boschen Mar 25 '17 at 22:03
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    $\begingroup$ it shouldn't matter @DanBoschen. they can do a complex division and have a complex frequency response from which magnitude and phase can be inferred. $\endgroup$ – robert bristow-johnson Mar 26 '17 at 2:48
  • $\begingroup$ Yes agreed, I was thinking about that after and concur $\endgroup$ – Dan Boschen Mar 26 '17 at 2:49
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Similar questions have been asked before, you may search. (I'm sorry but I cannot refer to one here)

You want to identify an analog LTI system from analysing its response $y(t)$ to a known input $x(t)$.

One method of analysis involves the use of continous Fourier transforms of the input and output signals. It can be shown that (for a stable system) $$H(j\Omega) = \frac{Y(j\Omega)}{X(j\Omega)} $$

Considering a continuous time analog LTI system, this method requires some convenient techniques to compute the Fourier transforms of the input and output signals. This method has practical limitations due to understandable reasons.

With the advent of digital computers and availability of dedicated hardwares, a more implementable approach has been developed during the rise of digital signal processing era, based on the sampling theorem to represent the continuous time signals by their samples, i.e., the continuous time analog system is represented in the form of an equivalent (and in a one to one relation to its continuous time counterpart) digital LTI system, as long as some necessary conditions are met.

An analysis based on the taken samples of input and output signals, however, should be handled with care, as the most critical element of success in this method relies on the fact that the continuous time system's frequency response function $H(j\Omega)$ be bandlimited to the half of the sampling frequency $\Omega_s = 2\pi F_s$, so that it can be digitally represented in an equivalent digital LTI filter with DTFT $H(e^{j\omega})$

If you know that this condition is satisfied, then there are other details such as the fact that Fourier transforms provide information on steady-state behaviour and not on the dynamic transients that will happen in any finite length measurements. (That can be avoided with longer recording durations, as you intent to do)

Also note that the LTI analog system must be stable so that its CTFT $H(e^{j\Omega})$ exists. Most often it will be stable I assume, but it's still important to consider though.

Finaly any noise and/or distortions on both the input and output signals not only prevent high precision computations but can even lead to misleading results.

So I suggest you to begin with a known setup first and apply your method to see if it can also identify that known LTI system. If yes, then you are ok.

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If your input is $x(t)$ and your output is $y(t)$ then you'll have in the Fourier domain $X(\omega)$ and $Y(\omega)$. The relation between the two will be your transfer function.

For example if you work only with the absolute values then you will have the the amplitudes of frequency response.

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