0
$\begingroup$

Suppose you have an arbitrary analog input signal $x_a(t)$ guaranteed to have frequencies within a bandwidth $[f_1,f_2]$ Hz.

Suppose your sampling frequency $F_s$Hz, and sample $x_a(t)$ to produce $x(n)=x_a(\frac{n}{F_s})$.

Then by the sampling theorem, you can successfully recover (or reconstruct) the signal so long as $f_2 < \frac{F_s}{2}$. Otherwise you will experience aliasing and cannot recover the original signal $x_a(t)$.

It appears (from this article) that there are "tricks" that can be employed that take advantage of aliasing scenario and can still recover the signal $x_a(t)$ completely.

The specific part of the article of concern is pasted below:

Using Nyquist aliasing as benefit

The trick is to use the aliasing (or frequency folding) to your advantage. By undersampling the data converter, higher-frequency content will be aliased into all of the lower Nyquist zones (see Figure 2). You will need to make absolutely sure that nothing ends up in the lower bands – any noise or frequency components in the lower zones will also be aliased into the first Nyquist. The good news is that the data rate from the data converter is only a fraction of the required RF input sample rate if this were a first Nyquist system. Under sampling greatly reduces the data rate of the samples supplied to the digital signal processor (DSP) or FPGA.

Could somebody explain this to me? Why is the Texas Instruments link not violating Nyquist's Theorem?

$\endgroup$
1
$\begingroup$

you can successfully recover (or reconstruct) the signal so long as $f_2<F_s/2$

Nope. The sampling theorem says that you need 2 samples per Hz of bandwidth, so in this case you'd need

$$Fs > 2*(f_2-f_1)$$

For more info on how this works google "bandpass sampling" or just read https://en.wikipedia.org/wiki/Undersampling.

The basic idea is that sampling creates a periodic repetition of the original spectrum with a period of $F_s$. As long as none of the repeated spectra falls on the same frequencies as the original spectrum, you don't get aliasing.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.