Multiband undersampling In Practical Signal Processing (Marc Owen), an exercise explore the topic of undersampling:


  [...] Suppose you know that an audio signal is a sine wave with a frequency that might be in one of two ranges, either between 120Hz and 130Hz or between 150Hz and 170Hz. What is the lowest sample rate you could use, and still be able to reconstruct the signal from (real-valued) samples? What if the second band extends from 150Hz to 172Hz?


The answers are actually provided at the end of the book (60Hz and 86.666Hz) but without further details.

I have apply some of the theory on undersampling and found the different ranges possible for each frequencies. For example, with $n=3$, I can see that the aliases possible for the range 120Hz-130Hz is $86.666 &lt; fs &lt; 120.0$. This theory applies well for one band but I can't find any explanation on the dependency created by another band. I cannot just take the intersection of both sets as it does not cover the case where the ranges are overlapping.

I also did a GNURadio example to see the influence of the sample rate on each range. By moving the sample rate slider around, I can see that 86.66Hz is indeed the solution (for the second case) but is there an analytical method to find this result?

Thanks
I found the answer in Akos, D.M., Stockmaster, M., Tsui, J.B., &amp; Caschera, J. (1999). Direct bandpass sampling of multiple distinct RF signals. IEEE Trans. Communications, 47, 983-988.

The extra condition between the 2 bands is: $|F_{IF1} - F_{IF2}| &gt;= \frac{BW_{I1} + BW_{I2}}{2}$ where $F_{IF}$ is the intermediate frequency and $BW$ the bandwidth.

There are then 4 equations that describe all these constraints (eq $(1)$ to $(4)$) in the paper. Unfortunately, the first constraint contains a non-linear combination which cannot be solved analytically (by something like mixed-integer programming).

This is confirm in the paper: "An iterative procedure, based on these equations, was developed to determine sampling frequencies that would meet the requirement, the results of which are displayed in Fig. 5.". I conclude there is no better solution than iteratively going through the frequencies under consideration.