Calculating dB: Understanding the Addition of Sound Intensities

[joe215]

Homework Statement

What is 80dB+75dB?

I found the individual intesities to be 0.0001 and .0000316228

Then I added them and:

dB=10log(1.31622x10^-4/1x10^-12)=81.19 dB

Is that right?

Thanks!

 

[jambaugh]

Whether you add values directly or add the intensities must be determined from the specific context. Without any I would assume you're adding numbers so 80dB + 75dB=155dB.

If for example you are considering decibel amplification or attenuation in stages then the sequential scaling of intensities will multiply and so again the decibel gains will add.

If you are superposing two sources with decibel intensities relative to some base level then your method would be correct.

But as you can see two contexts yields two different methods. Do you have more info?

 

[skeptic2]

Decibels are used in so many different ways it would help to know how the decibels are being used to be sure of getting the right answer. Generally decibels represent a ratio or more precisely 10 times the logarithm of a ratio. When used to represent an intensity, they usually have another letter to indicate what they are representing such as dBm or dBu. When the ratio is between two voltages, decibels are usually calculated using 20 times the logarithm instead of 10 times so the dB value represents the power ratio, not the voltage ratio. Note: when decibels are used for voltages, all the voltages must be measured at the same impedance.

Usually when decibels are added, they are simply added which is equivalent to multiplying the base ratios. Note that decibel values greater than zero represent ratios or values greater than one and values less than zero represent numbers between zero and one. Adding decibels is equivalent to multiplying the base values and subtracting is equivalent to dividing.

If I understand the problem correctly, you have several errors in your example resulting in a wrong answer.