7 ms·

How does 24/192 help with dynamic range?

akvadrako 9y ago
That's the 24-bit part, which provides more play room then 16-bit. Normally 16 is plenty, by consider an example track which is heavily compressed so only the top 1% of range is used: 16-bits = 2^16 * 0.01 = 655 discrete levels 24-bits = 2^24 * 0.01 = 168000 discrete levels
- Mediterraneo10 9y ago
  No, that has nothing to do with it. See my reply above.
  - akvadrako 9y ago
    Your reply is not the whole picture. Compression reduces the effective number of bits of resolution.
    - hashmal 9y ago
      If you had an RGB (24 bits) image that is mostly very dark, would you say that somehow because the image doesn't use the full range of possible values, the image quality suffered? Would you say that changing the format/bit-depth would actually lead to a perceived increase in quality when looking at the image?
      - akvadrako 9y ago
        I don't understand why you said I was wrong in the other comment, but you give here an appropriate analogy which makes the effect obvious. Because we only have 8 bits per channel, we all can see banding in (dynamically) compressed images. That's what increasing the depth improves.
- djaychela 9y ago
  I think you have the wrong end of the stick there. Even if a track is heavily multiband compressed, it's still using the full range of levels, as audio will oscillate between -ve and +ve for each cycle, potentially using most of the 65,536 levels available at 16-bit (and the 16777216 levels at 24 bit). The 0.01 is not relevant.
  - akvadrako 9y ago
    No, I don't have it wrong. Compression means it's effectively using less of the range and the dynamics could be represented with fewer bits in an optimal encoding. All the levels may be used in track, but it's more biased towards the high end. To help you understand, imagine 1-bit audio. What would it sound like? For each frequency, you can only have a single volume, i.e. it's maximally compressed.
    - hashmal 9y ago
      I can confirm, you have it wrong.
      - dom0 9y ago
        But also correct here: > Compression means it's effectively using less of the [dynamic] range and the dynamics [range] could be represented with fewer bits in an optimal encoding. Lower DNR can indeed be encoded with fewer bits per sample.
    - djaychela 9y ago
      I still think you have it wrong; (Dynamic Audio) Compression does mean it may be using less of the dynamic range when taken overall, but the waveform will still take up the entire span of the available bit depth. Granted that with an optimal encoding this could be biased towards the "high" end (but bear in mind that would have to be both +ve and -ve), but I still think this would miss the point, and I think it would also create distortion in the 'zero' region of the audio. If you open up heavily compressed music in an audio editor (pick any recent EDM track), it will still cross the zero point, and have maximum and minimum levels present. It will also have 'zero' (i.e. 32768 if the range is 0-65536 in 16-bit audio) present - often at the beginning and end of the track, and the audio will be centred about that in the vast majority of tracks. I'm not sure the 1-bit extreme helps with understanding this, particularly because the audio we'd be dealing with will be a mix of many frequencies; the waveform is an aggregate of the overall 'signal' at each frequency present in the mix; this is one of the reasons why multiband rather than single-band compression has become so popular (as it allows you to get the 'max' in each frequency band rather than one band dominating the overall result of the compression). I think there's a difference to take into account when considering any given momentary sample and the overall effect - yes, compression does reduce the dynamic range of the music, but you would need some sort of variable bit depth which was linked to the instantaneous amount of compression being applied to get any kind of workable lower-bit-depth encoding scheme, which seems like a lot of complexity for no significant gain (to me?).
- copperx 9y ago
  As Barbie would say: Audio engineering is hard! Let's play with the calculator.
  - Blahah 9y ago
    Barbie has piloted commercial airliners, been on a scientific space mission, and has an M.D., so I don't think she'd be likely to say that.
    - finnthehuman 9y ago
      https://en.wikipedia.org/wiki/Teen_Talk_Barbie https://en.wikipedia.org/wiki/Teen_Talk_Barbie The meme "Math class is hard, let's go shopping" is only slightly apocryphal. Two of the voicebox lines were "Math class is tough" and "Want to go shopping?"
Mediterraneo10 9y ago
It has nothing to do with the 24/192 values themselves, but rather the treatment given to the recording during the mastering. Because 24/192 and SACD are "audiophile formats", the engineers who master the recordings expect them to be listened to in serious listening environments that are relatively noise-insulated so that very soft parts can be heard clearly, and there are no complaining neighbours so the loud parts can really blast. Consequently, an ample dynamic range is preserved. "Remastered" reissues on the other formats – CD or lower-bitrate downloads – are nowadays expected to be listened to through cheap earbuds or speakers and perhaps in noisy urban environments. So, the engineer applies "loudness wars" treatment, compressing their dynamic range so the listener can still hear the quiet parts even with all the noise around them.
- aw3c2 9y ago
  You are proving the point. Treating the audio well, not giving in to the loudness, has only correlation with 24/192 releases (if you say so), there is no causation.
  - Mediterraneo10 9y ago
    Sure, there is no causation of "higher bitrate necessarily makes for better sound", but the correlation of "higher bitrate format just happens to have better mastering" is often strong enough to make these the go-to formats. There are a large number of classic recordings that are difficult to obtain in the digital era without harsh compression being applied to them, so the 24/192 or SACD release is the most convenient way of hearing them with decent dynamic range.
  - traviscj 9y ago
    In other words, it is not a technical advantage, but a social/political one.
  - JohnBooty 9y ago
    Right, absolutely. 16 bit samples give 100db+ of perceived dynamic range, more than would ever be needed. The other poster is not saying otherwise, though. They're just saying that "hi-def" formats, while technically unnecessary for end-user listening, are often the only way to obtain a decently-mastered recording. There's no technical reason for things to be that way. But that's how things are. It's sort of like buying a new car. You want the more powerful engine? Well then you also have to buy the package with heated seats, the moonroof, and the fog lights. There's no technical reason for it, but that's the only way they'll sell it to you.

That's the 24-bit part, which provides more play room then 16-bit. Normally 16 is plenty, by consider an example track which is heavily compressed so only the top 1% of range is used: 16-bits = 2^16 * 0.01 = 655 discrete levels 24-bits = 2^24 * 0.01 = 168000 discrete levels

Mediterraneo10 9y ago
No, that has nothing to do with it. See my reply above.
- akvadrako 9y ago
  Your reply is not the whole picture. Compression reduces the effective number of bits of resolution.
  - hashmal 9y ago
    If you had an RGB (24 bits) image that is mostly very dark, would you say that somehow because the image doesn't use the full range of possible values, the image quality suffered? Would you say that changing the format/bit-depth would actually lead to a perceived increase in quality when looking at the image?
    - akvadrako 9y ago
      I don't understand why you said I was wrong in the other comment, but you give here an appropriate analogy which makes the effect obvious. Because we only have 8 bits per channel, we all can see banding in (dynamically) compressed images. That's what increasing the depth improves.
djaychela 9y ago
I think you have the wrong end of the stick there. Even if a track is heavily multiband compressed, it's still using the full range of levels, as audio will oscillate between -ve and +ve for each cycle, potentially using most of the 65,536 levels available at 16-bit (and the 16777216 levels at 24 bit). The 0.01 is not relevant.
- akvadrako 9y ago
  No, I don't have it wrong. Compression means it's effectively using less of the range and the dynamics could be represented with fewer bits in an optimal encoding. All the levels may be used in track, but it's more biased towards the high end. To help you understand, imagine 1-bit audio. What would it sound like? For each frequency, you can only have a single volume, i.e. it's maximally compressed.
  - hashmal 9y ago
    I can confirm, you have it wrong.
    - dom0 9y ago
      But also correct here: > Compression means it's effectively using less of the [dynamic] range and the dynamics [range] could be represented with fewer bits in an optimal encoding. Lower DNR can indeed be encoded with fewer bits per sample.
  - djaychela 9y ago
    I still think you have it wrong; (Dynamic Audio) Compression does mean it may be using less of the dynamic range when taken overall, but the waveform will still take up the entire span of the available bit depth. Granted that with an optimal encoding this could be biased towards the "high" end (but bear in mind that would have to be both +ve and -ve), but I still think this would miss the point, and I think it would also create distortion in the 'zero' region of the audio. If you open up heavily compressed music in an audio editor (pick any recent EDM track), it will still cross the zero point, and have maximum and minimum levels present. It will also have 'zero' (i.e. 32768 if the range is 0-65536 in 16-bit audio) present - often at the beginning and end of the track, and the audio will be centred about that in the vast majority of tracks. I'm not sure the 1-bit extreme helps with understanding this, particularly because the audio we'd be dealing with will be a mix of many frequencies; the waveform is an aggregate of the overall 'signal' at each frequency present in the mix; this is one of the reasons why multiband rather than single-band compression has become so popular (as it allows you to get the 'max' in each frequency band rather than one band dominating the overall result of the compression). I think there's a difference to take into account when considering any given momentary sample and the overall effect - yes, compression does reduce the dynamic range of the music, but you would need some sort of variable bit depth which was linked to the instantaneous amount of compression being applied to get any kind of workable lower-bit-depth encoding scheme, which seems like a lot of complexity for no significant gain (to me?).
copperx 9y ago
As Barbie would say: Audio engineering is hard! Let's play with the calculator.
- Blahah 9y ago
  Barbie has piloted commercial airliners, been on a scientific space mission, and has an M.D., so I don't think she'd be likely to say that.
  - finnthehuman 9y ago
    https://en.wikipedia.org/wiki/Teen_Talk_Barbie https://en.wikipedia.org/wiki/Teen_Talk_Barbie The meme "Math class is hard, let's go shopping" is only slightly apocryphal. Two of the voicebox lines were "Math class is tough" and "Want to go shopping?"

Mediterraneo10 9y ago

It has nothing to do with the 24/192 values themselves, but rather the treatment given to the recording during the mastering. Because 24/192 and SACD are "audiophile formats", the engineers who master the recordings expect them to be listened to in serious listening environments that are relatively noise-insulated so that very soft parts can be heard clearly, and there are no complaining neighbours so the loud parts can really blast. Consequently, an ample dynamic range is preserved. "Remastered" reissues on the other formats – CD or lower-bitrate downloads – are nowadays expected to be listened to through cheap earbuds or speakers and perhaps in noisy urban environments. So, the engineer applies "loudness wars" treatment, compressing their dynamic range so the listener can still hear the quiet parts even with all the noise around them.

aw3c2 9y ago
You are proving the point. Treating the audio well, not giving in to the loudness, has only correlation with 24/192 releases (if you say so), there is no causation.
- Mediterraneo10 9y ago
  Sure, there is no causation of "higher bitrate necessarily makes for better sound", but the correlation of "higher bitrate format just happens to have better mastering" is often strong enough to make these the go-to formats. There are a large number of classic recordings that are difficult to obtain in the digital era without harsh compression being applied to them, so the 24/192 or SACD release is the most convenient way of hearing them with decent dynamic range.
- traviscj 9y ago
  In other words, it is not a technical advantage, but a social/political one.
- JohnBooty 9y ago
  Right, absolutely. 16 bit samples give 100db+ of perceived dynamic range, more than would ever be needed. The other poster is not saying otherwise, though. They're just saying that "hi-def" formats, while technically unnecessary for end-user listening, are often the only way to obtain a decently-mastered recording. There's no technical reason for things to be that way. But that's how things are. It's sort of like buying a new car. You want the more powerful engine? Well then you also have to buy the package with heated seats, the moonroof, and the fog lights. There's no technical reason for it, but that's the only way they'll sell it to you.