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This is my first question on a StackExchange.
When the basis functions to represent a signal are chosen as $e^{j\omega t}$ such as in a continuous-time Fourier transform then the sample rate $f_\text{s}$ must be more than twice the maximum frequency taken by the chosen continuous time signal. Two questions:

  1. Is it possible to get a Nyquist sampling rate less (I doubt) or more than this by choosing a different basis set? I understand that "frequency" itself may no longer be well defined, so basically: would it be possible to choose a set where one could afford to sample at a higher/lower rate without loss.

  2. If yes, is there some reasoning behind how the set was chosen? Any theory/references would be appreciated.

Would be glad to clarify any doubts/inconsistencies.

robert bristow-johnson
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Television
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1 Answers1

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In the most general case, if you want to sample a continuous-time signal without loss of information, the minimum sampling rate is independent of any choice basis functions. The faster the signal changes with respect to the independent variable (which doesn't need to be time), the faster you have to sample. And if the signal is not band-limited or if it can't be parameterized in some other way, sampling will be lossy.

Note, however, that the (Nyquist-Shannon) sampling theorem is only a sufficient criterion for sampling a continuous-time signal without loss of information. In general you have to choose the sampling frequency greater than twice the highest frequency component in the signal. However, under certain conditions you can get away with sampling at a lower rate.

One example are bandpass signals, which can often be sampled at rates much lower than their highest frequency, especially if their bandwidth is small compared to their center frequency. But note that it is generally not sufficient to sample at a rate equal to twice the signal's bandwidth. See this answer for more details on bandpass sampling.

Another special case are sparse signals. The sparsity can be exploited to represent these signals with fewer samples than required by the sampling theorem. The corresponding technique is called compressed sensing.

In general, if a given signal has a "finite rate of innovation (FRI)" (such signals are called "FRI-signals"), they can be sampled at or above their rate of innovation. This subject treated in this important paper.

I also recommend reading this overview paper: Sampling - 50 Years After Shannon.

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Matt L.
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  • Would bandpass signals be considered sparse in the frequency domain? But you can also have signals that are sparse in some other domain? – endolith Jun 15 '16 at 14:49
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    @endolith: Yes (to both questions). Sparsity in other domains can be taken advantage of. As long as the signal has a "finite rate of innovation" (FRI signals), they can be sampled (at or above their rate of innovation). – Matt L. Jun 15 '16 at 15:14
  • What's the simplest example of a different domain (not time or frequency) that a signal can be sparse in? (I've been trying and failing to understand compressed sensing for a while.) – endolith Jun 15 '16 at 15:37
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    @endolith: In the general sense I think "sparse" is most naturally interpreted as "parametric", e.g., a sparse signal can be given by $$s(t)=\sum_ka_k\phi(t-kT)$$ where $\phi(t)$ is orthogonal to its shifted versions (by multiples of $T$). – Matt L. Jun 15 '16 at 16:10
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    "FRI signals". that's a semantic i hadn't heard of before. – robert bristow-johnson Jun 16 '16 at 03:59
  • Thanks for your answer. Spin-offs from the original question: 1. Would it be possible to choose a basis set that has a descending (in some region) sampling rate vs. loss profile? (N.B. I am using the term "loss" without the knowledge of its exact mathematical representation) 2. Given a continuous time signal and a predefined sampling rate (lower than the rate of innovation), how would one decide which basis set to choose in order to minimize losses, reasoning behind the same. – Television Jun 16 '16 at 05:00
  • @NivedRajaraman: I think that these question can't be answered in comments. You could formulate them as a new question, but I'm not sure if anybody can come up with a satisfactory answer. – Matt L. Jun 16 '16 at 06:21
  • I understand. I'm looking to go deeper into the subject and do some research anyways, So if an answer turns up, I'll be happy to post it separately as a question with the answer and link it here. Thanks again! – Television Jun 16 '16 at 06:24