Sound Pressure Level Attenuation Equation

[mart34]

Hi. I’m working on an equation to find Sound Pressure Level attenuation over distance. Equation 1 allows me to find the distance at which SPL drops to a desired SPL.

Finding Distance (r₂), when -
Reference SPL (L₁) is known
Reference Distance (r₁) is known
Desired SPL (L₂) is known

For instance, I know the SPL (L₁) is 87dB at Distance (r₁) of 1meter.
I need to find the Distance (r₂), at which the SPL falls to the Desired SPL (L₂) of 66dB.
Equation 1 allows me to find the answer, which is 11.19meters. This just follows the 6dB loss per doubling of distance rule.

However, I’m also trying to include an additional amount of dB loss per meter (let’s call it “f”).
Therefore, I have a formulated Equation number 2 to take into account the “f”dB loss per meter, aswell as the original 6dB loss per doubling of distance.

My problem is that I now have Distance (r₂) on both sides of the equation. Can this be solved, to isolate “r₂ “ by itself on one side of the equal sign ?

Any help would be very much appreciated. Many thanks, Martin

 

[HallsofIvy]

That equation cannot be solved in terms of "elementary" functions but can be solved using "Lambert's W function". The W function is defined as the inverse function to \(\displaystyle f(x)= xe^x\). That is, \(\displaystyle W(xe^x)= x\) for all x.

The first thing I would do is write that 10 to the power as a product: \(\displaystyle 10^{\frac{L_1- fr_2- L_2}{20}}= (10^{\frac{(L_1-L_2)}{20}}10^{-\frac{fr_2}{20}}\) so that the equation becomes \(\displaystyle r_2= r_110^{\frac{(L_1- L_2)}{20}}10^{-\frac{fr_2}{20}}\) so that \(\displaystyle r_210^{\frac{fr_2}{20}}= r_110^{\frac{(L_1- L_2)}{20}}\).

Now, we need to convert that power of 10 to a power of e: \(\displaystyle 10^a= e^{ln(10^a)}= e^{a ln(10)}\). So we can write the equation as \(\displaystyle r_2e^{\frac{fr_2}{20} ln(10)}= r_1 10^{\frac{(L_1- L_2)}{20}}\).

Let \(\displaystyle x= \frac{fr_2}{20} ln(10)\) so that \(\displaystyle r_2= \frac{20x}{f ln(10)}\). The equation becomes \(\displaystyle \frac{20x}{f ln(10)}e^x= r_1 10^{\frac{(L_1- L_2)}{20}}\)

Then \(\displaystyle xe^x= \frac{f ln(10)}{20} r_1 10^{\frac{(L_1- L_2)}{20}}\)

Now, we can say that \(\displaystyle x= W\left(\frac{f ln(10)}{20} r_1 10^{\frac{(L_1- L_2)}{20}}\right)\) and then solve for \(\displaystyle r_2\).

 

[Ishuda]

What it sounds like you are trying to do is add in an attenuating media or possibly 'rough boundary' affect. That is, the actual sound pressure level is decreased not only by the spherical spreading but also by an attenuation effect due to the media. If the effect is linear, it is generally reported as a dB per distance, i.e. pressure P is given by
P ~ P₀ r^-1 10^{-\(\displaystyle \alpha\) r/20}
or
SPL = 20 log(P) = SPL₀ - [20 log(r) + \(\displaystyle \alpha r\)]

If you want something different as an attenuating function, i.e. your f(r), just replace the \(\displaystyle \alpha\) r/20 by your f(r).

Edit to add: Oh, and that SPL₀ = 20 log(P₀). It is generally the SPL at a reference distance such as 1 metre.

 

[mart34]

That equation cannot be solved in terms of "elementary" functions but can be solved using "Lambert's W function". The W function is defined as the inverse function to \(\displaystyle f(x)= xe^x\). That is, \(\displaystyle W(xe^x)= x\) for all x.

The first thing I would do is write that 10 to the power as a product: \(\displaystyle 10^{\frac{L_1- fr_2- L_2}{20}}= (10^{\frac{(L_1-L_2)}{20}}10^{-\frac{fr_2}{20}}\) so that the equation becomes \(\displaystyle r_2= r_110^{\frac{(L_1- L_2)}{20}}10^{-\frac{fr_2}{20}}\) so that \(\displaystyle r_210^{\frac{fr_2}{20}}= r_110^{\frac{(L_1- L_2)}{20}}\).

Now, we need to convert that power of 10 to a power of e: \(\displaystyle 10^a= e^{ln(10^a)}= e^{a ln(10)}\). So we can write the equation as \(\displaystyle r_2e^{\frac{fr_2}{20} ln(10)}= r_1 10^{\frac{(L_1- L_2)}{20}}\).

Let \(\displaystyle x= \frac{fr_2}{20} ln(10)\) so that \(\displaystyle r_2= \frac{20x}{f ln(10)}\). The equation becomes \(\displaystyle \frac{20x}{f ln(10)}e^x= r_1 10^{\frac{(L_1- L_2)}{20}}\)

Then \(\displaystyle xe^x= \frac{f ln(10)}{20} r_1 10^{\frac{(L_1- L_2)}{20}}\)

Now, we can say that \(\displaystyle x= W\left(\frac{f ln(10)}{20} r_1 10^{\frac{(L_1- L_2)}{20}}\right)\) and then solve for \(\displaystyle r_2\).

I can't thank you enough for this. Although I'm a bit out of my depth here. I'm embarrassed to ask, but could you please apply some numbers to show this providing an answer for r2 ? Say -
87 for "L1"
66 for "L2"
1 for "r1"
1.3 for "f"

Now there is an "x" and a"W", I'm confused, as I don't know which values to replace these with.

Thanks so much for your help. Apollogies for asking for more help.

 

[Ishuda]

Start with
SPL = SPL₀ - [20 log(r) + \(\displaystyle \alpha\) (r-1)]
where r is in meters, SPL₀ is the reference SPL, i.e.
L₁ = SPL₀ - [20 log(1) + \(\displaystyle \alpha\) (r-1)]
where we have included the attenuation suffered in being 1 meter from the source in SPL₀.

Note that the difference in levels between any two distances r₁ and r₂ is
L2 - L1 = SPL₀ - [20 log(r₂) + \(\displaystyle \alpha\) (r₂-1)] - SPL₀ + [20 log(r₂) + \(\displaystyle \alpha\) (r₂-1)] = 20 log(r₁/r₂) - \(\displaystyle \alpha\) (r₂-r₁)
So if r₂ is twice the distance of r₁ and [\tex]\alpha[/tex] = 0, the difference is -20 log(2) ~ -6 dB.

Now suppose L1 = 87 dB, r2 = 2 meters, and L2 = 80 dB, then
L2 - L1 = 80 - 87 = -7 = 20 log(1/2) - \(\displaystyle \alpha\) (r₂-r₁) ~ - 6 - \(\displaystyle \alpha\)
or
\(\displaystyle \alpha\) ~ 1 dB / meter