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Down sampling, bandpass sampling theorem, downconversion

  1. Nov 29, 2011 #1
    I'm trying to demodulate simple AM by using an ADC and the bandpass sampling theorem (as I understand it.) The way I understand the theorem is that by sampling a bandpass signal of frequency f[itex]_{0}[/itex] and bandwidth B, where f[itex]_{0}[/itex] >> B, as long as I use a sampling frequency of >2B I can reproduce the signal even though the sampling frequency is much less than f[itex]_{0}[/itex].
    From what I've read, to accomplish the bandwidth of the ADC must be at least f[itex]_{0}[/itex]. I'm not sure exactly what this means. Does it only mean the input to the ADC must not attenuate a signal of f[itex]_{0}[/itex]? Or are they referring to the sampling aperture. I see the formula use to demonstrate the concept is m[n] = M(t)δ (t-nTs). But i realize that δ is instantaneous where as a ADC is not, so I'm wondering if the width of my sample needs to be around 1/f[itex]_{0}[/itex]?

    If anyone can lend some insight I'd appreciate the help.
  2. jcsd
  3. Dec 4, 2011 #2


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    Gold Member

    Yes, that is correct.
    I don't understand your question.

    Bandpass sampling is somewhat sophisticated, and requires different filtration than regular baseband sampling to achieve a good SNR. I fear that a couple of short answers on this forum won't give you a deep enough understanding of sampling, filtration and DSP. Have you had a course in DSP? Do you have a DSP text to read?
    Last edited: Dec 4, 2011
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