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Let's consider any physical quantity depending on the frequency. For example, the impedance of a certain electrical component: $Z(f)$.

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Now, imagine to measure it in a continuous interval of frequencies. You get a graph. Now, let's take some samples separated by $∆f$ (uniform sampling).

My questions are:

  1. If I compute the Fourier Transform of $Z(f)$, what will I get? I think it is a signal like $Z(t)$, but it seems strange to me that from a frequency dependent signal, it will appear a time - dependence only by calculating the Fourier Transform.
  2. Which is the mathematical condition to apply to $∆f$ to avoid aliasing? I'd say that it should be greater of the maximum frequency of $Z(f)$, which is a graph on frequency. Is this frequency variation of Z with respect to frequency related to time?
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  • $\begingroup$ Re 1.: "Fourier transform" (not: transformer) of a frequency signal pretty much is a time signal, with a twist (just write down the Fourier Transform formula for a time signal $x(t)$, and the Fourier Transform of that, and you'll notice how the exponent doesn't fully cancel). Why is that surprising? That's exactly what a Fourier Transform does. (also, there's actually plenty of questions here dealing with "Fourier Transform of Fourier Transform of signal", do a bit of searching around here :) ) $\endgroup$ – Marcus Müller Oct 6 at 9:46
  • $\begingroup$ Re 2: Your statement makes no sense, please draw a spectrum of a signal with a maximum frequency $f_\text{max}$, and then sample that in the spectral domain with a resolution $\Delta f > f_\text{max}$! $\endgroup$ – Marcus Müller Oct 6 at 9:48
  • $\begingroup$ Yes, I have added now an example of spectrum $\endgroup$ – Kinka-Byo Oct 6 at 14:00
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If I compute the Fourier Transform of Z(f), what will I get?

If you apply the inverse Fourier Transform, you get the impulse response of the system which is indeed a time domain signal. You can also apply the forward transform and would get the impulse response time reversed and scaled since the forward and inverse transform are quite similar.

Which is the mathematical condition to apply to Δf to avoid aliasing?

$\Delta f < 1/L$, where L is the length of the impulse response. The sampling theorem is the same in both domains: in order to sample in time the signal needs to be limited in frequency and in order to sample in frequency the signal needs to be limited in time. Note: if $\Delta f$ is too big you get time domain aliasing, not frequency domain aliasing.

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  • $\begingroup$ Thank you for the answer. It is clear now. Only a question: should be a factor of 2 in your last statement? $\endgroup$ – Kinka-Byo Oct 6 at 15:11
  • $\begingroup$ No. At least not for a causal system. In the frequency domain we typically select the valid frequency range from $[-f_N/2,+f_N/2]$ where $f_N$ is the Nyquist frequency. That's why the total bandwidth is $f_N$ and NOT $f_N/2$. This is of more a convention of convenience: the signals are periodic anyway, so technically you can set the start and end point wherever you want as long as you cover one entire period $\endgroup$ – Hilmar Oct 6 at 15:30

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