Let's say that we want to digitally sample an analog signal and have a sampling frequency of 8000 samples/second. Usually the maximum frequency that can be captured is the half of the sampling frequency; that is the maximum frequency that can be captured is approximately 4KHz for the example above. But what is the mathematical relationship between the two? Thank you.
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2$\begingroup$ It's a very common misconception that the sampling rate limits the maximum frequency when in fact it limits the bandwidth of the signal. An 8kHz sampling rate will allow you to reconstruct a signal with a bandwidth of 8kHz. Therefore, a real signal with a maximum frequency of 4kHz can always be reconstructed as a signal with a bandwidth of 8kHz, with frequencies ranging from -4kHz to +4kHz. $\endgroup$– JazzmaniacApr 20, 2015 at 12:01
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2$\begingroup$ Do you know and understand the sampling theorem? Because this should answer your question. Otherwise please formulate a more specific question. $\endgroup$– Matt L.Apr 20, 2015 at 12:01
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3$\begingroup$ it's small potatoes, but for a real signal being sampled, in fact the sampling frequency must exceed twice the maximum frequency. it is not (even in theory) sufficient to sample a sinusoid at 4 kHz with a sample rate of 8 kHz (but, in theory, you can sample 3.99999 kHz sufficiently with 8 kHz sample rate). $\endgroup$– robert bristow-johnsonApr 20, 2015 at 12:17
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The maximum frequency is also limited by how long you sample at a given sample rate. If you only sample for a finite length of time at 8000 sps, the bandwidth will be less than 4kHz. A 4000 Hz signal (or any integer multiple thereof) will be ambiguous, or simply disappear.
A signal somewhat below the Nyquist frequency (depending on how long you sample at a given S/N or accuracy) can be "captured", as well as any signals between two successive integer multiples of Fs/2 given a-priori knowledge of no aliasing from any other spectra inside any other folds.
