Compute dynamic range in a linear quantization system

Question

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:

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

@MattL. Thanks, but that answer is about SNR that is different from the dynamic range... — sleepwalking, Sep 09 '17 at 14:27

argentum2f · Answer 1 · 2018-08-01T01:50:05.807

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

edited Aug 01 '18 at 01:50

answered Jul 30 '18 at 16:34

argentum2f

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Also see this answer here https://dsp.stackexchange.com/questions/41497/snr-of-a-16-bit-dsp-with-12-bit-adc-40-bit-accumulator/41509#41509 which may help give further insight. – Dan Boschen Jul 31 '18 at 01:40
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And this one: https://dsp.stackexchange.com/questions/40259/what-are-advantages-of-having-higher-sampling-rate-of-a-signal/40261#40261 – Dan Boschen Jul 31 '18 at 11:11

robert bristow-johnson · Answer 2 · 2018-09-30T06:01:27.250

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.

Compute dynamic range in a linear quantization system

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