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Why is it that spectral images appear? I know they appear for $F_s$, but why those replicas appear, I don't know.

E.g., sampling $x(t) = sin(1000 * 2\pi t)$ with a sample frequency of 230 Hz. I know it will alias, due to spectral replicas, but why do those appear at all?

enter image description here

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  • $\begingroup$ The 1000 Hz tone will alias to 80 Hz in your situation. I'm not sure what "spectral replicas" you are asking about. $\endgroup$ – John Jan 7 '14 at 14:00
  • $\begingroup$ I mean these replicas.. spectral images.. what are they called snag.gy/Kj5Vg.jpg $\endgroup$ – user7277 Jan 7 '14 at 14:21
  • $\begingroup$ They are called "aliases". There is a decent article on Wikipedia under the topic of undersampling. $\endgroup$ – John Jan 7 '14 at 15:24
  • $\begingroup$ no, @John, they're called "images". "aliases are frequency components higher than $f_s/2$ that are folded over, due to sampling. in the figure above, there are no aliases, but there are images. $\endgroup$ – robert bristow-johnson Jan 10 '14 at 0:56
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the reason that they appear is that the sampling function is periodic and has these Fourier sinusoidal components all with coefficients of 1.

$$ \begin{align} s(t) & = T \sum_{n=-\infty}^{+\infty} \delta(t - nT) \\ & = \sum_{k=-\infty}^{+\infty} e^{j 2 \pi (k/T) t } \\ \end{align} $$

so when you multiply $x(t)$ by $s(t)$ to sample it, you are multiplying by all of these sinusoids, with frequencies of $k/T$ and that bumps up your original baseband spectrum from something centered at DC to something centered at $k/T$. and then it adds it all up.

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One way to represent AD conversion mathematically is to think of it as a multiplication of the signal by an impulse train, which is a sequence of dirac delta distributions spaced by the sampling period $T_s$.

With this in mind and remembering that multiplicating in the time domain is equivalent to doing convolution in the frequency domain, it's possible to see that the frequency representation of the digitalized signal will be the convolution of the frequency representation of the sampled analog signal with the frequency representation of the impulse train.

As it turns out, the frequency representation of an impulse train spaced by $T_s$ is an impulse train spaced by the sampling frequency $f_s$. Also, the convolution of a dirac delta with any function $g$ is just $g$ itself. By displacing the delta function by $f_s$, we also displace the result of the convolution by $f_s$.

So, the spectrum of the digitalized signal will be an infinite sum of copies of the spectrum of the analog signal distant of each other by $f_s$. Since the spectrum is bilateral, if your analog signal has frequency components above $f_s/2$, those components will end up summed from two different replicas in the spectrum of the digital signal, and this is in fact the aliasing effect.

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