# Aliasing at $f_0 + kf_s$

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.

• there's nothing special here. You can pick any $f_0$! – Marcus Müller Oct 12 '20 at 21:54
• @MarcusMüller My mean was aliasing occurs only at $f_0 + kf_s$. Obviously we can pick any $f_0$. – S.H.W Oct 12 '20 at 21:57
• well, that's explained in the question you've linked to (and in hundreds of sources on aliasing) – Marcus Müller Oct 12 '20 at 21:58
• @MarcusMüller I couldn't find a visual explanation for that special case of aliasing. Could you give me some references, please? – S.H.W Oct 12 '20 at 22:01
• I don't understand – this is really not a special case. – Marcus Müller Oct 13 '20 at 21:16

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.