> Randy Yates wrote:
>
> > On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
> > so the precise answer to your question depends on the impulse response
> > of that lowpass filter and the initial conditions of the system. The
> > output is the convolution of the lowpass filter inpulse response and
> > the sequence of 1's and 0's from the DSD bitstream.
>
> Randy, isn't that low pass filter really acting as an integrator, or
> analog counter?
Nope. An integrator would look like a low pass filter that only
passes DC unmodified. An brickwall audio filter would look more
like a sinc function. In reality, it would be something in-between
(in the digital domain perhaps some optimized FIR likely looking very
similar to a windowed sinc.)
IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M
Reply by Ron N.●January 26, 20062006-01-26
Carey Carlan wrote:
> Here's what I understand:
>
> Fact: DSD has a sampling rate of 1 bit * 64 * 44100 per second.
> Assumption: Each sample raises or lowers the volume one "bit", one unit.
> Conclusion: A 22.05 kHz tone gets 128 samples and has a peak to peak
> height of -32 to +31 to -32 again (6 bits) and a 44.1 kHz tone only -16 to
> +15.
Here's part of the fallacy in the above thinking. You assume
that the 22 kHz tone gets only 128 samples. That's sort of a
rectangular window in the time domain, which is a sinc window in
the frequency domain. In fact, what you want is something that
looks like a smoothed out rectangular window in the frequency
domain, which looks something like a windowed sinc in the
time domain... a windowed sinc a lot wider than 128 samples.
Say it's thousands of samples wide, but with weightings of other
than 1.0 for each sample (in fact some of the weightings of a
sinc function are negative.) This allows the volume to raise or
lower by thousands of units even at 22 kHz.
IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M
Reply by Jerry Avins●January 26, 20062006-01-26
Mark wrote:
...
> I think the DSD is analogous to a Class D amplifier switching at about
> 2 MHz.
Precisely. Well put.
> The low pass filter at the output is required to remove the
> very high frequency 2 MHz ripple but the cutoff is above the audio band
> so it does not behave as an integrator in the audio band and it does
> not create slew rate limitations in the audio band.
I think that frequencies lower than 2 MHz are introduced by asymmetries
in the chopped waveform needed to reproduce intermediate amplitudes and
frequencies. They remain high, however.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Mark●January 26, 20062006-01-26
William Sommerwerck wrote:
> >> On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
> >> so the precise answer to your question depends on the impulse response
> >> of that lowpass filter and the initial conditions of the system. The
> >> output is the convolution of the lowpass filter inpulse response and
> >> the sequence of 1's and 0's from the DSD bitstream.
>
> > Randy, isn't that low pass filter really acting as an integrator, or
> > analog counter?
>
> That's the way I see it.
>
> > It's from looking at it from this point of view that one can question
> > the dynamic range available at the higher frequencies before slew rate
> > limiting sets in due to the limited bit rate.
>
> Exactly. That was the question I was asking.
An integrator would be a low pass filter with a cutoff that starts at
say 20 Hz (yes I mean Hz not kHz) ) and has a 6 dB per octave slope all
through the audio band and beyond.
I think the low pass filter used for DSD the cutoff is well above 20
kHz (yes kHz not Hz) so there is no integration effect or slew rate
limiting effect in the audio band. Yes there may be slew rate
limitations above the cutoff of the filter but I expect that would be
above 100 kHz. The filter is required to remove the 2 MHz component
which a cutoff of 100 kHz will do without slew rate limiting in the
audio band.
I think the DSD is analogous to a Class D amplifier switching at about
2 MHz. The low pass filter at the output is required to remove the
very high frequency 2 MHz ripple but the cutoff is above the audio band
so it does not behave as an integrator in the audio band and it does
not create slew rate limitations in the audio band.
thanks for the very good discussion..
Mark
Reply by William Sommerwerck●January 26, 20062006-01-26
>> On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
>> so the precise answer to your question depends on the impulse response
>> of that lowpass filter and the initial conditions of the system. The
>> output is the convolution of the lowpass filter inpulse response and
>> the sequence of 1's and 0's from the DSD bitstream.
> Randy, isn't that low pass filter really acting as an integrator, or
> analog counter?
That's the way I see it.
> It's from looking at it from this point of view that one can question
> the dynamic range available at the higher frequencies before slew rate
> limiting sets in due to the limited bit rate.
Exactly. That was the question I was asking.
Reply by Bob Cain●January 26, 20062006-01-26
Jerry Avins wrote:
> All of that is characteristic of delta modulation, not sigma-delta
> conversion. It's an unfortunate but understandable confusion.
Aha! As always, thanks Jerry. Time to do some more reading.
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
Reply by Bob Cain●January 26, 20062006-01-26
Randy Yates wrote:
> On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
> so the precise answer to your question depends on the impulse response
> of that lowpass filter and the initial conditions of the system. The
> output is the convolution of the lowpass filter inpulse response and
> the sequence of 1's and 0's from the DSD bitstream.
Randy, isn't that low pass filter really acting as an integrator, or
analog counter?
It's from looking at it from this point of view that one can question
the dynamic range available at the higher frequencies before slew rate
limiting sets in due to the limited bit rate.
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
Reply by Jerry Avins●January 25, 20062006-01-25
Mark wrote:
> Randy Yates wrote:
>
>>Carey Carlan <gulfjoe@hotmail.com> writes:
>>
>>>[...]
>>>Randy Yates <yates@ieee.org> wrote in news:k6cofda8.fsf@ieee.org:
>>>
>>>>The reason for this has to do with the way DSD operates, and you'd
>>>>need to understand what a delta sigma modulator is in order to fully
>>>>explain. If you're interested, you can find some info on Wikipedia on
>>>>it. We'd be happy to answer more detailed questions here if you
>>>>desire.
>>>
>>>I read the Wikipedia article before coming here. Either I
>>>misunderstand or it's not complete enough for this non-intuitive
>>>thinker.
>>
>>Did you mean the Wikipedia article on delta sigma modulation
>>or the Wikipedia article on DSD? I meant the former,
>>
>> http://en.wikipedia.org/wiki/Sigma_delta
>>
>>and it is quite detailed, so I'm thinking you meant the latter,
>>which is very brief.
>>
>>
>>>The specific question is "On playback, how much change occurs in the
>>>signal for each of a series of '1' values.?"
>>
>>On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
>>so the precise answer to your question depends on the impulse response
>>of that lowpass filter and the initial conditions of the system. The
>>output is the convolution of the lowpass filter inpulse response and
>>the sequence of 1's and 0's from the DSD bitstream.
>>
>>
>>>If I can get an understandable answer to that, I think I can make it the
>>>rest of the way or else fuel another question.
>>
>>Carey, why are you asking? What is the higher-level information you
>>want to know?
>>--
>>% Randy Yates % "With time with what you've learned,
>>%% Fuquay-Varina, NC % they'll kiss the ground you walk
>>%%% 919-577-9882 % upon."
>>%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
>>http://home.earthlink.net/~yatescr
>
>
> Randy,
>
> I think the fundamental question is the belief being addressed is that
> the SACD DSD system is slew rate limited such that it cannot reproduce
> high frequency signal (5 kHz to 20 kHz) at full scale (16 bit) the way
> a straight PCM system can..
>
> this is based on the belief that the DSD system operates such that each
> bit indicates a 1 LSB increase or a 1 LSB decrease in amplitude
> relative to the present amplitude and at ~2 Mbps this leads to a rather
> limited slew rate.
>
> or another (incorrect) way of expressing it is the belief "that the
> DSD system frequency response is level dependent" and as the tone
> frequency goes up, the max level that can be reproduced goes down....
All of that is characteristic of delta modulation, not sigma-delta
conversion. It's an unfortunate but understandable confusion.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Mark●January 25, 20062006-01-25
Randy Yates wrote:
> Carey Carlan <gulfjoe@hotmail.com> writes:
> > [...]
> > Randy Yates <yates@ieee.org> wrote in news:k6cofda8.fsf@ieee.org:
> >> The reason for this has to do with the way DSD operates, and you'd
> >> need to understand what a delta sigma modulator is in order to fully
> >> explain. If you're interested, you can find some info on Wikipedia on
> >> it. We'd be happy to answer more detailed questions here if you
> >> desire.
> >
> > I read the Wikipedia article before coming here. Either I
> > misunderstand or it's not complete enough for this non-intuitive
> > thinker.
>
> Did you mean the Wikipedia article on delta sigma modulation
> or the Wikipedia article on DSD? I meant the former,
>
> http://en.wikipedia.org/wiki/Sigma_delta
>
> and it is quite detailed, so I'm thinking you meant the latter,
> which is very brief.
>
> > The specific question is "On playback, how much change occurs in the
> > signal for each of a series of '1' values.?"
>
> On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
> so the precise answer to your question depends on the impulse response
> of that lowpass filter and the initial conditions of the system. The
> output is the convolution of the lowpass filter inpulse response and
> the sequence of 1's and 0's from the DSD bitstream.
>
> > If I can get an understandable answer to that, I think I can make it the
> > rest of the way or else fuel another question.
>
> Carey, why are you asking? What is the higher-level information you
> want to know?
> --
> % Randy Yates % "With time with what you've learned,
> %% Fuquay-Varina, NC % they'll kiss the ground you walk
> %%% 919-577-9882 % upon."
> %%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
> http://home.earthlink.net/~yatescr
Randy,
I think the fundamental question is the belief being addressed is that
the SACD DSD system is slew rate limited such that it cannot reproduce
high frequency signal (5 kHz to 20 kHz) at full scale (16 bit) the way
a straight PCM system can..
this is based on the belief that the DSD system operates such that each
bit indicates a 1 LSB increase or a 1 LSB decrease in amplitude
relative to the present amplitude and at ~2 Mbps this leads to a rather
limited slew rate.
or another (incorrect) way of expressing it is the belief "that the
DSD system frequency response is level dependent" and as the tone
frequency goes up, the max level that can be reproduced goes down....
Mark
Reply by Randy Yates●January 25, 20062006-01-25
Carey Carlan <gulfjoe@hotmail.com> writes:
> [...]
> Randy Yates <yates@ieee.org> wrote in news:k6cofda8.fsf@ieee.org:
>> The reason for this has to do with the way DSD operates, and you'd
>> need to understand what a delta sigma modulator is in order to fully
>> explain. If you're interested, you can find some info on Wikipedia on
>> it. We'd be happy to answer more detailed questions here if you
>> desire.
>
> I read the Wikipedia article before coming here. Either I
> misunderstand or it's not complete enough for this non-intuitive
> thinker.
Did you mean the Wikipedia article on delta sigma modulation
or the Wikipedia article on DSD? I meant the former,
http://en.wikipedia.org/wiki/Sigma_delta
and it is quite detailed, so I'm thinking you meant the latter,
which is very brief.
> The specific question is "On playback, how much change occurs in the
> signal for each of a series of '1' values.?"
On playback, the 1-bit DSD signal is simply applied to a lowpass filter,
so the precise answer to your question depends on the impulse response
of that lowpass filter and the initial conditions of the system. The
output is the convolution of the lowpass filter inpulse response and
the sequence of 1's and 0's from the DSD bitstream.
> If I can get an understandable answer to that, I think I can make it the
> rest of the way or else fuel another question.
Carey, why are you asking? What is the higher-level information you
want to know?
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr