Date:         Mon, 8 Jan 2007 01:46:51 +0100
Reply-To:     Association for Recorded Sound Discussion List
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From:         George Brock-Nannestad <[log in to unmask]>
Subject:      Re: square waves....Re: [ARSCLIST] Libraries disposing of records
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From: Patent Tactics, George Brock-Nannestad

Steven Barr(x) wrote:

> ----- Original Message ----- 
> From: "phillip holmes" <[log in to unmask]>
> > A square wave on a record doesn't look like a square because of the 
> > velocity involved--or something--that I can't explain.  A square wave is
> > built up of stacked "odd integer harmonics" and contains multiple 
> > harmonics.  Somewhere around 12 harmonics, it really looks like a square
> > wave.  The more harmonics present, the more perfectly formed the square 
> > wave becomes.  A perfect square wave would require infinite bandwidth 
> > and infinitely fast electronics (forget the stylus, we can't even 
> > generate a perfect square wave, only a mathematical representation of 
> > one). 
> >
> Now...let's assume we have a DC battery and a switch...so the output
> of the combination is either "current" (switch closed, DC voltage
> present) or "no current" (switch open, no DC voltage). We physically
> flip the switch between the two, say, five times per second. Since
> the voltage rises from zero to X as soon as the switch makes contact,
> don't we get a five Hertz "square wave?" Assume we have nice straight
> wiring (no inductance)...what, if anything, would distort the wave
> shape of our output?

----- nothing - your electrical voltage signal would be near perfect, but it 
would have a DC component at half the battery voltage, and its amplitude 
would be half the battery voltage. However that is an electrical signal 
without inertia. Once you drive an electromagnetic loudspeaker you have the 
inertia in the cone, the self-inductance in the coil, and the two are 
coupled, so you have a resonating circuit. That will be excited by some of 
the components of the square wave, and you will no longer have a square wave 
current, nor square wave acoustical output. But loudspeakers are different, 
and some present a more ohmic load to the generator than others, and this 
improves matter.

----- if you want at least a square wave _electrical_ signal from a 
mechanical record, you will have to have a waveform on that groove that will 
generate a square wave (with some upper frequency limitation, i.e. rounded 
corners) when it is traced by a stylus with a rate-of-change-sensitive 
transducer, such as a magnetic pickup. That groove shape is triangular. The 
slope of that groove is what decides the rate of change of the number of 
magnetic lines that induce the voltage in the coil. So a steep slope one way 
gives a high constant voltage of one polarity, and the same steepness, only 
the other way, gives a high voltage of the opposite polarity, with a huge 
acceleration where the two steepnesses join. That is the precise description 
of a square wave. Less slope means less amplitude of the square wave. The 
slope is negotiated in a given time, and if you double the time, you halve 
(minus 6dB) the square wave voltage. That is what happens in half-speed 
mastering, you avoid overload. Once you send that square wave to 
loudspeakers, you have the situation above again

It is the acceleration that causes most wear and groove overload. A further 
matter: a tape and tonehead act precisely the same way, because a tonehead is 
also a rate-of-change device. Not so with certain digital heads or Hall 
effect heads, where it is the absolute magnetization that is detected.

Kind regards,


George