# Down sampling, bandpass sampling theorem, downconversion

1. Nov 29, 2011

### FrankJ777

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

2. Dec 4, 2011

### marcusl

Yes, that is correct.