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In this question, it has been proved that $x(t) = \sin(2\pi f_0 t)$ and $x_k(t) = \sin(2\pi (f_0 + k f_s) t)$ have the same sample points. So sinusoids with frequencies $f_0$ and $f_0 + kf_s$ will identify as the same signals. In other words, we keep sampling frequency constant and increase frequency by $kf_s$. Apparently, only frequencies $f_0 + kf_s$ lead to aliasing. Why this frequency is special? I tried to see this result in the frequency domain but it wasn't useful. I know by sampling in time domain we get shifted replicas of the original spectrum which is two delta functions at $f_0$ and $-f_0$ but how this corresponds to $f_s$ and aliasing at $f_0 + kf_s$? What I'm trying to get here is a visual explanation of mentioned result. In the case of the sampling theorem it's really easy to see aliasing occurs when there is an overlap.

enter image description here

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  • $\begingroup$ there's nothing special here. You can pick any $f_0$! $\endgroup$ Oct 12 '20 at 21:54
  • $\begingroup$ @MarcusMüller My mean was aliasing occurs only at $f_0 + kf_s$. Obviously we can pick any $f_0$. $\endgroup$
    – S.H.W
    Oct 12 '20 at 21:57
  • $\begingroup$ well, that's explained in the question you've linked to (and in hundreds of sources on aliasing) $\endgroup$ Oct 12 '20 at 21:58
  • $\begingroup$ @MarcusMüller I couldn't find a visual explanation for that special case of aliasing. Could you give me some references, please? $\endgroup$
    – S.H.W
    Oct 12 '20 at 22:01
  • $\begingroup$ I don't understand – this is really not a special case. $\endgroup$ Oct 13 '20 at 21:16
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You've made an error in this statement: "I know by sampling in time domain we get shifted replicas of the original spectrum which is two delta functions at $f_0$ and $−f_0$ but how this corresponds to $f_s$ and aliasing at $f_0+kf_s$?"

The two delta functions at $-f_0$ and $f_0$ are the result of taking the Fourier transform of a sinusoid $x(t)$:

$$ x(t) = \sin(2{\pi}f_0t)$$

$$\mathcal{F}(x(t)) = \frac{1}{2i} \big(\delta(f - f_0) + \delta(f + f_0) \big)$$

We haven't introduced sampling yet, this is simply the result of taking the Fourier transform. When we start sampling, then we introduce replicas of the signal's spectrum.

Let's assume we sample the $x(t)$ above at a frequency of $f_s$. We will have copies of that signal's spectrum at $f_0 + kf_s$ for all integer values of $k$. Assuming we met Nyquist, there's no problem here. The problem now comes when we have the sinusoid

$$ s(t) = \sin(2{\pi}(f_0 + f_s)t)$$

$$\mathcal{F}(s(t)) = \frac{1}{2i}\big( \delta(f - [f_0 + f_s]) + \delta(f + [f_0 + f_s]) \big)$$

When sampled, we have the spectrum copies centered at $(f_0 + f_s) + kf_s$ for all integer values of $k$. If we choose $k = -1$ then we have a copy at

$$ (f_0 + f_s) + kf_s|_{k = -1} = f_0$$

So if we had a combination of $x(t)$ and $s(t)$ and sampled it, a spectrum copy of $s(t)$ would step on $x(t)$'s copy at $f_0$. This is aliasing.

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