1
$\begingroup$

I'm trying to simulate the behavior of my FFT spectrum analyzer, I am using Mathematica, but my questions are more conceptual than technical implementation of code...I hope.

If we consider a simple cosine wave transient signal $y(t) = A \cos(2 \pi \nu_0 t)$, where $A = 1$ and $\nu_{0} = 24$ $\rm{Hz}$. This signal is sampled at some time interval $dt$ and this gives me 2048 points (2048 because of my device), I want to resolve this data to some frequency resolution, $\Delta f$, and from the Fourier limit: $t_{sample} = 1/\Delta{f}$ so my sample rate, $S_{r} = 1/dt$ is adjusted accordingly.

I then take the absolute value of the FFT of these 2048 points and reconstruct the frequency component using $\Delta f$ (in this example $\Delta f = 0.0625$ $\rm{Hz}$, this value is because of my device and the features I want to resolve) and the index of the resultant FFT, and get this:

enter image description here

I understand that the symmetric appearance of an FFT comes from the real and imaginary components having the same response to signals -- the absolute value of their responses will be the same. But the phase of the real and imaginary parts is shifted by $\pi / 4$. This gives us our mirrored appearance.

FIRST QUESTION: Usually I would just throw the right hand frequencies away and multiply the amplitude of what is left by $2$. Is this the correct approach or should I reverse the order of one half of these 2048 points and then add them together (intuitively I would say no, because both halves should be identical so multiplying by $2$ should suffice)? What is the best method of reconstructing the frequency component/axis?

SECOND QUESTION: This is something I just don't understand. If I increase the frequency of my transient signal $\nu_{0}$ then the two mirrored peaks move closer to each other, converging in the center. This makes sense. If I keep increasing $\nu_{0}$ of the transient arbitrarily (past $124$ $\rm{Hz}$ which corresponds to the limit of frequencies) however I still see a peak, in fact I see this oscillating behavior of the peaks just bouncing back and forth as I further increase the frequency -- Why?! Is this natural or because of the way I define my frequencies.

$\endgroup$
2
$\begingroup$

It depends on your spectrum analyzer.

If you want to analyze a one channel real baseband signal, the mirrored frequencies are redundant. you should keep the first N/2+1 points.

For communications signals, there can be an I and Q (two reals as a single complex), you want to keep them all. the frequencies will not be (in general) symmetric.

If you want to calculate the cross spectrum of two reals, there will be mirrored response, but if you want to do something like taking the inverse of the cross spectrum, you should keep both, but only display half.

It really depends on the actual application. There really isn't a right or wrong way. There are conventions and exceptions. There are niche applications. Spectrum analyzers that are commercial standalone test equipment, usually have lots of buttons, lots of modes.

for your second question, this is what spectral aliasing looks like if you don't apply an antialiasing filter prior to the DFT.

$\endgroup$
0
$\begingroup$

The most precise way to deal with them is to compare them using the algorythm used in the Wigner timer frequency distribution.

It's a processing intensive way of comparing the real and imaginary results of the FFT, I am completely ignorant of the other types of FFT implementations.

The wigner time frequency distribution uses both products to construct a quantum uncertainty distribution, a bell curve of some kind to describe where the amplitude probably was, for the time and frequency domain, a bit like electrons and photons can be in multiple places at once, the time-frequency graph has an unknown and stochastic basis that can be implemented into a graph. It results the most precise FFT frequency graph according to the professors.

It is one way of processing the real and imaginary signals, there are probably more popular ones, I am a bit ignorant, even a 3rd year maths or physics student can't easily program it.

https://en.wikipedia.org/wiki/Wigner_distribution_function#Cross_term_property

They say In the ancestral physics Wigner quasi-probability distribution has useful physics consequences, required for precision. By contrast, the short-time Fourier transform does not have this feature. Negative features of the WDF are reflective of the Gabor limit of the classical signal and physically unrelated to any possible underlay of quantum structure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.