Is there a condition for bandpass sampling?
It's generally not true that a band pass signal can be sampled and recovered without error if $f_s&gt;2B$ is satisfied, where $B=f_h-f_l$ is the signal's bandwidth. This condition is just necessary but not sufficient.

You have to make sure that the aliased spectra do not overlap. This results in the following condition on the sampling frequency:

$$\frac{2f_h}{n+1}&lt;f_s&lt;\frac{2f_l}{n}\tag{1},\quad n=0,1,\ldots$$

If Eq. $(1)$ can only be satisfied for $n=0$, you get the familiar Nyquist sampling theorem where $f_s$ must be greater than twice the highest frequency of the signal: $f_s&gt;2f_h$. The lowest possible sampling frequency is obtained for the largest integer $n$ such that $(1)$ is still satisfied. This maximum value of $n$ is given by

$$n_{max}=\left\lfloor\frac{f_l}{B}\right\rfloor\tag{2}$$

where $B=f_h-f_l$ is the bandwidth. Eq. $(2)$ is obtained from $(1)$ by computing the largest value of $n$ such that the left-most term is still less than the right-most term.

As an example, take a band pass signal with $f_l=10\text{ kHz}$ and $f_h=30\text{ kHz}$. From $(2)$, Eq. $(1)$ can only be satisfied for $n=0$, so you have to choose $f_s&gt;2f_h=60\text{ kHz}$, which is a stricter condition than just requiring $f_s&gt;2B=40\text{ kHz}$. However, if $f_l=10\text{ kHz}$ and $f_h=14\text{ kHz}$, the maximum value of $n$ for which $(1)$ is satisfied is $n=2$, which gives you a range $2\cdot 14/3=9.33&lt;f_s&lt;10$ of possible sampling frequencies, all of which of course satisfy $f_s&gt;2B=8\text{ kHz}$.