Help to confirm this simple math calculation with dB's

Hi guys,

Hope everything is going well.

I'd like to ask you some help here regarding a simply math calculation. I was discussing with a guy who was giving a course on a company and at certain time I saw in her slides this:

60dB + 60dB = 63dB.

In the end of the class I spoke with him telling him that that statement that he had on his slides was wrong and he was debating that it is well and there is no mistake. Because I didn't remember the rules and some other stuffs I went home and did some calculations.

Her point of view was that each time we doble a number (in this case he was refering to the sound pressure in Pa) what we are doing is adding +3dB to the pressure in dB. What I want to ask is:

I can understand his point of view, but that is valid for linear units. He can't say that 60dB + 60dB = 63dB because 2 x 60dB in logarithm units is not the same as in the linear units.

Am I right? I told him this but he said that it is not like I was saying, that I was wrong lol. WTF, the logarithm scale was made to ease the calculations, instead of being multiplying and dividing we just need to sum or subtract.

If not, please explain.

Thanks in advance.

Regards.

In terms of math, you are correct.

However look at this scenario. Suppose you have one cat meowing at 60 dB. We hear 60 dB sound level.

Now add another identical cat meowing at 60 dB. What decibel level do we hear?

It is not 120 dB. It's more like 63 or 66 dB. Could that be the approach which your colleague was using?

It might be a good trick question to ask someone.

And here the decibel scale was supposed to be a handy way to quantify our subjective hearing perception...

might help to look at using the anti-logs then the log

just as an example

2 is the log of LOG(100)


100 + 100 = 200

BUT LOG(200) is 2.3

Rhode and Schwarz has an application note called "Db or not DB?": https://www.rohde-schwarz.com/en/applications/db-or-not-db-application-note_56280-15534.html
The math is very simply explained.
They also have a calculator (software) for download.

Addition in log scale is multiplication in normal scale:
LOG(100)= 2
100*100 = 10000
LOG(10000) = 4 = LOG(100) + LOG(100)

kam1787 said:

might help to look at using the anti-logs then the log

just as an example

2 is the log of LOG(100)


100 + 100 = 200

BUT LOG(200) is 2.3

Click to expand...

@AMSA84
Perhaps you can't add the dB values since this means that the normal values are multiplied.
Probably you have to add the normal values. This would mean:
Two sources of 60dB (60dB = 10^6):
10^6+10^6 = 20^6
10*log(2*10^6) = 63dB.

Example:
Two gain stages in series - each with a gain of A1=A2=60 dB:

That menas:
A1=1000
A2=1000
Series connection: A=A1*A2=1E6 equivalent to 120 dB

Hence: 60dB+60dB=120 dB.

Anything wrong?

No.
But I think the adding of two sound sources is not the same as two gain stages. Because the sound sources doesn't do a multiplication of each other since the sound pressure (in Pa) is added.

Probably something like to voltage sources in series:
U1 = 100V and U2 = 100V
U1 = 20dBV, U2 = 20dBV

U = U1 + U2 = 200V = 23dBV

LvW said:

Example:
Two gain stages in series - each with a gain of A1=A2=60 dB:

That menas:
A1=1000
A2=1000
Series connection: A=A1*A2=1E6 equivalent to 120 dB

Hence: 60dB+60dB=120 dB.

Anything wrong?

Click to expand...

This is why I was making some confusion. I did the analogy thinking in this and for example, measuring the attenuation of a cable in dBm. In those we sum the values.

- - - Updated - - -

MRahimi said:

Rhode and Schwarz has an application note called "Db or not DB?": https://www.rohde-schwarz.com/en/applications/db-or-not-db-application-note_56280-15534.html
The math is very simply explained.
They also have a calculator (software) for download.

Click to expand...

Rahimi, taking into account the application note that you have shared, in page 12 we can read:

30 dBm + 30 dBm = 60 dBm? Of course not! If we convert these power levels to linear values, it is obvious that 1 W + 1 W = 2 W. This is 33 dBm and not 60 dBm. But this is true only if the power levels to be added are uncorrelated. Uncorrelated means that the instantaneous values of the power levels do not have a fixed phase relationship with one another.

So this means that, if we think that the subject that the guy who was giving that class was related to noise in working environment (work conditions on the company) this means that we are referring to sound and in this case sound can be considered uncorrelated. (page 12)

I sent an e-mail to a person who is the owner of this website http://www.sengpielaudio.com/ asking if he could explain to me this thing. He was not so explicit and I couldn't understand the explanation that he gave Me and some links:

"Sound calculations - Total level adding of incoherent sound sources":
http://www.sengpielaudio.com/calculator-leveladding.htm
Here is clearly told why 60 dB + 60 dB = 63 for level adding of incoherent sound sources.

Here comes more help if it is needed.

"Adding Amplitudes and Levels of coherent ad incoherent signals":
http://www.sengpielaudio.com/AddingAmplitudesAndLevels.pdf

You find here logarithm units in decibels and also linear units as vectors.

Click to expand...

Adding coherent signals:
Linear: 1 + 1 = 2 (vectors in line)
means the same as
logarithmic: 0 dB + 0 dB = +6 dB.
(or 60 dB + 60 dB = 66 dB coherent signals)

Adding incoherent signals:
Linear: sqrt (1² + 1²) = 1.414 (Pythagoras)
means the same as
logarithmic: 0 dB + 0 dB = + 3 dB.
(or 60 dB + 60 dB = 63 dB incoherent signals)

linear 1 means logarithmic 0 dB
linear 1.414 means logarithmic +3 dB
linear 2 means logarithmic +6 dB

Help: "Adding amplitudes and levels of coherent ad incoherent signals":
http://www.sengpielaudio.com/AddingAmplitudesAndLevels.pdf

"Relative level - decibel chart":
http://www.sengpielaudio.com/dB-chart.htm

Click to expand...

This person uses vectors. WTF? I never was taugh in this way regarding power levels, logarithm scale, etc!

I am totally confused.

Continuing with the application note. On the same page (12), looking at that example they apply the same principle. However I am confused when we consider the example that LvW gave and got more confused with that person who owns the http://www.sengpielaudio.com/ website!

For example, measuring the attenuation on a cable. We apply a signal to the power meter, for example, 1dBm. We watch the value one that power meter and we take a note on the paper. After that we insert the cable and we measure the signal at the end of the cable. We sum the values and the result is the attenuation. Speaking roughly.

Can you guys understand my point?

One more thing, I insist that we can't write, mathematicaly, this crap! 60dB + 60dB = 63dB! Don't you guys agree?

I never spoke about this in school!

Hi,

Remember that dB is a unit for expressing ratios, it's not unit for absolute values.
Absolute values can be expressed in dB"something", i.e. relative to a reference level, like:

dBm (power, rel. to 1 mW),
dBu (voltage, rel. to 0.775 V),
dBW/Hz (power spectral density, rel. to 1W/Hz)
dBHz (frequency, rel. to 1Hz)

Nevertheless, in the case of sound pressure level, it is usual to use "dB" for absolute values referred to a conventional level (near the human threshold). In order to avoid confusions, it's better to express it as dB(SPL), or dBA if the sound is A-weighted.

The rules of manipulate magnitudes in dB (adding) are, for instance:

60 dB + 60 dB = 120 dB [the gain of two cascaded stages, each one with gain 60 dB]
60 dBm + 60 dB = 120 dBm (A terrible power!!!) [the power at the otput of an amplifier with 60 dB gain applying a signal of 60 dBm at its input]
60 dBm + 60 dBm = ... makes no sense in usual practice
-90 dBW/Hz + 60 dBHz = -30 dBW [the power of a noise of power spectral density -90 dBW/Hz over a bandwidth of 1 MHz]

The statement 60dB + 60dB = 63dB is incorrect, although we understand what it tries to express.
The correct one could be something like this:

P(60 dBm) + P(60 dBm) = P(63 dBm)

that would read "a signal with power of 60 dBm added to another signal with power of 60 dBm gives a signal with power of 63 dBm" (if the two signals are uncorrelated).

Regards

Z

flanello said:

No.
But I think the adding of two sound sources is not the same as two gain stages. Because the sound sources doesn't do a multiplication of each other since the sound pressure (in Pa) is added.

Probably something like to voltage sources in series:
U1 = 100V and U2 = 100V
U1 = 20dBV, U2 = 20dBV

U = U1 + U2 = 200V = 23dBV

Click to expand...

Independent on the kind of source (sound or something else) adding db values is equivalent to multiplication of absolute values.
(By the way: This is one of the advantages the usage of log quantities has).
This was shown very nicely by the examples as presented by zorro.

Hi Zorro and thank you very much for your post. I understood everything you wrote and in part I knew that already. However the only thing here that was confusing Me was that statement that I think and most of the people here agrees is written in a wrong fashion. Mathematically.

To finish, allow Me to ask another thing: and what about the bode plot? Is in dB's and when we want to construct the final bode plot we add all the components. What about this?

Regards.

If you have several stages or blocks in cascade, the total transfer function is the product of the individual functions.
In this way, expressing them in dB, at any frequency the total transfer function in dB is the sum of the individual functions (always in dB) at that frequency.
I hope it is clear.
Regards

Z

AMSA84 said:

Hi guys,

Hope everything is going well.

I'd like to ask you some help here regarding a simply math calculation. I was discussing with a guy who was giving a course on a company and at certain time I saw in her slides this:

60dB + 60dB = 63dB.

Click to expand...

I think there was a misnomer .
dB vs dBm

dB is a ratio , and dBm is a power referenced to 1mW.

So 60dB + 60 dB is not a clear statement. Unless you mean a factor of 1000 + another factor of 1000 ?...

60dBm + 60dBm should mean adding 1,000 W with 1,000 W which =2,000 W & 2,000 W = 63dBm !

Or if you said:
60dBm + 60dB --> = 120dBm = 1,000,000,000Watts !


Cheers