On Fri, 24 Aug 2012 03:38:38 -0700, "William Sommerwerck"
wrote:

First off you have to find/have/create an amplitude-
responding pickup. Do that first then re-enter the discussion.

They're commonly found in cheap turntables, especially of the USB ilk.

I seriously doubt that. Those pickups are either magnetic or
piezoelectric, both velocity response.

Second, a triangle wave is the integral of a square wave,
or the same relation between amplitude and velocity response.

Yes, of course. That's simple calculus, which I took in high school almost
50 years ago.


But they will claim that the z-transform make it so when
challenged. Just as you argue about velocity- versus
amplitude-responding pickups. Find me a phonograph
pickup that is amplitude responding. Then we have
something to discuss.

First of all, the issue has nothing whatever to do with whether
amplitude-responding pickups exist, ever have existed, or ever could exist.
We are talking about a manufacturer claiming that a test record shows how a
pickup responds to (mechanical) square-wave modulation, when the disk
doesn't have such modulation.

But it IS a square wave in velocity. Also, is there any cutter that you
can imagine that cut a square wave in amplitude or a stylus that can track
it?

Piezoelectric transducers are basically amplitude-sensitive. Crystal and
ceramic pickups were manufactured for decades, but gradually disappeared as
magnetic pickups grew less expensive and tracked at lower forces. There have
been "good" ceramic pickups (Sonotone and Weathers, for example), but they
were rare. However, you can still find ceramic pickups in cheap turntables.

http://www.knowzy.com/computers/audi...comparison.htm

(The author's claim that anti-skating is an absolute necessity is debatable,
to say the least.)

There are plenty of other techniques have equivalent geometry to the
cutting lathe to be sure.

If you want to get super-ultra picky about it, many ceramic pickups are
mechanically equalized to compensate for the ~12dB shelf in response when
playing RIAA LPs. This doesn't change the fact that the pickup is,
fundamentally, an amplitude-responding device.

If you get a strain sensor pickup they are amplitude responding. Rare and
very expensive. Also there are laser types that are "amplitude"
responding; even more expensive.


I constantly fight with other engineers that claim that just because it is
on some manufacturers' data sheet it is right/true. e.g. that the name of
the standard is RS-232 when the cover of the current 15+ year old (1997)
version of standard says TIA-232. Likewise network connectors which are
IEC 60603-7-x (See page 13 of TIA-568.2) or 8P8C modular instead of RJ-45
(which not even phone companies use any more). I will bet that you have
seen such issues but rarely dug into it to the finish.

It appears that we are equally annoyed by the spread of misinformation.

First off you have to find/have/create an amplitude-
responding pickup. Do that first then re-enter the
discussion.

They're commonly found in cheap turntables, especially
of the USB ilk.

I seriously doubt that. Those pickups are either magnetic
or piezoelectric, both velocity response.

Sorry, piezo pickups are amplitude-responding. Where did you ever get the
idea they were velocity-responding? If you have any doubts, get a ouija
board and contact Pierre Curie.

I detest "appealing to authority", but here are six (technically, five)...

http://www.enjoythemusic.com/cartridgehistory.htm (This article is 54
years old.)

http://www.audioregenesis.com/downlo...%20Process.pdf

http://www.epanorama.net/links/audiohifi.html#vinyl

http://www.epanorama.net/circuits/phono.html

http://www.audiobanter.com/showthrea...t=53419&page=2

http://www.roger-russell.com/sonopg/sonopc.htm (Note, in particular, the
response curves.)

These are among the top listings when Googling "ceramic pickup amplitude
response" (sans quotes). Note that in the fifth one, the writer shows a
fundamental misunderstanding of why recording EQ is used and how it is
implemented.


(The author's claim that anti-skating is an absolute necessity
is debatable, to say the least.)

There are plenty of other techniques have equivalent geometry
to the cutting lathe to be sure.

Anti-skating has nothing to do with matching the geometry of the cutting
lathe's path. However, there have been numerous attempts at tone arms that
either track linearly, or which have zero tracking error. Unfortunately,
there's never been one on an $80 hunk-o'-plastic.


If you want to get super-ultra picky about it, many ceramic pickups
are mechanically equalized to compensate for the ~12dB shelf in
response when playing RIAA LPs. This doesn't change the fact
that the pickup is, fundamentally, an amplitude-responding device.

If you get a strain sensor pickup they are amplitude responding.
Rare and very expensive. Also there are laser types that are
"amplitude" responding; even more expensive.

See preceding.

In article ,
"William Sommerwerck" wrote:

First off you have to find/have/create an amplitude-
responding pickup. Do that first then re-enter the
discussion.

They're commonly found in cheap turntables, especially
of the USB ilk.

I seriously doubt that. Those pickups are either magnetic
or piezoelectric, both velocity response.

Sorry, piezo pickups are amplitude-responding. Where did you ever get the
idea they were velocity-responding? If you have any doubts, get a ouija
board and contact Pierre Curie.

I detest "appealing to authority", but here are six (technically, five)...

http://www.enjoythemusic.com/cartridgehistory.htm (This article is 54
years old.)

http://www.audioregenesis.com/downlo...%20Process.pdf

http://www.epanorama.net/links/audiohifi.html#vinyl

http://www.epanorama.net/circuits/phono.html

http://www.audiobanter.com/showthrea...t=53419&page=2

http://www.roger-russell.com/sonopg/sonopc.htm (Note, in particular, the
response curves.)

These are among the top listings when Googling "ceramic pickup amplitude
response" (sans quotes). Note that in the fifth one, the writer shows a
fundamental misunderstanding of why recording EQ is used and how it is
implemented.


(The author's claim that anti-skating is an absolute necessity
is debatable, to say the least.)

There are plenty of other techniques have equivalent geometry
to the cutting lathe to be sure.

Anti-skating has nothing to do with matching the geometry of the cutting
lathe's path. However, there have been numerous attempts at tone arms that
either track linearly, or which have zero tracking error. Unfortunately,
there's never been one on an $80 hunk-o'-plastic.

Linear tracking doesn't work out well in practice in any case, because
the signal sent to the cutter is distorted (or pre-altered, or modified,
or call-it-what-you-will), to benefit playback with a pickup traversing
the disk in an arc.

Isaac

Linear tracking doesn't work out well in practice in any case,
because the signal sent to the cutter is distorted (or pre-altered,
or modified, or call-it-what-you-will), to benefit playback with
a pickup traversing the disk in an arc.

This is absolutely untrue. If it were true -- what would the point of linear
tracking?

It is theoretically possible (in much the same way that Dynagroove distorted
the groove to compensate for the finite diameter of the stylus) to do this,
but it would require a standard for the tracking error -- that is, every arm
would have to have "mis-track" in the same way.

There is no such standard, and never has been.

In article ,
"William Sommerwerck" wrote:

Linear tracking doesn't work out well in practice in any case,
because the signal sent to the cutter is distorted (or pre-altered,
or modified, or call-it-what-you-will), to benefit playback with
a pickup traversing the disk in an arc.

This is absolutely untrue. If it were true -- what would the point of linear
tracking?

My point exactly. There is no point to it, for playback -- well, except
to sell strange contraptions to the uninformed.

It is theoretically possible (in much the same way that Dynagroove distorted
the groove to compensate for the finite diameter of the stylus) to do this,
but it would require a standard for the tracking error -- that is, every arm
would have to have "mis-track" in the same way.

There is no such standard, and never has been.

A "standard", no, but alteration of the cutting waveform to minimize
several forms of "distortion", certainly. Each record company had their
own "recipe".

Again I recommend you consult one of the older editions of Eargle's
"Handbook of Recording Engineering".

Isaac

Linear tracking doesn't work out well in practice in any case,
because the signal sent to the cutter is distorted (or pre-altered,
or modified, or call-it-what-you-will), to benefit playback with
a pickup traversing the disk in an arc.

This is absolutely untrue. If it were true -- what would the point
of linear tracking?

My point exactly. There is no point to it, for playback -- well, except
to sell strange contraptions to the uninformed.

No, that was not your point. You stated something that was absolutely
incorrect. No one has ever cut a phonograph record that was predistorted to
compensate for tracking error.

Linear tracking is theoretically correct. Whether a particular arm
implements it in a way that's simple and reliable, and doesn't introduce its
own problems, is another matter.


It is theoretically possible (in much the same way that Dynagroove
distorted the groove to compensate for the finite diameter of the
stylus) to do this, but it would require a standard for the tracking
error -- that is, every arm would have to have "mis-track" in the same
way.

There is no such standard, and never has been.

A "standard", no, but alteration of the cutting waveform to minimize
several forms of "distortion", certainly. Each record company had their
own "recipe".

Tracking distortion was not one of them.

On Aug 27, 6:47*am, "William Sommerwerck"
wrote:
Linear tracking doesn't work out well in practice in any case,
because the signal sent to the cutter is distorted (or pre-altered,
or modified, or call-it-what-you-will), to benefit playback with
a pickup traversing the disk in an arc.
This is absolutely untrue. If it were true -- what would the point
of linear tracking?
My point exactly. There is no point to it, for playback -- well, except
to sell strange contraptions to the uninformed.

No, that was not your point. You stated something that was absolutely
incorrect. No one has ever cut a phonograph record that was predistorted to
compensate for tracking error.

Linear tracking is theoretically correct. Whether a particular arm
implements it in a way that's simple and reliable, and doesn't introduce its
own problems, is another matter.

It is theoretically possible (in much the same way that Dynagroove
distorted the groove to compensate for the finite diameter of the
stylus) to do this, but it would require a standard for the tracking
error -- that is, every arm would have to have "mis-track" in the same
way.
There is no such standard, and never has been.
A "standard", no, but alteration of the cutting waveform to minimize
several forms of "distortion", certainly. Each record company had their
own "recipe".

Tracking distortion was not one of them.

This discussion has morphed into something very interesting, however
and regardless of what anyone may think of the validity, or even the
sanity for that matter of my original post, the stuff of which has
somehow disappeared from this thread, it doesn't solve my problem as
to where I might find a replacement for an old mechanical three speed
automatic turntable with a simple high output ceramic cartridge. Lenny

This discussion has morphed into something very interesting, however
and regardless of what anyone may think of the validity, or even the
sanity for that matter of my original post, the stuff of which has
somehow disappeared from this thread, it doesn't solve my problem as
to where I might find a replacement for an old mechanical three speed
automatic turntable with a simple high output ceramic cartridge.

I said nothing about this, because it seems unlikely you can find a /direct/
replacement. These changers usually sit in product-specific cutout.

You might try looking for an early KLH with the Garrard changer you could
mount different pickups in. (I might have one I could sell you cheap.) The
problem is... How are you going to mount the unit?

You can buy cheap phono preamps. If you can find an easy way to power one,
you could substitute a 'table with a magnetic pickup.

Generally to mount it you hopefully have a standalone case. You cut the original cutout out usually with a jigsaw or something, even a sawsall, being careful of course. Then you take the top of the case and make it fit in that spot. I used to only remove the material that needed to be removed which was determined by laying the new piece over the old and marking it.

I forsee trouble making a ceramic cartridge fit in an arm designed for magnetic. It might almost be better to just find a standalone preamp. That adds a little bit of cost but will save alot of time and trouble.

What it does also boil down to is whether the customre is willing to pay enough to make any money off this. Fixing the original starts looking attractive when all of what the changeover entails is considered.

Plus we are talking about a turntable that works IIRC, just not automatically. Automatic turntables that take magnetic cartridges are available but less common than full manual ones. Audiophiles usually shunned "record changers", only accepting "auto return" for obvious reasons.

J

klem kedidelhopper wrote:

This discussion has morphed into something very interesting, however
and regardless of what anyone may think of the validity, or even the
sanity for that matter of my original post, the stuff of which has
somehow disappeared from this thread, it doesn't solve my problem as
to where I might find a replacement for an old mechanical three speed
automatic turntable with a simple high output ceramic cartridge. Lenny


Thrift stores? Flea markets? I've seen a lot of them collecting dust.
How about the radio & phono newsgroup?
news:rec.antiques.radio+phono

On Aug 28, 9:14*pm, "Michael A. Terrell"
wrote:
klem kedidelhopper wrote:

This discussion has morphed into something very interesting, however
and regardless of what anyone may think of the validity, or even the
sanity for that matter of my original post, the stuff of which has
somehow disappeared from this thread, it doesn't solve my problem as
to where I might find a replacement for an old mechanical three speed
automatic turntable with a simple high output ceramic cartridge. Lenny

* *Thrift stores? Flea markets? I've seen a lot of them collecting dust.
How about the radio & phono newsgroup?
news:rec.antiques.radio+phono

Those are some good ideas guys. I wouldn't even bother with this but
he's an old guy, (even older than me) and seems very attached to his
"record player", so I thought I'd try to help him out. I never thought
it would be so difficult to find one of these things new. I guess I
should have. The problem is I've been doing this for so long I've lost
track of how many years ago it was that I could just walk into my
jobber's store and pick up one for 12.00 or so. I typically never had
the luxury of having a proper mounting board though. I would have to
"cut and paste" to make the replacement changer fit but it always
worked out. Funny thing is I just realized that I actually have an old
changer with a ceramic cartridge in a plastic cabinet. My wife used
this cheapie in her Pre School until she recently retired and it went
up into the closet with some of my other relics I've no doubt also
lost track of. I'll have to get it out and check it over. I would not
play my records on it and so we'll never use it again. I have two
Thorens machines and a Benjamin Miracord all with magnetic cartridges.
Lenny

On Sat, 25 Aug 2012 03:23:30 -0700, "William Sommerwerck"
wrote:


Sorry, piezo pickups are amplitude-responding. Where did you ever get the
idea they were velocity-responding? If you have any doubts, get a ouija
board and contact Pierre Curie.

I detest "appealing to authority", but here are six (technically, five)...

http://www.enjoythemusic.com/cartridgehistory.htm (This article is 54
years old.)

http://www.audioregenesis.com/downlo...20Process..pdf

http://www.epanorama.net/links/audiohifi.html#vinyl

http://www.epanorama.net/circuits/phono.html

http://www.audiobanter.com/showthrea...t=53419&page=2

http://www.roger-russell.com/sonopg/sonopc.htm (Note, in particular, the
response curves.)

These are among the top listings when Googling "ceramic pickup amplitude
response" (sans quotes). Note that in the fifth one, the writer shows a
fundamental misunderstanding of why recording EQ is used and how it is
implemented.


Only one link gives any support to amplitude response. Nor do i trust
that one.

If you check the physics it is force responding, which is neither
amplitude nor velocity responding.

http://en.wikipedia.org/wiki/Piezoelectricity

?-)

Mr Barrett, I urge you to do your own research on this, rather than assuming
you know what you're talking about. You don't.

A ceramic pickup is, indeed, an amplitude-responding device. This is fact. I
gave multiple references but you don't believe them. I can only refer you to
the work of Pierre Curie. (You know him. He was Monsieur Madame Curie.)

http://en.wikipedia.org/wiki/Pierre_Curie

Hot-lookin' guy. (No wonder Marie was attracted to him.) Certainly
better-looking than Walter Pidgeon.


If you check the physics it is force-responding, which
is neither amplitude nor velocity responding.

http://en.wikipedia.org/wiki/Piezoelectricity

The displacement of a piezo device is proportional to the force applied.
(Ever hear of Hook's Law?) Ergo...

http://en.wikipedia.org/wiki/Hook%27s_law

I'll sum it up this way... An LP is cut with constant-amplitude response. *
If you play a test record with a ceramic pickup, you will get flat response.

Case closed. I don't have time to discuss this further. Take Pope's advice
to heart: "A little learning is a dangerous thing."

* Between the turnover and rolloff frequencies. Between them, LPs are
constant-velocity.

That should have been...

* Above and below the turnover and rolloff frequencies.
Between them, LPs are constant-velocity.

On Tue, 4 Sep 2012 06:05:21 -0700, "William Sommerwerck"
wrote:

Mr Barrett, I urge you to do your own research on this, rather than assuming
you know what you're talking about. You don't.

I will do some if i remember.

A ceramic pickup is, indeed, an amplitude-responding device. This is fact. I
gave multiple references but you don't believe them. I can only refer you to
the work of Pierre Curie. (You know him. He was Monsieur Madame Curie.)

http://en.wikipedia.org/wiki/Pierre_Curie

Hot-lookin' guy. (No wonder Marie was attracted to him.) Certainly
better-looking than Walter Pidgeon.


If you check the physics it is force-responding, which
is neither amplitude nor velocity responding.

http://en.wikipedia.org/wiki/Piezoelectricity

The displacement of a piezo device is proportional to the force applied.
(Ever hear of Hook's Law?) Ergo...

Well we agree on that.

http://en.wikipedia.org/wiki/Hook%27s_law

I'll sum it up this way... An LP is cut with constant-amplitude response. *
If you play a test record with a ceramic pickup, you will get flat response.

Case closed. I don't have time to discuss this further. Take Pope's advice
to heart: "A little learning is a dangerous thing."

* Between the turnover and rolloff frequencies. Between them, LPs are
constant-velocity.

I do believe you have that backwards. It is amplitude only between 1000
Hz and 2122 Hz (barely over an octave), and velocity otherwise. It has to
do with bandpass changes before the cutter head gets into the act. And
yes, the cutter head is an amplitude device (much like a speaker), the
signal that feeds it results in a mostly velocity groove. Try looking at
it with a microscope.

?-)

I'll sum it up this way... An LP is cut with constant-amplitude response. *
If you play a test record with a ceramic pickup, you will get flat
response.

* Between the turnover and rolloff frequencies. Between them, LPs are
constant-velocity.

Got it backwards. I corrected this error in a following post several days
ago.

On Sat, 8 Sep 2012 02:03:45 -0700, "William Sommerwerck"
wrote:

I'll sum it up this way... An LP is cut with constant-amplitude response. *
If you play a test record with a ceramic pickup, you will get flat
response.

* Between the turnover and rolloff frequencies. Between them, LPs are
constant-velocity.

Got it backwards. I corrected this error in a following post several days
ago.


Get out your microscope and look at the grooves and compare to the
playback waveforms. Complex waveforms are ok but something like a square
wave groove would be best.

?-)

Get out your microscope and look at the grooves and compare
to the playback waveforms. Complex waveforms are ok but
something like a square wave groove would be best.

Do you know of a disk that has a square wave cut on it?

I have a lot of test disks, but I don't remember one. I'll have to look.

In article ,
"William Sommerwerck" wrote:

Get out your microscope and look at the grooves and compare
to the playback waveforms. Complex waveforms are ok but
something like a square wave groove would be best.

Do you know of a disk that has a square wave cut on it?

I have a lot of test disks, but I don't remember one. I'll have to look.

Think about what the stylus would have to do. That's why you won't find
a test disk with a square-wave groove on it.

Isaac

Get out your microscope and look at the grooves and compare
to the playback waveforms. Complex waveforms are ok but
something like a square wave groove would be best.

Do you know of a disk that has a square wave cut on it?
I have a lot of test disks, but I don't remember one. I'll have to look.

Think about what the stylus would have to do. That's why you won't find
a test disk with a square-wave groove on it.

I assume you mean the cutting stylus. It is possible to send a signal to a
cutting head that would, in principle, produce a square wave on the blank.
For example, you could cut a 200Hz square wave with 36dB of pre-emphasis at
12.8kHz, which is not unreasonable.

What the groove would look like is another matter. Obviously, the cutter
head cannot move instantaneously, and there would probably be ringing from
system resonances. Still, I expect that the groove modulation would be
readily identifiable as a square wave, and not some other waveform.

In article ,
"William Sommerwerck" wrote:

Get out your microscope and look at the grooves and compare
to the playback waveforms. Complex waveforms are ok but
something like a square wave groove would be best.

Do you know of a disk that has a square wave cut on it?
I have a lot of test disks, but I don't remember one. I'll have to look.

Think about what the stylus would have to do. That's why you won't find
a test disk with a square-wave groove on it.

I assume you mean the cutting stylus.

No; I mean the playback stylus. In the groove, the rising and falling
edges of the waveform would be perfectly perpendicular to the motion of
the disk, and so there would be no possible way for the stylus to follow
it. And if they were not perpendicular, then it wouldn't be a square
wave.

It is possible to send a signal to a
cutting head that would, in principle, produce a square wave on the blank.
For example, you could cut a 200Hz square wave with 36dB of pre-emphasis at
12.8kHz, which is not unreasonable.

That still would not be truly a "square" wave, because the rise and fall
times would not be infinitely fast. You could do it by stopping the
rotation of the disk while the rising and falling edges were being cut,
for example, but no matter; the groove still could not be tracked by a
mechanical stylus.

Isaac

It is possible to send a signal to a cutting head that would,
in principle, produce a square wave on the blank. For example,
you could cut a 200Hz square wave with 36dB of pre-emphasis
at 12.8kHz, which is not unreasonable.

That still would not be truly a "square" wave, because the rise and fall
times would not be infinitely fast. You could do it by stopping the
rotation of the disk while the rising and falling edges were being cut,
for example, but no matter; the groove still could not be tracked by a
mechanical stylus.

I don't know what the logical fallacy is involved here (other than
intellectual petulance), but it's impossible to have zero rise time on any
square wave. There is no infinite bandwidth in either the electrical or
mechanical realms.

You're saying that because we cannot generate perfect ("infinite-bandwidth")
square waves that there is no such thing as a square wave. I suggest you
look at the output of a square-wave generator on a 'scope, and tell us
whether a reasonable person would consider the waveform "square".

In practical terms, if the highest harmonic in a square wave is
significantly higher than the bandwidth of the DUT, then the DUT should
behave as if a "real" square wave were being applied.

In article ,
William Sommerwerck wrote:

You're saying that because we cannot generate perfect ("infinite-bandwidth")
square waves that there is no such thing as a square wave. I suggest you
look at the output of a square-wave generator on a 'scope, and tell us
whether a reasonable person would consider the waveform "square".

In practical terms, if the highest harmonic in a square wave is
significantly higher than the bandwidth of the DUT, then the DUT should
behave as if a "real" square wave were being applied.

That's a very good working approximation.

And, again, in practical terms, once you make the "angle" in the
(near) square wave you have cut/pressed sharp enough, a mechanical
stylus just won't track it. The angle of the groove's motion
side-to-side, the V-shape of the groove, and the shape of the stylus
(typically an elliptical cone) will be such that most of the
mechanical force generated by the pressure of the stylus on the groove
will be upwards rather than sideways. The stylus tip will accelerate
upwards more rapidly than it will accelerate sideways, and it'll pop
out of the groove entirely.

The Sheffield Direct to Disc pressing of the 1812 Overture was
notorious for this... the "fire the cannon" passage would pop almost
any cartridge out of the groove. It wasn't quite at the point where
it was literally impossible to track... but it was a *really* severe
test of a cartridge's ability to follow a groove.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

And, again, in practical terms, once you make the "angle" in the
(near) square wave you have cut/pressed sharp enough, a mechanical
stylus just won't track it. The angle of the groove's motion
side-to-side, the V-shape of the groove, and the shape of the stylus
(typically an elliptical cone) will be such that most of the
mechanical force generated by the pressure of the stylus on the groove
will be upwards rather than sideways. The stylus tip will accelerate
upwards more rapidly than it will accelerate sideways, and it'll pop
out of the groove entirely.

The Sheffield Direct to Disc pressing of the 1812 Overture was
notorious for this... the "fire the cannon" passage would pop almost
any cartridge out of the groove. It wasn't quite at the point where
it was literally impossible to track... but it was a *really* severe
test of a cartridge's ability to follow a groove.

I think you're thinking of the Telarc disk. (I don't remember a Sheffield
"1812".) I don't know what the waveform "looked like", but I doubt it
approximated a square wave.

I do remember it knocking the pickup out of the groove. Not only did you
need a good-tracking pickup, but the arm-mass / stylus-compliance resonant
had to be very low -- so low that it would not normally be considered a
reasonable resonant point.

In article ,
"William Sommerwerck" wrote:

It is possible to send a signal to a cutting head that would,
in principle, produce a square wave on the blank. For example,
you could cut a 200Hz square wave with 36dB of pre-emphasis
at 12.8kHz, which is not unreasonable.

That still would not be truly a "square" wave, because the rise and fall
times would not be infinitely fast. You could do it by stopping the
rotation of the disk while the rising and falling edges were being cut,
for example, but no matter; the groove still could not be tracked by a
mechanical stylus.

I don't know what the logical fallacy is involved here (other than
intellectual petulance)

There's no fallacy. That is, in fact, one way to cut a groove in a vinyl
record that is a true, accurate square wave. Basically, you run the
lathe like a circular plotter. It's sort of the limit case for the
practice of half-speed mastering (which, done right, is a Very Good
Thing to do).

but it's impossible to have zero rise time on any
square wave. There is no infinite bandwidth in either the electrical or
mechanical realms.

Yup. Mathematics, now, can handle them just fine. But if you want to
claim that a band-limited square wave still should be called a "square
wave", I disagree. If the wave's fundamental is, say, 15 kHz, then what
gets carved into the vinyl will be pretty close to a sine wave. And I
think not a lot of folks would describe that as a "square wave".

You're saying that because we cannot generate perfect ("infinite-bandwidth")
square waves that there is no such thing as a square wave. I suggest you
look at the output of a square-wave generator on a 'scope, and tell us
whether a reasonable person would consider the waveform "square".

Having compensated my share of "high bandwidth" 'scope probes (and built
a few special-purpose probes along the way) I'd have to say, nope, those
things with the horns on the edges (Gibbs' phenomenon) don't look very
square to me.


In practical terms, if the highest harmonic in a square wave is
significantly higher than the bandwidth of the DUT, then the DUT should
behave as if a "real" square wave were being applied.

Generally, that is true. There are situations, however, where the
"functional bandwidth" of the DUT seems to predict one thing, while the
"trouble-causing bandwidth" causes something rather different to happen.
Ever see an audio amp that works (and measures) fine with the expected
(and designed in) rolloff above the audible range, but then the gain
pops back up again at a few megahertz (or more), which cause the thing
to go all unstable under certain conditions?

One reason for testing with the fastest-edged waves you can get is to
discover things like that.

Isaac

In article ,
"William Sommerwerck" wrote:

And, again, in practical terms, once you make the "angle" in the
(near) square wave you have cut/pressed sharp enough, a mechanical
stylus just won't track it. The angle of the groove's motion
side-to-side, the V-shape of the groove, and the shape of the stylus
(typically an elliptical cone) will be such that most of the
mechanical force generated by the pressure of the stylus on the groove
will be upwards rather than sideways. The stylus tip will accelerate
upwards more rapidly than it will accelerate sideways, and it'll pop
out of the groove entirely.

The Sheffield Direct to Disc pressing of the 1812 Overture was
notorious for this... the "fire the cannon" passage would pop almost
any cartridge out of the groove. It wasn't quite at the point where
it was literally impossible to track... but it was a *really* severe
test of a cartridge's ability to follow a groove.

I think you're thinking of the Telarc disk. (I don't remember a Sheffield
"1812".) I don't know what the waveform "looked like", but I doubt it
approximated a square wave.

I do remember it knocking the pickup out of the groove. Not only did you
need a good-tracking pickup, but the arm-mass / stylus-compliance resonant
had to be very low -- so low that it would not normally be considered a
reasonable resonant point.

I also think it was a Telarc disk. And the problem with it wasn't fast
rise time, it was that the groove made a huge excursion -- so great
that, on most pickups, the stylus just couldn't move that far. Usually,
the ones that could track it did so because (as you point out) the arm
mass was so low that the whole arm moved sideways, which cut the signal
down, but kept the needle in the groove.

Isaac

but it's impossible to have zero rise time on any square wave.
There is no infinite bandwidth in either the electrical or
mechanical realms.

Yup. Mathematics, now, can handle them just fine. But if you want to
claim that a band-limited square wave still should be called a "square
wave", I disagree. If the wave's fundamental is, say, 15 kHz, then what
gets carved into the vinyl will be pretty close to a sine wave. And I
think not a lot of folks would describe that as a "square wave".

In the real world, ALL SQUARE WAVES ARE BAND-LIMITED. All complex waveforms
are band-limited. Period.

Do you argue merely for the sake of arguing? We are talking about a
particular situation -- the use of a "real" square wave on an LP to get an
idea of the pickup's mechanical characteristics.

If it's of any interest, I'm working on a fantasy screen play in which
Josiah Gibbs appears.

I think you're thinking of the Telarc disk. (I don't remember a Sheffield
"1812".) I don't know what the waveform "looked like", but I doubt it
approximated a square wave.

I do remember it knocking the pickup out of the groove. Not only did you
need a good-tracking pickup, but the arm-mass / stylus-compliance
resonant
had to be very low -- so low that it would not normally be considered a
reasonable resonant point.

I also think it was a Telarc disk. And the problem with it wasn't fast
rise time, it was that the groove made a huge excursion -- so great
that, on most pickups, the stylus just couldn't move that far. Usually,
the ones that could track it did so because (as you point out) the arm
mass was so low that the whole arm moved sideways, which cut the
signal down, but kept the needle in the groove.

Stylus, stylus, stylus!

I worked in an audio store at the time, and the best arm for this disk was
the Dynavector, which had very high lateral mass, with relatively low
vertical mass.