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?

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    $\begingroup$ Decimation requires a low-pass filter. Averaging is a rather bad quality low-pass filter. $\endgroup$ – Juancho Oct 1 '18 at 20:00
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    $\begingroup$ 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. $\endgroup$ – user28715 Oct 1 '18 at 20:45
  • $\begingroup$ 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. $\endgroup$ – Aleksander Alekseev Oct 2 '18 at 10:51
  • $\begingroup$ Is this a duplicate of dsp.stackexchange.com/questions/40259/… ? $\endgroup$ – Dan Boschen Oct 4 '18 at 3:26
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    $\begingroup$ Possible duplicate of What are advantages of having higher sampling rate of a signal? $\endgroup$ – lennon310 Oct 4 '18 at 11:46

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