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I know the dynamic range is the ratio between the maximum amplitude and the minimum amplitude of a signal, calculating the dynamic range in dB I get a difference of logarithms:

dynamic_range = maxAmp/minAmp

But why the dynamic range of a N-bit digital system with linear quantization is 6N dB?

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    $\begingroup$ Take a look at this answer over at Electronics.SE. $\endgroup$ – Matt L. Sep 9 '17 at 14:21
  • $\begingroup$ @MattL. Thanks, but that answer is about SNR that is different from the dynamic range... $\endgroup$ – beth Sep 9 '17 at 14:27
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    $\begingroup$ It's exactly the same calculation, have a look at this. $\endgroup$ – Matt L. Sep 9 '17 at 14:30
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Just to summarize the comments, for the next person who comes looking:

decibels are $20 \log_{10}{x}$ (or $10 \log_{10}{\left( x^2 \right) }$ when power levels used instead of the amplitude levels)

When looking at bit noise limited N-bit system, there are $2^{N}$ measurement levels, and the smallest distinguishable difference is $1$. So

DR$_{dB}$ = $20 \log_{10}{\frac{2^N}{1} } = N*20\log_{10}{2}\approx 6 N$

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For an A/D converter (or an equivalent quantizer with both finite precision and range of measured amplitude), I would define Dynamic Range in dB of that A/D to be the sum of dB of Signal-to-Noise ratio (S/N) plus dB of Headroom plus dB of Crest Factor. For a fixed crest factor, as the signal level increases (against a constant quantization noise floor), the S/N increases, but the headroom dB decreases by about the same amount.

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