CSD313L_SoundIntensityPressureNarration

# Reference decibel sound pressure decibel db il 10

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Unformatted text preview: l – Sound Pressure deciBel dB IL = 10 * log10(Ix / Ir) Since Ix = px2 / Z and Ir = pr2 / Z Since Using px2 / Z to replace Ix and Ir = pr2 / Z to replace Ir, So – dB IL = 10 * log10(Ix / Ir) = 10* log10((px2/Z) / (pr2/Z)) = 10* )) log10(px2/pr2) = 10* log10(px/pr)2 (law 3 of logarithm) = log 10*2* log10(px/pr) = 20 * log10(px/pr) = dB SPL 10*2* log – Thus, dB SPL = 20 * log10(px/pr), in which pr = 2 x 10-5 Thus, log Pa Pa – Very important, dB IL = dB SPL dB SPL - Example dB Given a sound pressure at 0.02 Pa, how Given much is dB SPL? How much is dB IL? much dB SPL = 20 * log10(px/pr) = 20 * log10(0.02/ dB log log (2 x 10-5)) = 20 * log10((2 x 10-2)/ (2 x 10-5)) (2 )) log 10 )) = 20 * log10(103) = 20 * 3 = 60 dB SPL log Since dB IL = dB SPL, this sound has Since intensity at 60 dB IL. intensity dB IL and dB SPL - Example dB 3 Given two sounds, A and B, IA = 10--3 watt/m2 watt/m and pB = 0.4 Pa, which one has higher and level? level? – For sound A, dB IL = 10 * log10(Ix / Ir) = 10*log10(10-3 / 10-12) = 10*log10(109) = 10*9 10*log 10*9 = 90 dB IL 90 – For sound B, dB SPL = 20 * log10(px/pr) = 20 * For log log10(0.4/ (2 x 10-...
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## This note was uploaded on 02/01/2014 for the course CSD 313L taught by Professor Liu during the Summer '08 term at University of Texas.

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