Sampling Rate Passband Signal
We can always use Nyquist theorem to decide the sampling rate. But in case of bandpass signal, undersampling (sampling rate less than Nyquist rate) can also do the job.
Let's assume we have a continuous input bandpass signal of bandwidth B. The bandpass signal is centered at \$f_c\$ Hz, and its sampled value spectrum is that shown in Figure.
We can sample that continuous signal at a rate, say \$f_{s'}\$ Hz, so the spectral replications of the positive and negative bands, P and Q, just butt up against each other exactly at zero Hz. This situation, depicted in Figure (a). With an arbitrary number of replications, say m, in the range of \$2f_c – B\$, we see that :
$$mf_{s'} = 2f_c-B\ \ \mathrm{or}\ \ f_{s'} = \frac{2f_c - B}{m}$$

Where, m can be any positive integer so long as \$f_{s'}\$ is never less than 2B.
If the sample rate \$f_{s'}\$ is increased, the original spectra (bold) do not shift, but all the replications will shift. At zero Hz, the P band will shift to the right, and the Q band will shift to the left. These replications will overlap and aliasing occurs. Thus for an arbitrary m, there is a frequency that the sample rate must not exceed, or
$$f_{s'} &lt; \frac{2f_c - B}{m}\tag1$$
If we reduce the sample rate below the \$f_{s'}\$, the spacing between replications will decrease in the direction of the arrows in Figure (b). Again, the original spectra do not shift when the sample rate is changed. At some new sample rate \$f_{s''}\$, where \$f_{s''} &lt; f_{s'}\$, the replication P will just butt up against the positive original spectrum centered at \$f_{c}\$ as shown in Figure (c). Decreasing sampling frequency below this will cause aliasing. So there is a lower limit given by
$$f_{s''} &gt; \frac{2f_c + B}{m+1}\tag2$$
From equation (1) and (2),
$$ \frac{2f_c - B}{m} &gt; f_{s} &gt; \frac{2f_c + B}{m+1}\tag3$$
Where m is an integer and \$f_s &gt; 2B\$.
Equation (3) gives the minimum and maximum sampling frequency so that there will be no aliasing.
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