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I don't understand the digital filter enough, so I'll think it over, and I will understand.
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Lets see if this helps ... Above you say you want to sample a signal out to 1.25MHz, with an 8-bit ADC, right?
An 8-bit ADC has a signal-to-noise of around 8 x 6dB = 48dB. To ensure that your 1.25MHz signal is not corrupted by signals that alias into the band after sampling, your analog filter needs to reject signals by at least 48dB for the signals that will alias into the band. If your signal was filtered with an analog filter that was a simple RC filter, then the -3dB bandwidth of the filter needs to be about 5MHz to keep your passband ripple below 0.3dB. That single pole RC filter has a response amplitude of about -20dB at 50MHz and -26dB at 100MHz. If your ADC samples to 8-bits at 100MHz, then the signal from 50MHz to 100MHz will alias into your sampled band. This means that the signal from 100MHz-1.25MHz to 100MHz will alias onto your band of interest (as will all other higher frequencies). Since the power in that region is only down by 26dB, you will corrupt your signal, and be left with a signal-to-noise of only 26dB, which is about 26/6 ~ 4.3bits, so you've totally defeated using an 8-bit ADC.
The analog filter would actually be designed with a much faster roll-off than a single-pole response. The number of poles to use depends on how fast you can operate your ADC, and how many bits you want to preserve in your signal. From this discussion you can see that you often need to operate your ADC much faster than the Nyquist bandwidth of the signal you want to sample. Once you have sampled the signal, you can then use a *digital filter* to eliminate the out-of-band signal (the analog filter band edges) and then decimate the result to the Nyquist bandwidth of the signal, or perhaps slightly more than the Nyquist bandwidth. Your digital filter design depends on your "system" design.
Cheers,
Dave
Honored Contributor
