Dear all, In a Delta-Sigma Modulator, after the quantizer, the output bit stream (+1 and -1) is fed back to the input and being subtract from the input(assuming first order loop). However, none of the references that I read mentioned how large this feedback signal should be. For example, if the input is limited within +-1, how larger this feedback signal should be? Also +-1? I did some experiments, it seems that the feedback signal must also be +-1 but I want to confirm it. Any references talk about this? Thanks.

# Delta-Sigma Modulator question

Started by ●November 29, 2004

Reply by ●November 29, 20042004-11-29

On 29 Nov 2004 01:46:23 -0800, vinma55@hotmail.com (Vincent Ma) wrote:>Dear all, > >In a Delta-Sigma Modulator, after the quantizer, the output bit stream >(+1 and -1) is fed back to the input and being subtract from the >input(assuming first order loop). However, none of the references that >I read mentioned how large this feedback signal should be. For >example, if the input is limited within +-1, how larger this feedback >signal should be? Also +-1? I did some experiments, it seems that the >feedback signal must also be +-1 but I want to confirm it. Any >references talk about this? Thanks.Yes, you are right. In the long term, the input values must lie within the range of values at the quantizer output. I can't see a good reason why in the short term the input values can't exceed that limit, provided that the subtractor and integrator aren't driven beyond their linear range. Regards, Allan

Reply by ●November 29, 20042004-11-29

On Mon, 29 Nov 2004 21:50:37 +1100, Allan Herriman>I can't see a good reason why in the short term the input values can't >exceed that limit, provided that the subtractor and integrator aren't >driven beyond their linear range.Usually the input signal is oversampled therefore only "the long term" is relevant since the signal should only vary slowly. However, if the input signal contains high frequency components then the modulator might be overloaded. A first order modulator will recover from this situation, higher order modulators may remain unstable. The maximum input signal level is dependent on the frequency with DC beeing the worst case. Unfortunately there is no signal norm I am aware of which allows to specify an upper bound for the maximum input signal level independent of the frequency. Regards Chris

Reply by ●November 30, 20042004-11-30

Christian Auner wrote:>On Mon, 29 Nov 2004 21:50:37 +1100, Allan Herriman > > > >>I can't see a good reason why in the short term the input values can't >>exceed that limit, provided that the subtractor and integrator aren't >>driven beyond their linear range. >> >> > >Usually the input signal is oversampled therefore only "the long term" >is relevant since the signal should only vary slowly. However, if the >input signal contains high frequency components then the modulator >might be overloaded. A first order modulator will recover from this >situation, higher order modulators may remain unstable. > >The maximum input signal level is dependent on the frequency with DC >beeing the worst case. Unfortunately there is no signal norm I am >aware of which allows to specify an upper bound for the maximum input >signal level independent of the frequency. > >Regards >Chris > >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. The effect of this overload will not be confined to the time in which the excessive analog input level is present. The integrator will be heading upwards all this time because there is a continuous positive eror. When the analog input finally falls to below 1.000 the error signal will go negative and modulator will continue to output 1s while the error signal drives the integrator in the down towards zero. We now have pulse-smearing in the reconstructed signal. Regards, John

Reply by ●November 30, 20042004-11-30

On Tue, 30 Nov 2004 21:42:58 +1100, John Monro <johnmonro@delete.optusnet.com.au> wrote:>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. > >The effect of this overload will not be confined to the time in which >the excessive analog input level is present. The integrator will be >heading upwards all this time because there is a continuous positive >eror. When the analog input finally falls to below 1.000 the error >signal will go negative and modulator will continue to output 1s while >the error signal drives the integrator in the down towards zero. We now >have pulse-smearing in the reconstructed signal.This consideration is true but limited to a constant input signal. For a time varying input signal the maximum value may exceed the value of the reference voltage. Of course this only holds for a first order modulator. Chris

Reply by ●November 30, 20042004-11-30

Christian Auner wrote:>On Tue, 30 Nov 2004 21:42:58 +1100, John Monro ><johnmonro@delete.optusnet.com.au> wrote: > > > >>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. >> >>The effect of this overload will not be confined to the time in which >>the excessive analog input level is present. The integrator will be >>heading upwards all this time because there is a continuous positive >>eror. When the analog input finally falls to below 1.000 the error >>signal will go negative and modulator will continue to output 1s while >>the error signal drives the integrator in the down towards zero. We now >>have pulse-smearing in the reconstructed signal. >> >> > >This consideration is true but limited to a constant input signal. For >a time varying input signal the maximum value may exceed the value of >the reference voltage. Of course this only holds for a first order >modulator. > >Chris > >Chris, Could you expand on this a little please? I was assuming a first-order modulator, but it seems to me that there is issue here may be independent of the order. For simplicity though, let's stick with the first-order modulator. The issue is: If the modulator outputs a stream of logic-1s all the while the input signal is at a level of 1.000, then what digital stream could the modulator produce to represent correctly an input level of more than 1.000? It seems to me that even if the input remains at this higher level for only a few milliseconds, the modulator will be typically producing hundreds of logic-1s during this time and when the signal is reconstructed the reconstructed level will be a few samples at a level of 1.000, the same as if the input had gone only to a level of 1.000. Regards, John

Reply by ●November 30, 20042004-11-30

On Wed, 01 Dec 2004 00:53:07 +1100, John Monro <johnmonro@delete.optusnet.com.au> wrote: Hi John!>Chris, >Could you expand on this a little please? I was assuming a first-order >modulator, but it seems to me that there is issue here may be >independent of the order. For simplicity though, let's stick with the >first-order modulator. >The issue is: If the modulator outputs a stream of logic-1s all the >while the input signal is at a level of 1.000, then what digital stream >could the modulator produce to represent correctly an input level of >more than 1.000? It seems to me that even if the input remains at this >higher level for only a few milliseconds, the modulator will be >typically producing hundreds of logic-1s during this time and when the >signal is reconstructed the reconstructed level will be a few samples at >a level of 1.000, the same as if the input had gone only to a level of >1.000.Assume that we restrict our input signal to a single frequency and we want to determine the amplitude of the input signal. Since we have a single input frequency it is sufficient to determine the integral of our input signal versus time in order to get a value proportional to the amplitude. This is exactly what the modulator does, it tries to approximate the area of the discrete-time version of our input signal. As long as this area can be represented by the modulator it will produce a proportional signal. For a sine wave the normalized area is 2/pi times the amplitude therefore a input signal with amplitude 1 will produce a bitstream that contains the number of ones which will approximate the ratio of 2/pi. Since this ratio is smaller than 1 there is still room for higher input levels. Chris

Reply by ●December 1, 20042004-12-01

Christian Auner wrote:>On Wed, 01 Dec 2004 00:53:07 +1100, John Monro ><johnmonro@delete.optusnet.com.au> wrote: > >Hi John! > > > >>Chris, >>Could you expand on this a little please? I was assuming a first-order >>modulator, but it seems to me that there is issue here may be >>independent of the order. For simplicity though, let's stick with the >>first-order modulator. >>The issue is: If the modulator outputs a stream of logic-1s all the >>while the input signal is at a level of 1.000, then what digital stream >>could the modulator produce to represent correctly an input level of >>more than 1.000? It seems to me that even if the input remains at this >>higher level for only a few milliseconds, the modulator will be >>typically producing hundreds of logic-1s during this time and when the >>signal is reconstructed the reconstructed level will be a few samples at >>a level of 1.000, the same as if the input had gone only to a level of >>1.000. >> >> > >Assume that we restrict our input signal to a single frequency and we >want to determine the amplitude of the input signal. Since we have a >single input frequency it is sufficient to determine the integral of >our input signal versus time in order to get a value proportional to >the amplitude. This is exactly what the modulator does, it tries to >approximate the area of the discrete-time version of our input signal. >As long as this area can be represented by the modulator it will >produce a proportional signal. For a sine wave the normalized area is >2/pi times the amplitude therefore a input signal with amplitude 1 >will produce a bitstream that contains the number of ones which will >approximate the ratio of 2/pi. Since this ratio is smaller than 1 >there is still room for higher input levels. > >Chris > >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. 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. 4. For a square-wave (or for a low-frequency sine wave at its peak value) the modulator will be producing an continuous string of 1s, which will be decimated into multi-bit samples representing 1.000. 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." Regards, John

Reply by ●December 2, 20042004-12-02

Christian Auner wrote:>On Wed, 01 Dec 2004 00:53:07 +1100, John Monro ><johnmonro@delete.optusnet.com.au> wrote: > >Hi John! > > > >>Chris, >>Could you expand on this a little please? I was assuming a first-order >>modulator, but it seems to me that there is issue here may be >>independent of the order. For simplicity though, let's stick with the >>first-order modulator. >>The issue is: If the modulator outputs a stream of logic-1s all the >>while the input signal is at a level of 1.000, then what digital stream >>could the modulator produce to represent correctly an input level of >>more than 1.000? It seems to me that even if the input remains at this >>higher level for only a few milliseconds, the modulator will be >>typically producing hundreds of logic-1s during this time and when the >>signal is reconstructed the reconstructed level will be a few samples at >>a level of 1.000, the same as if the input had gone only to a level of >>1.000. >> >> > >Assume that we restrict our input signal to a single frequency and we >want to determine the amplitude of the input signal. Since we have a >single input frequency it is sufficient to determine the integral of >our input signal versus time in order to get a value proportional to >the amplitude. This is exactly what the modulator does, it tries to >approximate the area of the discrete-time version of our input signal. >As long as this area can be represented by the modulator it will >produce a proportional signal. For a sine wave the normalized area is >2/pi times the amplitude therefore a input signal with amplitude 1 >will produce a bitstream that contains the number of ones which will >approximate the ratio of 2/pi. Since this ratio is smaller than 1 >there is still room for higher input levels. > >Chris > >Chris, The modulator would produce an output of 2/pi only if one output sample represented the whole of one half-cycle of the input signal, and this would occur only if the input signal frequency was half the output sample rate. I think my original claim still holds: "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." Regards, John

Reply by ●December 2, 20042004-12-02

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