FrankJ777
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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_{0} and bandwidth B, where f_{0} >> 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_{0}.
From what I've read, to accomplish the bandwidth of the ADC must be at least f_{0}. I'm not sure exactly what this means. Does it only mean the input to the ADC must not attenuate a signal of f_{0}? 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_{0}?
If anyone can lend some insight I'd appreciate the help.
Thanks
From what I've read, to accomplish the bandwidth of the ADC must be at least f_{0}. I'm not sure exactly what this means. Does it only mean the input to the ADC must not attenuate a signal of f_{0}? 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_{0}?
If anyone can lend some insight I'd appreciate the help.
Thanks