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I am reading this document and in the section "Aliasing and Anti-Aliasing", the author make the following statement:

Consider a signal with two frequency components: $f_1 = 10\,\text{Hz}$ which is our desired signal and $f_2 = 20\,\text{Hz}$ which is noise. Let's say that we sample the signal at $f_s = 30\,\text{Hz}$. The first frequency component, $f_1$, will generate the following frequency components at the output of the multiplier, $10\,\text{Hz}, 20\,\text{Hz}, 40\,\text{Hz}, 50\,\text{Hz}, 70\,\text{Hz}$ and so on. The second frequency component $f_2 = 20\,\text{Hz}$ will generate the following frequency components at the output of the multiplier, $20\,\text{Hz}, 10\,\text{Hz}, > 50\,\text{Hz}, 40\,\text{Hz}, 80\,\text{Hz}$ and so on.

It's probably obvious to most but could someone explain where those generated frequency components come from. Thanks.

The background is that I've started reading Richard Lyons' DSP text and I really like it so far except that now I hit the bandpass sampling section and I didn't follow it. So, I've been looking for other material. If anyone knows of some good material ( books or papers ) on bandpass sampling ( I've printed out the dsp.stackexchange material and will check that out next ), it's appreciated.

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  • $\begingroup$ The problem here is that your sampling frequency is set at 30Hz which is sufficient for your 10Hz component but not for the 20Hz frequency component (because the Nyquist rule says that you should sample at at least twice the highest frequency, i.e at 40Hz). Because of this you'll get aliasing. In order to prevent aliasing, you should first run your signal through a low-pass filter (with a cutoff frequency of say 15Hz and a relatively steep rolloff) $\endgroup$ – dsp_user Nov 28 '17 at 7:51
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Just as a reminder, which could help you solve the confusion: The mathematical equation used to model the sampling process is the impulse train modulation, which could be stated in the simplest sense as:

$$ x_s(t) = \delta_{T_s}(t) \cdot x_c(t) = \left( \sum_{k=-\infty}^{\infty} \delta(t-kT) \right) x_c(t) $$ where $x_c(t)$ is the bandlimited continuous time function and $x_s(t)$ is the sampled version which is still a continuous time signal (not yet converted into a sequence of samples).

The spectrum of the sampled signal $x_s(t)$ can be derived based on a number of ways for which the most standard notation is simplified as the following: $$ X_s(f) =\mathcal{FT} \{ x_s(t) \} = \mathcal{FT} \{ \delta_{T_s}(t) \cdot x_c(t) \} \implies \left( \sum_{k=-\infty}^{\infty} \delta(f - k f_s) \right) \star X_c(f) $$

and therefore it follows that: $$ X_s(f) = K \sum_{k=-\infty}^{\infty} X_c(f - k f_s) $$ where $K$ is some linear scaling. So this shows why the spectrum of teh sampled signal contains shifted copies of the original spectrum at multiple shifts of sampling frequency $f_s$. In your example $f_s = 30$ Hz. So the spectrum will shift by $30$ Hz hoppings.

Your example involves two ideal (infinitely long) sinusoidal components whose Fourier transforms are $$ x_1(t) = \cos(2\pi 10 t) \implies X_1(f) = 0.5\delta(f-10) + 0.5\delta(f+10) $$ and $$ x_2(t) = \cos(2\pi 20 t) \implies X_2(f) = 0.5\delta(f-20) + 0.5\delta(f+20) $$

Note that each component's spectrum has two impulses located in frequency at $\pm f_0$ Hertz respectively. Now it's up to you to sketch those impulses shifted by $\pm k 30 $ Hz hops and to see the resulting spectrum...

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    $\begingroup$ will read carefully. my guess is that your explanation will be better than the things I've been reading. it's much appreciated. $\endgroup$ – mark leeds Nov 28 '17 at 21:31
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    $\begingroup$ That was perfect and beautiful. But, for people who want to understand the bandpass sampling concept in general, go to the link below and read the explanation of Fat32 there. It is by far the clearest explanation of bandpass sampling that I have seen. and I've looked around a lot. dsp.stackexchange.com/questions/29773/… $\endgroup$ – mark leeds Nov 29 '17 at 4:44
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The frequencies are $\left|10\,\mathrm{Hz}+k\cdot30\,\mathrm{Hz}\right|, k\in \mathbb{Z}$ and the same for $20\,\mathrm{Hz}$.

Note that real signals always have (complex conjugate) signal components for positive and negative frequencies, so a $10\,\mathrm{Hz}$ signal has components at $\pm10\,\mathrm{Hz}$. Consequently aliasing in a $30\,\mathrm{Hz}$ grid will also reach $\mp 20\,\mathrm{Hz}$.

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  • $\begingroup$ @user32274: just read it again more carefully and have one more question. where do the 20 and 50 for the first and the 10 and the 40 for the second come from ? thanks. $\endgroup$ – mark leeds Nov 28 '17 at 18:04

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