Subject: Re: two sine tones simultaneously within one critical band From: beaucham <beaucham(at)MANFRED.MUSIC.UIUC.EDU> Date: Fri, 7 Oct 2005 08:55:58 -0500Bob, Recall the trig identity sin(A) + sin(B) = 2*cos(.5(A-B))*sin(.5(A+B)) Let A = 99*2pi*t and B = 101*2pi*t . Then you get Reinhart's formula, which is well known, except that he forgot to include the 2pi's in the final version. Jim You wrote: >From: Bob Masta <audio(at)DAQARTA.COM> >Date: Fri, 7 Oct 2005 08:44:59 -0400 >To: AUDITORY(at)LISTS.MCGILL.CA >Subject: Re: two sine tones simultaneously within one critical band > >On 7 Oct 2005 at 12:40, Reinhart Frosch wrote: > >> Sound pressure of first sine-tone: >> >> p_1(t) = p_0 * sin(99 * 2pi * t); >> >> sound pressure of second sine-tone: >> >> p_2(t) = p_0 * sin(101 * 2pi * t). >> >> [ * = multiplication sign; t = time in seconds.] >> >> Total sound pressure: >> >> p(t) = p_1(t) + p_2(t) = 2p_0 * cos(t) * sin(100 * t). >> >> That last formula implies a 100-hertz sine-tone >> amplitude-modulated so that there are two beats per second. >> >> The 1-mm-long basilar membrane piece strongly excited by a >> soft 99-hertz sine-tone and that strongly excited by >> a soft 101-hertz sine-tone overlap almost completely. >> > >Reinhart: > >I am completely at a loss to understand how you arrived >at your last formula. It appears that you have not simply >added the pressure waves, but multiplied them. This is >not what happens in air at normal sound levels, where >there is essentially no nonlinearity. In air the two original >tones are linearly summed and spectral analysis of >the waveform output from a (linear) microphone shows >that only those components are present. The beat >tones are only in the head of the listener. > > >Best regards, > >Bob Masta > >audioATdaqartaDOTcom