FPGA Central

Hi,

Consider a sigma delta adc which first oversamples, does
noise-shaping and then filters and decimates to give final output
at Nyquist rate and with 16 bits per sample.

For this final output, note that quantisation stepsize is uniform.

Now, is it possible to get this same output using a normal adc
for some sampling phase and uniform sampling of the analog signal ?

Or will it be correct to say that the final output is a
non-uniformly sampled version of the original analog signal ?

( I understood the functionality in freq domain, and am trying
to figure out in time domain . )

shankar

[email protected] (kbc) writes:

> Hi,
>
> Consider a sigma delta adc which first oversamples, does
> noise-shaping and then filters and decimates to give final output
> at Nyquist rate and with 16 bits per sample.

OK, so this is a standard delta sigma ADC. Fine.

> For this final output, note that quantisation stepsize is uniform.

Hmmm. I really wouldn't say that. At the final output, there is no
such thing as a quantization stepsize since a delta sigma ADC is NOT a
uniform quantizer. Instead, the signal, which was quantized to N bits
(N typically is 1) inside the modulator, has been modified with
feedback in the modulator and the result filtered. We're far away from
a simple M-bit quantizer at the decimator output (M >> N due to the
oversampling and noise-shaping action).

> Now, is it possible to get this same output using a normal adc
> for some sampling phase and uniform sampling of the analog signal ?

No.

> Or will it be correct to say that the final output is a
> non-uniformly sampled version of the original analog signal ?

It is uniformly time quantized, but not uniformly amplitude quantized.

> ( I understood the functionality in freq domain, and am trying
> to figure out in time domain . )

You're thinking - this is a good thing. If this isn't clear, keep
asking questions.
--
% Randy Yates % "Bird, on the wing,
%% Fuquay-Varina, NC % goes floating by
%%% 919-577-9882 % but there's a teardrop in his eye..."
%%%% <[email protected]> % 'One Summer Dream', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Randy Yates <[email protected]> wrote in message news:<[email protected]>...
> [email protected] (kbc) writes:
>
> > Hi,
> >
> > Consider a sigma delta adc which first oversamples, does
> > noise-shaping and then filters and decimates to give final output
> > at Nyquist rate and with 16 bits per sample.
>
> OK, so this is a standard delta sigma ADC. Fine.
>
> > For this final output, note that quantisation stepsize is uniform.
>
> Hmmm. I really wouldn't say that. At the final output, there is no
> such thing as a quantization stepsize since a delta sigma ADC is NOT a
> uniform quantizer. Instead, the signal, which was quantized to N bits
> (N typically is 1) inside the modulator, has been modified with
> feedback in the modulator and the result filtered. We're far away from
> a simple M-bit quantizer at the decimator output (M >> N due to the
> oversampling and noise-shaping action).
>
> > Now, is it possible to get this same output using a normal adc
> > for some sampling phase and uniform sampling of the analog signal ?
>
> No.
>
> > Or will it be correct to say that the final output is a
> > non-uniformly sampled version of the original analog signal ?
>
> It is uniformly time quantized, but not uniformly amplitude quantized.
>

I think you are wrong. It is uniformly amplitude quantized also.

I think we have effectively dithered the analog signal and then
passed it thru a normal adc.

Now, why dither helps, I am trying to understand.

shankar

[email protected] (kbc) writes:

> Randy Yates <[email protected]> wrote in message news:<[email protected]>...
> > [email protected] (kbc) writes:
> >
> > > Hi,
> > >
> > > Consider a sigma delta adc which first oversamples, does
> > > noise-shaping and then filters and decimates to give final output
> > > at Nyquist rate and with 16 bits per sample.
> >
> > OK, so this is a standard delta sigma ADC. Fine.
> >
> > > For this final output, note that quantisation stepsize is uniform.
> >
> > Hmmm. I really wouldn't say that. At the final output, there is no
> > such thing as a quantization stepsize since a delta sigma ADC is NOT a
> > uniform quantizer. Instead, the signal, which was quantized to N bits
> > (N typically is 1) inside the modulator, has been modified with
> > feedback in the modulator and the result filtered. We're far away from
> > a simple M-bit quantizer at the decimator output (M >> N due to the
> > oversampling and noise-shaping action).
> >
> > > Now, is it possible to get this same output using a normal adc
> > > for some sampling phase and uniform sampling of the analog signal ?
> >
> > No.
> >
> > > Or will it be correct to say that the final output is a
> > > non-uniformly sampled version of the original analog signal ?
> >
> > It is uniformly time quantized, but not uniformly amplitude quantized.
> >
>
> I think you are wrong. It is uniformly amplitude quantized also.

If you've already made up your mind as to what you believe about
the matter, why did you ask?
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
[email protected], 919-472-1124

Randy Yates <[email protected]> wrote in message news:<[email protected]>...
> [email protected] (kbc) writes:
>

> > > > Now, is it possible to get this same output using a normal adc
> > > > for some sampling phase and uniform sampling of the analog signal ?
> > >
> > > No.
> > >
> > > > Or will it be correct to say that the final output is a
> > > > non-uniformly sampled version of the original analog signal ?
> > >
> > > It is uniformly time quantized, but not uniformly amplitude quantized.
> > >
> >
> > I think you are wrong. It is uniformly amplitude quantized also.
>
> If you've already made up your mind as to what you believe about
> the matter, why did you ask?


Does it not occur to you that I might have understood the connection
between dithering and sigma-delta after i posted my question here ?

kbc wrote:

> Randy Yates <[email protected]> wrote in message news:<[email protected]>...
>
>>[email protected] (kbc) writes:
>>
>
>
>>>>>Now, is it possible to get this same output using a normal adc
>>>>>for some sampling phase and uniform sampling of the analog signal ?
>>>>
>>>>No.
>>>>
>>>>
>>>>>Or will it be correct to say that the final output is a
>>>>>non-uniformly sampled version of the original analog signal ?
>>>>
>>>>It is uniformly time quantized, but not uniformly amplitude quantized.
>>>>
>>>
>>>I think you are wrong. It is uniformly amplitude quantized also.
>>
>>If you've already made up your mind as to what you believe about
>>the matter, why did you ask?
>
>
>
> Does it not occur to you that I might have understood the connection
> between dithering and sigma-delta after i posted my question here ?

Do you mean that you didn't really want or care about the content of an
answer? Surely, you don't declare yourself to be a troll!

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

[email protected] (kbc) writes:

> Randy Yates <[email protected]> wrote in message news:<[email protected]>...
>> [email protected] (kbc) writes:
>>
>
>> > > > Now, is it possible to get this same output using a normal adc
>> > > > for some sampling phase and uniform sampling of the analog signal ?
>> > >
>> > > No.
>> > >
>> > > > Or will it be correct to say that the final output is a
>> > > > non-uniformly sampled version of the original analog signal ?
>> > >
>> > > It is uniformly time quantized, but not uniformly amplitude quantized.
>> > >
>> >
>> > I think you are wrong. It is uniformly amplitude quantized also.
>>
>> If you've already made up your mind as to what you believe about
>> the matter, why did you ask?
>
>
> Does it not occur to you that I might have understood the connection
> between dithering and sigma-delta after i posted my question here ?

No, it did not occur to me because it is clear you do not understand. Delta
sigma quantization is not uniform amplitdue quantization. I attempted to
provide you with some insight on why this is so and you have responded
with rudeness. If you want to discuss it further with me, you will need
to modify your behavior by humbling yourself a bit and practicing a
modicum of politeness.
--
% Randy Yates % "Bird, on the wing,
%% Fuquay-Varina, NC % goes floating by
%%% 919-577-9882 % but there's a teardrop in his eye..."
%%%% <[email protected]> % 'One Summer Dream', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

> >
> > Does it not occur to you that I might have understood the connection
> > between dithering and sigma-delta after i posted my question here ?
>
> No, it did not occur to me because it is clear you do not understand. Delta
> sigma quantization is not uniform amplitdue quantization. I attempted to
> provide you with some insight on why this is so and you have responded
> with rudeness. If you want to discuss it further with me, you will need
> to modify your behavior by humbling yourself a bit and practicing a
> modicum of politeness.


I was not rude.

So, the output of the sigma delta adc cannot to be fed to a normal dac ??

How does a dac know how the amplitude quanti. has happened ?? Is there a
table kept ?

[email protected] (kbc) writes:

>> >
>> > Does it not occur to you that I might have understood the connection
>> > between dithering and sigma-delta after i posted my question here ?
>>
>> No, it did not occur to me because it is clear you do not understand. Delta
>> sigma quantization is not uniform amplitdue quantization. I attempted to
>> provide you with some insight on why this is so and you have responded
>> with rudeness. If you want to discuss it further with me, you will need
>> to modify your behavior by humbling yourself a bit and practicing a
>> modicum of politeness.
>
>
> I was not rude.

Well, it sure came across that way to me.

> So, the output of the sigma delta adc cannot to be fed to a normal dac ??

Yes, it can.

> How does a dac know how the amplitude quanti. has happened ?? Is there a
> table kept ?

It "knows" because of the shape of the quantization noise in the digital
signal. The quantization noise is not flat in the digital signal that
results from a delta sigma ADC, even when the input signal is "complex",
i.e., even when an input signal that would result in a flat quantization
noise from a uniform ADC is fed into the delta sigma ADC.

Logically you can put it this way. If a signal is uniformly quantized,
then the quantization noise spectrum is flat (given that the input
signal is sufficiently complex). The contrapositive of that is, If the
quantization noise spectrum is not flat, then the signal is not
uniformly quantized.
--
% Randy Yates % "Ticket to the moon, flight leaves here today
%% Fuquay-Varina, NC % from Satellite 2"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <[email protected]> % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr