[Dick Pierce]

On Jun 22, 9:13*pm, "Arny Krueger" wrote:
"Sonnova" wrote in message
Would you (or someone) like to explain to me how
one would go about cutting a square wave into a
record groove?

You use a cutting lathe with a power amplifier and
appropriate test signal which is contrived to produce
a square wave with an ideal cartridge and preamp,
if the preamp is equalized.

However, the question is misstated because quality *
cartridges as a rule have velocity-sensitive response,
and require a non-square wave cut into the record
groove in order to produce a square wave at either
the output of the cartridge, or as it is more commonly
done, at the output of a RIAA preamp.

No, not correct either.

You are right: the output of the phono cartridge is a
function of the instantaneous velocity of the stylus:
a sqquare wave physically cut into the groove will
result in an alternating train of positive- and negative-
going unit impulses. In that sense, the cartridge is
acting as a differentiator.

But you forget that the frequency response of the
phono preamp has TWO components to it: one of
them is the "RIAA EQ" shelf response with a zero
at 318 uS and a pole at 75 uS. This is what gives
the little shelf between 500 and 2 kHz, roughly.

But the overall response is dominated by the
-6 dB per octave slope across the entire band.
This is an integration function: it will take a train
of alternating unit impulse responses and turn it
back into a train of square waves, e.g.:

integral(derivative(f(x)) = f(x), w.r.t.t.

Thus, ignoring the little shelf around 1 kHz, a
square wave on the disk result in a square wave
coming out of the of the phono preamp.

Which points even more ot the effective
impossibility of cutting anything remotely
resembling a real square wave on a disk: it's
really an compromises, non-optimum exercise
in black magic.