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Decibel Calculation

finagle

Apr 27, 2017
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Hello all
I am a new member but I hope someone can help me

I have this equation
DbChange = Log(Vout/Vin) x 0.43429 x 10 x 2

which seem to work fine for me to calculate the Db change of an input to output voltage
but I would also like to input Vin and Db change and calculate Vout

I got this from a friend of mine (who knows more math than I do)

Vout = Vin/ 10 ^ (DbChange/0.43429)

but it doesn't seem to work

Appreciate any help on this
 

davenn

Moderator
Sep 5, 2009
14,254
Joined
Sep 5, 2009
Messages
14,254
Hello all
I am a new member but I hope someone can help me

I have this equation
DbChange = Log(Vout/Vin) x 0.43429 x 10 x 2

which seem to work fine for me to calculate the Db change of an input to output voltage
but I would also like to input Vin and Db change and calculate Vout

I got this from a friend of mine (who knows more math than I do)

Vout = Vin/ 10 ^ (DbChange/0.43429)

but it doesn't seem to work

Appreciate any help on this

because you have an error in your formula

dB gain/loss = 20log(Vout/Vin)


Dave
 

AnalogKid

Jun 10, 2015
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As above, your equation has an extra term in it (0.43429). It might be a conversion factor related to wherever the equation came from, but without more information there is no way to know.

Is this homework?

ak
 

finagle

Apr 27, 2017
4
Joined
Apr 27, 2017
Messages
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The Equation
dB gain/loss = 20log(Vout/Vin)

didn't not work in Microsoft access
I did plug it into Microsoft Excel and it worked but I am working in Microsoft Access and it didn't until I added
the 0.43429

not homework my (new) job requires me to work on Powerline carrier equipment in the Electric transmission system
and some of this equipment has a rated decibel loss/gain so I was hoping to see if I could measure the input voltage and calculate the expected output voltage based on the rated loss/gain
 

AnalogKid

Jun 10, 2015
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The Equation
dB gain/loss = 20log(Vout/Vin)
didn't not work in Microsoft access
Why would it?

That conversion factor is wrong. If it makes the results correct, then there is a problem somewhere in the calculation engine. Access is a database program, not even remotely related to something to use for calculations. I haven't run the numbers, but Access might be performing a natural log instead of base 10, and the 0.43 number just happens to correct that mistake for one set in values.

Gardening, jet engine repair, and network calculations all follow the same rule set:
1. Pick the right tool for the job.
2. Let the tool do the work.

ak
 

davenn

Moderator
Sep 5, 2009
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didn't not work in Microsoft access
I did plug it into Microsoft Excel and it worked but I am working in Microsoft Access and it didn't until I added
the 0.43429

of course not
as Analogkid said .... USE the correct tools

I have given you the correct formula
 

finagle

Apr 27, 2017
4
Joined
Apr 27, 2017
Messages
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I believe you are correct about the natural log but the equation as it stands in access does seem to work I have tried it for many different input variables (as I inferred in the original post my math skills are not good) I also know the limitation of Access vs Excel but it is easier to build a nice user interface with access. and I can through VBA programming input an equation into it. Having said that at this point I would be happy If someone could take the equation
dB gain/loss = 20log(Vout/Vin) that works in Excel
and solve for Vout
 

davenn

Moderator
Sep 5, 2009
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I believe you are correct about the natural log but the equation as it stands in access does seem to work I have tried it for many different input variables

I don't know what else to say, 1000's of people have been using it for many, many years without problems ;)
 

AnalogKid

Jun 10, 2015
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I agree. A 43.429% error is a frustrating problem.

I understand your frustration, but the problem here is that we actually do know what we're talking about, and we are not refusing to answer your question because we're pedantic jerks. The mathematical definition of a bel has been around for way over 100 years, and is very well understood. Because your extra term is in fact an error, there is no way to guess how it should figure into the standard equation permutations.

ak
 
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