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Digital Volume Control

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Anonymous
June 8, 2005 1:50:13 AM

Archived from groups: rec.audio.tech (More info?)

I believe that a digital volume control (for example on a CD player)
sacrifices 1 bit of resolution for every 6dB of attenuation. This raises for
me a couple of questions.

If the original signal is 16/44 and it is upsampled to 24/96 before the
digital volume control, would 6dB of attenuation result in 15-bit resolution
or 23-bit resolution? Any explanation would be most helpful.

Wadia claim to have a digital volume control on their CD player that does
not reduce the bit resolution. The claim is that redundant information
(where?) is used for the volume control and so there is no loss of
resolution (as my workmate/Wadia-owner explained it). How does this work?

More about : digital volume control

Anonymous
June 8, 2005 1:50:14 AM

Archived from groups: rec.audio.tech (More info?)

On Tue, 7 Jun 2005 21:50:13 +0930, "nowater" <nowater@grantsellek.com>
wrote:

>I believe that a digital volume control (for example on a CD player)
>sacrifices 1 bit of resolution for every 6dB of attenuation.

Most digital VC's use bit shifting between the LSB (least significant bit)
and the MSB (most significant bit). The signal shifs towards the LSB for
attenuation, thus reducing resolution.

>If the original signal is 16/44 and it is upsampled to 24/96 before the
>digital volume control, would 6dB of attenuation result in 15-bit resolution
>or 23-bit resolution? Any explanation would be most helpful.

An upsampled signal doesn't contain more information than the original.

Some manufacturers claim that their interpolation schemes boost resolution
but what goes out isn't what goes in: information is *created* by the
system. A 24/96 system with an attenuation of 6 dB results in the loss of
one bit of resolution on the original signal.

>Wadia claim to have a digital volume control on their CD player that does
>not reduce the bit resolution. The claim is that redundant information
>(where?) is used for the volume control and so there is no loss of
>resolution (as my workmate/Wadia-owner explained it). How does this work?

Wadia uses 22+ bits converters coupled with bit shifting and interpolation.
According to them, a 36 dB attenuation doesn't result in the loss of
information (22 - 6 = 6 x 6). This is pure BS: signal bits are thrown away.

OTOH, you can design a sliding system based on a 32-bit processor coupled
with 24-bit converters: a typical 16-bit signal can thus be slided down by 8
bits, or 48 dB, w/o any significant loss (monotonicity et al. kick in when
it comes to the lower bits). This architecture is typical of pro gear or
well designed consumer equipment.
Anonymous
June 8, 2005 1:50:14 AM

Archived from groups: rec.audio.tech (More info?)

nowater wrote:

> I believe that a digital volume control (for example on a
CD player)
> sacrifices 1 bit of resolution for every 6dB of
attenuation.

Most analog volume controls also lose 6 dB (=1 bit) of
resolution for every 6 dB of attenuation, more or less.

> This raises for me a couple of questions.

> If the original signal is 16/44 and it is upsampled to
24/96 before
> the digital volume control, would 6dB of attenuation
result in 15-bit
> resolution or 23-bit resolution?

16 bits or less. Upsampling doesn't increase the resolution
of a signal, it just makes it take up more data space.

If you take a 16 bit signal and upsample it to 24 bits and
then attenuate it in the 24 bit domain, you can apply up to
8 bits of attenuation (about 46 dB) before you start losing
resolution in the digital attenuator. Thing is that the
signal doesn't stay in the digital domain, but instead
looses additional amounts of resolution in the conversion to
the analog domain, and other processes that follow.

The whole discussion lacks practical relevance since the
source material, prior to being put on the CD, has 75 or
less dB dynamic range due to noise upstream in the live
performance, initial recording process, and production
processing.

The commonly-stated bugabear in digital attenuation is
low-level nonlinear distortion, which does not actually
happen in a properly-dithered digital attenuator.

> Wadia claim to have a digital volume control on their CD
player that
> does not reduce the bit resolution.

IME, Wadia claim lots of things that are kinda strange,
and/or have limited to nonexistent practical benefits.

> The claim is that redundant
> information (where?) is used for the volume control and so
there is
> no loss of resolution (as my workmate/Wadia-owner
explained it).

You've got to be a true believer to believe a lot of what
Wadia says. The very act of paying Wadia prices for an
optical player that if perfect, would sonically
indistinguishable from a good $60 DVD player seems to take a
toll on the objectivity of Wadia equipment.

> How does this work?

See above. It can work in theory, but just in the digital
domain.
Related resources
Anonymous
June 8, 2005 1:50:14 AM

Archived from groups: rec.audio.tech (More info?)

"nowater" wrote ...
>I believe that a digital volume control (for example on a CD player)
> sacrifices 1 bit of resolution for every 6dB of attenuation. This
> raises for
> me a couple of questions.
>
> If the original signal is 16/44 and it is upsampled to 24/96 before
> the
> digital volume control, would 6dB of attenuation result in 15-bit
> resolution
> or 23-bit resolution? Any explanation would be most helpful.
>
> Wadia claim to have a digital volume control on their CD player that
> does
> not reduce the bit resolution. The claim is that redundant information
> (where?) is used for the volume control and so there is no loss of
> resolution (as my workmate/Wadia-owner explained it). How does this
> work?

When you "attenuate" the signal by 6dB, you aren't even *using*
that bit of resolution anymore. So "sacrafice" seems like a mis-
leading characterization. I reject the concept that attenuating a
signal "sacrafices" the digital resolution. It is a silly semantic
argument. Doesn't hold any water for me.
Anonymous
June 8, 2005 1:50:14 AM

Archived from groups: rec.audio.tech (More info?)

Arny Krueger wrote:
> nowater wrote:
> > snip...
> Most analog volume controls also lose 6 dB (=1 bit) of
> resolution for every 6 dB of attenuation, more or less.

OK I never thought about that ... kind of invalidates the argument
against digital VC!

> > This raises for me a couple of questions.
>
> > If the original signal is 16/44 and it is upsampled to
> 24/96 before
> > the digital volume control, would 6dB of attenuation
> result in 15-bit
> > resolution or 23-bit resolution?
>
> 16 bits or less. Upsampling doesn't increase the resolution
> of a signal, it just makes it take up more data space.

Whoops I knew that... should have worded my question more carefully
(see below).

> If you take a 16 bit signal and upsample it to 24 bits and
> then attenuate it in the 24 bit domain, you can apply up to
> 8 bits of attenuation (about 46 dB) before you start losing
> resolution in the digital attenuator.

Thanks FYLG and AK for helpful replies. I think this answers the
question I really meant, namely: "If 16/44 is upsampled (say to 24
bits) prior to digital attenuation, can we then apply up to 48 dB of
attenuation without losing any of the original 16 bits of resolution?"
The answer seems to be Yes.

This seems to be a neat solution to the "loss of bits" argument against
digital volume control. It is also what I have implemented at home
(16-bit CD => 24-bit upsampling => digital VC), one reason for my
original question, and I now feel reassured.

In fact, if I read AK correctly, it is a better solution than an analog
attenuator - any analog attenuator.

> Thing is that the
> signal doesn't stay in the digital domain, but instead
> looses additional amounts of resolution in the conversion to
> the analog domain, and other processes that follow.

Is this a caveat on the use of digital volume controls, or just a
general observation about digital-analog hybrid systems, with and
without digital volume control?

>
> The whole discussion lacks practical relevance since the
> source material, prior to being put on the CD, has 75 or
> less dB dynamic range due to noise upstream in the live
> performance, initial recording process, and production
> processing.

I disagree about lack of practical relevance. If you have a CD player
with internal digital volume control operating at 16 bits (like my
brother in law's Denon), and you connect it directly to a sensitive
high-power power amp, in order to get normal listening levels it may
be necessary to dial back say 48dB of digital attenuation, taking 8
bits off your 16-bit signal and only having 8 bits of resolution (48dB)
available.

If I read AK correctly, the above scenario only allows about 20 dB of
digital attenuation from the 16-bit (say 96dB resolution) before the
attenuator results in less resolution than the 75dB of dynamic range in
the recorded music signal prior to being put on CD.

> > Wadia claim to have a digital volume control on their CD
> > player that does not reduce the bit resolution.
<<snip>>
> > How does this work?
>
> See above. It can work in theory, but just in the digital
> domain.

OK so I now see that the Wadia system works as claimed. They use 21-bit
upsampling, so their digital attenuator would permit up to 30dB of
attenuation with full 16-bit signal resolution.

Thanks to posters for clarifying. Please correct my conclusions if I
misunderstand you.
Anonymous
June 8, 2005 1:50:15 AM

Archived from groups: rec.audio.tech (More info?)

François Yves Le Gal wrote:
> On Tue, 7 Jun 2005 21:50:13 +0930, "nowater"
<nowater@grantsellek.com>
> wrote:
>
>> I believe that a digital volume control (for example on a
CD player)
>> sacrifices 1 bit of resolution for every 6dB of
attenuation.
>
> Most digital VC's use bit shifting between the LSB (least
significant
> bit) and the MSB (most significant bit). The signal shifs
towards the
> LSB for attenuation, thus reducing resolution.
>
>> If the original signal is 16/44 and it is upsampled to
24/96 before
>> the digital volume control, would 6dB of attenuation
result in
>> 15-bit resolution or 23-bit resolution? Any explanation
would be
>> most helpful.
>
> An upsampled signal doesn't contain more information than
the
> original.
>
> Some manufacturers claim that their interpolation schemes
boost
> resolution but what goes out isn't what goes in:
information is
> *created* by the system. A 24/96 system with an
attenuation of 6 dB
> results in the loss of one bit of resolution on the
original signal.
>
>> Wadia claim to have a digital volume control on their CD
player that
>> does not reduce the bit resolution. The claim is that
redundant
>> information (where?) is used for the volume control and
so there is
>> no loss of resolution (as my workmate/Wadia-owner
explained it). How
>> does this work?
>
> Wadia uses 22+ bits converters coupled with bit shifting
and
> interpolation. According to them, a 36 dB attenuation
doesn't result
> in the loss of information (22 - 6 = 6 x 6). This is pure
BS: signal
> bits are thrown away.
>
> OTOH, you can design a sliding system based on a 32-bit
processor
> coupled with 24-bit converters: a typical 16-bit signal
can thus be
> slided down by 8 bits, or 48 dB, w/o any significant loss
> (monotonicity et al. kick in when it comes to the lower
bits). This
> architecture is typical of pro gear or well designed
consumer
> equipment.

Hi François,

I was looking at this very nice post of yours, wondering
what happened to the days when you and I used to fight like
cats and dogs? ;-)
Anonymous
June 8, 2005 1:50:16 AM

Archived from groups: rec.audio.tech (More info?)

On Tue, 7 Jun 2005 08:50:18 -0400, "Arny Krueger" <arnyk@hotpop.com> wrote:

>I was looking at this very nice post of yours, wondering
>what happened to the days when you and I used to fight like
>cats and dogs?

Well Arny, when you behave like a schmuck, I still treat you like a schmuck,
don't I?

BTW, despite what you may believe, I've never been on the wee-wee side of
audiophilia.

The difference between you and me is that you're still a Luddite somewhere,
stuck into, say, 16/44 and refusing to admit - despite numerous objective
tests - that this obsolete format doesn't allow for full signal resolution,
either in bandwidth or dynamics.

Ditto for amps: if models of similar approaches and topologies used within
their performance envelope essentially sound the same, there are some huge
measurable as well as perceptible differences between different amps. And so
on.
Anonymous
June 8, 2005 1:50:17 AM

Archived from groups: rec.audio.tech (More info?)

On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger" <arnyk@hotpop.com> wrote:

>I never said that 16/44 allows for full signal resolution,
>either in bandwidth or dynamics. Obviously it doesn't.

Ahem. here's just an example: "we find that 16/44 is a pretty good match to
the limits of the ears and typical recording (or even exceptional
recording ) and playback circumstances. Its even overkill. That means
that more resolution and higher sample rates are sonically moot."
<9qi0mi01le5@enews3.newsguy.com> in RAHE

Another one? OK: "The CD format makes it entirely possible to make a
recording that has just about any content that can be recorded and heard".
<lyUU9.477$PH6.24600791@newssvr15.news.prodigy.com> right here.

And so on.
Anonymous
June 8, 2005 1:50:18 AM

Archived from groups: rec.audio.tech (More info?)

"François Yves Le Gal" <flegal@aingeal.com> wrote in message
news:4ngba1pg34uv4qqj78cpm2ahqqb1j9qoom@4ax.com...
> On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger" <arnyk@hotpop.com>
> wrote:
>
>>I never said that 16/44 allows for full signal resolution,
>>either in bandwidth or dynamics. Obviously it doesn't.
>
> Ahem. here's just an example: "we find that 16/44 is a pretty good
> match to
> the limits of the ears and typical recording (or even exceptional
> recording ) and playback circumstances. Its even overkill. That means
> that more resolution and higher sample rates are sonically moot."
> <9qi0mi01le5@enews3.newsguy.com> in RAHE
>
> Another one? OK: "The CD format makes it entirely possible to make a
> recording that has just about any content that can be recorded and
> heard".
> <lyUU9.477$PH6.24600791@newssvr15.news.prodigy.com> right here.
>
> And so on.

Ahem. Perhaps you missed the part about "the question I like
to deal with is practical sufficiency." But if you like to argue
just for argument's sake, go for it. I'll go back to ignoring such
silly time-wasters.
Anonymous
June 8, 2005 1:50:18 AM

Archived from groups: rec.audio.tech (More info?)

François Yves Le Gal wrote:

> On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger"
<arnyk@hotpop.com>
> wrote:
>
>> I never said that 16/44 allows for full signal
resolution,
>> either in bandwidth or dynamics. Obviously it doesn't.
>> However, the question I like to deal with is practical
>> sufficiency.

> Ahem. here's just an example: "we find that 16/44 is a
pretty good
> match to the limits of the ears and typical recording (or
even
> exceptional recording ) and playback circumstances. Its
even
> overkill. That means
> that more resolution and higher sample rates are sonically
moot."
> <9qi0mi01le5@enews3.newsguy.com> in RAHE

> Another one? OK: "The CD format makes it entirely possible
to make a
> recording that has just about any content that can be
recorded and
> heard".
<lyUU9.477$PH6.24600791@newssvr15.news.prodigy.com> right
> here.

> And so on.

Easily explained by the sentence about practical
sufficiency.

Where's the beef?
Anonymous
June 8, 2005 1:50:19 AM

Archived from groups: rec.audio.tech (More info?)

Richard Crowley wrote:
> "François Yves Le Gal" <flegal@aingeal.com> wrote in
message
> news:4ngba1pg34uv4qqj78cpm2ahqqb1j9qoom@4ax.com...
>> On Tue, 7 Jun 2005 09:54:03 -0400, "Arny Krueger"
<arnyk@hotpop.com>
>> wrote:
>>
>>> I never said that 16/44 allows for full signal
resolution,
>>> either in bandwidth or dynamics. Obviously it doesn't.
>>> However, the question I like to deal with is practical
>>> sufficiency.

>> Ahem. here's just an example: "we find that 16/44 is a
pretty good
>> match to
>> the limits of the ears and typical recording (or even
exceptional
>> recording ) and playback circumstances. Its even
overkill. That means
>> that more resolution and higher sample rates are
sonically moot."
>> <9qi0mi01le5@enews3.newsguy.com> in RAHE
>>
>> Another one? OK: "The CD format makes it entirely
possible to make a
>> recording that has just about any content that can be
recorded and
>> heard".
>> <lyUU9.477$PH6.24600791@newssvr15.news.prodigy.com> right
here.
>>
>> And so on.

> Ahem. Perhaps you missed the part about "the question I
like
> to deal with is practical sufficiency."

Since it was in my OP, seems like putting that sentence back
in to François' quote is reasonable.

> But if you like to argue
> just for argument's sake, go for it. I'll go back to
ignoring such
> silly time-wasters.

Agreed.
Anonymous
June 8, 2005 1:50:19 AM

Archived from groups: rec.audio.tech (More info?)

On Tue, 7 Jun 2005 09:13:39 -0700, "Richard Crowley" <rcrowley7@xprt.net>
wrote:

>I'll go back to ignoring such
>silly time-wasters.

Sure, whatever you say.
Anonymous
June 8, 2005 1:50:19 AM

Archived from groups: rec.audio.tech (More info?)

On Tue, 7 Jun 2005 12:21:32 -0400, "Arny Krueger" <arnyk@hotpop.com> wrote:

>Easily explained by the sentence about practical
>sufficiency.

What did I say? The schmuck is back...
Anonymous
June 8, 2005 6:39:56 AM

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> writes:
> [...]
> If you take a 16 bit signal and upsample it to 24 bits and
> then attenuate it in the 24 bit domain, you can apply up to
> 8 bits of attenuation (about 46 dB) before you start losing
> resolution in the digital attenuator.

Wrong.

Even if the signal stays 24 bits all the way to the DAC, this is not
correct. You lose resolution immediately. If you had a true 24-bit
DAC, you'd have a usable SNR of about 144 dB relative to a full-scale
level. When you attentuate X dB, you shave X dB off that SNR just as
you would with a 16-bit DAC or a 9-bit DAC or whatever.

The only advantage is that, with a high-resolution DAC, you can afford
to throw away some of your SNR, so in that sense it does make a
digital volume control more viable.

What seems to be going on in your mind (and others) is that if you
don't "lose any of the original bits" you don't lose anything. That
is not correct since the noise floor of the DAC stays constant,
and thus any attenuation of the maximum signal power degrades
the SNR of the signal.
--
% Randy Yates % "My Shangri-la has gone away, fading like
%% Fuquay-Varina, NC % the Beatles on 'Hey Jude'"
%%% 919-577-9882 %
%%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO
http://home.earthlink.net/~yatescr
Anonymous
June 8, 2005 7:53:21 AM

Archived from groups: rec.audio.tech (More info?)

nowater wrote:

> Arny Krueger wrote:

>> nowater wrote:

>> Most analog volume controls also lose 6 dB (=1 bit) of
>> resolution for every 6 dB of attenuation, more or less.

> OK I never thought about that ... kind of invalidates the
argument
> against digital VC!

>>> This raises for me a couple of questions.

>>> If the original signal is 16/44 and it is upsampled to
>> 24/96 before
>>> the digital volume control, would 6dB of attenuation
>> result in 15-bit
>>> resolution or 23-bit resolution?

>> 16 bits or less. Upsampling doesn't increase the
resolution
>> of a signal, it just makes it take up more data space.

> Whoops I knew that... should have worded my question more
carefully
> (see below).

>> If you take a 16 bit signal and upsample it to 24 bits
and
>> then attenuate it in the 24 bit domain, you can apply up
to
>> 8 bits of attenuation (about 46 dB) before you start
losing
>> resolution in the digital attenuator.

> Thanks FYLG and AK for helpful replies. I think this
answers the
> question I really meant, namely: "If 16/44 is upsampled
(say to 24
> bits) prior to digital attenuation, can we then apply up
to 48 dB of
> attenuation without losing any of the original 16 bits of
resolution?"
> The answer seems to be Yes.

Agreed.

> This seems to be a neat solution to the "loss of bits"
argument
> against digital volume control. It is also what I have
implemented at
> home (16-bit CD => 24-bit upsampling => digital VC), one
reason for my
> original question, and I now feel reassured.

> In fact, if I read AK correctly, it is a better solution
than an
> analog attenuator - any analog attenuator.

....other than the fact that eventually the output of any
practical digital volume control will have to be converted
to analog, which tends to wipe out the superiority of the
digital volume control over the analog volume control.

>> Thing is that the
>> signal doesn't stay in the digital domain, but instead
>> looses additional amounts of resolution in the conversion
to
>> the analog domain, and other processes that follow.

> Is this a caveat on the use of digital volume controls, or
just a
> general observation about digital-analog hybrid systems,
with and
> without digital volume control?

It's just a general observation about digital-analog hybrid
systems.

>> The whole discussion lacks practical relevance since the
>> source material, prior to being put on the CD, has 75 or
>> less dB dynamic range due to noise upstream in the live
>> performance, initial recording process, and production
>> processing.

> I disagree about lack of practical relevance. If you have
a CD player
> with internal digital volume control operating at 16 bits
(like my
> brother in law's Denon), and you connect it directly to a
sensitive
> high-power power amp, in order to get normal listening
levels it may
> be necessary to dial back say 48dB of digital attenuation,
taking 8
> bits off your 16-bit signal and only having 8 bits of
resolution
> (48dB) available.

If you connect a volume control to a highly sensitive power
amp (e.g., one whose input voltage spec is significantly
less than the output voltage spec of the DAC) both good
practice and the specifics of this application suggest that
you first attenuate the output of the DAC with an
appropriate analog attenuator.

The input gain control of the power amp may be a readily
availalable tool for the purpose of matching its sensitivity
to the maximum output of the DAC. Thus, the full resolution
of the DAC is available for the purpose at hand, which is
providing a controllable attenuator that exploits all of the
hardware at hand.

> If I read AK correctly, the above scenario only allows
about 20 dB of
> digital attenuation from the 16-bit (say 96dB resolution)
before the
> attenuator results in less resolution than the 75dB of
dynamic range
> in the recorded music signal prior to being put on CD.

As long as this 20 dB of attenuation by the digital
attenuator results in 20 dB attenuation of the final SPL,
there is no problem. This will happen if any excess gain of
the power amp is essentually absorbed by an analog
attenuator between the DAC and the power amp.

The noise floor of most power amps is around 80-95 dB below
their maximum output. If the full output of a 16 bit DAC is
just barely sufficient to drive the power amp to full
output, then any residual noise at the output of the DAC
will be at, near, or below the residual noise of the power
amp.

There are exceptional power amps with residual noise on the
order of 110 dB below full output. To fully exploit such a
power amp, a digital attenuator with > 16 bit resolution
will be required. However, room noise etc, may allow this
requirement to be relaxed. For example, a typical listening
room will have residual acoustical noise on the order of
35-45 dB. 96 dB over this is 131-151 dB SPL which is at or
beyond the threshold of pain. A real-world audio system will
probably need to develop no more than 120 dB SPL, with 115
dB being a more practical maximum. A 16 bit (96 dB) digital
attenuator will have residual noise on the order of 19 to 24
dB SPL which will only be noticable in a typical residential
listening room if you put your head close to the speakers
and don't actually play a recording.
Anonymous
June 8, 2005 7:58:49 AM

Archived from groups: rec.audio.tech (More info?)

Randy Yates wrote:
> "Arny Krueger" <arnyk@hotpop.com> writes:
>> [...]
>> If you take a 16 bit signal and upsample it to 24 bits
and
>> then attenuate it in the 24 bit domain, you can apply up
to
>> 8 bits of attenuation (about 46 dB) before you start
losing
>> resolution in the digital attenuator.
>
> Wrong.
>
> Even if the signal stays 24 bits all the way to the DAC,
this is not
> correct. You lose resolution immediately. If you had a
true 24-bit
> DAC, you'd have a usable SNR of about 144 dB relative to a
full-scale
> level.

No you wouldn't. The dynamic range of the input signal,
which comes from a 16 bit CD, limits the total usable
dynamic range of the system to about 96 dB, assuming a
perfectly noiseless power amp.
Anonymous
June 9, 2005 4:17:53 AM

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> writes:

> Randy Yates wrote:
>> "Arny Krueger" <arnyk@hotpop.com> writes:
>>> [...]
>>> If you take a 16 bit signal and upsample it to 24 bits
> and
>>> then attenuate it in the 24 bit domain, you can apply up
> to
>>> 8 bits of attenuation (about 46 dB) before you start
> losing
>>> resolution in the digital attenuator.
>>
>> Wrong.
>>
>> Even if the signal stays 24 bits all the way to the DAC,
> this is not
>> correct. You lose resolution immediately. If you had a
> true 24-bit
>> DAC, you'd have a usable SNR of about 144 dB relative to a
> full-scale
>> level.
>
> No you wouldn't. The dynamic range of the input signal,
> which comes from a 16 bit CD, limits the total usable
> dynamic range of the system to about 96 dB, assuming a
> perfectly noiseless power amp.

If the dynamic range of the system was 96 dB, then the attenuator
would degrade it to <96 dB.

I see your view. It's wrong.

The 16-bit CD signal is represented by the four-digit hexadecimal
number 0xklmn. This is converted to a 24-bit number "left justified"
so that it becomes 0xklmn00. The attenuator operates on that input and
produces a value 0xopqrst. That value is free to range over almost 144
dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss
of 0.000133 dB from a true 24-bit system).

Another way to look at it is this. The attenuation procedure computes
the following product: a*x, where "x" is the 16-bit input signal and
"a" is a positive, real number between 0 and 1. Since "a" has infinite
precision, then so does a*x. The final "noise" in the system is
related to the number of bits to which the product a*x is quantized.
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% <yates@ieee.org> % 'Waterfall', *Face The Music*, ELO
http://home.earthlink.net/~yatescr
Anonymous
June 9, 2005 4:17:54 AM

Archived from groups: rec.audio.tech (More info?)

"Randy Yates" wrote ...
> The 16-bit CD signal is represented by the four-digit hexadecimal
> number 0xklmn. This is converted to a 24-bit number "left justified"
> so that it becomes 0xklmn00. The attenuator operates on that input and
> produces a value 0xopqrst. That value is free to range over almost 144
> dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss
> of 0.000133 dB from a true 24-bit system).

What's the point in the real world? You can't add dynamic range merely
by shifting a 16-bit sample into a 24-bit word. You act as if there were
some magic imputed by converting to 24-bit before attenuating.
Anonymous
June 9, 2005 4:17:55 AM

Archived from groups: rec.audio.tech (More info?)

A point that folks seem to be missing:

The point of 24-bit sampling is HEADROOM. It allows you to turn up the
input and get a reliable 16-or-so bits without having the signal clip
unexpectedly if your input levels were too high. It allows you to mix
multiple 16-bit-or-so samples without the cumulative signal becoming too
loud to handle, and without having to attenuate them prematurely and
lose part of the original signal before you have to.


More bits can be useful IF you have a specific use for them. But "go
faster stripes" on a car don't really make it go faster, and extra bits
for their own sake don't necessarily buy you anything.
Anonymous
June 9, 2005 4:17:55 AM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <richard.7.crowley@intel.com> wrote in
message news:D 882vc$g32$1@news01.intel.com...
> "Randy Yates" wrote ...
> > The 16-bit CD signal is represented by the four-digit
hexadecimal
> > number 0xklmn. This is converted to a 24-bit number
"left justified"
> > so that it becomes 0xklmn00. The attenuator operates on
that input and
> > produces a value 0xopqrst. That value is free to range
over almost 144
> > dB of a 24-bit system (except for the 8 LSBs, which
amounts to a loss
> > of 0.000133 dB from a true 24-bit system).
>
> What's the point in the real world? You can't add dynamic
range merely
> by shifting a 16-bit sample into a 24-bit word.

You know that, I know that but somehow this has escaped the
attention of Mr. yates.

>You act as if there were
> some magic imputed by converting to 24-bit before
attenuating.

Indeed.
Anonymous
June 9, 2005 4:17:56 AM

Archived from groups: rec.audio.tech (More info?)

"Joe Kesselman" wrote ...
>A point that folks seem to be missing:
>
> The point of 24-bit sampling is HEADROOM.

Only if the original sample was 24-bits. If the original was 16-bit,
then none of what follows makes any sense....

> It allows you to turn up the input and get a reliable 16-or-so bits
> without having the signal clip unexpectedly if your input levels were
> too high. It allows you to mix multiple 16-bit-or-so samples without
> the cumulative signal becoming too loud to handle, and without having
> to attenuate them prematurely and lose part of the original signal
> before you have to.
>
>
> More bits can be useful IF you have a specific use for them. But "go
> faster stripes" on a car don't really make it go faster, and extra
> bits for their own sake don't necessarily buy you anything.
Anonymous
June 9, 2005 4:22:24 AM

Archived from groups: rec.audio.tech (More info?)

François Yves Le Gal <flegal@aingeal.com> writes:

> The difference between you and me is that you're still a Luddite somewhere,
> stuck into, say, 16/44 and refusing to admit - despite numerous objective
> tests - that this obsolete format doesn't allow for full signal resolution,
> either in bandwidth or dynamics.

Can you cite or comment on these "numerous objective tests"?
--
% Randy Yates % "My Shangri-la has gone away, fading like
%% Fuquay-Varina, NC % the Beatles on 'Hey Jude'"
%%% 919-577-9882 %
%%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO
http://home.earthlink.net/~yatescr
Anonymous
June 9, 2005 6:53:51 AM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <richard.7.crowley@intel.com> writes:

> "Randy Yates" wrote ...
>> The 16-bit CD signal is represented by the four-digit hexadecimal
>> number 0xklmn. This is converted to a 24-bit number "left justified"
>> so that it becomes 0xklmn00. The attenuator operates on that input and
>> produces a value 0xopqrst. That value is free to range over almost 144
>> dB of a 24-bit system (except for the 8 LSBs, which amounts to a loss
>> of 0.000133 dB from a true 24-bit system).
>
> What's the point in the real world? You can't add dynamic range merely
> by shifting a 16-bit sample into a 24-bit word.

That much is true. However, the operation you describe is only the
first of several that occur in a digital attenuator, and it is
the subsequent operations that provide the extra dynamic range.

> You act as if there were some magic imputed by converting to 24-bit
> before attenuating.

If you consider a little arithmetic magic, then I guess so.
--
% Randy Yates % "My Shangri-la has gone away, fading like
%% Fuquay-Varina, NC % the Beatles on 'Hey Jude'"
%%% 919-577-9882 %
%%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO
http://home.earthlink.net/~yatescr
Anonymous
June 9, 2005 6:53:52 AM

Archived from groups: rec.audio.tech (More info?)

"Randy Yates" <yates@ieee.org> wrote in message
news:is0odyv4.fsf@ieee.org...
> "Richard Crowley" <richard.7.crowley@intel.com> writes:
>
>> "Randy Yates" wrote ...
>>> The 16-bit CD signal is represented by the four-digit hexadecimal
>>> number 0xklmn. This is converted to a 24-bit number "left justified"
>>> so that it becomes 0xklmn00. The attenuator operates on that input
>>> and
>>> produces a value 0xopqrst. That value is free to range over almost
>>> 144
>>> dB of a 24-bit system (except for the 8 LSBs, which amounts to a
>>> loss
>>> of 0.000133 dB from a true 24-bit system).
>>
>> What's the point in the real world? You can't add dynamic range
>> merely
>> by shifting a 16-bit sample into a 24-bit word.
>
> That much is true. However, the operation you describe is only the
> first of several that occur in a digital attenuator, and it is
> the subsequent operations that provide the extra dynamic range.
>
>> You act as if there were some magic imputed by converting to 24-bit
>> before attenuating.
>
> If you consider a little arithmetic magic, then I guess so.

Unless you can actually show how you think this works, sounds
like you're just relying on magic that you don't really understand.
Anonymous
June 9, 2005 12:11:38 PM

Archived from groups: rec.audio.tech (More info?)

> Unless you can actually show how you think this works, sounds
> like you're just relying on magic that you don't really understand.

See other thread. No magic, just leaving room for accumulation and error.
Anonymous
June 9, 2005 12:19:48 PM

Archived from groups: rec.audio.tech (More info?)

"Joe Kesselman" <keshlam-nospam@comcast.net> wrote in message
news:ZPSdnaQFb_VsrDXfRVn-1g@comcast.com...
> > Unless you can actually show how you think this works, sounds
> > like you're just relying on magic that you don't really understand.
>
> See other thread. No magic, just leaving room for accumulation and error.

In an attenuator? (Refer to the subject line if you have lost track of
the discussion.)
Anonymous
June 9, 2005 7:30:37 PM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <rcrowley7@xprt.net> wrote in message
news:11afer3a4ioabe1@corp.supernews.com...
> "Joe Kesselman" wrote ...
> > The point of 24-bit sampling is HEADROOM.
>
> Only if the original sample was 24-bits. If the original was 16-bit,
> then none of what follows makes any sense....


I suggest you look up the meaning of "headroom".

MrT.
Anonymous
June 9, 2005 7:30:38 PM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" wrote ...
>
> "Richard Crowley" wrote ...
>> "Joe Kesselman" wrote ...
>> > The point of 24-bit sampling is HEADROOM.
>>
>> Only if the original sample was 24-bits. If the original was 16-bit,
>> then none of what follows makes any sense....
>
>
> I suggest you look up the meaning of "headroom".

And I suggest you look up the meaning of "magic".
Unless you have devised some method of recreating the
missing 8 bits of data it doesn't matter WHERE you put
the 16-bits of real data within the 24-bit word. You've
got only 16-bits of dynamic range. Period.
Anonymous
June 9, 2005 7:30:38 PM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" <MrT@home> wrote in message
news:42a7d407$0$30957$afc38c87@news.optusnet.com.au...
>
> "Richard Crowley" <rcrowley7@xprt.net> wrote in message
> news:11afer3a4ioabe1@corp.supernews.com...
> > "Joe Kesselman" wrote ...
> > > The point of 24-bit sampling is HEADROOM.
> >
> > Only if the original sample was 24-bits. If the original
was 16-bit,
> > then none of what follows makes any sense....
>
>
> I suggest you look up the meaning of "headroom".

Your typical 16 -> 24 bit conversion adds no headroom.
Anonymous
June 9, 2005 9:12:36 PM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <rcrowley7@xprt.net> wrote in message
news:11afp2ek8rjbo41@corp.supernews.com...
> "Mr.T" wrote ...
> > "Richard Crowley" wrote ...
> >> "Joe Kesselman" wrote ...
> >> > The point of 24-bit sampling is HEADROOM.
> >>
> >> Only if the original sample was 24-bits. If the original was 16-bit,
> >> then none of what follows makes any sense....
> >
> > I suggest you look up the meaning of "headroom".
>
> And I suggest you look up the meaning of "magic".

I see no mention of that in *my* post?

> Unless you have devised some method of recreating the
> missing 8 bits of data it doesn't matter WHERE you put
> the 16-bits of real data within the 24-bit word. You've
> got only 16-bits of dynamic range. Period.

Exactly, and 24 bits gives you HEADROOM to avoid clipping. I assume you have
never done any *real* recording.

MrT.
Anonymous
June 9, 2005 9:12:37 PM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" wrote ...
> "Richard Crowley" wrote ...
>> "Mr.T" wrote ...
>> > "Richard Crowley" wrote ...
>> >> "Joe Kesselman" wrote ...
>> >> > The point of 24-bit sampling is HEADROOM.
>> >>
>> >> Only if the original sample was 24-bits. If the original was
>> >> 16-bit,
>> >> then none of what follows makes any sense....
>> >
>> > I suggest you look up the meaning of "headroom".
>>
>> And I suggest you look up the meaning of "magic".
>
> I see no mention of that in *my* post?

Your methodology of thinking that putting 16-bits worth
of data into a 24-bit word somehow gives you extra
"headroom" can only be described as "magic". There
is cerainly no technical explanation for it. At least you
and your magical 24-bit friends have offered none.

>> Unless you have devised some method of recreating the
>> missing 8 bits of data it doesn't matter WHERE you put
>> the 16-bits of real data within the 24-bit word. You've
>> got only 16-bits of dynamic range. Period.
>
> Exactly, and 24 bits gives you HEADROOM to avoid
> clipping. I assume you have never done any *real*
> recording.

I've been recording long before digital was invented.
I have recorded hundreds (thousands?) of hours back in
the analog era, and hundreds more in 16-bit. I now
record mostly 24-bit and also produce digital video.
I know not only the analog implications of headroom,
but also practice both digital and analog circuit design
as well as writing code for everything from micro-
controllers to supercomputers. I have possibly been
recording since before you were born and have been
involved in the digital world for many years before it
was ever practical to apply to audio.

Unless you can explain how these empty 8 bits expand
your "headroom" you appear to be talking through your
hat.
Anonymous
June 9, 2005 9:12:37 PM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" <MrT@home> wrote in message
news:42a7ebee$0$29336$afc38c87@news.optusnet.com.au...
>
> "Richard Crowley" <rcrowley7@xprt.net> wrote in message
> news:11afp2ek8rjbo41@corp.supernews.com...
> > "Mr.T" wrote ...
> > > "Richard Crowley" wrote ...
> > >> "Joe Kesselman" wrote ...
> > >> > The point of 24-bit sampling is HEADROOM.
> > >>
> > >> Only if the original sample was 24-bits. If the
original was 16-bit,
> > >> then none of what follows makes any sense....
> > >
> > > I suggest you look up the meaning of "headroom".
> >
> > And I suggest you look up the meaning of "magic".
>
> I see no mention of that in *my* post?
>
> > Unless you have devised some method of recreating the
> > missing 8 bits of data it doesn't matter WHERE you put
> > the 16-bits of real data within the 24-bit word. You've
> > got only 16-bits of dynamic range. Period.
>
> Exactly, and 24 bits gives you HEADROOM to avoid clipping.
I assume you have
> never done any *real* recording.

Repeat: Your typical 16-> 24 bit conversion adds no
headroom.

In a digital volume control, what matters is more like "foot
room".
Anonymous
June 9, 2005 9:12:38 PM

Archived from groups: rec.audio.tech (More info?)

Richard Crowley wrote:
> Your methodology of thinking that putting 16-bits worth
> of data into a 24-bit word somehow gives you extra
> "headroom" can only be described as "magic".

I think we're still talking at cross purposes and applying different
assumptions about what's going to be done with the data.


If all you're doing is record-and-playback, then of course you can't
play back more information than you originally recorded.

BUT: even assuming that you have 16-bit starting samples, as soon as you
start mixing them together you either need more bits or you lose
information. Add two waves that peak at the top of the 16-bit range, and
you either need another bit or you lose information off the bottom. This
*is* a headroom consideration -- without that headroom, your mixer
becomes lossy.

No magic, just engineering.

(And as I pointed out, originally recording at 24 increases the odds of
your getting a good reliable 16-meaningful. Actually, I really like
Sony's (?) patent for floating-point digitization, but I haven't seen
any affordable hardware or software which uses it yet. I'm still kicking
myself for not having filed that patent when I first came up with the idea!)
Anonymous
June 9, 2005 10:07:48 PM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <rcrowley7@xprt.net> wrote in message
news:11afrsddd6iih62@corp.supernews.com...
> >> > I suggest you look up the meaning of "headroom".
> >>
> >> And I suggest you look up the meaning of "magic".
> >
> > I see no mention of that in *my* post?
>
> Your methodology of thinking that putting 16-bits worth
> of data into a 24-bit word somehow gives you extra
> "headroom" can only be described as "magic".

Not, that's exactly what headroom is for when doing REAL recording. One
doesn't know exactly what the peak levels will be ahead of time.
A couple of bits of headroom is a nice thing to have.
(assumes the hardware can achieve better than 16 bits of course, but that's
common these days)

>There
> is cerainly no technical explanation for it.

Yes there is. Look up recording headroom like I told you.

>At least you
> and your magical 24-bit friends have offered none.

You simply refuse to look.

> > Exactly, and 24 bits gives you HEADROOM to avoid
> > clipping. I assume you have never done any *real*
> > recording.
>
> I've been recording long before digital was invented.
> I have recorded hundreds (thousands?) of hours back in
> the analog era, and hundreds more in 16-bit. I now
> record mostly 24-bit and also produce digital video.
> I know not only the analog implications of headroom,
> but also practice both digital and analog circuit design
> as well as writing code for everything from micro-
> controllers to supercomputers. I have possibly been
> recording since before you were born and have been
> involved in the digital world for many years before it
> was ever practical to apply to audio.

All that so called experience and you still have no idea what headroom is.
Hint: its the *unused* part of the recorders dynamic range *above* the
actual signal peak.
Simply allows you to avoid clipping when you don't know ahead of time what
the peak level will be.

It's obvious you are too busy trying to big-note yourself to see what *I*
have actually said.

> Unless you can explain how these empty 8 bits expand
> your "headroom" you appear to be talking through your
> hat.

It's quite simple, IF the recorder has less than 16 bits resolution, then
you achieve nothing by going to 24 bits.
*IF* your recorder *CAN* achieve better than 16 bits (17, 18 bits etc) then
the extra bits can be used for recording HEADROOM.
IF *you* don't require the extra headroom, fine.

MrT.
Anonymous
June 9, 2005 10:07:49 PM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" <MrT@home> wrote in message
news:42a7f8de$0$30960$afc38c87@news.optusnet.com.au...

> It's quite simple, IF the recorder has less than 16 bits
resolution, then
> you achieve nothing by going to 24 bits.

Not necessarily. You can achieve what I called "foot room"
which is relevant to digital attenuators.

> *IF* your recorder *CAN* achieve better than 16 bits (17,
18 bits etc) then
> the extra bits can be used for recording HEADROOM.

Only if you turn the levels down.

> IF *you* don't require the extra headroom, fine.

What you always get with a 16->24 bit conversion is the
potential for better dynamic range in follow-on processing.
Anonymous
June 9, 2005 10:07:49 PM

Archived from groups: rec.audio.tech (More info?)

> *IF* your recorder *CAN* achieve better than 16 bits (17, 18 bits etc) then
> the extra bits can be used for recording HEADROOM.

Bingo. And generally that's a better use for the extra bits -- in the
recording stage -- than turning up the preamp and putting the extra bits
on the bottom would be.

I really think we've got two separate arguments going on -- whether more
than 16 bits buys you anything significant in finished output, and
whether it has practical uses before that point.

I think the answers are "probably not" and "hell yes". The latter's
unquestionable and easily demonstrable. The former's an audiophile
argument and probably impossible to separate from placebo effect and
opinion.
Anonymous
June 9, 2005 10:07:49 PM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" <MrT@home> wrote in message
news:42a7f8de$0$30960$afc38c87@news.optusnet.com.au...
>
> "Richard Crowley" <rcrowley7@xprt.net> wrote in message
> news:11afrsddd6iih62@corp.supernews.com...
> > >> > I suggest you look up the meaning of "headroom".
> > >>
> > >> And I suggest you look up the meaning of "magic".
> > >
> > > I see no mention of that in *my* post?
> >
> > Your methodology of thinking that putting 16-bits worth
> > of data into a 24-bit word somehow gives you extra
> > "headroom" can only be described as "magic".
>
> Not, that's exactly what headroom is for when doing REAL recording. One
> doesn't know exactly what the peak levels will be ahead of time.
> A couple of bits of headroom is a nice thing to have.
> (assumes the hardware can achieve better than 16 bits of course, but
that's
> common these days)

We agree completely that recording in 24 bits is highly desirable
and will most certainly increase the dynamic range i.e. provide
significant headroom.

What you fail to realize is that adding 8 more bits to a pre-recorded
16 bit sample buys you NOTHING.
Anonymous
June 9, 2005 11:27:27 PM

Archived from groups: rec.audio.tech (More info?)

On Thu, 9 Jun 2005 08:18:58 -0700, "Richard Crowley"
<richard.7.crowley@intel.com> wrote:

>What you fail to realize is that adding 8 more bits to a pre-recorded
>16 bit sample buys you NOTHING.

You get 8 more bits, so for instance you can slide down your signal, or edit
it without any loss.
Anonymous
June 9, 2005 11:27:28 PM

Archived from groups: rec.audio.tech (More info?)

"François Yves Le Gal" wrote...
> "Richard Crowley" wrote:
>>What you fail to realize is that adding 8 more bits to a pre-
>> recorded 16 bit sample buys you NOTHING.
>
> You get 8 more bits, so for instance you can slide down
> your signal, or edit it without any loss.

If you want to discuss editing and other kinds of processing,
then start a new conversation. Please refer back to the subject
line of THIS conversation and remember that we are talking
about a "Digital Volume Control" (i.e. attenuator).

By definition, attenuators do nothing except REDUCE dynamic
range (and REDUCE the number of active bits). Any expansion
of word-width to accomplish *attenuation* is just ignorant of the
principles.
Anonymous
June 10, 2005 3:17:55 AM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <rcrowley7@xprt.net> writes:

> "Randy Yates" <yates@ieee.org> wrote in message
> news:is0odyv4.fsf@ieee.org...
>> "Richard Crowley" <richard.7.crowley@intel.com> writes:
>>
>>> "Randy Yates" wrote ...
>>>> The 16-bit CD signal is represented by the four-digit hexadecimal
>>>> number 0xklmn. This is converted to a 24-bit number "left justified"
>>>> so that it becomes 0xklmn00. The attenuator operates on that input
>>>> and
>>>> produces a value 0xopqrst. That value is free to range over almost
>>>> 144
>>>> dB of a 24-bit system (except for the 8 LSBs, which amounts to a
>>>> loss
>>>> of 0.000133 dB from a true 24-bit system).
>>>
>>> What's the point in the real world? You can't add dynamic range
>>> merely
>>> by shifting a 16-bit sample into a 24-bit word.
>>
>> That much is true. However, the operation you describe is only the
>> first of several that occur in a digital attenuator, and it is
>> the subsequent operations that provide the extra dynamic range.
>>
>>> You act as if there were some magic imputed by converting to 24-bit
>>> before attenuating.
>>
>> If you consider a little arithmetic magic, then I guess so.
>
> Unless you can actually show how you think this works, sounds
> like you're just relying on magic that you don't really understand.

Sure. Define dynamic range (DR) as the ratio of the loudest sound to
the softest sound, or, in decibels, the difference between the loudest
sound and softest sound. Let a 16-bit system have a loudest sound of L16
dB and a softest sound of S16 dB, so that its dynamic range DR16 is

DR16 = L16 - S16 [dB]
= 96 [db]. (note 1)

If we input the loudest sound at L16 dB into a "24-bit left-justified"
digital volume control at a gain of 0 dB, we have at the output of the
digital volume control a signal level LDV dB, where LDV = L16.

The "24-bit left-justified" digital volume control can implement a
maximum attenuation of about 48 dB perfectly (i.e., without
introducing any extra noise) by right shifting the left-justified
input by 8 bits. Thus if we input the softest 16-bit sound at S16 dB
into the device with the gain at this setting, the output level is

SDV = S16 - 48.

Then the dynamic range of the digital volume control, DRDV, is

DRDV = LDV - SDV
= L16 - (S16 - 48)
= L16 - S16 + 48
= 96 + 48
= 144 [dB],

which was to be shown.

--RY

Notes:
1. The dynamic range isn't EXACTLY 96 dB but we consider it so
in this example.
--
% Randy Yates % "Midnight, on the water...
%% Fuquay-Varina, NC % I saw... the ocean's daughter."
%%% 919-577-9882 % 'Can't Get It Out Of My Head'
%%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra
http://home.earthlink.net/~yatescr
Anonymous
June 10, 2005 3:17:56 AM

Archived from groups: rec.audio.tech (More info?)

"Randy Yates" <yates@ieee.org> wrote in message
news:3brrce71.fsf@ieee.org...
> "Richard Crowley" <rcrowley7@xprt.net> writes:
>
>> "Randy Yates" <yates@ieee.org> wrote in message
>> news:is0odyv4.fsf@ieee.org...
>>> "Richard Crowley" <richard.7.crowley@intel.com> writes:
>>>
>>>> "Randy Yates" wrote ...
>>>>> The 16-bit CD signal is represented by the four-digit hexadecimal
>>>>> number 0xklmn. This is converted to a 24-bit number "left
>>>>> justified"
>>>>> so that it becomes 0xklmn00. The attenuator operates on that input
>>>>> and
>>>>> produces a value 0xopqrst. That value is free to range over almost
>>>>> 144
>>>>> dB of a 24-bit system (except for the 8 LSBs, which amounts to a
>>>>> loss
>>>>> of 0.000133 dB from a true 24-bit system).
>>>>
>>>> What's the point in the real world? You can't add dynamic range
>>>> merely
>>>> by shifting a 16-bit sample into a 24-bit word.
>>>
>>> That much is true. However, the operation you describe is only the
>>> first of several that occur in a digital attenuator, and it is
>>> the subsequent operations that provide the extra dynamic range.
>>>
>>>> You act as if there were some magic imputed by converting to 24-bit
>>>> before attenuating.
>>>
>>> If you consider a little arithmetic magic, then I guess so.
>>
>> Unless you can actually show how you think this works, sounds
>> like you're just relying on magic that you don't really understand.
>
> Sure. Define dynamic range (DR) as the ratio of the loudest sound to
> the softest sound, or, in decibels, the difference between the loudest
> sound and softest sound. Let a 16-bit system have a loudest sound of
> L16
> dB and a softest sound of S16 dB, so that its dynamic range DR16 is
>
> DR16 = L16 - S16 [dB]
> = 96 [db]. (note 1)
>
> If we input the loudest sound at L16 dB into a "24-bit left-justified"
> digital volume control at a gain of 0 dB, we have at the output of the
> digital volume control a signal level LDV dB, where LDV = L16.
>
> The "24-bit left-justified" digital volume control can implement a
> maximum attenuation of about 48 dB perfectly (i.e., without
> introducing any extra noise) by right shifting the left-justified
> input by 8 bits. Thus if we input the softest 16-bit sound at S16 dB
> into the device with the gain at this setting, the output level is
>
> SDV = S16 - 48.
>
> Then the dynamic range of the digital volume control, DRDV, is
>
> DRDV = LDV - SDV
> = L16 - (S16 - 48)
> = L16 - S16 + 48
> = 96 + 48
> = 144 [dB],
>
> which was to be shown.
>
> --RY
>
> Notes:
> 1. The dynamic range isn't EXACTLY 96 dB but we consider it so
> in this example.

Fine. But you have completely lost track of the original topic of
this thread. Please refer back to the Subject Line. Expanding the
available computational range is completely useless (and IMHO
ignorant) when all you are doing is attenuation (i.e. *reduction*
of the dynamic range)
Anonymous
June 10, 2005 5:35:04 AM

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:Ao2dnTlieeI-sjXfRVn-gw@comcast.com...
> Your typical 16 -> 24 bit conversion adds no headroom.

Of course not, why would you think it would?

MrT.
Anonymous
June 10, 2005 5:35:05 AM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" <MrT@home> wrote in message
news:42a861b1$0$10320$afc38c87@news.optusnet.com.au...
>
> "Arny Krueger" <arnyk@hotpop.com> wrote in message
> news:Ao2dnTlieeI-sjXfRVn-gw@comcast.com...
> > Your typical 16 -> 24 bit conversion adds no headroom.
>
> Of course not, why would you think it would?

That was the (mis)statement that started this whole discussion.
Anonymous
June 10, 2005 5:35:59 AM

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:45Kdnfp_tc1FsjXfRVn-1Q@comcast.com...

> Repeat: Your typical 16-> 24 bit conversion adds no
> headroom.

Of course not, where did I say it did?

MrT.
Anonymous
June 10, 2005 5:40:19 AM

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:sfCdnYyXv6XbrTXfRVn-vA@comcast.com...
> "Mr.T" <MrT@home> wrote in message
> news:42a7f8de$0$30960$afc38c87@news.optusnet.com.au...
>
> > It's quite simple, IF the recorder has less than 16 bits
> resolution, then
> > you achieve nothing by going to 24 bits.
>
> Not necessarily. You can achieve what I called "foot room"
> which is relevant to digital attenuators.
>
> > *IF* your recorder *CAN* achieve better than 16 bits (17,
> 18 bits etc) then
> > the extra bits can be used for recording HEADROOM.
>
> Only if you turn the levels down.

Duh!

> > IF *you* don't require the extra headroom, fine.
>
> What you always get with a 16->24 bit conversion is the
> potential for better dynamic range in follow-on processing.

Yes, but if you wish to reply to something other than what I said, why tack
it onto my post? Why not the person who said it?

MrT.
Anonymous
June 10, 2005 5:42:35 AM

Archived from groups: rec.audio.tech (More info?)

"Joe Kesselman" <keshlam-nospam@comcast.net> wrote in message
news:ZPSdnaUFb_UgrDXfRVn-1g@comcast.com...
> I really think we've got two separate arguments going on -- whether more
> than 16 bits buys you anything significant in finished output, and
> whether it has practical uses before that point.
>
> I think the answers are "probably not" and "hell yes". The latter's
> unquestionable and easily demonstrable. The former's an audiophile
> argument and probably impossible to separate from placebo effect and
> opinion.

Agreed.

MrT.
Anonymous
June 10, 2005 5:55:58 AM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <richard.7.crowley@intel.com> wrote in message
news:D 89ml2$90a$1@news01.intel.com...
> We agree completely that recording in 24 bits is highly desirable
> and will most certainly increase the dynamic range i.e. provide
> significant headroom.
>
> What you fail to realize is that adding 8 more bits to a pre-recorded
> 16 bit sample buys you NOTHING.

Why do you think I fail to realise that, since I never mentioned it?

As others have pointed out though, it *does* buy you less computational
error when mathematically manipulating the data in any way that causes
rounding errors.

MrT.
Anonymous
June 10, 2005 5:55:59 AM

Archived from groups: rec.audio.tech (More info?)

"Mr.T" <MrT@home> wrote in message
news:42a86697$0$30957$afc38c87@news.optusnet.com.au...
>
> "Richard Crowley" <richard.7.crowley@intel.com> wrote in message
> news:D 89ml2$90a$1@news01.intel.com...
> > We agree completely that recording in 24 bits is highly desirable
> > and will most certainly increase the dynamic range i.e. provide
> > significant headroom.
> >
> > What you fail to realize is that adding 8 more bits to a pre-recorded
> > 16 bit sample buys you NOTHING.
>
> Why do you think I fail to realise that, since I never mentioned it?
>
> As others have pointed out though, it *does* buy you less computational
> error when mathematically manipulating the data in any way that causes
> rounding errors.

In an *attenuator*?
(Refer to the subject line if you have lost track of the discussion.)
Anonymous
June 10, 2005 5:56:00 AM

Archived from groups: rec.audio.tech (More info?)

On Thu, 9 Jun 2005 09:14:27 -0700, "Richard Crowley"
<richard.7.crowley@intel.com> wrote:

>In an *attenuator*?

Try to design an attenuator using a single bit DSD signal as a source.
Anonymous
June 10, 2005 6:02:12 AM

Archived from groups: rec.audio.tech (More info?)

"Richard Crowley" <richard.7.crowley@intel.com> wrote in message
news:D 89od6$9op$1@news01.intel.com...
>
> "Mr.T" <MrT@home> wrote in message
> news:42a861b1$0$10320$afc38c87@news.optusnet.com.au...
> >
> > "Arny Krueger" <arnyk@hotpop.com> wrote in message
> > news:Ao2dnTlieeI-sjXfRVn-gw@comcast.com...
> > > Your typical 16 -> 24 bit conversion adds no headroom.
> >
> > Of course not, why would you think it would?
>
> That was the (mis)statement that started this whole discussion.

Maybe, but not mine.

MrT.
!