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What is the maximum/minimum possible sampling rate for the signal to be full recovered for a passband signal?

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  • 1
    \$\begingroup\$ Ever heard of the Nyquist frequency? \$\endgroup\$ – Matt Young Mar 14 '15 at 16:43
  • \$\begingroup\$ passband signal not bandpass signal \$\endgroup\$ – user36685 Mar 14 '15 at 16:52
  • \$\begingroup\$ That makes no difference. \$\endgroup\$ – Matt Young Mar 14 '15 at 16:59
  • \$\begingroup\$ @MattYoung, it actually does. This is called undersampling of a bandpass aignal and if proper conditions are met it allows the recovery of the signal. en.m.wikipedia.org/wiki/Undersampling \$\endgroup\$ – Roger C. Mar 14 '15 at 17:47
  • \$\begingroup\$ @user36685, I think that you mean in your comment that it is a bandpass signal and not a "baseband" signal. Bandpass or passband would be pretty much the same I guess! \$\endgroup\$ – Roger C. Mar 14 '15 at 17:53
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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 :

$$mf_{s'} = 2f_c-B\ \ \mathrm{or}\ \ f_{s'} = \frac{2f_c - B}{m}$$

enter image description here

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

$$f_{s'} < \frac{2f_c - B}{m}\tag1$$

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''} < 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

$$f_{s''} > \frac{2f_c + B}{m+1}\tag2$$

From equation (1) and (2),

$$ \frac{2f_c - B}{m} > f_{s} > \frac{2f_c + B}{m+1}\tag3$$

Where m is an integer and \$f_s > 2B\$.

Equation (3) gives the minimum and maximum sampling frequency so that there will be no aliasing.

Content copied from this Source.

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I don't have practical experience with this, but I did look into it a bit at one time. So take this with a grain of salt.

Let's say you are sampling at 10 MHz. The classic arrangement is to insure that the signal is band limited to the range 0 to 5 MHz. However, you can also band limit the signal to the range 5-10 MHz, or 10-15MHz, etc. In any of these cases, there will be no ambiguity, because the signal BW is known beforehand.

In principle, you can use any 5MHz swath of spectrum, but the way the signal frequency gets remapped into the frequency domain will be kind of weird if you don't stick to even multiples of Nyquist.

Please note that some ADC's have an analog bandwidth limit on their input, and may not be able to reproduce signals higher than Fs/2. So look into that before you get too far along.

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A minimum of twice the frequency of the baseband signal (as given by the Nyquist Sampling Theorem) is to be used. As for the maximum, anything beyond the above mentioned sampling frequency will still result in correct reconstruction. However, an extremely large sampling frequency is inefficient, in terms of bandwidth usage.

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