THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

or in terms of complex exponential functionsp 1 = Re [ P 1 e i(ω 1t + φ 1 ) ]p 2 = Re [ P i( ω 2 t+ φ 2 )]2where3.16 Correlated and Uncorrelated Sound 59p = instantaneous sound pressureP = amplitude of sound pressureω = angular frequencyφ = phase angle(3.33c)(3.33d)The instantaneous sound pressure resulting from the superimposition of the twowaves is given by the sum of the two instantaneous sound pressures, and theroot-mean-square sound pressure of the combined waves can be found fromp 2 rms = 1 T∫ T0(p 1 + p 2 ) 2 dt (3.34)where T represents the averaging time, which should be an integer number ofperiods at both frequencies. In real measurements it suffices to have the averagingtime cover many periods so that contributions from fractional periods becomeinsignificant. This condition is met if T ≫ 1/f lower , where f lower is the lower ofthe two frequencies. Inserting Equations (3.33a) and (3.33b) or Equations (3.33c)and (3.33d) into Equation (3.34) and integrating, we obtainprms 2 = P12 + P2 2= prms1 2 2+ prms2 2 for ω 1 ≠ ω 2 (3.35)p 2 rms = P12 + P2 2+ P1 P2 cos(φ21 φ 2 ) for ω 1 = ω 2 (3.36)Consider the case of two 4-kHz signals that are in phase at the receiving point.Each of the signals has a sound pressure level of 60 dB. From Equation (3.36) forsound waves of the same frequency, the root-mean-square pressure is given byprms 2 = P2 1 + P2 2+ P 1 P 2 cos 0 = 2P1 2 2= 4 prms2The increase in sound pressure level as the result of adding an identical in-phasepure tone is( p 2L P − L P1 = 10 log rmsp 2 ref= 10 log 4 ≈ 6dB)− 10 log( p 2) (rms14p2)= 10 log rms1p 2 refp 2 rms1The combined signals result in a 4-Hz signal with an SPL of 66 dB. If the amplitudeswould be out of phase by π radians (or 180 ◦ ) and the frequencies are equal, thesound pressure would theoretically be zero at the observation point.