3 ms·
Perhaps it would be helpful to consider the path of a phonon going through a barrier. The odds of that phonon making it through is 1 chance in N. If there are t
by javajosh 3y ago
Perhaps it would be helpful to consider the path of a phonon going through a barrier. The odds of that phonon making it through is 1 chance in N. If there are two barriers, the odds of a phonon making it through both in series is 1 chance in N^2. That is, the odds multiply, and you are absolutely correct that dB's add (since they are basically log(N) in this case and log(N^2) = 2log(N)).
- kazinator 3y agoThis probability model doesn't seem to add anything to just straightforward reasoning about how each barrier attenuates the intensity of the sound (or other signal) by a certain fraction. E.g. 50% is -3 dB, then 25% is another -6 dB for at total of -9 dB. 1> (log10 .5) -0.301029995663981 2> (* 10 (log10 .5)) -3.01029995663981 3> (* 10 (log10 .25)) -6.02059991327962 4> (* 10 (log10 (* .5 .25))) -9.03089986991944 Just logs turning multiplication into addition, like your grandpa's slide rule. In electronics we use a factor of 20 for voltage cuts/boosts. Half the voltage is -6 dB, because power/intensity (Watts) is voltage squared: half the Volts is a quarter the Watts (P = V^2/R).