Re: two sine tones simultaneously within one critical band

[beaucham <beaucham@xxxxxxxxxxxxxxxxxxxxxx>]

Bob,

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@xxxxxxxxxxx>
>Date:         Fri, 7 Oct 2005 08:44:59 -0400
>To: AUDITORY@xxxxxxxxxxxxxxx
>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