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Given a sampling frequency Fs lets say we plot the magnitude of the fft of a temporal signal $x$, for different frequencies above the Nyquist frequency, to show the effect of aliasing:

Fs=10;
t=0:1/Fs:1-(1/Fs);
f=6;%7,8,9, ... to see aliasing starting a 6 since 6 is just above the Nyquist freq
x=sin(2*pi*f*t);
freq_range=Fs*((0:length(x)-1)/length(x));
plot(freq_range,abs(fft(x)));

Can someone help to formulate mathematically to which lower frequencies a given freqency f > Fs/2 will be aliased to?

Experimentally, we can see that (for integer frequencies, dont know if it really makes sense to use fractionnal ones?):

6 -> 4

7 -> 3

8 -> 2

9 -> 1

10 -> 0 <- should be zero? but the spectrum goes crazy and there are peaks everywhere.

Why? Then one would think that this would be periodic i.e. f=11 -> f_aliased=1, etc. back to zero in decreasing order, then again some strange things happens at 15 then it goes on cyclically etc. What is the relation?

Also, w.r.t. spectral resolution ($\Delta f = F_s/N$, let $N$ be the nb of samples, i.e. length(x)), one could ask if the next frequency abouve the Nyquist is in fact $f_{Nyq}+\Delta f = (F_s/2)+\Delta f$ ? But in this example the spectral resolution $\Delta f$ = 1 right?

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  1. If you do spectral analysis is typically better to use WAY more points then 10 in the time domain
  2. Look at your y axis and make sure your graphs are properly scaled. $f=10$ aliases indeed back to $f=0$. All you see is numerical noise and your y-scale is $10^{-15}$. This would be different if you choose a a different phase, cosine instead of sine.
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