Dave Platt wrote:
I'm sure many of you have experienced the frustration in different playback
levels when playing CDs in a CD carousel or a CD jukebox, especially in
shuffle or random mode. The reason for this may be "technological
advances" as most of the seemingly "low- level" CDs are early issues.
Anyway, large differences in levels tend to make variations in music
programming less than optimum.

In my experience, the "technological advances" you cite are precisely
backwards.

Most modern CDs (popular music in particular) are being mixed/mastered
at a very high level, with a very great deal of compression being
applied to the signal. The goal of this appears to be a deliberate
desire to have the music sound "loud" or "punchy" - that is,
attention-getting. It's purely market-and-marketing driven.

The compression is often so great that there are two very negative
side effects:

- The music often has very little dynamic range left - the
distance between loudest and softest passages is often only
a few dB.


see, for example, the march of progress on display he

http://forums.lukpac.org/viewtopic.php?t=643


--

-S.

"They've got God on their side. All we've got is science and reason."
-- Dawn Hulsey, Talent Director

"Laurence Payne" wrote in
message
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

6 dB, way too much.

I leave 1 dB headroom. This number is based on testing a ton of digital
audio gear.

"Laurence Payne" wrote in
message
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

6 dB, way too much.

I leave 1 dB headroom. This number is based on testing a ton of digital
audio gear.

"Laurence Payne" wrote in
message
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

6 dB, way too much.

I leave 1 dB headroom. This number is based on testing a ton of digital
audio gear.

"Laurence Payne" wrote in
message
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

6 dB, way too much.

I leave 1 dB headroom. This number is based on testing a ton of digital
audio gear.

Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

Laurence Payne wrote in message . ..
On Tue, 16 Mar 2004 07:12:14 -0000, (Dave Platt)
wrote:

Leaving a small fraction of a dB of digital headroom seems like good
practice,

What fraction of a dB would correspond to one bit under maximum?

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne

What fraction of a dB would correspond to one bit under maximum?

When you ask about "1 bit" are you asking about ..

THE most significant bit (msb) (perhaps referenced to a "signed" 32767)
or
THE least significant bit (lsb)
or
some other bit
or
some particular bit based upon and arbitary level.

There's quite a difference to the way the question should be answered.

On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne

What fraction of a dB would correspond to one bit under maximum?

When you ask about "1 bit" are you asking about ..

THE most significant bit (msb) (perhaps referenced to a "signed" 32767)
or
THE least significant bit (lsb)
or
some other bit
or
some particular bit based upon and arbitary level.

There's quite a difference to the way the question should be answered.

On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne

What fraction of a dB would correspond to one bit under maximum?

When you ask about "1 bit" are you asking about ..

THE most significant bit (msb) (perhaps referenced to a "signed" 32767)
or
THE least significant bit (lsb)
or
some other bit
or
some particular bit based upon and arbitary level.

There's quite a difference to the way the question should be answered.

On Tue, 16 Mar 2004 23:33:58 +0000, Laurence Payne

What fraction of a dB would correspond to one bit under maximum?

When you ask about "1 bit" are you asking about ..

THE most significant bit (msb) (perhaps referenced to a "signed" 32767)
or
THE least significant bit (lsb)
or
some other bit
or
some particular bit based upon and arbitary level.

There's quite a difference to the way the question should be answered.

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed. If one were directly outputing the data from an A/D, you could
point the finger to this stage, but usually there is some (a crapload?)
digital processing to the signal before D/Aing. It is hard to ensure
that all DSP algorithms result in an equivalent analog waveform that
doesn't exceed the D/A limits. In fact, it can be shown (although it
is a pathological case) that a digital signal can be generated which
would result in an infinite output level if a theoretical brick-wall
reconstruction filter is applied. Of course such a filter is not
physically realizable, but long filter are and the potential for
overflow is real.

I believe that the usual working definition of a
"digital over" is two or more consecutive samples at the positive or
negative limit.

Leaving a small fraction of a dB of digital headroom seems like good
practice, because (as you note) some CD players will show clipping
artifacts at the digital limit, and I understand that some can
actually show clipping problems even if the limit isn't reached. CD
players with digital oversampling filters (many/most of 'em) can have
problems with clipping during the oversampling process - even if two
adjacent samples in the original 16-bit data stream are below the
limit, the reconstructed/oversampled waveform calculated by the
filters can go over the limit and be clipped.

As long as the digital signal is maintained below the "equivalent analog
limit," oversampling shouldn't do this. If it does, the designer screwed
up a gain somewhere.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed. If one were directly outputing the data from an A/D, you could
point the finger to this stage, but usually there is some (a crapload?)
digital processing to the signal before D/Aing. It is hard to ensure
that all DSP algorithms result in an equivalent analog waveform that
doesn't exceed the D/A limits. In fact, it can be shown (although it
is a pathological case) that a digital signal can be generated which
would result in an infinite output level if a theoretical brick-wall
reconstruction filter is applied. Of course such a filter is not
physically realizable, but long filter are and the potential for
overflow is real.

I believe that the usual working definition of a
"digital over" is two or more consecutive samples at the positive or
negative limit.

Leaving a small fraction of a dB of digital headroom seems like good
practice, because (as you note) some CD players will show clipping
artifacts at the digital limit, and I understand that some can
actually show clipping problems even if the limit isn't reached. CD
players with digital oversampling filters (many/most of 'em) can have
problems with clipping during the oversampling process - even if two
adjacent samples in the original 16-bit data stream are below the
limit, the reconstructed/oversampled waveform calculated by the
filters can go over the limit and be clipped.

As long as the digital signal is maintained below the "equivalent analog
limit," oversampling shouldn't do this. If it does, the designer screwed
up a gain somewhere.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed. If one were directly outputing the data from an A/D, you could
point the finger to this stage, but usually there is some (a crapload?)
digital processing to the signal before D/Aing. It is hard to ensure
that all DSP algorithms result in an equivalent analog waveform that
doesn't exceed the D/A limits. In fact, it can be shown (although it
is a pathological case) that a digital signal can be generated which
would result in an infinite output level if a theoretical brick-wall
reconstruction filter is applied. Of course such a filter is not
physically realizable, but long filter are and the potential for
overflow is real.

I believe that the usual working definition of a
"digital over" is two or more consecutive samples at the positive or
negative limit.

Leaving a small fraction of a dB of digital headroom seems like good
practice, because (as you note) some CD players will show clipping
artifacts at the digital limit, and I understand that some can
actually show clipping problems even if the limit isn't reached. CD
players with digital oversampling filters (many/most of 'em) can have
problems with clipping during the oversampling process - even if two
adjacent samples in the original 16-bit data stream are below the
limit, the reconstructed/oversampled waveform calculated by the
filters can go over the limit and be clipped.

As long as the digital signal is maintained below the "equivalent analog
limit," oversampling shouldn't do this. If it does, the designer screwed
up a gain somewhere.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed. If one were directly outputing the data from an A/D, you could
point the finger to this stage, but usually there is some (a crapload?)
digital processing to the signal before D/Aing. It is hard to ensure
that all DSP algorithms result in an equivalent analog waveform that
doesn't exceed the D/A limits. In fact, it can be shown (although it
is a pathological case) that a digital signal can be generated which
would result in an infinite output level if a theoretical brick-wall
reconstruction filter is applied. Of course such a filter is not
physically realizable, but long filter are and the potential for
overflow is real.

I believe that the usual working definition of a
"digital over" is two or more consecutive samples at the positive or
negative limit.

Leaving a small fraction of a dB of digital headroom seems like good
practice, because (as you note) some CD players will show clipping
artifacts at the digital limit, and I understand that some can
actually show clipping problems even if the limit isn't reached. CD
players with digital oversampling filters (many/most of 'em) can have
problems with clipping during the oversampling process - even if two
adjacent samples in the original 16-bit data stream are below the
limit, the reconstructed/oversampled waveform calculated by the
filters can go over the limit and be clipped.

As long as the digital signal is maintained below the "equivalent analog
limit," oversampling shouldn't do this. If it does, the designer screwed
up a gain somewhere.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

Randy Yates writes:

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed.

Actually, let me correct myself here. It is possible for an analog
waveform to exceed the reference voltage of an A/D and still not clip
digitally. It *will* be clipped when it is reproduced over a D/A, but
it *hasn't* been clipped by the A/D.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Randy Yates writes:

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed.

Actually, let me correct myself here. It is possible for an analog
waveform to exceed the reference voltage of an A/D and still not clip
digitally. It *will* be clipped when it is reproduced over a D/A, but
it *hasn't* been clipped by the A/D.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Randy Yates writes:

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed.

Actually, let me correct myself here. It is possible for an analog
waveform to exceed the reference voltage of an A/D and still not clip
digitally. It *will* be clipped when it is reproduced over a D/A, but
it *hasn't* been clipped by the A/D.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Randy Yates writes:

(Dave Platt) writes:

In article ,
Barry Mann wrote:

- The sound is often pushed to such a high volume level that
it's hitting the maximum-positive/negative number boundaries
of the 16-bit digital number system used by CDs. In other
words, it's clipping - quite hard, often very harshly, and
often for a prolonged period.

I don't have a problem with the CD touching the top level at one or
more spots. This is good management of dynamic range. It's just a
simple computer trick to do this. Redbook CD's have a 16 bit range.
"Clipping" implies a loss of information because the signal would have
gone beyond the limit if it could have. "Touching" means the signal
reached the limit and didn't need to go past it.

Touching the limit, for a single sample, usually doesn't cause
technical problems. Two or more samples, touching the limit, usually
means that the original signal tried to go past the limit, and has in
fact been clipped.

Agreed.

Actually, let me correct myself here. It is possible for an analog
waveform to exceed the reference voltage of an A/D and still not clip
digitally. It *will* be clipped when it is reproduced over a D/A, but
it *hasn't* been clipped by the A/D.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.
--
% Randy Yates % "Rollin' and riding and slippin' and
%% Fuquay-Varina, NC % sliding, it's magic."
%%% 919-577-9882 %
%%%% % 'Living' Thing', *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn wrote in message . ..
(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is:

B = lg (P_out/P_in) where P stands for Power.

And since P(ower) is proportinal to the square of V(oltage) then

B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in)

and expressed in dB gives: 20*lg(V_out/V_in)

A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6
and a halfing: 20*lg(1/2)= -6 (approx)

Therefor you have smaller dB-steps at higher voltage than you have in
lower voltage in a linear PCM system (as the CD format is). This is a
(kind of)drawback in audio PCM since you use more bits than you
necessarily must use to obtain a suitable HiFi-level. I guess they
(the constructor of CD) choosed linear PCM when designing CD because
at that time it was easier to construct a reasonable good sounding
linear D/A-converter than a logarithmic one.

In the analog days you more or less learned that high volumes produced
more distortion than low volumes. In PCM you have the opposite. High
volumes that does not exceeds the headroom limit have lesser
distortion than low volumes. But since you have a lots of bits the
overall distortion at normal levels are very low.

O boy. It is a bit difficult to explain in a language that is not my
mother tounge. Hope you all understand my points. It was also very
long time ago a studied PCM (back in 1977/78).

BTW. Just recently I read how dts was coded and boy that was
complicated (but beautiful) using Adaptive Differentiell PCM with
prediction and splitting up the signal into sub frequency bands. But I
am impressed. Back in 77 we (at school) just touched the technique of
ADPCM.

Best Regards from Sweden

/Kent

Norbert Hahn wrote in message . ..
(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is:

B = lg (P_out/P_in) where P stands for Power.

And since P(ower) is proportinal to the square of V(oltage) then

B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in)

and expressed in dB gives: 20*lg(V_out/V_in)

A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6
and a halfing: 20*lg(1/2)= -6 (approx)

Therefor you have smaller dB-steps at higher voltage than you have in
lower voltage in a linear PCM system (as the CD format is). This is a
(kind of)drawback in audio PCM since you use more bits than you
necessarily must use to obtain a suitable HiFi-level. I guess they
(the constructor of CD) choosed linear PCM when designing CD because
at that time it was easier to construct a reasonable good sounding
linear D/A-converter than a logarithmic one.

In the analog days you more or less learned that high volumes produced
more distortion than low volumes. In PCM you have the opposite. High
volumes that does not exceeds the headroom limit have lesser
distortion than low volumes. But since you have a lots of bits the
overall distortion at normal levels are very low.

O boy. It is a bit difficult to explain in a language that is not my
mother tounge. Hope you all understand my points. It was also very
long time ago a studied PCM (back in 1977/78).

BTW. Just recently I read how dts was coded and boy that was
complicated (but beautiful) using Adaptive Differentiell PCM with
prediction and splitting up the signal into sub frequency bands. But I
am impressed. Back in 77 we (at school) just touched the technique of
ADPCM.

Best Regards from Sweden

/Kent

Norbert Hahn wrote in message . ..
(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is:

B = lg (P_out/P_in) where P stands for Power.

And since P(ower) is proportinal to the square of V(oltage) then

B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in)

and expressed in dB gives: 20*lg(V_out/V_in)

A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6
and a halfing: 20*lg(1/2)= -6 (approx)

Therefor you have smaller dB-steps at higher voltage than you have in
lower voltage in a linear PCM system (as the CD format is). This is a
(kind of)drawback in audio PCM since you use more bits than you
necessarily must use to obtain a suitable HiFi-level. I guess they
(the constructor of CD) choosed linear PCM when designing CD because
at that time it was easier to construct a reasonable good sounding
linear D/A-converter than a logarithmic one.

In the analog days you more or less learned that high volumes produced
more distortion than low volumes. In PCM you have the opposite. High
volumes that does not exceeds the headroom limit have lesser
distortion than low volumes. But since you have a lots of bits the
overall distortion at normal levels are very low.

O boy. It is a bit difficult to explain in a language that is not my
mother tounge. Hope you all understand my points. It was also very
long time ago a studied PCM (back in 1977/78).

BTW. Just recently I read how dts was coded and boy that was
complicated (but beautiful) using Adaptive Differentiell PCM with
prediction and splitting up the signal into sub frequency bands. But I
am impressed. Back in 77 we (at school) just touched the technique of
ADPCM.

Best Regards from Sweden

/Kent

Norbert Hahn wrote in message . ..
(Kega) wrote:

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

We're talking about 16 bit systems, don't we? So a step from 0001 to
0002 will again be 20*lg(1/65536) as it is when going from 65536 to
65535. All steps are of the same size as linear PCM is used.

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Norbert

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Decibel is a logaritmic scale and is a tenth of a Bel. A bel (B) is:

B = lg (P_out/P_in) where P stands for Power.

And since P(ower) is proportinal to the square of V(oltage) then

B = lg(V_out**2/V_in**2) = lg((V_out/V_in)**2) = 2 * lg(V_out/V_in)

and expressed in dB gives: 20*lg(V_out/V_in)

A double in voltage gives: 20*lg(2/1) is approx 20*0.3=6
and a halfing: 20*lg(1/2)= -6 (approx)

Therefor you have smaller dB-steps at higher voltage than you have in
lower voltage in a linear PCM system (as the CD format is). This is a
(kind of)drawback in audio PCM since you use more bits than you
necessarily must use to obtain a suitable HiFi-level. I guess they
(the constructor of CD) choosed linear PCM when designing CD because
at that time it was easier to construct a reasonable good sounding
linear D/A-converter than a logarithmic one.

In the analog days you more or less learned that high volumes produced
more distortion than low volumes. In PCM you have the opposite. High
volumes that does not exceeds the headroom limit have lesser
distortion than low volumes. But since you have a lots of bits the
overall distortion at normal levels are very low.

O boy. It is a bit difficult to explain in a language that is not my
mother tounge. Hope you all understand my points. It was also very
long time ago a studied PCM (back in 1977/78).

BTW. Just recently I read how dts was coded and boy that was
complicated (but beautiful) using Adaptive Differentiell PCM with
prediction and splitting up the signal into sub frequency bands. But I
am impressed. Back in 77 we (at school) just touched the technique of
ADPCM.

Best Regards from Sweden

/Kent

Randy Yates wrote:

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.

The number you use here is what we see on a PC, it is not what is on a CD.
The previous poster, however, ...

Laurence Payne wrote in message ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits. Thus I made
my example using the representation on the CD.

Norbert

Randy Yates wrote:

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.

The number you use here is what we see on a PC, it is not what is on a CD.
The previous poster, however, ...

Laurence Payne wrote in message ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits. Thus I made
my example using the representation on the CD.

Norbert

Randy Yates wrote:

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.

The number you use here is what we see on a PC, it is not what is on a CD.
The previous poster, however, ...

Laurence Payne wrote in message ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits. Thus I made
my example using the representation on the CD.

Norbert

Randy Yates wrote:

Norbert Hahn writes:

Your zero value is at 32768, thus going from 32768 to 32769 doubles
the output voltage, but the reference level remains 65536.

Or am I grossly mistaken?

Almost all digital systems use "signed two's complement"
representation of the data. For a 16-bit system, this results in
integers from -32768 (full-scale negative) to +32767 (full-scale
positive), with 0 corresponding to 0.

The number you use here is what we see on a PC, it is not what is on a CD.
The previous poster, however, ...

Laurence Payne wrote in message ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

Note that one step close to the maximum level (FFFF) is very small in
decibel compared to a step at the minumum level. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits. Thus I made
my example using the representation on the CD.

Norbert

(Kega) wrote:

Norbert Hahn wrote in message . ..
(Kega) wrote:

Or am I grossly mistaken?

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Well, it all depends on the selection of a reference value. IHMO the
reference value for a 16 bit audio signal should be either 32768 or
65536. I don't think it will help for the discussion to base your
example 1 ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

on 65536 and example 2 ...

.. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

on quite a differenct level. When those numbers are translated into
voltages by the D/A process - using 5,0000 V as reference level and
a symmetric output voltage, hence 10 V voltage swing

65535 - 5,000000 V 65534 - 4,9999237 while
00000 - -5,000000 V 00001 - -4,9999237

clearly showing that a step from 00000 to 00001 is not 6 dB.

And what is going on he

32768 - 0,000000 V
32769 - 0,000305 V

Again, those numbers are taken from the representation we find on the CD.

I hope I clarified my view somewhat.

Norbert

(Kega) wrote:

Norbert Hahn wrote in message . ..
(Kega) wrote:

Or am I grossly mistaken?

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Well, it all depends on the selection of a reference value. IHMO the
reference value for a 16 bit audio signal should be either 32768 or
65536. I don't think it will help for the discussion to base your
example 1 ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

on 65536 and example 2 ...

.. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

on quite a differenct level. When those numbers are translated into
voltages by the D/A process - using 5,0000 V as reference level and
a symmetric output voltage, hence 10 V voltage swing

65535 - 5,000000 V 65534 - 4,9999237 while
00000 - -5,000000 V 00001 - -4,9999237

clearly showing that a step from 00000 to 00001 is not 6 dB.

And what is going on he

32768 - 0,000000 V
32769 - 0,000305 V

Again, those numbers are taken from the representation we find on the CD.

I hope I clarified my view somewhat.

Norbert

(Kega) wrote:

Norbert Hahn wrote in message . ..
(Kega) wrote:

Or am I grossly mistaken?

Yes I think you are. A double of output voltage is always approx 6 dB.
When counting with dB you always count with the ratio between output
and input (or between to levels).

Well, it all depends on the selection of a reference value. IHMO the
reference value for a 16 bit audio signal should be either 32768 or
65536. I don't think it will help for the discussion to base your
example 1 ...

Well if you mean you go from the value FFFF (hexadecimal) (=65536) to
FFFE (65535) then the decibel change will be (20*lg(65535/65536))
where lg is the logarithm using 10 as base. A rather small value
indeed.

on 65536 and example 2 ...

.. A step between 0001
to 0002 is aprox 6 dB. You have doubbled to voltage value. The
quantization is linear but the ear's sensitivity is logarithmic.

on quite a differenct level. When those numbers are translated into
voltages by the D/A process - using 5,0000 V as reference level and
a symmetric output voltage, hence 10 V voltage swing

65535 - 5,000000 V 65534 - 4,9999237 while
00000 - -5,000000 V 00001 - -4,9999237

clearly showing that a step from 00000 to 00001 is not 6 dB.

And what is going on he

32768 - 0,000000 V
32769 - 0,000305 V

Again, those numbers are taken from the representation we find on the CD.

I hope I clarified my view somewhat.

Norbert