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post #31 of 54 Old 12-17-2008, 08:45 PM
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Originally Posted by ryrules1 View Post

Exactly thats what in trying to tell there fools, Im not a genius but I know what sounds better. 24bit-96Khz encoded using FLACer about 200mb a song.

I'm assuming that's per channel?

If it sounds good to you, no reason to change, but as an FYI, modern DACs introduce about 3 times more noise when converting a 92kHz signal than when converting a 44.1kHz signal. We're not exactly talking about earth shattering sound quality differences as THD+N goes from about 0.0005% to 0.0015%, but like I mentioned earlier there is really no audible advantage to encode at anything higher than 44.1/24.
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post #32 of 54 Old 12-17-2008, 09:16 PM
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I had the same plan as the OP.

So I have a Denon AV receiver that I will be using for my "amp." Thus, it has many optical inputs.

If this is the case, do I need to get an external USB DAC or does the optical cable from the line-out of the macbook pro work just fine?

This is new to me.
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post #33 of 54 Old 12-17-2008, 09:46 PM
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If this is the case, do I need to get an external USB DAC or does the optical cable from the line-out of the macbook pro work just fine?

it'll work just fine. It's a mini-toslink to toslink cable. But for $100 you can free your MBP from your amp.
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post #34 of 54 Old 12-17-2008, 09:48 PM
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Originally Posted by Bing View Post

it'll work just fine. It's a mini-toslink to toslink cable. But for $100 you can free your MBP from your amp.

"Free it" as in get an external usb dac? I won't mind just plugging in a mini-toslink to the computer when I need it, as it seems easy.
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post #35 of 54 Old 12-17-2008, 10:12 PM
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Quote:
Originally Posted by Bing View Post

it'll work just fine. It's a mini-toslink to toslink cable. But for $100 you can free your MBP from your amp.

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Originally Posted by ViperVenom18 View Post

"Free it" as in get an external usb dac? I won't mind just plugging in a mini-toslink to the computer when I need it, as it seems easy.

As pointed out, the optical connection to your receiver will work just fine. Your receiver's DACs will then be used.

I'm not certain, but I think he is referring to buying an AirPort Express, which should be LESS than $100. The AirPort Express has the same mini-toslink optical out as the MBP. With it, you will be able to transmit your music from your MBP to the receiver wirelessly.

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post #36 of 54 Old 12-17-2008, 10:20 PM
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Originally Posted by sivadselim View Post

As pointed out, the optical connection to your receiver will work just fine. Your receiver's DACs will then be used.

I'm not certain, but I think he is referring to buying an AirPort Express, which should be LESS than $100. The AirPort Express has the same mini-toslink optical out as the MBP. With it, you will be able to transmit your music from your MBP to the receiver wirelessly.

Awesome. So since my receiver accepts optical input, I'll just go that route!
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post #37 of 54 Old 12-17-2008, 11:09 PM
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I'm not certain, but I think he is referring to buying an AirPort Express, which should be LESS than $100. The AirPort Express has the same mini-toslink optical out as the MBP. With it, you will be able to transmit your music from your MBP to the receiver wirelessly.

Yeah that's what I meant.
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post #38 of 54 Old 12-18-2008, 03:17 AM
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Originally Posted by cansp6 View Post

Actually it would be oversampled, thus guaranteeing that it would sound exactly the same for sources 48kHz or less. According to Nyquist's Theorem, a digital signal can be used to encode an analog wave without loss if the digital signal is sampled at twice the rate of the analog. By going above and beyond that sampling rate (from 44.1k to 48k), you are simply being less efficient in your encoding. No signal loss occurs during encoding.

It is possible to do a good job resampling a 44.1KHz signal to 48KHz. Good enough that I think the difference could be inaudible. But it isn't quite perfect, since it sort of compounds the quantization error. IMO, you could call that signal loss. I would not want such a resampler in my hifi system.

In reality, the upsampling isn't as easy to do as it seems like it should be. I think it is very likely that a $15 sound card would not do it very well, and that it would introduce audible noise as a result.

-Max
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post #39 of 54 Old 12-18-2008, 04:03 AM
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Originally Posted by maxcooper View Post

Are you sure that the laptop doesn't already do digital out? My old MacBookPro had a headphone jack that could also be used as an optical digital output using a little adapter thing. Yes, I know that may sound nuts, but I know what I am talking about and the same jack could be used for headphones or toslink. I think mine even came with the adapter.

Update: Apple says MacBook already has digital out:
http://support.apple.com/kb/HT1562

What exactly does it output? Is it PCM, or does my receiver need to know how to decode the lossless format? Would it output a DRM file or not? I don't have a minitoslink - toslink handy, or I'd just plug it in and see.
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post #40 of 54 Old 12-18-2008, 08:25 AM
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What exactly does it output?

PCM I suppose. The MBP has a cd/dvd drive so neither format has lossless codecs.
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post #41 of 54 Old 12-18-2008, 08:27 AM
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I don't have a minitoslink - toslink handy, or I'd just plug it in and see.

????? It won't even fit. The output is 1/8" headphone jack.


Edit: Nevermind my post........I goofed...
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post #42 of 54 Old 12-18-2008, 09:43 AM
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^ the output jack can be used as a mini-headphone jack OR mini-toslink. This article shows one connector that will work, but the one I have is even smaller -- just a little adapter that plugs in and has a normal optical toslink connector on the other side.

http://digitalprosound.digitalmedian...fterinter=true

-Max
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post #43 of 54 Old 12-18-2008, 10:14 AM
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The Dac Magic by cambridge has a USB option so ur laptop doesnt even need to have a optical out. No drivers are needed either just plug and play, plus it sounds fantastic, i use it for my laptop and my cd player
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post #44 of 54 Old 12-19-2008, 03:44 AM
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Originally Posted by Bing View Post

PCM I suppose. The MBP has a cd/dvd drive so neither format has lossless codecs.

FLAC, Apple Lossless?
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post #45 of 54 Old 12-19-2008, 02:54 PM
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Quote:
Originally Posted by maxcooper View Post

It is possible to do a good job resampling a 44.1KHz signal to 48KHz. Good enough that I think the difference could be inaudible. But it isn't quite perfect, since it sort of compounds the quantization error. IMO, you could call that signal loss. I would not want such a resampler in my hifi system.

In reality, the upsampling isn't as easy to do as it seems like it should be. I think it is very likely that a $15 sound card would not do it very well, and that it would introduce audible noise as a result.

-Max

I might be hitting my head against a wall here, but we go. Quantization errors are due to encoding of the waveform in some sort of PCM scheme. It has absolutely nothing to do with the sampling frequency. The 44.1kHz or 48kHz is simply a measure of how many samples (of the waveform) you are taking each second. There is mathematical proof, based on our current knowledge of the universe, that in order to be lossless the sample rate needs to be double the frequency of the wave. Since human ears are incapable of hearing anything about 20kHz or so, there is no audible reason to sample more than 40,000 times per second. Let's be really sure and use 44,100 data points.

So now we have 44,100 data points for each second of music. We then need to package these data points in some sort of way so that we can exchange this information between the player and the receiver or whatever. CDs use 44.1/24 which means that each data point is encoded using 24bits, which is enough to record values anywhere from zero to 16,777,216 if we use the absolute basic, first grade level of PCM.

Now I suppose you could argue that 16.7 million is not enough to properly represent the height of the wave at each of the data points. That somehow some information is missed by using this constrained quantum (range). I suppose you could argue about quantization errors. In theory, it's possible. On the other hand, in theory anything is possible.

But if we're going to talk about theory, then we really need to remember that no modern day ADC is going to use such a simple and inefficient PCM algorithm. Modern day ADCs use QPCM, which encodes the full height only for the very first data point and thereafter encodes differences between this data point and the next. I don't think I need to explain how this method allows for a significantly higher range to be encoded.

Now getting back to the 44.1 to 48kHz conversion, a very basic approach to create the missing 3900 data points is simply to make them up. You slot the missing data so that each new point is in between 2 existing points. Then, you calculate the average value of the 2 existing points and assign it to the newly added point. Using the same proof as before, we can guaranteed that not only is no new information added, but also that no old information is lost. Thus, you get a lossless conversion from 44.1 to 48kHz.

Do I win the internets now?
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post #46 of 54 Old 12-19-2008, 03:35 PM
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Quote:
Originally Posted by cansp6 View Post

CDs use 44.1/24 which means that each data point is encoded using 24bits...........

Whether it is relevant to the point you are trying to make or not, CDs are 16bit. Not 24bit.

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post #47 of 54 Old 12-19-2008, 07:48 PM
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Quote:
Originally Posted by cansp6 View Post

I might be hitting my head against a wall here, but we go. Quantization errors are due to encoding of the waveform in some sort of PCM scheme. It has absolutely nothing to do with the sampling frequency. The 44.1kHz or 48kHz is simply a measure of how many samples (of the waveform) you are taking each second. There is mathematical proof, based on our current knowledge of the universe, that in order to be lossless the sample rate needs to be double the frequency of the wave. Since human ears are incapable of hearing anything about 20kHz or so, there is no audible reason to sample more than 40,000 times per second. Let's be really sure and use 44,100 data points.

So now we have 44,100 data points for each second of music. We then need to package these data points in some sort of way so that we can exchange this information between the player and the receiver or whatever. CDs use 44.1/24 which means that each data point is encoded using 24bits, which is enough to record values anywhere from zero to 16,777,216 if we use the absolute basic, first grade level of PCM.

Now I suppose you could argue that 16.7 million is not enough to properly represent the height of the wave at each of the data points. That somehow some information is missed by using this constrained quantum (range). I suppose you could argue about quantization errors. In theory, it's possible. On the other hand, in theory anything is possible.

But if we're going to talk about theory, then we really need to remember that no modern day ADC is going to use such a simple and inefficient PCM algorithm. Modern day ADCs use QPCM, which encodes the full height only for the very first data point and thereafter encodes differences between this data point and the next. I don't think I need to explain how this method allows for a significantly higher range to be encoded.

Now getting back to the 44.1 to 48kHz conversion, a very basic approach to create the missing 3900 data points is simply to make them up. You slot the missing data so that each new point is in between 2 existing points. Then, you calculate the average value of the 2 existing points and assign it to the newly added point. Using the same proof as before, we can guaranteed that not only is no new information added, but also that no old information is lost. Thus, you get a lossless conversion from 44.1 to 48kHz.

Do I win the internets now?

Let's be cool and keep it informational rather than confrontational. This stuff is pretty interesting, though I am sure we are boring most of the readers. I looked a bunch of this stuff up as a result of our exchange. I looked it up because I suspected that you were oversimplifying things and because I wanted to understand how resampling worked and what its limitations are. I remain convinced that I don't want my hifi system to upsample from 44.1 to 48KHz because it definitely changes the audio, for the worse, even with a high-quality resampler. That change may or may not be audible, but I'd rather not take the chance. And it seems likely that the $15 USB sound card doesn't use a high-quality resampler, so I stand by my suggestion not to use it.

Let's say that upsampling from 44.1 to 48 requires about one new sample for each 12 samples in the original file. So imagine the following 12 samples from the original file:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Your proposal is to squeeze in an extra sample with a value that is the average of the two adjacent samples. So here is the resampled version, with the new sample added in the middle (another 6):

0, 1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11

These two series of samples will play with precisely the same duration. Notice that the audio produced will not be exactly the same. The output of the resampled version becomes 1 at just 1/13th of the way into playback, where the original does not go to 1 until 1/12th of the way through. The resampled version also plays two 6s in a row, while the original progresses linearly from 0 to 11. The resampled version is inferior to the original.

This is the simplest example I could come up with. And though the resampling algorithm you proposed is not a good one, I think it does as well as any could do on this set of samples. No upsampling algorithm is perfect when the new sample rate is not an integer multiple of the original sample rate.

If 13 samples were taken from the original analog audio waveform (and not the 12 samples we started with in the example), the extra sample could end up anywhere in the sequence. Let's say that it should have occurred at sample 2. That means all the samples between sample 2 and where we actually added the new sample are off by 1. Perhaps calling this an increase in quantization error isn't quite right (or maybe it is?), but the effect is quite similar, since it is caused by the limited resolution of the samples.

Here are some links about Nyquist's Theorem and resampling that you may find interesting:
http://www.wescottdesign.com/article.../sampling.html
http://ccrma.stanford.edu/~jos/resample/
http://www.mega-nerd.com/SRC/

-Max
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post #48 of 54 Old 12-20-2008, 12:57 PM
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Quote:
Originally Posted by maxcooper View Post

Let's say that upsampling from 44.1 to 48 requires about one new sample for each 12 samples in the original file. So imagine the following 12 samples from the original file:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Your proposal is to squeeze in an extra sample with a value that is the average of the two adjacent samples. So here is the resampled version, with the new sample added in the middle (another 6):

0, 1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11

These two series of samples will play with precisely the same duration. Notice that the audio produced will not be exactly the same. The output of the resampled version becomes 1 at just 1/13th of the way into playback, where the original does not go to 1 until 1/12th of the way through. The resampled version also plays two 6s in a row, while the original progresses linearly from 0 to 11. The resampled version is inferior to the original.

-Max

This would be true if the clock was kept at 44.1kHz, but with a 48kHz encoding the clock is now running at 48kHz as well. So it's not quite accurate to say that the 2nd 6 (which is actually a 6.5 as that is the average between 6 and 7) will now be played 8/13 of the way through as opposed to 7/12 way through of the original. If you look at it your way, the 2nd 6 will be played at 7.5/12 way through due to the clock change. And even that is not quite correct because the actual wave reproduced and sent to the speakers will be a an approximation of the encoding anyway with its accuracy to the encoding determined by the amp technology used.

I stand by my earlier assertion that a 44.1kHz signal re-encoded in 48kHz produces an identical analog waveform as the original encoding.

As for reliance on price as some sort of determinant in quality, it should be noted that a $13 chip that does not only this simple conversion but also decodes DD, DTS, matrix the signal and perform post-processing like THX is commonly found in entry level AVRs.
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post #49 of 54 Old 12-20-2008, 01:17 PM
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Originally Posted by sivadselim View Post

Whether it is relevant to the point you are trying to make or not, CDs are 16bit. Not 24bit.

I actually had to look this up. You are correct, according to the spec (IEC 60908) CDs are encoded with a signed 16-bit word size. This was actually kind of surprising since most studios use 24-bit recorders for both original capture and post-processing.
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post #50 of 54 Old 12-20-2008, 02:17 PM
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Quote:
Originally Posted by cansp6 View Post

I actually had to look this up. You are correct, according to the spec (IEC 60908) CDs are encoded with a signed 16-bit word size. This was actually kind of surprising since most studios use 24-bit recorders for both original capture and post-processing.

Yeah, it's dithered down to 16bit.

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post #51 of 54 Old 12-20-2008, 04:01 PM
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Quote:
Originally Posted by cansp6 View Post

This would be true if the clock was kept at 44.1kHz, but with a 48kHz encoding the clock is now running at 48kHz as well.

The complete 12-sample original will play with exactly the same duration as the 13-sample version, since the clock has been increased on the resampled version. I thought I was pretty clear about that. To make this simple, let's say the sampling rate for the first one is 12Hz and the sampling rate for sequence #2 is 13Hz. They will both have a duration of 1 second in real time.

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Originally Posted by cansp6 View Post

So it's not quite accurate to say that the 2nd 6 (which is actually a 6.5 as that is the average between 6 and 7) will now be played 8/13 of the way through as opposed to 7/12 way through of the original. If you look at it your way, the 2nd 6 will be played at 7.5/12 way through due to the clock change.

You can't have a 6.5. This is digital. The numbers are integers. And the word size is the same in the resampled version. I made the numbers small to make the example simple to understand. This is not theory, this is real data, resampled. And the resampled version is different because of the discrete time slices and integer math.

I made no statement about when the second 6 would play. I didn't need to. I chose a simpler illustration of the problem. I said that the resampled version would "go to 1" before the original does, near the start of the playback. That is a difference. My main intent was to show that the waveforms would be different.

If you consider that the analog signal sampled at the higher rate may have resulted in a 0 for the second sample in the 13-sample version, instead of a 1, you will see that playing the 12-sample digital is closer to the real waveform than playing the resampled 13-sample version in that area. IOW, the resampled version is inferior to the digital version we started with, in a way that is very much like having a larger quantization error.

These guys even tell you how much difference there will be on the home page of their resampling library:
http://www.mega-nerd.com/SRC/

"SRC provides a small set of converters to allow quality to be traded off against computation cost. The current best converter provides a signal-to-noise ratio of 97dB with -3dB passband extending from DC to 96% of the theoretical best bandwidth for a given pair of input and output sample rates."

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Originally Posted by cansp6 View Post

And even that is not quite correct because the actual wave reproduced and sent to the speakers will be a an approximation of the encoding anyway with its accuracy to the encoding determined by the amp technology used.

I stand by my earlier assertion that a 44.1kHz signal re-encoded in 48kHz produces an identical analog waveform as the original encoding.

I have shown above that differences in the digital representation are clearly present. To convince me that the differences in the digital representation are completely obscured by the rest of the playback chain, you will need to provide proof. I doubt that is true, and really can't even imagine how it could be true. Until then, I will assume that the differences, somewhat but not completely diminished, do make it to the listeners ears, after which point I admit they may be eaten up by the fuzz of our brains . However, I am aware of a lot of people spending time bypassing the Windows kmixer in computer audio, which resamples audio to 48KHz, so I have a moderately positive suspicion that the effects of such resampling may be audible, at least with some resampler implementations.

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Originally Posted by cansp6 View Post

As for reliance on price as some sort of determinant in quality, it should be noted that a $13 chip that does not only this simple conversion but also decodes DD, DTS, matrix the signal and perform post-processing like THX is commonly found in entry level AVRs.

The reason I think price matters is that it isn't easy to resample 44.1KHz to 48KHz with high quality. You keep saying it is simple, as if it was just a matter of doing the math right. But it isn't. There is no algorithm that does that resample "perfectly", so the implementation quality does matter. Doing it real-time probably puts even more limits on the potential quality. It is conceivable that some dedicated engineer at a chip company could have lobbied to produce a chip that does the resample with very high quality. But I think that chip would cost more. And would thus not be selected for use in the budget-priced device that started this discussion. And I still don't want ANY 44.1-to-48KHz upsampling in my music playback system!

-Max
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post #52 of 54 Old 12-20-2008, 11:12 PM
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Let's try another approach. Here's a graph of a 1000 Hz signal converted from 44.1 to 48 kHz encoding. The top is the signal at 44.1, the middle is the signal at 48, and the last graph is the error of the 48 kHz encoding. Notice that the error is bound by 0.0000004%.

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post #53 of 54 Old 12-21-2008, 11:39 AM
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Quick recap:
1. So we clearly agree now that the resampling changes the ideal waveform represented by the digital format.
2. I am also arguing that the resampling operation can be done in more than one way with significantly different results; a point which you haven't seemed to acknowledge yet.
3. And that the $15 sound card probably doesn't use a high quality resampling algorithm, but that is just speculation on my part and I don't expect that we will arrive at a definitive answer.
4. It is still up for debate whether those differences are still present after the DAC, amp, and speakers. I say yes, you say no.
5. And then also if the difference is audible, which I contend it will be with a low-quality resampling algorithm, but may or may not be with a high-quality algorithm. You say that the differences will be totally inaudible because they won't be present to hear.

I just looked at that graph again. What it actually shows is:
top: ideal 1000 Hz sine wave
middle: error for 1000 Hz sine wave using linear resampling algorithm from 44.1KHz to 48KHz
bottom: error for 1000 Hz sine wave using sinc_best resampling algorithm from 44.1KHz to 48KHz

This proves point #1 above. And also proves point #2.

On point #3, I think it is very likely that the $15 USB sound card does something like linear interpolation for its resampling. And the error shown in the graph is as much as 15% with a 1KHz sine wave. I have a feeling that music played through that resampler would have audible degradation.

And I'm still pretty sure that even sinc_best will have a lot of trouble at higher frequencies when doing a 44.1 to 48KHz resample.

On point #4, the error above is 15%. I think it is safe to assume that that would make it through the rest of the playback system. And even with small errors, since error compounds those errors will change the playback. You don't buy an amp that is +/-2dB and assume it won't make any difference because your speakers are +/-3dB. The result could be as bad as +/-5dB; it's not like the speakers "hide" the 2dB error in the amp. Whether the difference is too small to be heard is a separate issue, but it should be clear that the differences caused by resampling 44.1 to 48KHz DO result in a change at the end of the playback system.

-Max
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post #54 of 54 Old 02-23-2009, 06:16 PM
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Quote:
Originally Posted by ViperVenom18 View Post

So will this work when just listening to music via iTunes, for example? Or is this for when surround sound is detected via fiber via the DVD player on the laptop?

No one has answered this yet, I don't think...

It is clear the the DVD player on the macbook has this option for optical out.

But if I am just playing music via iTunes to my Receiver via toslink, will the macbook recognize that there is a digital connection via toslink there?

This is NOT using the DVD layer.
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