Thread: Natural sound frequency filtering

1. Natural sound filtering

Hi!

This is my theory of why high frequent sound waves have a higher ability to be filtered out by obstacles in the environment such as trees and stuff, and then even by walls.

There is many paths for the sound to travel from the source to the target which will result in a "local minima" in travel time, and the sound will take every one of these paths. If each of those would have taken equally long time to travel, the sound from the different paths would have made perfect resonance with each other. However, if the different paths differ a little bit in time from each other, the sound will begin to decrease their resonance with each other, and the total sound will be less loud.

Say that there are $n \rightarrow \infty$ paths for the sound to travel, and that the travel time for the different paths is evenly distributed between $a$ and $b$, for simplicity. Also say that the amplitude in air pressure for each of the wavefronts is $\frac{Y}{n}$. Assume that the sound has the frequency $f$. Now, if every path would have taken equally long time, the amplitude of the soundwaves put together at the target point, would have been $Y$. But now that's not the case.

Say that the formula for how much the pressure differ from the normal air pressure at each wavefront is $\sin(t\cdot f\cdot 2\pi)\cdot\frac{Y}{n}$, where $t$ is the time, then the difference in air pressure at the target point would be

$Y\cdot\int^1_{0}\sin((t-(a+(a-b)\cdot x))\cdot f\cdot 2\pi)dx\ =$
$=\ Y\cdot\frac{\cos((t-b)\cdot f\cdot 2\pi)-\cos((t-a)\cdot f\cdot 2\pi)}{(b-a)\cdot f\cdot 2\pi}\ =$
$=\ Y\cdot\sin\left(\left(t-\frac{b+a}{2}\right)\cdot f\cdot 2\pi\right)\cdot\frac{\sin\left(\displaystyle{\fra c{b-a}{2}}\cdot f\cdot 2\pi\right)}{\displaystyle{\frac{b-a}{2}}\cdot f\cdot 2\pi}$

So, as you can se, we first have the factor $Y$ which has to do with the amplitude of the wave, then we have the $\sin\left(\left(t-\frac{b+a}{2}\right)\cdot f\cdot 2\pi\right)$ factor, which is responsible for the frequency of the wave (which is still $f$).

Then we have the $\frac{\sin\left(\displaystyle{\frac{b-a}{2}}\cdot f\cdot 2\pi\right)}{\displaystyle{\frac{b-a}{2}}\cdot f\cdot 2\pi}$ factor will be constant over time, and also has to do with the amplitude of the wave. And as the frequency is small, this factor will be close to 1. But as the frequency grows, or as the difference between $a$ and $b$ grows, the factor will decrease and eventually converge to 0 as the frequency or the difference between $a$ and $b$ grows.

So, as the frequency gets higher, or the more obstacles there is in the environment so the travel time will differ more from different paths, the more the sound will be muted and filtered out by he obstacles. Also, if the sound speed is increased (maybe because of increased air temperature), the differece in traveling time for the different paths will be decreased so that the sound will be less muted. But if we keep the same wavelength instead of the same frequency, the frequency is increased, and that will conpensate for what I previoulsy said. So at the end, what really matters is the wavelength of the sound.

What do you say about it? Any comments? Hard to follow, to poorley explained? Incorrect?

Also, I believe that walls and other "sound-absorbing" medias, has a special ability to absorbe certain wavelengths (especially high) when the sound is moving through it. Then the frequency would mater too.

2. Originally Posted by TriKri
If each of those would have taken equally long time to travel, the sound from the different paths would have made perfect resonance with each other, and the total sound will be less loud.
Multiple paths of the same travel time -> constructive interference ->
louder that if there was but one path.

However, if the different paths differ a little bit in time from each other, the sound will begin to decrease their resonance with each other.
Incoherent summation -> less loud than coherent summation as above.

RonL