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Why is 16-bit undithered a "rough" wave, while 24-bit is smooth?

I like to think that I understand how digital audio works pretty well, but please help me understand this:

I was reading these measurements from Stereophile:


Figure 6 shows an undithered 16-bit tone, whereas figure 7 shows the same with 24 bit:
"[T]he M51's reproduction of an undithered 16-bit tone at exactly –90.31dBFS was essentially perfect (fig.6), with a symmetrical waveform and the Gibbs Phenomenon "ringing" on the waveform tops well defined. With 24-bit data, the M51 produced a superbly defined sinewave (fig.7). "


Figure 6:

View attachment 340354


Figure 7:

View attachment 340355

Why are the two waves so different between 16-bit and 24-bit?

They look different because of resolution. OTOH this is not a big deal and it is mostly exploited because of marketing.
 
The graphs are somewhat misleading because at the last bit of 24-bit, it should look similar to the last bit of 16-bits. Below that is quantization noise. But it looks like the reviewer is testing something specific about sending an un-dithered signal through the device that I don't quite understand.

edit: Oy I forgot that there are many more steps in 24-bit than 16-bit.

In practice, the difference I've found between 16-bit and 24-bit is that with 24-bit, you can record audio at line level (around -18dbFS RMS) and not worry about quantization distortion creeping into the mix as multiple recorded elements are layered, compressed and loudened during mixing, and then loudness maximized during mastering. At 24-bit, preamp noise is going to be the bigger concern :)
 
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Why are the two waves so different between 16-bit and 24-bit?

At such a low level, there's almost no resolution left at 16-bit. The measurement shows that the DAC is working well.

This is what an undithered 16-bit 1 kHz sinewave looks like in Adobe Audition, at -90.31 dB:

1 khz 16-bit audition.png


1. Generate a 0.006s 1 kHz sinewave, volume 0dB (max), at 32-bit, 44.1 kHz.

2. Use effects, amplitude, amplify/fade, constant amplification and -90.31 dB.

3. Edit, Convert Sample Type, 16-bit, dither disabled.

4. Convert sample type back to 32-bit.

5. Amplify it by 85 dB, and cut it to match the Stereophile measurement.

6. Use Effects, Filters, FFT filter and apply a brickwall filter at 20500 Hz (to match the M51 measurements).
 
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I like to think that I understand how digital audio works pretty well, but please help me understand this:

I was reading these measurements from Stereophile:


Figure 6 shows an undithered 16-bit tone, whereas figure 7 shows the same with 24 bit:
"[T]he M51's reproduction of an undithered 16-bit tone at exactly –90.31dBFS was essentially perfect (fig.6), with a symmetrical waveform and the Gibbs Phenomenon "ringing" on the waveform tops well defined. With 24-bit data, the M51 produced a superbly defined sinewave (fig.7). "


Figure 6:

View attachment 340354


Figure 7:

View attachment 340355

Why are the two waves so different between 16-bit and 24-bit?
A belated response: I have been performing this test for a long time, because with the 16-bit data it reveals DAC linearity problems.

In the twos-complement encoding used by 16-bit digital audio, –1 least significant bit (LSB) is represented by 1111 1111 1111 1111, digital zero by 0000 0000 0000 0000, and +1 LSB by 0000 0000 0000 0001. If the waveform at exactly -90.31dBFS is symmetrical, this indicates that changing all 16 bits in the digital word gives exactly the same change in the analog output level as changing just the LSB.

With 24-bit data, the test reveals if the DAC's analog noisefloor is low enough for the sinewave not to be obscured by noise.

John Atkinson
Technical Editor, Stereophile
 
A belated response: I have been performing this test for a long time, because with the 16-bit data it reveals DAC linearity problems.

In the twos-complement encoding used by 16-bit digital audio, –1 least significant bit (LSB) is represented by 1111 1111 1111 1111, digital zero by 0000 0000 0000 0000, and +1 LSB by 0000 0000 0000 0001. If the waveform at exactly -90.31dBFS is symmetrical, this indicates that changing all 16 bits in the digital word gives exactly the same change in the analog output level as changing just the LSB.

With 24-bit data, the test reveals if the DAC's analog noisefloor is low enough for the sinewave not to be obscured by noise.

John Atkinson
Technical Editor, Stereophile
So sorry in advance for this stupidly basic question, John:
if every dac uses the digital map captured as input for a reconstruction filter ( sincx function or similar), according to Shannon the reconstructed analog wave should almost perfectly resemble the original sine wave, as long as the sampling is 2x tha max frecuency captured ( which is the case in the graph), no matter the bit rate, only slightly disturbed by the noise floor implicit in the bit rate used. Then, why this is not seen in the 16 bit graph and only in 24 bits? Just because of the short distance between the signal @-90 db and the noise floor @-98 db? If so, isn’t that noise random, thus leading to random jagged reconstruction instead of the symmetrical gibbs ringing shown in these graphs?
 
If so, isn’t that noise random, thus leading to random jagged reconstruction instead of the symmetrical gibbs ringing shown in these graphs?
Generally speaking, quantization noise is not at all random! It can be highly correlated to the signal.

This is why low-level signals generally tend to require dither to help them make it over this nonlinearity, similar to what high-frequency AC bias does for tape. This is what then turns quantization noise into uncorrelated noise. If the signal is left undithered, this does not happen.

BTW, you can generate some high-level test signals that have the neat property of being "self-dithering", i.e. if you remove the signal you will actually get a residual resembling white noise (which will be lower in amplitude than when adding dither). For example, in a sine wave, the frequency needs to be chosen such that each cycle is being sampled at slightly different points for as long as possible before the pattern starts to repeat.
1 kHz straight makes a very bad choice because the pattern repeats every 441 samples at 44.1 kHz or even every 48 samples at 48 kHz. (And periodicity in time domain equates to sampling in frequency domain, so quantiztation noise ends up forming a bunch of spikes every 44100/441 = 100 Hz.)
With prime multiples of a fraction of 1 Hz in the 1 kHz vicinity we have so far routinely made it to 10-20 second pattern lengths at 16 bit, even into the low hundreds of seconds at times. For example, 999.91 Hz or 1000.03 Hz fit the bill, but some prime fractions of 1 kHz have also shown good results. The finite resolution ultimately limits pattern lengths to below what you would theoretically expect (which would potentially be into the millions of years) as slightly different values end up being rounded to be the same. Finding the actual maximum would make an interesting mathematical problem.
 
So sorry in advance for this stupidly basic question, John:
if every dac uses the digital map captured as input for a reconstruction filter ( sincx function or similar), according to Shannon the reconstructed analog wave should almost perfectly resemble the original sine wave, as long as the sampling is 2x tha max frecuency captured ( which is the case in the graph), no matter the bit rate, only slightly disturbed by the noise floor implicit in the bit rate used. Then, why this is not seen in the 16 bit graph and only in 24 bits? Just because of the short distance between the signal @-90 db and the noise floor @-98 db? If so, isn’t that noise random, thus leading to random jagged reconstruction instead of the symmetrical gibbs ringing shown in these graphs?
With undithered 16-bit data representing a signal at exactly -90.31dBFS, there are only 3 DC voltage levels in the reconstructed analog signal: +1, 0. -1. By definition,these cannot represent a sinewave. However, with the NAD M66's very low noisefloor you can see the leading edges of the transitions between these levels overlaid with the minimum-phase impulse response of the reconstruction filter. See figs.12 & 15 at https://www.stereophile.com/content/nad-m66-streaming-preamplifier-measurements-page-2.

John Atkinson
Technical Editor, Stereophile
 
So the steps are likely much smaller...
The "steps" are the same size whether 16-bit or 24-bit sample size; 24-bit just has more of them. A -90 dB sine wave, while at the limit of 16-bit accuracy, would have many more bits to work with if captured in 24-bit. If you measured a -140 dB sine wave in 24-bit, you'd see a similar distortion as in the original post.
 
The "steps" are the same size whether 16-bit or 24-bit sample size; 24-bit just has more of them. A -90 dB sine wave, while at the limit of 16-bit accuracy, would have many more bits to work with if captured in 24-bit. If you measured a -140 dB sine wave in 24-bit, you'd see a similar distortion as in the original post.
That depends upon the output level (actually the lsb level, since we're being pedantic). If the DAC's maximum output is 4 V for 16 or 24 bits, assuming the same buffer gain, then the 24-bit DAC will have much smaller steps. In practice, at least in my decades of design, that is most often the case, as creating a DAC with 256 times the output voltage is usually impractical. The analogy falls apart somewhat with delta-sigma designs, but no matter the architecture the effective lsb size is usually much smaller for a 24-bit DAC vs. a 16-bit DAC. At least in my experience.
 
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