Towards the end of the paper that I'm providing a link to, the author, referring to the intensities of first and second sound waves before summing, states that-

"For example, the direct sum of 0 dB and -20 dB is only 0.8 dB."

From my calculation, the intensity of the summation would be +0.04 dB.

Is the author correct?

https://factualaudio.com/post/sum/

Regards,
Pete

 

Towards the end of the paper that I'm providing a link to, the author, referring to the intensities of first and second sound waves before summing, states that-

"For example, the direct sum of 0 dB and -20 dB is only 0.8 dB."

From my calculation, the intensity of the summation would be +0.04 dB.

Is the author correct?

https://factualaudio.com/post/sum/

Regards,
Pete

It depends.
Are the sounds coherent? If so, add the pressures. If not, add the powers.

 

I get the sum being equal to 0.83 dB, so I'm with the author.

\( log(1.1)\;=\;0.04139 \)

\( 20\times0.04139\;=\;0.82785 \text{ dB} \)

 

From my calculation, the intensity of the summation would be +0.04 dB.

Do you mean +0.4dB?

I get the sum being equal to 0.83

Yes, that using the voltage equation for dB.
The power equation will give you half that, or 0.4dB, which I believe is appropriate for sound intensity.

 

Do you mean +0.4dB?
Yes, that using the voltage equation for dB.
The power equation will give you half that, or 0.4dB, which I believe is appropriate for sound intensity.

Since the TS mentions "summing" two signals, this would usually be done with a voltage device like an operational amplifier. Summing does not usually involve currents, although that is possible, and it seems unlikely that the TS is talking about summing power.

Here is the quote from the paper referenced by the TS:

I’ll leave you with one last rule of thumb: decibels being a logarithmic unit, the contribution of a given term to the sum falls quickly as the decibels decrease. For example, the direct sum of 0 dB and -20 dB is only 0.8 dB. This can be used to dismiss the contribution of the smaller terms as negligible. ​

I think it is pretty clear that the "author" is talking about the voltage equation. We can argue about what he should or should not be talking about with respect to "sound".

 

It depends.
Are the sounds coherent? If so, add the pressures. If not, add the powers.

My assumption was that he was considered what the author calls unrelated signals, that is, incoherent or signals of differing frequency. But I think he doesn't make clear which he was referring to.

 

I get the sum being equal to 0.83 dB, so I'm with the author.

\( log(1.1)\;=\;0.04139 \)

\( 20\times0.04139\;=\;0.82785 \text{ dB} \)

Yes, I think that you are correct. My mistake was thinking that he was referring to an intensity of -20 dB, but looking at the article again, he is referring to voltage amplitude. So -20 dB of amplitude would be one-tenth and summing that with 0 dB of amplitude, with the waves in phase would sum to an amplitude of 1.1. Thanks. {Edit} That is, 1.1 times the lower amplitude.

 

Do you mean +0.4dB?
Yes, that using the voltage equation for dB.
The power equation will give you half that, or 0.4dB, which I believe is appropriate for sound intensity.

The case that I'm interested in understanding is how to correctly calculate the sum where the two waves are sound waves of differing frequency and intensity.

If he were referring to intensity rather than amplitude, then intensity of the 0 dB wave would be a hundred times the intensity of the -20 dB wave,

10*log(100) = 20 dB

Summing the intensities of the 0 dB wave and the -20 dB wave gives an intensity of 101 times the intensity of the -20 dB wave.
Therefore

Intensity sum = 10*log (101*I2/ Io)

where Io is threshold intensity and I2 is intensity of the lower level wave, -20 dB.

intensity sum = 10*log(I2 / Io) + 10*log(101)

intensity sum = -20 dB +20.04 dB =+ .04 dB

This way of calculating I found in a book on acoustics by John Backus. Hopefully I'm applying his method correctly.

 

It depends.
Are the sounds coherent? If so, add the pressures. If not, add the powers.

Sorry, I think that I misstated that he was referring to intensities. In fact I think he was referring to amplitude.

 

Is it assumed that these waves are in phase?
2*pi*f*t+phase

 

Sorry, I think that I misstated that he was referring to intensities. In fact I think he was referring to amplitude.

so the figure is 10*log10( 10^0+10^(-20/10))
=0.04dB
if the noises are not coherent.
and
so the figure is 20*log10( 10^0+10^(-20/20))
=0.8dB
if they are.

 

The website claims to be evidenced backed. Most home audio enthusiast do not have sufficient equipment and credentials.

Electronic simulation is making a transition in buying and testing components. It can be more informative than a datasheet
by allowing a user to modify parameters. On the bench we can measure and compare what the semiconductor company is
claiming. A manufacturer is more accountable and is there is more incentive to meet those specifications.
The test bench is merging with the pc.

Some audio is also following these newer methods.
I can see a hint of that on the website, but too early to tell if they can filter simultaneous sound waves well enough.
Based on what is presented the evidence backing is in the right direction and not meant to be a credentialed sound lab.
All that could change rapidly, some of the micro controller home based audio enthusiast might
attain that capability and more spin off ventures, some of the sound engineers have been busy.

 

Is it assumed that these waves are in phase?
2*pi*f*t+phase

The paper considers both cases of in-phase and not in-phase.

 

so the figure is 10*log10( 10^0+10^(-20/10))
=0.04dB
if the noises are not coherent.
and
so the figure is 20*log10( 10^0+10^(-20/20))
=0.8dB
if they are.

Is that reasonable to you? That seems as if there is too great of a difference in the summations. If you are stating that that is correct, can you explain why in terms of coherence or not?

 

Working with decibels can be confusing in electronics because dB is used for both power and voltage.

Also dB is a relative measure and has no units, i.e. it represents a ratio of two quantities.

If the author is measuring intensity in dB then one can assume that it is comparing power.

In the audio recording industry, a reference level 1 mW into 600Ω is used to reference 0 dB, written as 0 dBm
-20 dBm would be 0.01 mW.

Now, you cannot just add 1 mW to 0.01 mW to get 1.01 mW. That is the mistake you are making.
You have to convert the power into voltage into a 600 Ω load and then add the voltages.
1 mW converts to 0.775 Vrms.
0.01 mW converts to 0.0775 Vrms.

Now you add the two voltages, 0.775 + 0.0775 = 0.8525 Vrms
Calculate the power into 600 Ω load = 1.2 mW
Convert this to dBm gives 0.8 dBm.

Hence, the author is correct.

 

Towards the end of the paper that I'm providing a link to, the author, referring to the intensities of first and second sound waves before summing, states that-

"For example, the direct sum of 0 dB and -20 dB is only 0.8 dB."

From my calculation, the intensity of the summation would be +0.04 dB.

Is the author correct?

https://factualaudio.com/post/sum/

Regards,
Pete

The article is fairly clear and thorough yet the pondering postscript lacks assumptions or context.

The statement assumes coherent addition of amplitudes (e.g., voltages), yielding ~0.82 dB ( , rounded to 0.8 dB. For incoherent power addition, the correct result is ~0.043 dB, indicating a context-specific assumption in the statement.

 

Now, you cannot just add 1 mW to 0.01 mW to get 1.01 mW. That is the mistake you are making.
You have to convert the power into voltage into a 600 Ω load and then add the voltages.
1 mW converts to 0.775 Vrms.
0.01 mW converts to 0.0775 Vrms.

that is exactly what you should do when dealing with uncorrelated noise, except that the noise power density will be in W/m^2.
If there is 1mW going through your square metre for one source, then along comes another 10uW from another source, then the total must be 1.01mW.

 

If you want to calculate the power of the sum of two in-phase waves, then you have to do what Papabravo did, which is extrapolate from the dB levels to amplitudes, add the amplitudes, and then solve for power of the sum.

Actually the ratio of two voltages or other amplitudes expressed in decibels is a statement of their relative power levels. That is,
expressing the ratio of two voltages in dB is solved according to

dB = 20 X log (V2/ V1)

where the multiplier 20 comes from the fact that the voltages are squared.