CSD313L_SoundIntensityPressureNarration

We will use db il 10 log10ix ir to figure out to ix

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: og10(Ix / Ir) to figure out to Ix. dB IL / 10 = log10(Ix / Ir) 10 dB IL/10 = Ix / Ir 10 Ix = Ir * 10 dB IL/10 – Example, for a sound at 70 dB IL, how much is Example, Ix ? Ix = Ir * 10 dB IL/10 = 10 -12 * 10 70/10 = 10 -12 * 10 7 10 +7= -5 10 -5 watts/m2 = 10 -12 Sound Pressure Sound Another way to measure sound level Force applied on a unit area – p = F/A – 1 Pa = 1 N/m2 Range of acoustic pressure for humans: 2 x Range 10-5 Pa (the softest sound that is audible) to 10 20 Pa (the sound level starting to cause 20 discomfort) Acoustic Intensity and Pressure I = p2 / Z – Acoustic intensity (I) is equal to pressure (p) square divided by acoustic impedance (Z) – Acoustic impedance is the total opposition to sound energy transfer. Absolute and Relative Measure of Sound Pressure Sound Absolute measure of sound pressure – 2 x 10-5 Pa – 3 Pa Relative measure of sound pressure – px / pr (ratio), in which px is the target sound pressure and pr is reference sound pressure and – Reference sound pressure: pr = 2 x 10-5 Pa – For a sound with a pressure at 10-3 Pa Relative (ratio) measure: px / pr = 10-3 Pa / 2x10-5 Pa = 50, meaning this sound has a pressure 50 times higher than the reference sound. reference deciBe...
View Full Document

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 at Austin.

Ask a homework question - tutors are online