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The general rule for aliasing. If my sampling frequency is Fs=800MHz signal frequency =120MHz.

For the 4th harmonics 480MHz its 80MHz above Nyquist frequency(400MHz) thus its mirrored 80MHz back to 320MHz.

What about 1120MHz the closest nyquist frequency to it is 2*400 so its supposed to be mirrored 800-320=480 Why we do double mirroring for it? Why it also mirrored to 320MHz?

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    $\begingroup$ How is 1120MHz harmonically related to anything? $\endgroup$
    – TimWescott
    Oct 2, 2020 at 18:32
  • $\begingroup$ i ask purely theoreticaly $\endgroup$
    – rocko445
    Oct 2, 2020 at 19:10
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    $\begingroup$ That's fine. The question would make sense if 1120MHz / 120MHz were an integer -- it's not, but you're calling it a "harmonic" -- that does not compute. Was it a typo, or do you misunderstand what a harmonic is, or what? $\endgroup$
    – TimWescott
    Oct 2, 2020 at 20:08

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Aliasing is more like circular wrapping around rather than just mirroring. Frequencies wrap circularly between -Fs/2 to Fs/2, an 800 Hz range from -400 to 400 at a sample rate of 800 sps.

120 is inside -Fs/2 to Fs/2, so doesn't need to wrap around at all.

But 480 wraps 80 past 400, and shows up at -320, which looks like 320 if you don't care about phase.

And 1120 wraps all the way around the circles range of 800 Hz, plus another 320, so it shows up at 320.

12000 Hz (12 kHz) would wrap around 15 times and show up identical to DC or 0 Hz.

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  • $\begingroup$ for 480 we have its pass by 80 the 400 line. So by wrapping we do 400-80=320 . with 1120 we have 800 line which we pass it by 320 its not the same thing they pass th nyquist line by diiferent amount. $\endgroup$
    – rocko445
    Oct 2, 2020 at 22:38
  • $\begingroup$ -400+80 = -320, 1120-800 = 320, -320 is different from 320 (if you don't ignore phase info) $\endgroup$
    – hotpaw2
    Oct 3, 2020 at 4:09
  • $\begingroup$ One way to think of it - By sampling at $Fs$, the frequency domain becomes periodic every $Fs$. Any frequencies that were in the original spectrum will still be there, i.e. at their original frequencies, but so will all the periodic extensions of those frequencies. Mathematically, if you have a sinusoid at $f_0$ you'll see tones at $f_N=f_0 \pm N\cdot Fs$, where $N=0,\pm 1, \pm 2, \ldots$ $\endgroup$
    – David
    Jul 25, 2022 at 13:29

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