Explaining a statement regarding aliasing in a document

Question

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.

Answer 1

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 $\mp 20\,\mathrm{Hz}$.

Answer 2

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...