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I have an input signal with bandwidth 2.5 MHz. Let's say it's filtered using analog Butterworth low-pass filter. I would like to sample the signal using AD9203 10-bit 40 Msps ADC. If I understand correctly, in this case 5 Msps is enough and I could use decimation to effectively increase number of bits per sample.

Is it enough to just add up every 8 samples @ 40 Msps on input and push resulting 13-bit sum @ 5 Msps to the output? Or a little more sophisticated processing is required?

Aleksander Alekseev
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    Decimation requires a low-pass filter. Averaging is a rather bad quality low-pass filter. – Juancho Oct 01 '18 at 20:00
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    what is the bandwidth of the butterworth filter? Without noise shaping, you pick up a bit of resolution for 4X sample rate. That is SNR gain over quantization noise. –  Oct 01 '18 at 20:45
  • The Butterworth filter doesn't pass anything above 2.5 Mhz (-50 dB at this frequency). The actual bandwidth (- 3 dB point) is a little narrower. – Aleksander Alekseev Oct 02 '18 at 10:51
  • Is this a duplicate of https://dsp.stackexchange.com/questions/40259/what-are-advantages-of-having-higher-sampling-rate-of-a-signal/40261#40261 ? – Dan Boschen Oct 04 '18 at 03:26
  • (See the question I linked above for full details but yes you stand to gain 1.5 bits (9 dB) by oversampling by 8, this is achieved digitally at the output of a 5 MHz filter that you would naturally have in this case. No decimation is required to do this, but given the bandwidth that you constrained it to, decimating it would only make sense to do once filtered. – Dan Boschen Oct 04 '18 at 03:31
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    Possible duplicate of What are advantages of having higher sampling rate of a signal? – lennon310 Oct 04 '18 at 11:46

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