Can you suggest me some books/webpages on Bandpass sampling?

I undestand that if the signal is restricted between $f_L$ and $f_H$, then the minimum bandwidth required is $2(f_H - f_L)$. But say the centre frequency is in GHz and the bandwidth is in MHz, then by this theory I would have the sampling frequency to be in a few MHz. So in case of GHz centre frequency signal(say 1 GHz), I collect the samples every microsecond. But the signal is changing every nanosecond. I did not understand this concept.


3 Answers 3


Yes, the signal is changing "every nanosecond" for any non DC signal. But if the bandwidth is really limited to 0.001 of the carrier frequency, then those changes are very predictable, as the changes result in a carrier waveform that is a nearly perfect sinusoid of a known frequency. The signal can't differ too much from that highly predictable sinusoidal waveform unless the bandwidth is increased. Assuming the bandwidth doesn't increase, any changes to the near perfect sinusoid are so small that one only needs to check the waveform every few hundred cycles of the sinusoid to see if it has drifted a bit from the expected point on the expected cycle. Thus under-sampling at somewhere above 0.002 cf can tell you all you need to know in addition to knowing that the bandwidth was absolutely constrained to 0.001 cf or below.

But the sampling jitter requirement has to be on the order of a very tiny fraction of a nanosecond, or else one will completely miss the expected point on the expected cycle of the 1 Ghz sinusoid.

  • $\begingroup$ Thank you sir. Supposing it is a random signal with frequency centered at 1 Ghz, and it spreads over a 1 MHz band of frequency. Then can we sample it at a 2 MHz for the complete information or should we ample it at 2 GHz ? For eg. Say it is a gaussian infrequency domain with a centre frequency of 1 GHz and spreads over a bandwidth ( which contains 99% of the signal (since Gaussian is an infinitely spreading curve). So in this case should we sample it at GHz frequency or is MHz enough? $\endgroup$ Mar 30, 2016 at 7:57
  • $\begingroup$ You have to sample a bit faster than 2 MHz. If the bandwidth exceeds 1 MHz with some probability, undersampling can fail with about that same order of probability. $\endgroup$
    – hotpaw2
    Mar 30, 2016 at 14:36

Bandpass sampling takes advantage of the empty bands within the signal spectrum so to reduce the minimum necessary sampling rate $\Omega_s$ from what's suggested by lowpass sampling theorem which considers the minimum $\Omega_s$ to be 2x the highest frequency in its spectrum, aka its bandwidth.

For a real, continuous-time bandpass signal whose spectrum (magnitude) as shown bleow is zero for $|\Omega| < \Omega_1$ and for $|\Omega| > \Omega_2$, the lowpass sampling theorem states that a necessary and sufficient condition for perfect reconstruction is $\Omega_s > 2 \Omega_2$. However, this may be reduced by applying bandpass sampling as follows.

Bandpass signal Spectra

A necessary and sufficient condition on the sampling frequency $\Omega_s$ for perfect recovery of original signal from its samples is that, shifted spectrums (due to impulse train modulation) do not overlap; i.e., there is no aliasing.

This is depicted from the figure below showing the shifted spectrums of $X_c(\Omega - k \Omega_s)$ for $k=0,1,...,m,$ and $k=m+1$.

sampled spectrum

from the figure, alias-free positioning is attained if the following conditions hold for some integer $m$:

$$ -\Omega_1 + m ~ \Omega_s < \Omega_1 \tag{1} $$ $$ -\Omega_2 + (m+1) ~ \Omega_s > \Omega_2 \tag{2} $$

Adding negative of Eq(1) to Eq(2) yields, a necessarry condition for the sampling rate $\Omega_s$ :

$$\Omega_s > 2(\Omega_2 - \Omega_1) $$

Given $\Omega_s$ that meets this necessary condition, then the maximum positive integer $m$ that satisifes Eq(1) is found to be:

$$m = \lfloor{ \frac{2 \Omega_1}{\Omega_s} }\rfloor $$

Then, finally, the second necessary condition on $\Omega_s$ to avoid any spectral overlapp is found from Eq.(2) for the $m$ found as above.

$$2\Omega_2 \leq (m+1) \Omega_s $$

Therefore for a given set of $\Omega_1$,$\Omega_2$ and $\Omega_s$ to ensure alias free bandpass sampling, $\Omega_s$ and integer $m$ satify :

$$ \Omega_s \geq 2 (\Omega_2 - \Omega_1) $$ $$0 \leq m = \lfloor{ \frac{2 \Omega_1}{\Omega_s} }\rfloor $$ $$2\Omega_2 \leq (m+1) \Omega_s $$


If you shift the signal so that the center frequency of the signal is at 0 Hz (i.e. multiply by $\exp(-j2\pi f_ct)$ and do a low-pass filter, then the bandwidth of this basebanded signal is $f_H-f_L$ and you can sample at slightly greater rate than $f_H-f_L$, but the samples are now complex - due to the loss of symmetry. Once you sample - your frequency spectrum then contains periodic repetitions of the original spectrum. This just shows that the signal can be represented with the lower samples rate - the information in the carrier or center frequency has been removed. Note this is just a representation of your signal - it doesn't equal the original signal. You can recover the original signal by doing D/A conversion and shifting the signal back up to the center frequency. To do this effectively there are requirements on the phase noise of the signal $\exp(-j2\pi f_ct)$ - a similar requirement is also required for the scheme below. This is traditional bandpass sampling.

Another technique is to just to sample the signal at $2(f_H-f_L)$. The sampling process makes the frequency domain periodic - so a bunch of repetitions of your signal occur in the frequency domain - even around 0 Hz. This assumes that there are no other signals outside the range $f_L$ to $f_H$ in the original - if there are, then these signals will also periodically repeat and overlap with your desired signal. The doubled sampling rate is required because for a real signal with complex conjugate symmetry in the frequency domain, the positive frequency components and the negative frequency components will both periodically repeat. Again this signal is not equivalent to your original, it is just a representation that allows you to recover your original signal (if desired).

  • $\begingroup$ @PeterK.I don't think so. You're shifting the positive frequency parts down, so they're centered at 0 Hz - These are the components you want to keep. What you are getting rid of is the original negative frequency components (I'm assuming the original signal was real) that were further shifted - so now they're centered on at $-2f_c$. This is the typical low-pass filter before sampling idea. Am I missing something? $\endgroup$
    – David
    Apr 29, 2016 at 14:46
  • $\begingroup$ Right! My mistake! I was thinking up, not down. $\endgroup$
    – Peter K.
    Apr 29, 2016 at 14:46
  • 1
    $\begingroup$ @PeterK.To clarify - the frequency shift is done in the analog domain. You can't really do a multiply by $\exp(-j2\pi f_ct)$ Instead you have generate the $\sin()$ and $\cos()$ components and keep track of the resulting real and imaginary components. Hope that's clear. $\endgroup$
    – David
    Apr 29, 2016 at 14:49

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