MQA technical analysis

[Abtr]

I haven't listened to it, but it's too low level to be audible (one could of course apply digital gain; there's some 90 dB headroom). Looking at the spectrogram, there is some correlation with the original.

Okay, so what do you think this means, if anything?

[mansr]

Okay, so what do you think this means, if anything?

Well, it shows that the low bits encode, somehow, the high-frequency content. Not exactly surprising, of course.

[Don Hills]

Well, it shows that the low bits encode, somehow, the high-frequency content. Not exactly surprising, of course.

In other words, what you appear to have done is removed the "base" 0-24 kHz audio samples and decoded the "folded" part of the audio. It shows that the original (pre MQA) 24-48 kHz audio has been encoded by the MQA process. Correct? If so, as you say it's not surprising, it's doing what the MQA patents say it does.

[Fokus]

Well, it shows that the low bits encode, somehow, the high-frequency content.

 

It shows that the folded part also contains information below Fs/2, possibly suggesting something about the filters used in the lossless spectral split and join at Fs/2.

 

(With Fs/2 meaning 24kHz.)

[Miska]

It shows that the folded part also contains information below Fs/2, possibly suggesting something about the filters used in the lossless spectral split and join at Fs/2.

 

You mean lossy split?

 

It is easy to make the split much better without overlap/aliasing around Fs/2...

[mansr]

For the next experiment, I replaced the top 15 bits of the MQA file with a 1 kHz sine wave (TPDF-dithered at 15 bits) and decoded the file. This is the resulting spectrum (input blue, output red):

 

[audiventory]

For the next experiment, I replaced the top 15 bits of the MQA file with a 1 kHz sine wave (TPDF-dithered at 15 bits) and decoded the file. This is the resulting spectrum (input blue, output red):

 

[ATTACH=CONFIG]33014[/ATTACH]

 

Blue signal in 15 bit resolution?

[mansr]

Blue signal in 15 bit resolution?

 

Yes.

[audiventory]

Yes.

 

If there 15 bit, why noise level -140 dB?

 

Must be about -90 ... -100 dB.

[mansr]

If there 15 bit, why noise level -140 dB?

 

Must be about -90 ... -100 dB.

 

You are confusing total noise with noise spectral density.

[audiventory]

You are confusing total noise with noise spectral density.

 

Spectrum 15 bit looks like 23...24 bit resolution.

[mansr]

Spectrum 15 bit looks like 23...24 bit resolution.

 

If you integrate across the frequency range, you get the level you're expecting. It's why doubling the sample rate lowers the noise floor equivalently to adding a bit. Same total noise spread across a wider frequency range.

[Abtr]

For the next experiment, I replaced the top 15 bits of the MQA file with a 1 kHz sine wave (TPDF-dithered at 15 bits) and decoded the file. This is the resulting spectrum (input blue, output red):

 

[ATTACH=CONFIG]33014[/ATTACH]

So what's the significance of what we see here?

[mansr]

So what's the significance of what we see here?

 

I don't know.

[bogi]

I wouldn't call it deblurring!

 

It's blurring to get better sound

[audiventory]

If you integrate across the frequency range, you get the level you're expecting. It's why doubling the sample rate lowers the noise floor equivalently to adding a bit. Same total noise spread across a wider frequency range.

 

Each expanding range 2 times give difference 6 dB for level (voltage) spectrum and 3 dB for power spectrum.

 

For sample rate 22 kHz (44 kHz sample rate) there -90 dB.

 

At the picture we see -140 dB. Difference is 50 dB=-90+140.

 

50 dB / 6 dB is about 8 times.

 

22 kHz * 8 times = 176 kHz band (352 kHz sample rate).

 

At the picture I see input band 22 ... 24 kHz, not 176 kHz.

 

1. What is analyzis software shown in the picture?

 

2. What is window applied?

[mansr]

Each expanding range 2 times give difference 6 dB for level (voltage) spectrum and 3 dB for power spectrum.

 

For sample rate 22 kHz (44 kHz sample rate) there -90 dB.

 

At the picture we see -140 dB. Difference is 50 dB=-90+140.

 

50 dB / 6 dB is about 8 times.

 

22 kHz * 8 times = 176 kHz band (352 kHz sample rate).

 

At the picture I see input band 22 ... 24 kHz, not 176 kHz.

 

Do you know what "integration" is?

 

1. What is analyzis software shown in the picture?

 

Octave. Matlab could also be used.

 

2. What is window applied?

 

Dolph-Chebychev.

[audiventory]

Do you know what "integration" is?

 

I suppose, the integration can't decrease level noise to 40 dB.

 

What is level (in dB) of the signal (by oscillogramm) in LFSU scale?

[bogi]

I read from more sources that the total amount of noise (integral through signal frequency range) is not changed by resampling or noise shaping, but the frequency range where that noise appears is changed by resampling and it's distribution within that range can be changed too, for example by noise shaping.

 

Therefore the average level of noise is lower after upsampling, but the noise distribution over signal frequency spectrum is also important.

 

Googling for delta sigma modulator and noise shaping results to many articles explaining it.

https://www.maximintegrated.com/en/app-notes/index.mvp/id/1870

[mansr]

I suppose, the integration can't decrease level noise to 40 dB.

 

Suppose a noise floor of -140 dB at 48 kHz sample rate. Multiply that by the 24 kHz bandwidth (i.e. integrate the constant level) and convert back to dB scale:

 

10 * log10(10^(-140/10) * 24000) = -96 dB

 

Now bear in mind that the window function affects the observed noise floor. Here's a comparison of 16-bit dither noise analysed with a few different window functions:

 

 

With the rectangle window we get exactly the level expected according the usual 6 dB per bit formula. The others lower the level around 6 dB, so the -140 dB level seen above with 15-bit dither is precisely where it should be.

 

What is level (in dB) of the signal (by oscillogramm) in LFSU scale?

 

The signal is mostly a 1 kHz tone at -3 dBFS, so looking at it like that isn't really helpful.

[audiventory]

Suppose a noise floor of -140 dB at 48 kHz sample rate. Multiply that by the 24 kHz bandwidth (i.e. integrate the constant level) and convert back to dB scale:

 

10 * log10(10^(-140/10) * 24000) = -96 dB

 

 

With the rectangle window we get exactly the level expected according the usual 6 dB per bit formula. The others lower the level around 6 dB, so the -140 dB level seen above with 15-bit dither is precisely where it should be.

 

1. Why you suppose noise floor -140 db?

 

2. -140/10 - whats here -140 and 10?

[mansr]

1. Why you suppose noise floor -140 db?

 

Just as an example.

 

2. -140/10 - whats here -140 and 10?

 

Do I really need to explain to you how the dB unit works?

[audiventory]

Do I really need to explain to you how the dB unit works?

 

Why -140 dB? Why not -110? Why not -200?

 

If you have time, could you show how you get the formula?

[mansr]

Why -140 dB? Why not -110? Why not -200?

 

As I already said, it was an example. The specific value has no significance.

 

If you have time, could you show how you get the formula?

 

I thought I already did, but lets take it step by step, again using -140 dB/Hz as the example noise floor:

 

1. Convert dB to linear units: -140 dB = 10 ^ (-140 / 10) = 1e-14

2. Multiply by the bandwidth in Hz: 1e-14 * 24000 = 2.4e-10

3. Convert linear to dB: 10 * log10(2.4e-10) = -96.2 dB

 

A noise floor of -140 dB/Hz over a 24 kHz bandwidth thus corresponds to a total noise level of -96.2 dB.

[witchdoctor]

Check out Bob Stuarts blog for technical MQA talk:

 

MQA Playback | Bob Talks