I have a 10-bit ADC with 8.5 bit ENOB. I have 2.5 quantization levels of thermal noise in front of the ADC. Is there a reason to process more than 8 bits from this ADC?
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How wide is the word width of the processor receiving these 10-bit values from the A/D converter? Wider than 8 bits?
What is there to gain by adding another layer of quantization error?
What you have here is a highly dithered signal that is dithered naturally. Sometimes in audio we actually go out of our way to dither a signal so that further processing and quantization is better.
If your processor is 8 bits wide and you absolutely do not want to do multi-byte arithmetic (with an Add-with-carry instruction) or do not have the computational bandwidth to do higher precision arithmetic, then perhaps it makes sense to quantize your noisy 10-bit value to an 8-bit value. Otherwise it does not.
answered Mar 20 '14 at 19:24
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$\begingroup$ The processor is an FPGA. 8-bit is preferred for simplicity. $\endgroup$ – John Mar 20 '14 at 20:48
