Talk:Effective number of bits

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ENOB > actual bits

This discussion refers to a previous revision of the article. The current revision does not mention the possibility of ENOB > actual bits. I have removed the {{Disputed}} banner which I assume referenced this discussion. --Kvng (talk) 15:27, 5 November 2011 (UTC)[reply]

I think there's a mistake: Typically the ENOB is less then the number of bits; i.e. ENOB=6 Bit by using a 3Bit DAC is not credible. —Preceding unsigned comment added by 153.96.69.66 (talk) 11:13, 13 August 2009 (UTC)[reply]

Yep, that's definitely wrong, ENOB must be <= total available bits. Side note - is deleting vandalism from a talk page considered 'good form'? (I have just deleted a line from the top of this page) GyroMagician (talk) 10:55, 12 December 2009 (UTC)[reply]

I disagree. Noise Shaping will increase the ENOB, within a band of interest, to far greater than the numebr of bits the DAC can represent. For example in audio DAC's it is possible to have 24bits of resolution between 0kHz and 20kHz with a 6bit noise shaped DAC at the expense of out of band noise.195.59.102.251 (talk) 10:55, 2 November 2010 (UTC)[reply]

How is it possible to increase the signal resolution by 18 bits, after the digitiser? I'm not saying you're wrong - just that I don't get it. GyroMagician (talk) 15:36, 2 November 2010 (UTC)[reply]

I also find it hard to understand how you can have a higher number of effective bits than the actual number of bits. A larger number of effective bits would imply a higher resolution capability. This would imply that if you change the signal amplitude by an amount equal to the LSB then you will see a change in the digital representation of the amplitude, but the digital representation of the signal can only have 2^n discrete states which sets the LSB equal to Vref/2^n. Please explain how you realize a signal representation that has an LSB with greater resolution than the actual number of bits.Rmonzello (talk) 18:02, 22 November 2010 (UTC)rmonzello[reply]

The actual number of bits is a function of the physical ADC hardware chosen. The effective number of bits is a function of the noise and signal ratios. If the ENOB is greater than the ADC's resolution, you have chosen an ADC which is not discriminate enough to fully describe the variations in the signal. If the ENOB is less than the ADC's resolution, you are using an ADC which discriminates enough or too much. If possible, you may wish to use a less accurate ADC to save cost. —Preceding unsigned comment added by 192.43.65.245 (talk) 17:22, 3 February 2011 (UTC)[reply]

Misleading example

The example is misleading. The worst case error occurs when the signal is between the quantization points. In other words, when the signal is 1.5 volts, what is the error? The error is approximately .0625 full scale, plus the measured amount for INL / DNL. Jay (talk) 17:34, 20 June 2011 (UTC)[reply]

Example does not match definition

The calculations in the Example section are done with different equations than those given in the Definition section. --Kvng (talk) 15:27, 5 November 2011 (UTC)[reply]

I removed the whole section. If you aren't using decibels then there's another way to write the formula which is:

\mathrm {ENOB} =N-\log _{2}\!\left({\frac {Q_{actual}}{Q_{ideal}}}\right)

You can play with it and come out with something like what the example has:

=\log _{2}\!\left(2^{N}\right)-\log _{2}\!\left({\frac {Q_{actual}}{Q_{ideal}}}\right)=\log _{2}\!\left({\dfrac {2^{N}}{\dfrac {Q_{actual}}{Q_{ideal}}}}\right)=\log _{2}\!\left({\frac {2^{N}Q_{ideal}}{Q_{actual}}}\right)

Before I flipped it, the example was:

D={\frac {V_{ref}}{2^{\mathrm {ENOB} }}}

Which is:

2^{\mathrm {ENOB} }D=V_{ref}

And:

2^{\mathrm {ENOB} }={\frac {V_{ref}}{D}}

And finally the more sensible but still wrong:

\mathrm {ENOB} =\log 2\!\left({\frac {V_{ref}}{D}}\right)

In the example, V_ref was 8 V which is 2^{3 bits} so this is basically like omitting the - 1.761 dB from the definition formula.

I've done some searching and I can't find anything that says anything other than what is in the definition. The example has been there since almost the beginning of the article but nobody's changed it except to tack on the comment-like caveats in the closing paragraph which aren't particularly encyclopedic themselves. As far as I can tell, all designers most certainly have the same definition and that includes a PDF I found on the MIT website as well as another paper that explicitly stated that DAC has the same definition as ADC. Another objecting statement appears to be pointing out what I just showed by saying something about "adding" the ideal error but that's also wrong because you subtract it. And beyond all of this, the ENOB formula is for a sine wave and the number set in the example is not a sine wave.

So it's wrong and now it's gone. Radiodef (talk) 00:32, 15 August 2013 (UTC)[reply]