"Mr.T" MrT@home wrote in message
u...

"Richard Crowley" wrote in message
...
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.)

"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.

"Richard Crowley" writes:

"Randy Yates" wrote in message
...
"Richard Crowley" 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'
%%%% % *El Dorado*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

"Randy Yates" wrote in message
...
"Richard Crowley" writes:

"Randy Yates" wrote in message
...
"Richard Crowley" 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)

"nowater" wrote in message
ups.com...
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.

I don't see how this is really any different than analog volume controls and
noise floor and headroom calculations. If you have components with
mismatched sensitivity, or you put the volume control in the wrong place in
the circuit, you will waste dynamic range and have either insufficient
headroom or excess residual noise or both.

In a properly designed digital audio system, digital attenuation should not
be any different. Remember, any time you re-quantize a digital audio signal
(including level shifting) you need to re-dither it (see VanDerKooy &
Lip****z) so that the noise floor and resolution remains constant.

"François Yves Le Gal" wrote ...
Are you really that thick or just yet another clueless troll?

Unless you can provide a coherent explanation of why
*additional* dynamic range is required to *reduce* the
amplitude of a signal, I will have to conclude that YOU
are the troll.

OTOH, forget it. This whole discussion has devolved
into a useless waste of time. Bye.

"François Yves Le Gal" wrote ...
Sure. Whatever you say gimp. Try to understand digital
attenuation next time, unless you fancy publicly demonstrating
*again* your abysmal stupidity.

That's what I thought. Thanks for confirming it. Fall back
to juvenile name-calling when your bluff is called. Now
that is what I call troll-like behavior.

"Richard Crowley" writes:

Unless you can provide a coherent explanation of why
*additional* dynamic range is required to *reduce* the
amplitude of a signal,

Richard,

Think about it for a moment. If your only method of controlling the
system gain is with a digital volume control, then the full-scale
digital output must produce an analog signal level (i.e., the preamp
line-level output signal to the power amp) that is strong enough to
drive your power amp to the loudest possible level you could use. That
system gain setting, call it Gs [dB], is fixed.

Then the digital noise floor is correspondingly raised by Gs dB
since the digital SNR is fixed.

If Gs is enough to produce, say, 130 dB SPL, and DRDV (continuing with
the notation I introduced in my previous post) is only 16 bits, or 96
dB, that places the digital noise at around 34 dB SPL. That is
relatively high, and when listening at low volume levels the noise
will probably be obvious.

If the digital noise floor is fixed (no matter what the digital volume
gain is) at a level that is audible, then such a system also has the
undesirable property of having a gain-dependend SNR. For example, when
the digital volume gain is 0 dB, the system SNR is the full digital
SNR. An alternate scenario is when the digital volume gain is low,
say, -60 dB, in which case your system SNR could be as low as 36 dB
(for a 16-bit system). In general, the system SNR, SNRs, as a function
of digital gain Gd is

SNRs = DRDV - Gd.

These system characteristics, along with the human ear's incredible
dynamic range, make it not unreasonable to require a DAC of resolution
greater than 16 bits in order to produce a digital volume control that
sounds good at all gain settings.
--
% Randy Yates % "And all that I can do
%% Fuquay-Varina, NC % is say I'm sorry,
%%% 919-577-9882 % that's the way it goes..."
%%%% % Getting To The Point', *Balance of Power*, ELO
http://home.earthlink.net/~yatescr

"Richard Crowley" wrote in message
...
(Refer to the subject line if you have lost track of the discussion.)

Refer to my previous posts if you are not sure of *my* points rather than
those made by others.

MrT.

"Richard Crowley" wrote in message
...
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.


I guess you have never heard "zipper" noise then?

MrT.

"Mr.T" wrote ...
I guess you have never heard "zipper" noise then?

Zipper noise can be found in poorly designed or malfunctioning
circuits of ANY sample width. Irrelevant to the question at hand.
Not surprising since nobody seems to remember what the topic
was despite the subject line at the top of every posting.

"Richard Crowley" wrote in message
...
Zipper noise can be found in poorly designed or malfunctioning
circuits of ANY sample width. Irrelevant to the question at hand.
Not surprising since nobody seems to remember what the topic
was despite the subject line at the top of every posting.

Very strange, I thought zipper noise WAS applicable to a digital volume
control when used in a linear mode.
Of course if you only want a step attenuator, that's another matter.

MrT.

Randy Yates writes:

"Richard Crowley" writes:

Unless you can provide a coherent explanation of why
*additional* dynamic range is required to *reduce* the
amplitude of a signal,

Richard,
[...]

Richard,

I've seen you fault others in this newsgroup for namecalling and
"troll-like behavior," yet my response to you on this subject, which
focuses on the technical issues and does no namecalling, goes
unanswered.

Is it the namecalling and trolling you would rather respond to or a
technical discussion?
--
% Randy Yates % "Ticket to the moon, flight leaves here today
%% Fuquay-Varina, NC % from Satellite 2"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

On Tue, 7 Jun 2005 08:45:32 -0400, "Arny Krueger"
wrote:

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.

Actually, it is very easy to build a digital volume control that
does not reduce resolution. You simply do it it the analog
domain, after the system DAC.

Feed your signal to the 'Ref' input of a DAC chip. Feed the
desired attenuation (as a digital number) into the DAC input.
Take the attenuated analog signal off the DAC output.
That's it.

Resolution of the attenuation factor is the same as the
resolution of the chosen DAC chip (8 bits is acceptable
for many/most applications. 10 bits are an overkill on
just about any). There is not resolution loss of the
signal, provided the chip is monotonic (reasonably
accurate is also good) -- easily and cheaply achievable
nowadays.

-- Ron





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.