miab

2009-11-13, 00:11

Will the Touch send out via SPDIF native formats of files on a central server or does it upsample? Almost all my music is 16/44.1 with no immediate plans for higher rates.

View Full Version : What will the Touch do to 16/44.1 over SPDIF

miab

2009-11-13, 00:11

Will the Touch send out via SPDIF native formats of files on a central server or does it upsample? Almost all my music is 16/44.1 with no immediate plans for higher rates.

radish

2009-11-13, 08:34

It sends out in whatever format it gets, so 16/44.1 for redbook. No upsampling.

JohnSwenson

2009-11-14, 13:27

For a 16/44.1 file you will get a 24/44.1 stream coming out of the S/PDIF (coax or optical). 16 bit originals are converted to 24 bit so there are more bits for volume control processing, this happens even if the volume is set to 100. All known S/PDIF receivers should be able to handle the 24 bit stream.

John S.

John S.

radish

2009-11-14, 15:53

You are of course correct John. I forgot about the bit depth expansion as it makes no difference to the signal. Thx for the correction.

Kellen

2009-11-14, 19:37

For a 16/44.1 file you will get a 24/44.1 stream coming out of the S/PDIF (coax or optical). 16 bit originals are converted to 24 bit so there are more bits for volume control processing, this happens even if the volume is set to 100. All known S/PDIF receivers should be able to handle the 24 bit stream.

John S.

Hi, does the Touch convert the 16 bit originals to 24 bits even if I connect an old external DAC (EAD 7000 in this case), that does 16 bit max, to it?

John S.

Hi, does the Touch convert the 16 bit originals to 24 bits even if I connect an old external DAC (EAD 7000 in this case), that does 16 bit max, to it?

JohnSwenson

2009-11-14, 23:52

all known S/PDIF receivers should receive the 24 bit data even if the DAC chips only do 16.

There should be no problems.

John S.

There should be no problems.

John S.

miab

2009-11-15, 01:05

all known S/PDIF receivers should receive the 24 bit data even if the DAC chips only do 16.

There should be no problems.

John S.

Hi John

This is probably a very stupid question. Does it just add bits on top of the 16 bit native files or is there some processing or manipulation to the original 16 bits. I really don't like my bits touched :~) no pun intended.

There should be no problems.

John S.

Hi John

This is probably a very stupid question. Does it just add bits on top of the 16 bit native files or is there some processing or manipulation to the original 16 bits. I really don't like my bits touched :~) no pun intended.

Themis

2009-11-15, 01:36

Hi John

This is probably a very stupid question. Does it just add bits on top of the 16 bit native files or is there some processing or manipulation to the original 16 bits. I really don't like my bits touched :~) no pun intended.

I doubt *any* processing would manipulate bit depth, as this is directly related to noise ratio. They all simply add non-significant information (zeros) to the bit depth (on top, as you say).

This is probably a very stupid question. Does it just add bits on top of the 16 bit native files or is there some processing or manipulation to the original 16 bits. I really don't like my bits touched :~) no pun intended.

I doubt *any* processing would manipulate bit depth, as this is directly related to noise ratio. They all simply add non-significant information (zeros) to the bit depth (on top, as you say).

Mnyb

2009-11-15, 08:48

I doubt *any* processing would manipulate bit depth, as this is directly related to noise ratio. They all simply add non-significant information (zeros) to the bit depth (on top, as you say).

would it not be added (padded) from the bottom beging with lsb ??

The lowest 8 bits get zeros and the rest is the 16bits delivered with the file, this makes headroom for the near transparent 24bit digital volume that we have.

I dont know but supects that the zeros are added from the bottom as otherwise my 24bit files should be very loud compared to the 16bit but they are not.

would it not be added (padded) from the bottom beging with lsb ??

The lowest 8 bits get zeros and the rest is the 16bits delivered with the file, this makes headroom for the near transparent 24bit digital volume that we have.

I dont know but supects that the zeros are added from the bottom as otherwise my 24bit files should be very loud compared to the 16bit but they are not.

radish

2009-11-15, 10:09

Yes, the extra 0 byte is added to the bottom (lsb).

batka

2009-11-15, 10:21

This is not clear to me.

Sorry if it's dumb question, but if it's converted to 24 bit, even if zeros added, how could it be bit-perfect then?

Sorry if it's dumb question, but if it's converted to 24 bit, even if zeros added, how could it be bit-perfect then?

JohnSwenson

2009-11-15, 18:09

Think of it this way:

12 = 12.0 = 12.00 = 12.000 = 12.00000000

The value is EXACTLY the same even though you add extra zeros to the right.

John S.

12 = 12.0 = 12.00 = 12.000 = 12.00000000

The value is EXACTLY the same even though you add extra zeros to the right.

John S.

pfarrell

2009-11-15, 18:15

JohnSwenson wrote:

> Think of it this way:

>

> 12 = 12.0 = 12.00 = 12.000 = 12.00000000

>

> The value is EXACTLY the same even though you add extra zeros to the

> right.

While I love how simple this is, and it greatly conveys the idea. There

are problems. Computers don't do floating point the way humans think.

The idea is right, the example, not so much in general

The good news is that SPDIF and all PCM audio uses integers, not

floating point, so the essence of what John is saying is true, even if

the details are not.

--

Pat Farrell

http://www.pfarrell.com/

> Think of it this way:

>

> 12 = 12.0 = 12.00 = 12.000 = 12.00000000

>

> The value is EXACTLY the same even though you add extra zeros to the

> right.

While I love how simple this is, and it greatly conveys the idea. There

are problems. Computers don't do floating point the way humans think.

The idea is right, the example, not so much in general

The good news is that SPDIF and all PCM audio uses integers, not

floating point, so the essence of what John is saying is true, even if

the details are not.

--

Pat Farrell

http://www.pfarrell.com/

JohnSwenson

2009-11-15, 23:58

I was trying to convey the concept, but if the details are important:

I'll still do the basic concept in decimal because that is what MOST humans are more comfortable in, then I'll go into the actual binary details.

DACs used normalised integers. The important concept is what is called Full Scale or FS for short. The official CD standard says that FS produces 2 volts on the output jacks. No matter how many digits are used to reperesnt the numbers, the largest one it can represent produces 2V.

For the sake of making things easy I'll do an example in decimal. Lets say you have two digits to play with, this gives a range of 0 - 99. To make this easy lets say full scale is defined as 100. Thus the maximum value you can get is 99/100 = .99FS.

Lets then go to a system with 3 digits, you now have 0-999 as possible values, BUT we also change the definition of FS to 1000, so the maximum value is now 999/1000 = .999FS. IF we take our original 99 and convert it to the second system we add a a zero to the right, which gives us 990, thus 990/1000 = .99FS. You will get exactly the same voltage from the output.

If you've followed this you have now realized tat the maximum number (999 in this example) isn't really the defined FS (1000). Thus you will never actually get exactly 2V from the output, but you can get very close. (The more digits the closer you can get)

Now to the REAL nitty gritty details. In reality twos complement integers are used. This is a way of representing both positive and negative numbers as integers. Here is a simple 4 bit example. Zero is represented as 0000, one is 0001, two is 0010 etc, just what you would expect from binary representation. The fun part is when it goes negative, minus 1 is 1111, minus 2 is 1110 etc. This is done so that if you add minus one and plus one you get zero. An interesting aspect of this is that you can go from (in this example) minus 8 to plus 7. This system would get mapped to minus 1 to plus 1 volts, its still a two volt range.

In this case FS gets slightly more complicated because you have to deal with negative numbers. For the 4 bit case its still common to say FS is 16, even though its really -8 to +8. You still have the issue that you can never quite hit FS on the top (but you can on the bottom!)

Adding bits works the same way as with the decimal example. The integers get bigger but so does FS. Adding zeros to the right does the same thing, the ratio of FS stays exactly the same so the voltage you get from the DAC does not change.

When adding zeros in this way its perfectly legitimate to call this "bit perfect" because the voltage output from the DAC does not change.

For completeness the numbers for a 16 bit system are:

FS = 65536

min = -32768

max = 32767

For 24 bit its:

FS = 16777216

min = -8388608

max = 8388607

In both cases that maps to -1 to +1 on the output of the DAC.

So there you have the details.

Just don't ask WHY you want to add zeros, THATS another post!

John S.

I'll still do the basic concept in decimal because that is what MOST humans are more comfortable in, then I'll go into the actual binary details.

DACs used normalised integers. The important concept is what is called Full Scale or FS for short. The official CD standard says that FS produces 2 volts on the output jacks. No matter how many digits are used to reperesnt the numbers, the largest one it can represent produces 2V.

For the sake of making things easy I'll do an example in decimal. Lets say you have two digits to play with, this gives a range of 0 - 99. To make this easy lets say full scale is defined as 100. Thus the maximum value you can get is 99/100 = .99FS.

Lets then go to a system with 3 digits, you now have 0-999 as possible values, BUT we also change the definition of FS to 1000, so the maximum value is now 999/1000 = .999FS. IF we take our original 99 and convert it to the second system we add a a zero to the right, which gives us 990, thus 990/1000 = .99FS. You will get exactly the same voltage from the output.

If you've followed this you have now realized tat the maximum number (999 in this example) isn't really the defined FS (1000). Thus you will never actually get exactly 2V from the output, but you can get very close. (The more digits the closer you can get)

Now to the REAL nitty gritty details. In reality twos complement integers are used. This is a way of representing both positive and negative numbers as integers. Here is a simple 4 bit example. Zero is represented as 0000, one is 0001, two is 0010 etc, just what you would expect from binary representation. The fun part is when it goes negative, minus 1 is 1111, minus 2 is 1110 etc. This is done so that if you add minus one and plus one you get zero. An interesting aspect of this is that you can go from (in this example) minus 8 to plus 7. This system would get mapped to minus 1 to plus 1 volts, its still a two volt range.

In this case FS gets slightly more complicated because you have to deal with negative numbers. For the 4 bit case its still common to say FS is 16, even though its really -8 to +8. You still have the issue that you can never quite hit FS on the top (but you can on the bottom!)

Adding bits works the same way as with the decimal example. The integers get bigger but so does FS. Adding zeros to the right does the same thing, the ratio of FS stays exactly the same so the voltage you get from the DAC does not change.

When adding zeros in this way its perfectly legitimate to call this "bit perfect" because the voltage output from the DAC does not change.

For completeness the numbers for a 16 bit system are:

FS = 65536

min = -32768

max = 32767

For 24 bit its:

FS = 16777216

min = -8388608

max = 8388607

In both cases that maps to -1 to +1 on the output of the DAC.

So there you have the details.

Just don't ask WHY you want to add zeros, THATS another post!

John S.

Phil Leigh

2009-11-16, 00:15

All of which is proven in practice by the fact that dts replay is perfect on max volume, which it wouldn't be if even the lowest bit of the 16 originals had been changed in any way, therefore it is "bit-perfect".

MrRalph

2009-11-17, 03:57

Looking forward to the Touch! I read the Touch can do 24/96 by itself. Does it also output higher rez to the S/PDIF out? So if I play 24/192 material, what happens?

1) The Touch itself gets downsampled 24/96 from the server (SOX) of does the Touch downsample itself for analoge out?

2) Can it feed my DAC 24/192?

Ralph

1) The Touch itself gets downsampled 24/96 from the server (SOX) of does the Touch downsample itself for analoge out?

2) Can it feed my DAC 24/192?

Ralph

Phil Leigh

2009-11-17, 04:45

Looking forward to the Touch! I read the Touch can do 24/96 by itself. Does it also output higher rez to the S/PDIF out? So if I play 24/192 material, what happens?

1) The Touch itself gets downsampled 24/96 from the server (SOX) of does the Touch downsample itself for analoge out?

2) Can it feed my DAC 24/192?

Ralph

Nope - 24/96 is what it outputs via s/pdif. Downsampling of > 96 happens on the server.

1) The Touch itself gets downsampled 24/96 from the server (SOX) of does the Touch downsample itself for analoge out?

2) Can it feed my DAC 24/192?

Ralph

Nope - 24/96 is what it outputs via s/pdif. Downsampling of > 96 happens on the server.

MrRalph

2009-11-17, 14:39

Nope - 24/96 is what it outputs via s/pdif. Downsampling of > 96 happens on the server.

Thanks for clarifying, helps a lot.

Ralph

Thanks for clarifying, helps a lot.

Ralph

bingcrosby

2009-12-03, 22:48

Does the touch alter a 24 bit input signal(with sampling rate no higher than 96 kHz) by adding bits? Also does it support any sampling rate between 32 kHz and 96 kHz?

I have a dac which is upto 24 bit 96 kHz compatible, but due to the noise floor of the analog some say the actual resolution of the dac is a bit lower than this ~ 20 bit. If the touch adds zeroes to a 16 bit CD to make it 24 bit, will I infact loose some of the CD data due a shifting in the position of the bits?

I have a dac which is upto 24 bit 96 kHz compatible, but due to the noise floor of the analog some say the actual resolution of the dac is a bit lower than this ~ 20 bit. If the touch adds zeroes to a 16 bit CD to make it 24 bit, will I infact loose some of the CD data due a shifting in the position of the bits?

Mnyb

2009-12-03, 23:33

Does the touch alter a 24 bit input signal(with sampling rate no higher than 96 kHz) by adding bits? Also does it support any sampling rate between 32 kHz and 96 kHz?

I have a dac which is upto 24 bit 96 kHz compatible, but due to the noise floor of the analog some say the actual resolution of the dac is a bit lower than this ~ 20 bit. If the touch adds zeroes to a 16 bit CD to make it 24 bit, will I infact loose some of the CD data due a shifting in the position of the bits?

No it adds bit from LSB "the bottom" so the 16 bits are shifted "upwards"

so that the 16bit MSB will be the 24bit MSB

Unless you fiddle with volume controll or replaygain.

Yes 24bits is 144dB signal to noise no practical device has that.

I have a dac which is upto 24 bit 96 kHz compatible, but due to the noise floor of the analog some say the actual resolution of the dac is a bit lower than this ~ 20 bit. If the touch adds zeroes to a 16 bit CD to make it 24 bit, will I infact loose some of the CD data due a shifting in the position of the bits?

No it adds bit from LSB "the bottom" so the 16 bits are shifted "upwards"

so that the 16bit MSB will be the 24bit MSB

Unless you fiddle with volume controll or replaygain.

Yes 24bits is 144dB signal to noise no practical device has that.

Phil Leigh

2009-12-03, 23:51

Does the touch alter a 24 bit input signal(with sampling rate no higher than 96 kHz) by adding bits? Also does it support any sampling rate between 32 kHz and 96 kHz?

I have a dac which is upto 24 bit 96 kHz compatible, but due to the noise floor of the analog some say the actual resolution of the dac is a bit lower than this ~ 20 bit. If the touch adds zeroes to a 16 bit CD to make it 24 bit, will I infact loose some of the CD data due a shifting in the position of the bits?

As MNYB said. I think you are confusing a few things.

Any SB device uses a 24-bit path internally, so if it gets 24 bits in it puts exactly the same 24-bits out - if volume is on MAX and Replaygain is disabled.

If it gets a 16-bit input, it adds zeroes to the end first, then behaves as it would with a 24-bit input.

The fact that your DAC (any DAC) can't resolve the lowest bits in the analogue domain because they are below the noise floor is irrelevant to the SB device and its digital output.

There is NO shifting of bits or loss of anything (until one is INSIDE the DAC itself).

Phil

I have a dac which is upto 24 bit 96 kHz compatible, but due to the noise floor of the analog some say the actual resolution of the dac is a bit lower than this ~ 20 bit. If the touch adds zeroes to a 16 bit CD to make it 24 bit, will I infact loose some of the CD data due a shifting in the position of the bits?

As MNYB said. I think you are confusing a few things.

Any SB device uses a 24-bit path internally, so if it gets 24 bits in it puts exactly the same 24-bits out - if volume is on MAX and Replaygain is disabled.

If it gets a 16-bit input, it adds zeroes to the end first, then behaves as it would with a 24-bit input.

The fact that your DAC (any DAC) can't resolve the lowest bits in the analogue domain because they are below the noise floor is irrelevant to the SB device and its digital output.

There is NO shifting of bits or loss of anything (until one is INSIDE the DAC itself).

Phil

bingcrosby

2009-12-04, 00:35

Thanks guys, it was as I thought. I was worried that the added bits would head the 16 bits.

Is the correct interpretation that my DAC gets the full 24 bits, however that the SNR of the DAC limits the lowest order bits from actually influencing the resulting analogue waveform? So that the DAC does a full 24 bit reproduction but then the extra information from the lowest 4 ish bits is lost in the SNR of the device?

Cheers.

Is the correct interpretation that my DAC gets the full 24 bits, however that the SNR of the DAC limits the lowest order bits from actually influencing the resulting analogue waveform? So that the DAC does a full 24 bit reproduction but then the extra information from the lowest 4 ish bits is lost in the SNR of the device?

Cheers.

Phil Leigh

2009-12-04, 00:51

Thanks guys, it was as I thought. I was worried that the added bits would head the 16 bits.

Is the correct interpretation that my DAC gets the full 24 bits, however that the SNR of the DAC limits the lowest order bits from actually influencing the resulting analogue waveform? So that the DAC does a full 24 bit reproduction but then the extra information from the lowest 4 ish bits is lost in the SNR of the device?

Cheers.

If the zeroes went to the 8 MSB's, that would be pretty disastrous!

Yes, the 24th bit (for example) is at -144dB which is way below the noise floor of all domestic audio equpiment. Not just your DAC analogue output stage, also your amps...

Is the correct interpretation that my DAC gets the full 24 bits, however that the SNR of the DAC limits the lowest order bits from actually influencing the resulting analogue waveform? So that the DAC does a full 24 bit reproduction but then the extra information from the lowest 4 ish bits is lost in the SNR of the device?

Cheers.

If the zeroes went to the 8 MSB's, that would be pretty disastrous!

Yes, the 24th bit (for example) is at -144dB which is way below the noise floor of all domestic audio equpiment. Not just your DAC analogue output stage, also your amps...

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