Hi,
I always thought that sampling at Fs >= 2Fmax is the thing to do when sampling a signal.
But I came across a school document that says if Fmin is also given, the sampling rate should be Fs >= 2(Fmax - Fmin).
It puzzles me because if Fmax and Fmin are very high and close, then Fs would be very small… So how can the original high frequency signal be recreated with a low sample rate ?
Disclaimer: It was a long time since I studied this and have never made practical use of it.
The key feature is that this critically depends on the signal being perfectly band-limited between the given frequencies. The reconstruction formula turns out differently and only creates higher frequency components. If I remember correctly it’s reasonably easy to derive the formula from the proof for the ordinary case.
The main difficulty is that real signals aren’t actually band-limited and the aliasing effects become quite severe when you reconstruct to higher frequencies.
What you’re after is called Band Pass Sampling which is common in RF.
The trick is to think the signal is shifted by some modulation (Image from Sampling Band Pass Signals):
This trick allows sampling very high frequency signals only by their real bandwidth.
The actual sampling process is composed of a Mixer and Base Band Sampling (Classic sampling).
There are some conditions for this kind of sampling (Given a Sampler: A/D Device). You may have a look at Mark Fowler - Bandpass Sampling.
By the way, instead of Off Topic It i better to use Signal and Image Processing category.
It’s basically what @RoyiAvital says, but I’ll try another explanation. Imagine you have a bandlimited signal with frequencies between [-f_{max} , f_{max}]. If you multiply it with a carrier signal exp(i2\pi f_ct) you get a new signal with the same information as the first one but with a shifted band.
The mapping between those signals is perfectly predictable, so if you have a signal with the [f_{min} , f_{max}] band you can easily capture the essential information by sampling with the ‘low’ sampling rate, downshifting with the center frequency (multiply by exp(-i\pi (f_{min}+ f_{max}))), reconstruct/resample, and then upshifting again.
But, yes, it’s the bandwidth that counts, not the max frequency.
And, yes, physical signals are rarely really bandlimited, but often you pass them through analog filters before sampling to get “close enough” to band-limitedness. Physical signals don’t need to be perfectly reconstructed, since they are anyway full of measurement noise. If the sampling noise doesn’t add significant extra noise, it’s ‘good enough’.
To add to RoyiAvital’s answer and graphic, you could use the higher sampling rate (2Fmax) to avoid information loss, and it would result in this repeating spectrum instead of (b):
When you sample in the time domain (which you must for digital processing of analog signals), you effectively repeat and add the original signal’s frequency spectrum. That indeed changes the signal, but as long as the repeats don’t overlap and thus irreversibly lose information in the addition, it can theoretically be used to recreate the original analog signal perfectly. Note that you would use different formulae to recreate the analog signal from (b) versus the 2Fmax-sampled spectrum; as you intuited, the wrong formula causes aliasing (high frequencies resulting in false lower frequencies).
