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multitone spectrum proposal

We are ignoring the noise and distortion also contributing to the 5W but ok i don't want to be pedantic.
Too late. ;)

So lets say its So 0 dBrA is 0.79 V RMS
Tahts far From Waht Amir is saying:

The voltage corresponding to 5 watts @ 4 ohm would be ~4.47V
But 32 Tones at 4.47V would add up to ~25V...
You are comparing a single tone producing 5 W to 32 tones producing the same 5 W average power. Your conclusion is not valid because the math is not linear.
 
Its not
You are comparing a single tone producing 5 W to 32 tones producing the same 5 W average power. Your conclusion is not valid because the math is not linear.
Thats not my conclusion this is quoting from Amir
I asked for "the Absolute level of 0dBrA" and he said "is the RMS voltage corresponding to 5 watts @ 4 ohm."
And no matter what the waveform is the RMS voltage corresponding to 5 watts @ 4 ohm is th 4.47V or so about.


But that's of cause wrong as this is not the value of 0dBrA
 
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We are ignoring the noise and distortion also contributing to the 5W but ok i don't want to be pedantic.
The contributions from the noise and distortions are totally negligible. You can see from the graph that the highest distortion/noise peak is -90 dB down, which means 10^-9 (a billionth) of the power of each of the tones, and the average is orders of magnitude lower. Even with 1 M (10^6) distortion components at -90 dB, their total power is still only 1/1000th of the power of each of the multitones.
 
Its not

Thats not my conclusion this is quoting from Amir
I asked for "the Absolute level of 0dBrA" and he said "is the RMS voltage corresponding to 5 watts @ 4 ohm."
And no matter what the waveform is the RMS voltage corresponding to 5 watts @ 4 ohm is th 4.47V or so about.

But tarts of cause wrong that this is the value of 0dBrA
A single sine wave of 4.47 Vrms yields 5 W into a 4 ohm load. If you look at the individual amplitudes of 32 tones they are much less, thus the lower number provided for that case, but the RMS value will still be 4.47 Vrms for the product of tones. Setting the display to 0 dB for individual tones is a common and useful way to display and compare multitone test results, though you are correct the peaks of each tone do not correspond to 5 W for each tone. But reducing the reference value is not really useful, since the whole graph will shift down, and just makes it harder for people to compare since 0 dB is a common and easy-to-see reference. Debating a common if somewhat arbitrary reference value does seem pedantic if not pointless...

In the RF world, two-tone testing usually shows the individual tones 3 dB down (in dBm), but multitone usually scales everything to 0 dB peaks with a note about the offset, again to make it easy to compare the performance through the signal chain.

Note dBrA simply means dB referenced to A, which is determined by the user.

Another way to think of this is that the average power of all 32 tones is 5 W. The graph is scaled so that each individual tone is at 0 dB. For 32 tones, 0 dB is not the same as for a single tone, since each tone has much lower amplitude. For a single tone, A corresponds to 4.47 Vrms, but for 32 tones corresponds to about 0.79 Vrms going by the earlier post (I did not do the math). I think that is what you are questioning?
 
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Ok, so the end conclusion is now:
the 5 watt is done with the combined rms signal of 4.47 Vrms into 5watt.
This means that the individual tones are at lower level ( the 0.79 Vrms mentioned earlier).
THIS is the 0dB level in the plot. So this is (0.79^2/4ohm) = 0.123 watt for a single tone.
All agree?
 
Can you pleases do this and upload it here to illustrate it :)
Here we go.

Dense multitone, periodic (512 samples), linear spacing of ~100Hz, spanning ~100Hz ... 12kHz, pink spectrum, RMS level of 0dB.

Spectrum:
1745227568687.png


Waveform for "Noise" (random start phases of the cosine oscillators), one period:
1745227768688.png

The crest factor (ratio of peak sample to RMS level) is directly readable from the peak sample, 2.7 in this case. Minimum for a sine-wave is 1.41.

Waveform for "Sweep" (so-called Newman start phases of the cosine oscillators):
1745227902174.png

The crest factor here happens to be higher with 3.0. Crest factor of Newman sweeps are very sensitive to the parameters and in this case, notably from the pink spectrum, it turned out to be higher than for noise case.
The point is, except for the overlap region this signal doesn't look like a multitone and will not produce a lot of IMD, only HD.

Waveform for "Pulse" (start phases of the cosine oscillators all being zero):
1745228238323.png

This one -- basically a pulse train, with some amount of ringing -- clearly does not look like anything resembling a multitone, yet it contains the exact same frequencies and levels as the other two. The distortion response from this one would be almost useless, and of course the crest factor went up to 13.

If we go to the extreme and make the spectrum as dense as possible (and white) then the three variants approach periodic noise, endless sweep with very low crest factor and a single pulse with extreme crest factor.

References: https://web.stanford.edu/~boyd/papers/pdf/multitone_low_crest.pdf

Note:
The discussed multitones are all linear-spaced, with all frequencies being integer multiples (aka harmonics) of a base frequency. However, for audio testing, we would prefer log-spaced harmonically unrelated frequencies, like the typical 1/10th decade spacing giving 31 tones from 20Hz to 20kHz. So far I have not figured out how to create a sweep-style multitone with arbitrary frequency selection and it could be it is even impossible, theoretically (or at least insanely impractical/time-consmuing to create with automated brute-force trial&error)

For our application, we are interested in a multitone that contains all selected frequencies "all at the same time" (random-like) but minimizes crest factor. Most often, simply using random start phases suffices is good enough. And zero phases produce the pulse-like pattern.
 
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