Klarinet Archive - Posting 000285.txt from 2001/02

From: Grant Green <gdgreen@-----.com>
Subj: Re: [kl] Combination tones......
Date: Thu, 8 Feb 2001 20:18:49 -0500

>Grant,
>
>Good - a useful statement of what I wanted to say. Sorry about the
>octave/harmonic confusion, although as you say, it doesn't matter for this
>purpose.
>
>I am not sure about your last statement, however. You are saying that if
>you start with two tones (say 1000 and 1001 Hz) you will hear a pulsation
>(a variation in amplitude) at a rate of 1 Hz, and that this 1Hz is a new
>tone? That is, it is really physically there? So that if I gradually

Yes, if the sources are close enough together. If you have a Mac,
you can visualize this by using the Graphing Calculator applet. For
the equation, type in (for example) "sin 4x = sin 6x": this will show
you two sine waves with a frequency ratio of 2:3 (a perfect fifth
interval). Now change the equation to read "sin 4x = sin 6x + sin
4x", and the graph will show the difference tone. The difference
tone is a somewhat more complex waveform, but it repeats exactly
twice for every 4 cycles of the other graph, which is the difference
frequency. The complex waveform indicates that the difference tone
has more timbre (audible higher harmonics) than the two generating
sine waves.

Now, change the equation to "sin 4x @-----. Again, you
get a complex difference tone, with a period equal to the difference
(1 cycle for every 4 cycles of the reference wave). If you then
change the equation to "sin 4x @-----.1x + sin 4x", you'll see a
complex wave that looks like the 4x wave varying from maximum volume
to zero volume to max and back to zero, at a rate of 40 sin 4x peaks
per fluctuation: beats. It is there in the math, regardless of the
system, as long as the sound waves add and cancel in the medium.

>increase the separation until the tones are 1000 and 1200, I will hear a
>200 Hz difference tone, EVEN IN A COMPLETELY LINEAR SYSTEM? I don't think

Yes, exactly. The difference tones arise from the cancelation and
reinforcement of two non-unison frequencies

>so. Returning to the radio world that Tom started from, this logic would
>imply that when my two local broadcast stations with carrier frequencies at
>550 and 1250 MHz are both on the air at once, there is a new carrier at 700
>MHz present all the time, even in the absence of non-linear elements. I
>certainly hope that this is not true, or the FCC is going to have one awful
>problem on its hands. (In fact, of course, there is no such problem, and
>the FCC is free to assign 700 MHz to a third station.)

I can't say I know that much about radio broadcasting, but I suspect
that (a) the FCC assigns carrier frequencies to avoid having two that
heterodyne at the same frequency as a third, and (b) radio receivers
are designed to reject spurious carriers. Come to think of it, the
FCC never assigns AM 700 (AFAIK), or any other multiple of 20 KHz:
they're all 510, 530, 570, etc., which means that any difference
frequencies will be some multiple of 20 KHz (e.g., 1470 KHz - 550 KHz
@-----.

Grant

++++++++++++++++++++++++++++++++++++++++++++++++
Grant Green gdgreen@-----.com
ecode:contrabass http://www.contrabass.com
Professional Fool -> http://www.mp3.com/ProFools
++++++++++++++++++++++++++++++++++++++++++++++++

---------------------------------------------------------------------
Unsubscribe from Klarinet, e-mail: klarinet-unsubscribe@-----.org
Subscribe to the Digest: klarinet-digest-subscribe@-----.org
Additional commands: klarinet-help@-----.org
Other problems: klarinet-owner@-----.org

   
     Copyright © Woodwind.Org, Inc. All Rights Reserved    Privacy Policy    Contact charette@woodwind.org