Delta-Sigma Modulator question

Christian Auner wrote:

>On Wed, 01 Dec 2004 22:38:46 +1100, John Monro
><johnmonro@delete.optusnet.com.au> wrote:
>
>John,
>  
>
>>Chris,
>>A few points:
>>
>>  1. The ratio is less than 1.000 only because the input instantaneous
>>     level is less than 1.000 for most of the cycle.
>>    
>>
>
>Of course, yes.
>
>  
>
>>  2. The modulator would generate a  value of 2/pi only under
>>     conditions where all the bits generated in the course of one
>>     half-cycle contributed to one single output-sample.  In other
>>     words, when the signal is right at the Nyquist limit 2fs.  (fs =
>>     output sample rate).
>>  3. The output could approach 1.000 if a square-wave analog signal was
>>     applied.
>>    
>>
>
>Yes, but the information is not lost. The information is not present
>in a single output-sample but in sequence of output-samples. Same is
>true if you consider a signal slightly below the reference voltage.
>The modulator can't represent that value instantaneously, however the
>information is stored in the integrator and will contribute to the
>following output-samples.
>
>[...]
>
>  
>
>>In view of all this, I think my original comment still stands:
>>
>>   "A problem I can see relates to the fact that the  modulator
>>   delivers all 1s  when the analog input is at a level of 1.000.    It
>>   follows that if the input ever exceeds 1.000 then the modulator can
>>   not represent this higher value because it can't do any better than
>>   continue outputting all 1s.  The result is clipping in the
>>   reconstructed waveform."
>>    
>>
>
>No, it is not always required to reconstruct the waveform. The output
>will always represent the area under the signal with the cumulative
>error beeing the state of the integrator.
>
>
>Chris
>  
>
Chris,
Apologies for an error in my posting.  Instead of 2fs for the Nyquist 
limit I should have written (1/2)fs.

Regarding your comments, we are discussing the multi-bit samples that 
come out of the decimator following the basic modulator.  One of these 
samples represents the average input-signal level during the time it 
takes the modulator to produce a large number of single-bit outputs..  
If the input waveform happens to have peaked at +1.000 for the whole of 
this time, the modulator will be producimg a string of single-bit values 
each having a value of logic 1.

There will be a cumulative error when the input signal is slightly less 
than 1.000.  This will cause the modulator to produce an occasional 
logic 0 among the logic 1s, but the cumulative error from one multi-bit 
output sample to the next has a maximum value of only one 
least-significant bit.  In no way is there any sort of summation over 
the whole of one cycle of the input signal, (except for very high input 
frequencies, as I noted.)

For an input of 1.000, the multi-bit sample that come out of the 
decimator will have a value equal to the highest number that can be 
represented with that number of bits.  It will not be able to exceed 
that value, no matter what the input signal does.

Regards,
John