In this presentation on slide 15 there is a corollary:

Rule of thumb: the dark noise must be larger than 0.5

Corollary: With a N bit digital signal you can deliver no more*) than N+1 bit dynamic range.

*) You can if you use loss-less compression

They give an example:

Example : A 102f camera with 11 bit dynamic range will deliver only 9 bit in Mono8 mode. Use Mono16!

Why N+1? How can an 8 bit signal deliver 9 bit dynamic range?


From taking a look at the presentation, it seems that they might be using a slightly different definition of dynamic range than is typical. Usually, as in the Wikipedia article on the topic, dynamic range is defined as:

$$ \text{dynamic range} = \frac{\text{largest possible representable value}}{\text{smallest possible representable value}} $$

For an $N$-bit (unsigned) signal, this is equal to:

$$ \text{dynamic range} = \frac{2^N-1}{1} = 2^N-1 $$

However, their discussion of dynamic range is interspersed with discussion of quantization noise. Therefore, I posit that they instead define dynamic range as:

$$ \text{dynamic range} = \frac{\text{largest possible representable value}}{\text{largest possible quantization error}} $$

For a uniformly-quantized quantity such as this, the maximum quantization error is equal to half of one bit. That leads to a dynamic range of:

$$ \text{dynamic range} = \frac{2^N-1}{0.5} = 2(2^N-1) $$

The extra factor of 2 gives you an approximate increase of 1 bit in "dynamic range" when measured this way. I assume that's what is meant by an $N$-bit digitized signal providing $N+1$ bits of dynamic range.

  • $\begingroup$ English isn't my first language. After actually looking up "corollary" in a dictionary, I think your answer makes the most sense. I think now I get that part where they say "toggeling -> mean = 0.5". You probably have to look at a series of images to be able to really have N+1 bits of DR? $\endgroup$ – LV3 Apr 15 '14 at 17:26

The values of dynamic range you're using seems to be base-2 logarithm (see Dynamic range article).

By the definition of dynamic range its best possible value for $N$-bit encoded signal is $2^N:1$ or just $N$ bits. You're right in that it's not possible to have more than $N$-bit DR for $N$-bit encoding (without any compression).

  • $\begingroup$ Thank you werediver, this is what I expected. Strange that Basler would write something like this and EMVA would publish it then, though. $\endgroup$ – LV3 Apr 15 '14 at 11:50

The definition of "dynamic range in dB" is the sum of the S/N dB plus the dB of headroom.

now if you assume the PDF of the signal has the same uniform distribution that the implied quantization noise (where one has quantized a longer word to these $N$ bits), then the dynamic range in dB is $20 \log_{10}(2^N) = 20 \log_{10}(2) N \approx 6.02 N$.

if you assume the signal is a sinusoid (so it has a sinusoidal PDF) then you add 1.76 dB, i think. the sine up to a level (and that level is the "rails" minus the dB headroom) is 1.76 dB louder than a uniform PDF (like a triangle wave would have) having the same maximum amplitude.


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