FPGA Central

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.

On 29 Nov 2004 01:46:23 -0800, [email protected] (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

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

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

On Tue, 30 Nov 2004 21:42:58 +1100, John Monro
<[email protected]> 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

Christian Auner wrote:

>On Tue, 30 Nov 2004 21:42:58 +1100, John Monro
><[email protected]> 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

On Wed, 01 Dec 2004 00:53:07 +1100, John Monro
<[email protected]> 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

Christian Auner wrote:

>On Wed, 01 Dec 2004 00:53:07 +1100, John Monro
><[email protected]> 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

Christian Auner wrote:

>On Wed, 01 Dec 2004 00:53:07 +1100, John Monro
><[email protected]> 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

On Wed, 01 Dec 2004 22:38:46 +1100, John Monro
<[email protected]> 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

Christian Auner wrote:

>On Wed, 01 Dec 2004 22:38:46 +1100, John Monro
><[email protected]> 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