(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

Norbert Hahn writes:

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.

Well, what is actually on a CD is a string of bits which, having been
interleaved and encoded using a parity-extended (32,28) Reed-Solomon code
along with eight-to-fourteen modulation, are pretty far from the actual
data samples.

HOWEVER, those data samples are indeed two's complement. See, e.g.,
the entry entitled "Quantization" in Table 1 on page 7 of
http://www.disctronics.co.uk/downloa...troduction.pdf

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits.

An "unsigned long integer" is a C data type and does not in itself specify
the encoding.
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

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.

Well, what is actually on a CD is a string of bits which, having been
interleaved and encoded using a parity-extended (32,28) Reed-Solomon code
along with eight-to-fourteen modulation, are pretty far from the actual
data samples.

HOWEVER, those data samples are indeed two's complement. See, e.g.,
the entry entitled "Quantization" in Table 1 on page 7 of
http://www.disctronics.co.uk/downloa...troduction.pdf

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits.

An "unsigned long integer" is a C data type and does not in itself specify
the encoding.
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

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.

Well, what is actually on a CD is a string of bits which, having been
interleaved and encoded using a parity-extended (32,28) Reed-Solomon code
along with eight-to-fourteen modulation, are pretty far from the actual
data samples.

HOWEVER, those data samples are indeed two's complement. See, e.g.,
the entry entitled "Quantization" in Table 1 on page 7 of
http://www.disctronics.co.uk/downloa...troduction.pdf

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits.

An "unsigned long integer" is a C data type and does not in itself specify
the encoding.
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Norbert Hahn writes:

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.

Well, what is actually on a CD is a string of bits which, having been
interleaved and encoded using a parity-extended (32,28) Reed-Solomon code
along with eight-to-fourteen modulation, are pretty far from the actual
data samples.

HOWEVER, those data samples are indeed two's complement. See, e.g.,
the entry entitled "Quantization" in Table 1 on page 7 of
http://www.disctronics.co.uk/downloa...troduction.pdf

using the representation of samples on the CD. The numbers there are
unsigned long integers using the Motorola order of bits.

An "unsigned long integer" is a C data type and does not in itself specify
the encoding.
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

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.

No, they had logarithmic converters in the 1980s, the telephone company used
them. But you can't easily process non-linear audio. Also, the noise level
varies with the signal level. Linear quantization is much better for HiFi.
Dither can linearize all quantization effects: Noise modulation, distortion,
giving digital signals an analog noise floor. All decent quality digital
systems use dither.

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.

If properly dithered, the distortion characteristics are as good or better
than any analog system. Distortion should be about the same at any level.

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

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.

No, they had logarithmic converters in the 1980s, the telephone company used
them. But you can't easily process non-linear audio. Also, the noise level
varies with the signal level. Linear quantization is much better for HiFi.
Dither can linearize all quantization effects: Noise modulation, distortion,
giving digital signals an analog noise floor. All decent quality digital
systems use dither.

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.

If properly dithered, the distortion characteristics are as good or better
than any analog system. Distortion should be about the same at any level.

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

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.

No, they had logarithmic converters in the 1980s, the telephone company used
them. But you can't easily process non-linear audio. Also, the noise level
varies with the signal level. Linear quantization is much better for HiFi.
Dither can linearize all quantization effects: Noise modulation, distortion,
giving digital signals an analog noise floor. All decent quality digital
systems use dither.

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.

If properly dithered, the distortion characteristics are as good or better
than any analog system. Distortion should be about the same at any level.

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

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.

No, they had logarithmic converters in the 1980s, the telephone company used
them. But you can't easily process non-linear audio. Also, the noise level
varies with the signal level. Linear quantization is much better for HiFi.
Dither can linearize all quantization effects: Noise modulation, distortion,
giving digital signals an analog noise floor. All decent quality digital
systems use dither.

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.

If properly dithered, the distortion characteristics are as good or better
than any analog system. Distortion should be about the same at any level.

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

"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.

The standard that is used by some in the recording industyr is 0.3 dB Below
Full Scale. -1dBFS is certainly safe and really a good idea and it isn't going
to be noticable except in direct comparison.

Some older machines really distort when presented withfull scale peaks.
Richard H. Kuschel
"I canna change the law of physics."-----Scotty

"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.

The standard that is used by some in the recording industyr is 0.3 dB Below
Full Scale. -1dBFS is certainly safe and really a good idea and it isn't going
to be noticable except in direct comparison.

Some older machines really distort when presented withfull scale peaks.
Richard H. Kuschel
"I canna change the law of physics."-----Scotty

"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.

The standard that is used by some in the recording industyr is 0.3 dB Below
Full Scale. -1dBFS is certainly safe and really a good idea and it isn't going
to be noticable except in direct comparison.

Some older machines really distort when presented withfull scale peaks.
Richard H. Kuschel
"I canna change the law of physics."-----Scotty

"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.

The standard that is used by some in the recording industyr is 0.3 dB Below
Full Scale. -1dBFS is certainly safe and really a good idea and it isn't going
to be noticable except in direct comparison.

Some older machines really distort when presented withfull scale peaks.
Richard H. Kuschel
"I canna change the law of physics."-----Scotty

On Thu, 18 Mar 2004 09:35:43 +1100, Bazza wrote:


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.


If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"

If that's not a valid analogy, are you inclined to helpfully explain
why?

My question was an honest one. Did you think otherwise?

On Thu, 18 Mar 2004 09:35:43 +1100, Bazza wrote:


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.


If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"

If that's not a valid analogy, are you inclined to helpfully explain
why?

My question was an honest one. Did you think otherwise?

On Thu, 18 Mar 2004 09:35:43 +1100, Bazza wrote:


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.


If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"

If that's not a valid analogy, are you inclined to helpfully explain
why?

My question was an honest one. Did you think otherwise?

On Thu, 18 Mar 2004 09:35:43 +1100, Bazza wrote:


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.


If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"

If that's not a valid analogy, are you inclined to helpfully explain
why?

My question was an honest one. Did you think otherwise?

On Fri, 19 Mar 2004 18:39:42 +0000, Laurence Payne
wrote:

If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"
Are you likely to ask such a question?
Why then bring it up? Lighten up.

If that's not a valid analogy, are you inclined to helpfully explain
why?
No. I don't see it as an analogy at all - hence, from my pov, irrelevant.


My question was an honest one. Did you think otherwise?
No sir. I did NOT think otherwise. And you're right. it is honest.
So think about it (the Q.) a bit more. Here it is again ....

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

Putting aside twos complement etc because we (I) don't want to get into side
arguments about terminologies and encodings or dB's used in CD production.

Assume that the digitalisation process has one aim - of representing a
specific voltage or current wherein 0 (zero E or I) is a binary count of 0
(00000000 h). I'm doing 8 bits, not 16 for illustration.

Assume further that the maximum level likely to be encountered (or indeed,
predefined as the maximum allowable - thus we have defined a bounded,
specific, dynamic range). The maximum will be 11111111 (255dec). You may
consider the physical system as having an 8 bit buss - indeed you *must*
because if we added only just 1 more bit it would be a 9 bit buss and adding
just one more 'system state' would 'tip' the binary definition into 100000000
(256dec).

So 11111111 is absolute max
So 01111111 is not max, (a)
So 11111110 is also not max (b)

___ Both (a) and (b) are altered by 1 bit___

Both situations fit your description insofar as there is 1 bit "short of the
maximum". You surely were not implying a 7 bit system (and in context - a 15
bit system)

This to MY mind means "a binary bit" turned off.
YOU probably meant the MSB but you did not state that.
Most replies to your question assumed that you meant the MSB

But ....
(a) is a significantly less value than max (in fact it's (127d) or 50%)
(b) on the otherhand is (254d) so that (254/255)*100 =99.6%

In the framework of being "just 1 bit" short of the 'absolute' maximum
I would contend that the ONLY bit state _just_ short of maximum is situation
(b) . Of course, it was my predilection to choose the definition of "_just_ 1
bit" but then again. that's what I was alluding to in my first answer -)

Firebreak in place. Hose ready ...

So perhaps the real answer according to a different viewpoint is 0.03413 dB

On Fri, 19 Mar 2004 18:39:42 +0000, Laurence Payne
wrote:

If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"
Are you likely to ask such a question?
Why then bring it up? Lighten up.

If that's not a valid analogy, are you inclined to helpfully explain
why?
No. I don't see it as an analogy at all - hence, from my pov, irrelevant.


My question was an honest one. Did you think otherwise?
No sir. I did NOT think otherwise. And you're right. it is honest.
So think about it (the Q.) a bit more. Here it is again ....

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

Putting aside twos complement etc because we (I) don't want to get into side
arguments about terminologies and encodings or dB's used in CD production.

Assume that the digitalisation process has one aim - of representing a
specific voltage or current wherein 0 (zero E or I) is a binary count of 0
(00000000 h). I'm doing 8 bits, not 16 for illustration.

Assume further that the maximum level likely to be encountered (or indeed,
predefined as the maximum allowable - thus we have defined a bounded,
specific, dynamic range). The maximum will be 11111111 (255dec). You may
consider the physical system as having an 8 bit buss - indeed you *must*
because if we added only just 1 more bit it would be a 9 bit buss and adding
just one more 'system state' would 'tip' the binary definition into 100000000
(256dec).

So 11111111 is absolute max
So 01111111 is not max, (a)
So 11111110 is also not max (b)

___ Both (a) and (b) are altered by 1 bit___

Both situations fit your description insofar as there is 1 bit "short of the
maximum". You surely were not implying a 7 bit system (and in context - a 15
bit system)

This to MY mind means "a binary bit" turned off.
YOU probably meant the MSB but you did not state that.
Most replies to your question assumed that you meant the MSB

But ....
(a) is a significantly less value than max (in fact it's (127d) or 50%)
(b) on the otherhand is (254d) so that (254/255)*100 =99.6%

In the framework of being "just 1 bit" short of the 'absolute' maximum
I would contend that the ONLY bit state _just_ short of maximum is situation
(b) . Of course, it was my predilection to choose the definition of "_just_ 1
bit" but then again. that's what I was alluding to in my first answer -)

Firebreak in place. Hose ready ...

So perhaps the real answer according to a different viewpoint is 0.03413 dB

On Fri, 19 Mar 2004 18:39:42 +0000, Laurence Payne
wrote:

If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"
Are you likely to ask such a question?
Why then bring it up? Lighten up.

If that's not a valid analogy, are you inclined to helpfully explain
why?
No. I don't see it as an analogy at all - hence, from my pov, irrelevant.


My question was an honest one. Did you think otherwise?
No sir. I did NOT think otherwise. And you're right. it is honest.
So think about it (the Q.) a bit more. Here it is again ....

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

Putting aside twos complement etc because we (I) don't want to get into side
arguments about terminologies and encodings or dB's used in CD production.

Assume that the digitalisation process has one aim - of representing a
specific voltage or current wherein 0 (zero E or I) is a binary count of 0
(00000000 h). I'm doing 8 bits, not 16 for illustration.

Assume further that the maximum level likely to be encountered (or indeed,
predefined as the maximum allowable - thus we have defined a bounded,
specific, dynamic range). The maximum will be 11111111 (255dec). You may
consider the physical system as having an 8 bit buss - indeed you *must*
because if we added only just 1 more bit it would be a 9 bit buss and adding
just one more 'system state' would 'tip' the binary definition into 100000000
(256dec).

So 11111111 is absolute max
So 01111111 is not max, (a)
So 11111110 is also not max (b)

___ Both (a) and (b) are altered by 1 bit___

Both situations fit your description insofar as there is 1 bit "short of the
maximum". You surely were not implying a 7 bit system (and in context - a 15
bit system)

This to MY mind means "a binary bit" turned off.
YOU probably meant the MSB but you did not state that.
Most replies to your question assumed that you meant the MSB

But ....
(a) is a significantly less value than max (in fact it's (127d) or 50%)
(b) on the otherhand is (254d) so that (254/255)*100 =99.6%

In the framework of being "just 1 bit" short of the 'absolute' maximum
I would contend that the ONLY bit state _just_ short of maximum is situation
(b) . Of course, it was my predilection to choose the definition of "_just_ 1
bit" but then again. that's what I was alluding to in my first answer -)

Firebreak in place. Hose ready ...

So perhaps the real answer according to a different viewpoint is 0.03413 dB

On Fri, 19 Mar 2004 18:39:42 +0000, Laurence Payne
wrote:

If i asked "What's one less than 200?" would you reply "Is that one
hundred, one ten or one unit?"
Are you likely to ask such a question?
Why then bring it up? Lighten up.

If that's not a valid analogy, are you inclined to helpfully explain
why?
No. I don't see it as an analogy at all - hence, from my pov, irrelevant.


My question was an honest one. Did you think otherwise?
No sir. I did NOT think otherwise. And you're right. it is honest.
So think about it (the Q.) a bit more. Here it is again ....

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

Putting aside twos complement etc because we (I) don't want to get into side
arguments about terminologies and encodings or dB's used in CD production.

Assume that the digitalisation process has one aim - of representing a
specific voltage or current wherein 0 (zero E or I) is a binary count of 0
(00000000 h). I'm doing 8 bits, not 16 for illustration.

Assume further that the maximum level likely to be encountered (or indeed,
predefined as the maximum allowable - thus we have defined a bounded,
specific, dynamic range). The maximum will be 11111111 (255dec). You may
consider the physical system as having an 8 bit buss - indeed you *must*
because if we added only just 1 more bit it would be a 9 bit buss and adding
just one more 'system state' would 'tip' the binary definition into 100000000
(256dec).

So 11111111 is absolute max
So 01111111 is not max, (a)
So 11111110 is also not max (b)

___ Both (a) and (b) are altered by 1 bit___

Both situations fit your description insofar as there is 1 bit "short of the
maximum". You surely were not implying a 7 bit system (and in context - a 15
bit system)

This to MY mind means "a binary bit" turned off.
YOU probably meant the MSB but you did not state that.
Most replies to your question assumed that you meant the MSB

But ....
(a) is a significantly less value than max (in fact it's (127d) or 50%)
(b) on the otherhand is (254d) so that (254/255)*100 =99.6%

In the framework of being "just 1 bit" short of the 'absolute' maximum
I would contend that the ONLY bit state _just_ short of maximum is situation
(b) . Of course, it was my predilection to choose the definition of "_just_ 1
bit" but then again. that's what I was alluding to in my first answer -)

Firebreak in place. Hose ready ...

So perhaps the real answer according to a different viewpoint is 0.03413 dB

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

A decibel is a measure of a _ratio of powers_,

dB = 10 * log(P2/P1),

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.

Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference. To
illustrate, let's not even go to the digital realm but stay in the
analog domain.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).

So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.
--
% 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

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

A decibel is a measure of a _ratio of powers_,

dB = 10 * log(P2/P1),

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.

Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference. To
illustrate, let's not even go to the digital realm but stay in the
analog domain.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).

So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.
--
% 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

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

A decibel is a measure of a _ratio of powers_,

dB = 10 * log(P2/P1),

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.

Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference. To
illustrate, let's not even go to the digital realm but stay in the
analog domain.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).

So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.
--
% 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

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

A decibel is a measure of a _ratio of powers_,

dB = 10 * log(P2/P1),

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.

Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference. To
illustrate, let's not even go to the digital realm but stay in the
analog domain.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).

So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.
--
% 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

On Sat, 20 Mar 2004 02:55:22 GMT, Randy Yates wrote:

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

Hope not. -)
I'm trying to keep out of the dB market by using a very broadly
explained concept of the digitisation as related to his original question.

A decibel is a measure of a _ratio of powers_,
dB = 10 * log(P2/P1),
Yes. A ratio if you would want it that way.

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.
No problem there.


Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference.
Either way is good for me. I'd rather keep things simple
without going into signed numbers etc. I used to design equipment
for converting BCD readouts from F counters to analog OP.
Just throwing in the fact to "establish" credentials, or at least,
familiarity.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

No no no. It's right at this spot that you lose me. I'm not saying you're
wrong. It's that your expressing of the terms is culturally different from my
own terms. What are you comparing again ? What do you mean when you say
"this signal"?. I fully understand that 1watt (DC in nature) or (RMS sine) can
be used as a reference - no problem with that.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).



So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

Given that I'm not closely following, allow me to step in here.
This is in relation to my point in the previous post for Laurence.

(a) I'll allow that more digital bits better describe discrete step levels
between any two arbitrary points be they defined as power. volts,
amperes, lux or acceleration. There can be reference(d) standards
(b) From the last sentence above, there's a philosophical aspect also
involved (aah, but can I explain it). We can only work, shall we say,
downwards from a max. self imposed system standard. There's little
to be gained from using only one bit in a 'dynamic' system.
The "light is on" versus "the light is off". Dynamic range = infinite

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.
If I read you aright, then we're in agreement.
But there can be no equivocation about what constitues "a bit" -)

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB. BUT when we get to a situation as extreme as (e.g. only 1 bit), terms like
AC and 'dynamic range' become meaningless.

The rotation of bits through a CPU will multiply/divide any number by 2 per
shift. The position of the bit's place (if nulled) in a sequence of bits
affects the extent of the resolved value.

Obviously, as I say, there's a limit to which this absurdity can go. With only
1 bit, the sound (or any level) is either "full on" or "full
off"(non-existent). (interpolate a note here - remember that early PC's used
to 'digitise' sounds through the game port ?) .

DC can still hold tho' -)

It's at this point that the old circular arguments re-appear in this NG


Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.

Noise? Now how did that get in here -)

On Sat, 20 Mar 2004 02:55:22 GMT, Randy Yates wrote:

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

Hope not. -)
I'm trying to keep out of the dB market by using a very broadly
explained concept of the digitisation as related to his original question.

A decibel is a measure of a _ratio of powers_,
dB = 10 * log(P2/P1),
Yes. A ratio if you would want it that way.

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.
No problem there.


Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference.
Either way is good for me. I'd rather keep things simple
without going into signed numbers etc. I used to design equipment
for converting BCD readouts from F counters to analog OP.
Just throwing in the fact to "establish" credentials, or at least,
familiarity.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

No no no. It's right at this spot that you lose me. I'm not saying you're
wrong. It's that your expressing of the terms is culturally different from my
own terms. What are you comparing again ? What do you mean when you say
"this signal"?. I fully understand that 1watt (DC in nature) or (RMS sine) can
be used as a reference - no problem with that.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).



So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

Given that I'm not closely following, allow me to step in here.
This is in relation to my point in the previous post for Laurence.

(a) I'll allow that more digital bits better describe discrete step levels
between any two arbitrary points be they defined as power. volts,
amperes, lux or acceleration. There can be reference(d) standards
(b) From the last sentence above, there's a philosophical aspect also
involved (aah, but can I explain it). We can only work, shall we say,
downwards from a max. self imposed system standard. There's little
to be gained from using only one bit in a 'dynamic' system.
The "light is on" versus "the light is off". Dynamic range = infinite

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.
If I read you aright, then we're in agreement.
But there can be no equivocation about what constitues "a bit" -)

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB. BUT when we get to a situation as extreme as (e.g. only 1 bit), terms like
AC and 'dynamic range' become meaningless.

The rotation of bits through a CPU will multiply/divide any number by 2 per
shift. The position of the bit's place (if nulled) in a sequence of bits
affects the extent of the resolved value.

Obviously, as I say, there's a limit to which this absurdity can go. With only
1 bit, the sound (or any level) is either "full on" or "full
off"(non-existent). (interpolate a note here - remember that early PC's used
to 'digitise' sounds through the game port ?) .

DC can still hold tho' -)

It's at this point that the old circular arguments re-appear in this NG


Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.

Noise? Now how did that get in here -)

On Sat, 20 Mar 2004 02:55:22 GMT, Randy Yates wrote:

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

Hope not. -)
I'm trying to keep out of the dB market by using a very broadly
explained concept of the digitisation as related to his original question.

A decibel is a measure of a _ratio of powers_,
dB = 10 * log(P2/P1),
Yes. A ratio if you would want it that way.

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.
No problem there.


Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference.
Either way is good for me. I'd rather keep things simple
without going into signed numbers etc. I used to design equipment
for converting BCD readouts from F counters to analog OP.
Just throwing in the fact to "establish" credentials, or at least,
familiarity.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

No no no. It's right at this spot that you lose me. I'm not saying you're
wrong. It's that your expressing of the terms is culturally different from my
own terms. What are you comparing again ? What do you mean when you say
"this signal"?. I fully understand that 1watt (DC in nature) or (RMS sine) can
be used as a reference - no problem with that.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).



So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

Given that I'm not closely following, allow me to step in here.
This is in relation to my point in the previous post for Laurence.

(a) I'll allow that more digital bits better describe discrete step levels
between any two arbitrary points be they defined as power. volts,
amperes, lux or acceleration. There can be reference(d) standards
(b) From the last sentence above, there's a philosophical aspect also
involved (aah, but can I explain it). We can only work, shall we say,
downwards from a max. self imposed system standard. There's little
to be gained from using only one bit in a 'dynamic' system.
The "light is on" versus "the light is off". Dynamic range = infinite

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.
If I read you aright, then we're in agreement.
But there can be no equivocation about what constitues "a bit" -)

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB. BUT when we get to a situation as extreme as (e.g. only 1 bit), terms like
AC and 'dynamic range' become meaningless.

The rotation of bits through a CPU will multiply/divide any number by 2 per
shift. The position of the bit's place (if nulled) in a sequence of bits
affects the extent of the resolved value.

Obviously, as I say, there's a limit to which this absurdity can go. With only
1 bit, the sound (or any level) is either "full on" or "full
off"(non-existent). (interpolate a note here - remember that early PC's used
to 'digitise' sounds through the game port ?) .

DC can still hold tho' -)

It's at this point that the old circular arguments re-appear in this NG


Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.

Noise? Now how did that get in here -)

On Sat, 20 Mar 2004 02:55:22 GMT, Randy Yates wrote:

Bazza writes:
[...]

You guys/gals are really confusing the issues here.

Hope not. -)
I'm trying to keep out of the dB market by using a very broadly
explained concept of the digitisation as related to his original question.

A decibel is a measure of a _ratio of powers_,
dB = 10 * log(P2/P1),
Yes. A ratio if you would want it that way.

where "log" is the base 10 logarithm and P1 and P2 are
the powers whose decibel ratio is being evaluated.
No problem there.


Those powers may be AC or DC, and the allowable voltage ranges
may be unipolar or bipolar, and both make a difference.
Either way is good for me. I'd rather keep things simple
without going into signed numbers etc. I used to design equipment
for converting BCD readouts from F counters to analog OP.
Just throwing in the fact to "establish" credentials, or at least,
familiarity.

Let's say you have a signal with a voltage swing of 0 to +10
volts. Assume that we are comparing this signal to a DC signal of 1
watt (grabbed from thin air). That is, P1 is 1 watt. Then the maximum
DC dynamic range relative to P1 is

D.R. = 10 * log((10^2/R)/(1 watt)).

If we assume R = 1, then this D.R. is 20 dB.

No no no. It's right at this spot that you lose me. I'm not saying you're
wrong. It's that your expressing of the terms is culturally different from my
own terms. What are you comparing again ? What do you mean when you say
"this signal"?. I fully understand that 1watt (DC in nature) or (RMS sine) can
be used as a reference - no problem with that.

Now, let's say we have the same voltage swing but instead it goes from
-5 volts to +5 volts. Then the maximum DC dynamic range is only 10 *
log(25/1), or about 14 dB - less than the 0 to 10 volt situation.

Now think about the AC dynamic range of the 0 to 10 volt system, where
again we choose an arbitrary AC reference power P1 of 1 watt. If you
utilize a full-scale square wave, the resulting ratio is again 10 *
log(25/1), or 14 dB, where I'm ignoring the DC power component because
we're talking about AC dynamic range. Thus the AC dynamic range is
less than the DC D.R. (and would be even smaller if we used a sine
wave).



So the two points I want to make a

1. If we're talking about DC dynamic range, the actual range of values
(and thus the number representation) matters.
2. AC and DC dynamic range differ for a given system.

Given that I'm not closely following, allow me to step in here.
This is in relation to my point in the previous post for Laurence.

(a) I'll allow that more digital bits better describe discrete step levels
between any two arbitrary points be they defined as power. volts,
amperes, lux or acceleration. There can be reference(d) standards
(b) From the last sentence above, there's a philosophical aspect also
involved (aah, but can I explain it). We can only work, shall we say,
downwards from a max. self imposed system standard. There's little
to be gained from using only one bit in a 'dynamic' system.
The "light is on" versus "the light is off". Dynamic range = infinite

To get back to the digital domain, since we can't hear DC, we are
concerned with the AC dynamic range in an audio system. Thus these
questions of "what constitutes a bit" should be answered in the
context of AC signals. Additionally I'd like to point out that, apart
from the above arguments, the number representation matters since it
determines the mapping between a "bit" and a range.
If I read you aright, then we're in agreement.
But there can be no equivocation about what constitues "a bit" -)

In a two's complement representation, the addition of a bit ALWAYS
results in a 6 dB increase in AC dynamic range. That is because a
two's complement system can be modeled as a sign bit and magnitude
bits, and the bits being added are added to the magnitude. Thus, e.g.,
if we have 8 total bits (1 sign bit and 7 magnitude bits) a square
wave can be as large as +/- 127 (2^7 - 1 = 127), and if a bit
is added, the magnitude doubles, and doubling the voltage (essentially
these numbers are like voltages) yields a 6 dB increase in dynamic
range.

Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB. BUT when we get to a situation as extreme as (e.g. only 1 bit), terms like
AC and 'dynamic range' become meaningless.

The rotation of bits through a CPU will multiply/divide any number by 2 per
shift. The position of the bit's place (if nulled) in a sequence of bits
affects the extent of the resolved value.

Obviously, as I say, there's a limit to which this absurdity can go. With only
1 bit, the sound (or any level) is either "full on" or "full
off"(non-existent). (interpolate a note here - remember that early PC's used
to 'digitise' sounds through the game port ?) .

DC can still hold tho' -)

It's at this point that the old circular arguments re-appear in this NG


Now, how you compute the AC noise power (the P1 power) for a digital
dynamic range computation is a whole other subject.

Noise? Now how did that get in here -)

Bazza writes:
[...]
Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB.

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% % 'Waterfall', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Bazza writes:
[...]
Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB.

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% % 'Waterfall', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Bazza writes:
[...]
Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB.

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% % 'Waterfall', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Bazza writes:
[...]
Precisely the terminological discussion points I was avoiding . Not saying
you're wrong, only that in a broad sense if 4 bits (or 8, or 16) represent an
arbitrary level (predefined or otherwise), then nullifying the MSB (removing
or setting it at 0) will halve the binary count and therefore reduce ANY level
previously represented by that count to a NEW level which will be down by 6
dB.

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% % 'Waterfall', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

On Sat, 20 Mar 2004 06:10:14 GMT, Randy Yates wrote:

Bazza writes:

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.


Aaaagh.
Who has mentioned mu-law?
Who has mentioned codecs?
Who has mentioned linearity relevance?

On Sat, 20 Mar 2004 06:10:14 GMT, Randy Yates wrote:

Bazza writes:

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.


Aaaagh.
Who has mentioned mu-law?
Who has mentioned codecs?
Who has mentioned linearity relevance?

On Sat, 20 Mar 2004 06:10:14 GMT, Randy Yates wrote:

Bazza writes:

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.


Aaaagh.
Who has mentioned mu-law?
Who has mentioned codecs?
Who has mentioned linearity relevance?

On Sat, 20 Mar 2004 06:10:14 GMT, Randy Yates wrote:

Bazza writes:

That is absolutely incorrect. It depends on the representation. For example,
mu-law codecs are not linear. Hence one of the major points - i.e., the
representation matters. Sorry you're having so much trouble getting it.


Aaaagh.
Who has mentioned mu-law?
Who has mentioned codecs?
Who has mentioned linearity relevance?

Norbert Hahn wrote:

(Kega=myself) 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

Hello all.

First I must appologize for confusing you all concerning linear PCM; I'm
so sorry.

My main point is that a loud sound (in a linear PCM system) has a finer
granularity than a faint one when it comes to how the ear percept the
sound. The decibel scale is a way to closer describe how the ear
functions.

Therefor it is of no use to talk about how many decibel a digital step
represents. It depends on the level.

Let me tired you with a some math. I hope I have calculated everything
correct:

Asume we have a signal ranging between -5V to +5V and that we use a
signed bit representation of 16 bits (where the most significant bit is
the sign bit).

Voltage Decimal(d) Binary(b) Hexadecimal(h)
+5 32767 0111 1111 1111 1111 7FFF
0 0 0000 0000 0000 0000 0000
-1 1111 1111 1111 1111 FFFF
-5 -32786 1000 0000 0000 0000 8000

The whole swing is divided into 2**16 parts. Each quantising part has
the same size (it is linear PCM) of 10/65536 V = approx 0.152588 mV

Case 1:

A rather load signal around 4.9 V peak. The bit representation will be
about (4.9/5)*32767 = 32112(d) = 7D70(h). The smallest increasing
change of that sound that could be represented in PCM is going from
7D70(h) to 7D71(h).

In decibel this will be:

20 lg((32113*5/32767)/(32112*5/32767)) = lg(32113/32112) = +0.000270 dB

Case 2:

A signal that lies about 80 dB below maximum 5V. Indeed very close to
the quantising noise floor. Minus 80 dB will be 0.5 mV. This extremely
low value will be represented by (0.5E-3/5) * 32767 = 3.2767 = 3(d) =
0003(h). The smallest increase of this low signal that could be
represented is the step from 3 to 4. In decibel this will be 20 lg(4/3)
= +2.498775 dB.

An interesting point is at what volume the error of the representaion
could be 1% or higher (In HiFi societies this is a rather high value. I
took the value 1% from the old HiFi DIN norm back in the 60:ies).

Let's say that half of the samples have 1% distortion or higher. This
will happen when the hex value is around 0032(h) = 50(d). In decibel
from the maximum this will occur at 20 lg (50/32767) = -56.33 dB.

If you only could accept 0.1% distortion then you only have the
threshold at 20 lg (500/32767) = -36 dB. Not much of a range when
talking abount high end HiFi.

In linear PCM system using 24 bits the maximum level is represented by
the value 8_388_607 ((2**23)-1). 1% threshold will be 20 lg (50/8388607)
= -104.5 dB. Now you have a dynamic range you may call true HiFi. 0.1%
level will be -84.5 dB. This may be an indication why 24 bits has been
choosen in professional equipments.

Note also that I have not in my reasing involved dithering and other
techniques to increase the quality in a 16 bit PCM at low signal values.

Regards Kent

--

Remove all characters 'c' before using mail address.

Norbert Hahn wrote:

(Kega=myself) 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

Hello all.

First I must appologize for confusing you all concerning linear PCM; I'm
so sorry.

My main point is that a loud sound (in a linear PCM system) has a finer
granularity than a faint one when it comes to how the ear percept the
sound. The decibel scale is a way to closer describe how the ear
functions.

Therefor it is of no use to talk about how many decibel a digital step
represents. It depends on the level.

Let me tired you with a some math. I hope I have calculated everything
correct:

Asume we have a signal ranging between -5V to +5V and that we use a
signed bit representation of 16 bits (where the most significant bit is
the sign bit).

Voltage Decimal(d) Binary(b) Hexadecimal(h)
+5 32767 0111 1111 1111 1111 7FFF
0 0 0000 0000 0000 0000 0000
-1 1111 1111 1111 1111 FFFF
-5 -32786 1000 0000 0000 0000 8000

The whole swing is divided into 2**16 parts. Each quantising part has
the same size (it is linear PCM) of 10/65536 V = approx 0.152588 mV

Case 1:

A rather load signal around 4.9 V peak. The bit representation will be
about (4.9/5)*32767 = 32112(d) = 7D70(h). The smallest increasing
change of that sound that could be represented in PCM is going from
7D70(h) to 7D71(h).

In decibel this will be:

20 lg((32113*5/32767)/(32112*5/32767)) = lg(32113/32112) = +0.000270 dB

Case 2:

A signal that lies about 80 dB below maximum 5V. Indeed very close to
the quantising noise floor. Minus 80 dB will be 0.5 mV. This extremely
low value will be represented by (0.5E-3/5) * 32767 = 3.2767 = 3(d) =
0003(h). The smallest increase of this low signal that could be
represented is the step from 3 to 4. In decibel this will be 20 lg(4/3)
= +2.498775 dB.

An interesting point is at what volume the error of the representaion
could be 1% or higher (In HiFi societies this is a rather high value. I
took the value 1% from the old HiFi DIN norm back in the 60:ies).

Let's say that half of the samples have 1% distortion or higher. This
will happen when the hex value is around 0032(h) = 50(d). In decibel
from the maximum this will occur at 20 lg (50/32767) = -56.33 dB.

If you only could accept 0.1% distortion then you only have the
threshold at 20 lg (500/32767) = -36 dB. Not much of a range when
talking abount high end HiFi.

In linear PCM system using 24 bits the maximum level is represented by
the value 8_388_607 ((2**23)-1). 1% threshold will be 20 lg (50/8388607)
= -104.5 dB. Now you have a dynamic range you may call true HiFi. 0.1%
level will be -84.5 dB. This may be an indication why 24 bits has been
choosen in professional equipments.

Note also that I have not in my reasing involved dithering and other
techniques to increase the quality in a 16 bit PCM at low signal values.

Regards Kent

--

Remove all characters 'c' before using mail address.

Norbert Hahn wrote:

(Kega=myself) 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

Hello all.

First I must appologize for confusing you all concerning linear PCM; I'm
so sorry.

My main point is that a loud sound (in a linear PCM system) has a finer
granularity than a faint one when it comes to how the ear percept the
sound. The decibel scale is a way to closer describe how the ear
functions.

Therefor it is of no use to talk about how many decibel a digital step
represents. It depends on the level.

Let me tired you with a some math. I hope I have calculated everything
correct:

Asume we have a signal ranging between -5V to +5V and that we use a
signed bit representation of 16 bits (where the most significant bit is
the sign bit).

Voltage Decimal(d) Binary(b) Hexadecimal(h)
+5 32767 0111 1111 1111 1111 7FFF
0 0 0000 0000 0000 0000 0000
-1 1111 1111 1111 1111 FFFF
-5 -32786 1000 0000 0000 0000 8000

The whole swing is divided into 2**16 parts. Each quantising part has
the same size (it is linear PCM) of 10/65536 V = approx 0.152588 mV

Case 1:

A rather load signal around 4.9 V peak. The bit representation will be
about (4.9/5)*32767 = 32112(d) = 7D70(h). The smallest increasing
change of that sound that could be represented in PCM is going from
7D70(h) to 7D71(h).

In decibel this will be:

20 lg((32113*5/32767)/(32112*5/32767)) = lg(32113/32112) = +0.000270 dB

Case 2:

A signal that lies about 80 dB below maximum 5V. Indeed very close to
the quantising noise floor. Minus 80 dB will be 0.5 mV. This extremely
low value will be represented by (0.5E-3/5) * 32767 = 3.2767 = 3(d) =
0003(h). The smallest increase of this low signal that could be
represented is the step from 3 to 4. In decibel this will be 20 lg(4/3)
= +2.498775 dB.

An interesting point is at what volume the error of the representaion
could be 1% or higher (In HiFi societies this is a rather high value. I
took the value 1% from the old HiFi DIN norm back in the 60:ies).

Let's say that half of the samples have 1% distortion or higher. This
will happen when the hex value is around 0032(h) = 50(d). In decibel
from the maximum this will occur at 20 lg (50/32767) = -56.33 dB.

If you only could accept 0.1% distortion then you only have the
threshold at 20 lg (500/32767) = -36 dB. Not much of a range when
talking abount high end HiFi.

In linear PCM system using 24 bits the maximum level is represented by
the value 8_388_607 ((2**23)-1). 1% threshold will be 20 lg (50/8388607)
= -104.5 dB. Now you have a dynamic range you may call true HiFi. 0.1%
level will be -84.5 dB. This may be an indication why 24 bits has been
choosen in professional equipments.

Note also that I have not in my reasing involved dithering and other
techniques to increase the quality in a 16 bit PCM at low signal values.

Regards Kent

--

Remove all characters 'c' before using mail address.