Hi, I've been chasing high dollar speaker cables for years. Finally, I
decided to make one myself.

However, I don't know how to measure the spec. of the cable.

Like how to use a handheld LCR meter to measure one speaker cable's
inductance and/or capacitance?

Could anyone kind enough to help me?

Thanks in advance!

Lawrence Leung

Lawrence:

Find out the frequency that the LCR meter uses to measure with.
Use a piece of cable -much- shorter than a wavelength of this frequency.
Measure the capacitance of an open circuited piece
Measure the inductance of a short circuted piece.
divide by the length of the piece to get cap per unit length
and inductance per unit length.
characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)
--------------------------------

More accurate measurements can be made taking into account the actual length
of the piece of cable.

Regards,

Cliff

"Lawrence Leung" wrote in message
54...
Hi, I've been chasing high dollar speaker cables for years. Finally, I
decided to make one myself.

However, I don't know how to measure the spec. of the cable.

Like how to use a handheld LCR meter to measure one speaker cable's
inductance and/or capacitance?

Could anyone kind enough to help me?

Thanks in advance!

Lawrence Leung

Lawrence:

Find out the frequency that the LCR meter uses to measure with.
Use a piece of cable -much- shorter than a wavelength of this frequency.
Measure the capacitance of an open circuited piece
Measure the inductance of a short circuted piece.
divide by the length of the piece to get cap per unit length
and inductance per unit length.
characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)
--------------------------------

More accurate measurements can be made taking into account the actual length
of the piece of cable.

Regards,

Cliff

"Lawrence Leung" wrote in message
54...
Hi, I've been chasing high dollar speaker cables for years. Finally, I
decided to make one myself.

However, I don't know how to measure the spec. of the cable.

Like how to use a handheld LCR meter to measure one speaker cable's
inductance and/or capacitance?

Could anyone kind enough to help me?

Thanks in advance!

Lawrence Leung

In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:

Lawrence:

Find out the frequency that the LCR meter uses to measure with.
Use a piece of cable -much- shorter than a wavelength of this frequency.
Measure the capacitance of an open circuited piece
Measure the inductance of a short circuted piece.
divide by the length of the piece to get cap per unit length
and inductance per unit length.

Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)

Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*. The same goes for propagation, which is
frequency-dependent. The way those parameters vary with frequency is
critically dependent upon the physical construction of the line.

For example, as frequency drops without limit, the propagation velocity
also drops *without limit*. Find out why early long-distance telephone
lines needed "loading coils". Then find out why they all were removed.

See the wonderful Schaum's Outline book on Transmission Lines for more
information on this.

More accurate measurements can be made taking into account the actual length
of the piece of cable.

Better still is to avoid the need for measurements at all.

Theoretical calculations will get you fairly close, for cables of any
length useful for speakers -- under a hundred feet, say; do it for a few
different sets of materials and different constructions if you like.
Learning enough to understand whether values anywhere near the ones you
calculated could have any possible effect on audible frequencies being
supplied by a decent, real-world amplifier to any rational set of
speakers will get you the rest of the way to true enlightenment.

Isaac

In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:

Lawrence:

Find out the frequency that the LCR meter uses to measure with.
Use a piece of cable -much- shorter than a wavelength of this frequency.
Measure the capacitance of an open circuited piece
Measure the inductance of a short circuted piece.
divide by the length of the piece to get cap per unit length
and inductance per unit length.

Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)

Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*. The same goes for propagation, which is
frequency-dependent. The way those parameters vary with frequency is
critically dependent upon the physical construction of the line.

For example, as frequency drops without limit, the propagation velocity
also drops *without limit*. Find out why early long-distance telephone
lines needed "loading coils". Then find out why they all were removed.

See the wonderful Schaum's Outline book on Transmission Lines for more
information on this.

More accurate measurements can be made taking into account the actual length
of the piece of cable.

Better still is to avoid the need for measurements at all.

Theoretical calculations will get you fairly close, for cables of any
length useful for speakers -- under a hundred feet, say; do it for a few
different sets of materials and different constructions if you like.
Learning enough to understand whether values anywhere near the ones you
calculated could have any possible effect on audible frequencies being
supplied by a decent, real-world amplifier to any rational set of
speakers will get you the rest of the way to true enlightenment.

Isaac

Isaac Wingfield wrote in message ...
In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:


Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)


Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*.



It is true there is no "characteristic impedance" for long coax
cables, at frequencies below several hundred KHz.

However, Zo = sqrt(L/C), will hold for *any* frequency if: R = 0 and G
= 0. (R the resistance of the cable and, G the conductance).

In the case of heavy gage (12 gage) audio cables , of a short length
(25 ft), R and G are very close to zero. For this case, the calculated
characterisitc impedance will hold down to audio frequenies.

Bob Stanton

Isaac Wingfield wrote in message ...
In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:


Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)


Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*.



It is true there is no "characteristic impedance" for long coax
cables, at frequencies below several hundred KHz.

However, Zo = sqrt(L/C), will hold for *any* frequency if: R = 0 and G
= 0. (R the resistance of the cable and, G the conductance).

In the case of heavy gage (12 gage) audio cables , of a short length
(25 ft), R and G are very close to zero. For this case, the calculated
characterisitc impedance will hold down to audio frequenies.

Bob Stanton

In article ,
(Bob-Stanton) wrote:

Isaac Wingfield wrote in message
...
In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:


Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)


Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*.

It is true there is no "characteristic impedance" for long coax
cables, at frequencies below several hundred KHz.

Not just "coax"; any transmission line at all.

However, Zo = sqrt(L/C), will hold for *any* frequency if: R = 0 and G
= 0. (R the resistance of the cable and, G the conductance).

Yes, but in practice they never are. That approximation only works in
the high frequency region of a line's operating range.

In the case of heavy gage (12 gage) audio cables , of a short length
(25 ft), R and G are very close to zero. For this case, the calculated
characterisitc impedance will hold down to audio frequenies.

In some sense, a "short" line doesn't have a "characteristic impedance",
and for a line long enough (in terms of wavelength) to have one, the
high-frequency approximation Zo = sqrt(L/C) does not hold at low
frequencies.

The point I think we are both making is that if the line is short enough
(in terms of the wavelengths it is carrying), then the line has no
particular characteristics (resistance aside) that could affect the
signal. That is certainly the case for domestic speaker cables.

Isaac

In article ,
(Bob-Stanton) wrote:

Isaac Wingfield wrote in message
...
In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:


Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)


Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*.

It is true there is no "characteristic impedance" for long coax
cables, at frequencies below several hundred KHz.

Not just "coax"; any transmission line at all.

However, Zo = sqrt(L/C), will hold for *any* frequency if: R = 0 and G
= 0. (R the resistance of the cable and, G the conductance).

Yes, but in practice they never are. That approximation only works in
the high frequency region of a line's operating range.

In the case of heavy gage (12 gage) audio cables , of a short length
(25 ft), R and G are very close to zero. For this case, the calculated
characterisitc impedance will hold down to audio frequenies.

In some sense, a "short" line doesn't have a "characteristic impedance",
and for a line long enough (in terms of wavelength) to have one, the
high-frequency approximation Zo = sqrt(L/C) does not hold at low
frequencies.

The point I think we are both making is that if the line is short enough
(in terms of the wavelengths it is carrying), then the line has no
particular characteristics (resistance aside) that could affect the
signal. That is certainly the case for domestic speaker cables.

Isaac

"Lawrence Leung" wrote ...
Hi, I've been chasing high dollar speaker cables for years. Finally, I
decided to make one myself.

What on earth for? Speaker cables are one of the most hyped subjects in the
entire audio world - as long as it's decent thick copper wire, doesn't go
open or short circuit, and has enough cores to carry the signal, anything
will do!
At a recent audio trade fair, a *very* reputable UK speaker manufacturer
was demonstating a new pair of speakers - all the "golden-eared" HiFi
reviewers were raving about the wonderful "clarity and transparency of the
treble, the fluid bass" (and rightly so). An unpretentious reporter noticed
some familiar-looking orange cable running across the room, behind the wall
drapes - "is that what it looks like?" he asked; "yeah, we forgot to bring
the cables, so I popped over the road to Homebase and picked up one of their
Black&Decker garden extension cables - they're good thick copper, make great
speaker cables"
Personally, I use QED 79-strand well terminated in sturdy connectors -
it's about as far as I want to go into expensive cables - and pay more
attention to obstructions and reflective surfaces in front of the speakers,
which make a hell of a lot more difference!

YMMV,
Dave H.
(The engineer formerly known as Homeless)

(ex senior engineer at a reputable pro-audio manufacturer, etc.)

"Lawrence Leung" wrote ...
Hi, I've been chasing high dollar speaker cables for years. Finally, I
decided to make one myself.

What on earth for? Speaker cables are one of the most hyped subjects in the
entire audio world - as long as it's decent thick copper wire, doesn't go
open or short circuit, and has enough cores to carry the signal, anything
will do!
At a recent audio trade fair, a *very* reputable UK speaker manufacturer
was demonstating a new pair of speakers - all the "golden-eared" HiFi
reviewers were raving about the wonderful "clarity and transparency of the
treble, the fluid bass" (and rightly so). An unpretentious reporter noticed
some familiar-looking orange cable running across the room, behind the wall
drapes - "is that what it looks like?" he asked; "yeah, we forgot to bring
the cables, so I popped over the road to Homebase and picked up one of their
Black&Decker garden extension cables - they're good thick copper, make great
speaker cables"
Personally, I use QED 79-strand well terminated in sturdy connectors -
it's about as far as I want to go into expensive cables - and pay more
attention to obstructions and reflective surfaces in front of the speakers,
which make a hell of a lot more difference!

YMMV,
Dave H.
(The engineer formerly known as Homeless)

(ex senior engineer at a reputable pro-audio manufacturer, etc.)

Isaac Wingfield wrote in message ...
In article ,
(Bob-Stanton) wrote:

Isaac Wingfield wrote in message
...
In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:


Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)


Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*.

It is true there is no "characteristic impedance" for long coax
cables, at frequencies below several hundred KHz.

Not just "coax"; any transmission line at all.

However, Zo = sqrt(L/C), will hold for *any* frequency if: R = 0 and G
= 0. (R the resistance of the cable and, G the conductance).

Yes, but in practice they never are. That approximation only works in
the high frequency region of a line's operating range.

In the case of heavy gage (12 gage) audio cables , of a short length
(25 ft), R and G are very close to zero. For this case, the calculated
characterisitc impedance will hold down to audio frequenies.

In some sense, a "short" line doesn't have a "characteristic impedance",
and for a line long enough (in terms of wavelength) to have one, the
high-frequency approximation Zo = sqrt(L/C) does not hold at low
frequencies.

The point I think we are both making is that if the line is short enough
(in terms of the wavelengths it is carrying), then the line has no
particular characteristics (resistance aside) that could affect the
signal. That is certainly the case for domestic speaker cables.

Isaac

You're right, I agree.

Bob Stanton

Isaac Wingfield wrote in message ...
In article ,
(Bob-Stanton) wrote:

Isaac Wingfield wrote in message
...
In article S7Gpb.99545$HS4.838304@attbi_s01,
"Cliff Curry" wrote:


Note:

characteristic impedance is sqrt(l/c)
propagation constant is sqrt(l*c)


Those approximations are not in general true below a few hundred
kilohertz or so. In the audio range, there is no "characteristic
impedance"; instead, at each frequency there is an impedance for *that
frequency only*.

It is true there is no "characteristic impedance" for long coax
cables, at frequencies below several hundred KHz.

Not just "coax"; any transmission line at all.

However, Zo = sqrt(L/C), will hold for *any* frequency if: R = 0 and G
= 0. (R the resistance of the cable and, G the conductance).

Yes, but in practice they never are. That approximation only works in
the high frequency region of a line's operating range.

In the case of heavy gage (12 gage) audio cables , of a short length
(25 ft), R and G are very close to zero. For this case, the calculated
characterisitc impedance will hold down to audio frequenies.

In some sense, a "short" line doesn't have a "characteristic impedance",
and for a line long enough (in terms of wavelength) to have one, the
high-frequency approximation Zo = sqrt(L/C) does not hold at low
frequencies.

The point I think we are both making is that if the line is short enough
(in terms of the wavelengths it is carrying), then the line has no
particular characteristics (resistance aside) that could affect the
signal. That is certainly the case for domestic speaker cables.

Isaac

You're right, I agree.

Bob Stanton

"Bob-Stanton" wrote in message
om...
Isaac Wingfield wrote in message
...
snip
The point I think we are both making is that if the line is short enough
(in terms of the wavelengths it is carrying), then the line has no
particular characteristics (resistance aside) that could affect the
signal. That is certainly the case for domestic speaker cables.

Isaac

You're right, I agree.

Bob Stanton

Aw, c'mon Bob - this is the net, you're not allowed to say that,
and even with the devil apostrophe in the right place!

Regards
Ian

;-)

"Bob-Stanton" wrote in message
om...
Isaac Wingfield wrote in message
...
snip
The point I think we are both making is that if the line is short enough
(in terms of the wavelengths it is carrying), then the line has no
particular characteristics (resistance aside) that could affect the
signal. That is certainly the case for domestic speaker cables.

Isaac

You're right, I agree.

Bob Stanton

Aw, c'mon Bob - this is the net, you're not allowed to say that,
and even with the devil apostrophe in the right place!

Regards
Ian

;-)

"Dave H." wrote in
:


"Lawrence Leung" wrote ...
Hi, I've been chasing high dollar speaker cables for years. Finally,
I decided to make one myself.

What on earth for? Speaker cables are one of the most hyped subjects
in the entire audio world - as long as it's decent thick copper wire,
doesn't go open or short circuit, and has enough cores to carry the
signal, anything will do!
At a recent audio trade fair, a *very* reputable UK speaker
manufacturer
was demonstating a new pair of speakers - all the "golden-eared" HiFi
reviewers were raving about the wonderful "clarity and transparency of
the treble, the fluid bass" (and rightly so). An unpretentious
reporter noticed some familiar-looking orange cable running across the
room, behind the wall drapes - "is that what it looks like?" he asked;
"yeah, we forgot to bring the cables, so I popped over the road to
Homebase and picked up one of their Black&Decker garden extension
cables - they're good thick copper, make great speaker cables"
Personally, I use QED 79-strand well terminated in sturdy
connectors -
it's about as far as I want to go into expensive cables - and pay more
attention to obstructions and reflective surfaces in front of the
speakers, which make a hell of a lot more difference!

YMMV,
Dave H.
(The engineer formerly known as Homeless)

(ex senior engineer at a reputable pro-audio manufacturer, etc.)




First of all, thank you so much for all the fellows for your inputs.

Well, I have to admit that, high dollar speaker cables really make the
difference, but the point is, rather the difference(s) worth the high
dollar.

I listened to a USD$0.30/ft cable and a USD$150.00/ft cable, they sound
REALLY difference, the high dollar cable REALLY sounds better, but
defintely not 500 times better than then cheap cable!!!

That is one of the reasons that I try to make speaker cable myself, I
just make one with double 12 AWG "normal" speaker cable (USD$0.40/ft),
terminated it real good, heat shrink it, put on foil, nylon sleeving the
whole nine yards, and guess what? I spent about USD$50.00 for that pair
of speaker cables and it sounds almost as good as my Analysis-Plus Oval 9
(which is about USD$500.00), although my cable lack a little bit detail
in the mid-range, the high pitch is not as clear. But, I blame it as new
cables, need another 100 hours to break-in.

What I want is to find out the characteristic of the Oval 9, like the Q,
L, and C. Then compare them with my home-made cable's Q, L, and C to see
if the "difference" can be quantified!!!

Hey, if I told you I make a good living, you might not have any idea what
"good living" mean, but if I told you I make 3 million dollars salary per
year, then you know what my good living mean!

We all live in a "number" world!

Once again, thanks for all the inputs, please keep feed me with
information and knowledge.

Lawrence Leung

"Dave H." wrote in
:


"Lawrence Leung" wrote ...
Hi, I've been chasing high dollar speaker cables for years. Finally,
I decided to make one myself.

What on earth for? Speaker cables are one of the most hyped subjects
in the entire audio world - as long as it's decent thick copper wire,
doesn't go open or short circuit, and has enough cores to carry the
signal, anything will do!
At a recent audio trade fair, a *very* reputable UK speaker
manufacturer
was demonstating a new pair of speakers - all the "golden-eared" HiFi
reviewers were raving about the wonderful "clarity and transparency of
the treble, the fluid bass" (and rightly so). An unpretentious
reporter noticed some familiar-looking orange cable running across the
room, behind the wall drapes - "is that what it looks like?" he asked;
"yeah, we forgot to bring the cables, so I popped over the road to
Homebase and picked up one of their Black&Decker garden extension
cables - they're good thick copper, make great speaker cables"
Personally, I use QED 79-strand well terminated in sturdy
connectors -
it's about as far as I want to go into expensive cables - and pay more
attention to obstructions and reflective surfaces in front of the
speakers, which make a hell of a lot more difference!

YMMV,
Dave H.
(The engineer formerly known as Homeless)

(ex senior engineer at a reputable pro-audio manufacturer, etc.)




First of all, thank you so much for all the fellows for your inputs.

Well, I have to admit that, high dollar speaker cables really make the
difference, but the point is, rather the difference(s) worth the high
dollar.

I listened to a USD$0.30/ft cable and a USD$150.00/ft cable, they sound
REALLY difference, the high dollar cable REALLY sounds better, but
defintely not 500 times better than then cheap cable!!!

That is one of the reasons that I try to make speaker cable myself, I
just make one with double 12 AWG "normal" speaker cable (USD$0.40/ft),
terminated it real good, heat shrink it, put on foil, nylon sleeving the
whole nine yards, and guess what? I spent about USD$50.00 for that pair
of speaker cables and it sounds almost as good as my Analysis-Plus Oval 9
(which is about USD$500.00), although my cable lack a little bit detail
in the mid-range, the high pitch is not as clear. But, I blame it as new
cables, need another 100 hours to break-in.

What I want is to find out the characteristic of the Oval 9, like the Q,
L, and C. Then compare them with my home-made cable's Q, L, and C to see
if the "difference" can be quantified!!!

Hey, if I told you I make a good living, you might not have any idea what
"good living" mean, but if I told you I make 3 million dollars salary per
year, then you know what my good living mean!

We all live in a "number" world!

Once again, thanks for all the inputs, please keep feed me with
information and knowledge.

Lawrence Leung

On Thu, 06 Nov 2003 23:34:15 GMT, Lawrence Leung
wrote:

Well, I have to admit that, high dollar speaker cables really make the
difference, but the point is, rather the difference(s) worth the high
dollar.

I listened to a USD$0.30/ft cable and a USD$150.00/ft cable, they sound
REALLY difference, the high dollar cable REALLY sounds better, but
defintely not 500 times better than then cheap cable!!!

Try listening to them again, when you don't *know* which cable is
connected. You'll find that all these 'REAL' differences vanish......
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Thu, 06 Nov 2003 23:34:15 GMT, Lawrence Leung
wrote:

Well, I have to admit that, high dollar speaker cables really make the
difference, but the point is, rather the difference(s) worth the high
dollar.

I listened to a USD$0.30/ft cable and a USD$150.00/ft cable, they sound
REALLY difference, the high dollar cable REALLY sounds better, but
defintely not 500 times better than then cheap cable!!!

Try listening to them again, when you don't *know* which cable is
connected. You'll find that all these 'REAL' differences vanish......
--

Stewart Pinkerton | Music is Art - Audio is Engineering

In , on 11/07/03
at 08:01 AM, (Stewart Pinkerton) said:

[ ... ]

Try listening to them again, when you don't *know* which cable is
connected. You'll find that all these 'REAL' differences vanish......

About half of the "magic" of tricky cables occurrs when you break down
the original connection (thus scraping off some of the oxides and crud)
and connect the new, wonderful cables.

I recommend that everyone break down their system once or twice a year
and reconnect everything. If you hear a difference after this exercise,
increase the frequency of these breakdowns.

-----------------------------------------------------------
SPAM:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, SPAMers are ruining my mailbox]
-----------------------------------------------------------

In , on 11/07/03
at 08:01 AM, (Stewart Pinkerton) said:

[ ... ]

Try listening to them again, when you don't *know* which cable is
connected. You'll find that all these 'REAL' differences vanish......

About half of the "magic" of tricky cables occurrs when you break down
the original connection (thus scraping off some of the oxides and crud)
and connect the new, wonderful cables.

I recommend that everyone break down their system once or twice a year
and reconnect everything. If you hear a difference after this exercise,
increase the frequency of these breakdowns.

-----------------------------------------------------------
SPAM:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, SPAMers are ruining my mailbox]
-----------------------------------------------------------

"Lawrence Leung" wrote in message
54

First of all, thank you so much for all the fellows for your inputs.

The best thanks is credibility.

Well, I have to admit that, high dollar speaker cables really make the
difference, but the point is, rather the difference(s) worth the high
dollar.

Not at all.

I listened to a USD$0.30/ft cable and a USD$150.00/ft cable, they
sound REALLY difference, the high dollar cable REALLY sounds better,
but defintely not 500 times better than then cheap cable!!!

I seriously doubt that you just listened when you made this comparison.
Typically comparisons like this are made under the worst of circumstances
from the standpoint of just listening. The levels aren't matched, the
listening opportunities aren't time-synched, and there are no bias controls.
Indeed the alleged listening test may be contemporaneous with a sales pitch
for the more expensive cable.

That is one of the reasons that I try to make speaker cable myself, I
just make one with double 12 AWG "normal" speaker cable (USD$0.40/ft),
terminated it real good, heat shrink it, put on foil, nylon sleeving
the whole nine yards, and guess what? I spent about USD$50.00 for
that pair of speaker cables and it sounds almost as good as my
Analysis-Plus Oval 9 (which is about USD$500.00), although my cable
lack a little bit detail in the mid-range, the high pitch is not as
clear. But, I blame it as new cables, need another 100 hours to
break-in.

Cable break-in is another one of those areas where much so-called evidence
is not obtained by means of just listening.

What I want is to find out the characteristic of the Oval 9, like the
Q, L, and C. Then compare them with my home-made cable's Q, L, and C
to see if the "difference" can be quantified!!!

Been there, done that. Slightly different context, but...

Hey, if I told you I make a good living, you might not have any idea
what "good living" mean, but if I told you I make 3 million dollars
salary per year, then you know what my good living mean!

Then you understand the pitfalls of unrestrained subjectivity.

We all live in a "number" world!

Not only a number world, but also a world where some numbers represent
differences that are irrelevant to human experience.

Once again, thanks for all the inputs, please keep feed me with
information and knowledge.

First off, learn how to "just listen". Free, online interactive tutorials
on that subject can be found at www.pcabx.com .

"Lawrence Leung" wrote in message
54

First of all, thank you so much for all the fellows for your inputs.

The best thanks is credibility.

Well, I have to admit that, high dollar speaker cables really make the
difference, but the point is, rather the difference(s) worth the high
dollar.

Not at all.

I listened to a USD$0.30/ft cable and a USD$150.00/ft cable, they
sound REALLY difference, the high dollar cable REALLY sounds better,
but defintely not 500 times better than then cheap cable!!!

I seriously doubt that you just listened when you made this comparison.
Typically comparisons like this are made under the worst of circumstances
from the standpoint of just listening. The levels aren't matched, the
listening opportunities aren't time-synched, and there are no bias controls.
Indeed the alleged listening test may be contemporaneous with a sales pitch
for the more expensive cable.

That is one of the reasons that I try to make speaker cable myself, I
just make one with double 12 AWG "normal" speaker cable (USD$0.40/ft),
terminated it real good, heat shrink it, put on foil, nylon sleeving
the whole nine yards, and guess what? I spent about USD$50.00 for
that pair of speaker cables and it sounds almost as good as my
Analysis-Plus Oval 9 (which is about USD$500.00), although my cable
lack a little bit detail in the mid-range, the high pitch is not as
clear. But, I blame it as new cables, need another 100 hours to
break-in.

Cable break-in is another one of those areas where much so-called evidence
is not obtained by means of just listening.

What I want is to find out the characteristic of the Oval 9, like the
Q, L, and C. Then compare them with my home-made cable's Q, L, and C
to see if the "difference" can be quantified!!!

Been there, done that. Slightly different context, but...

Hey, if I told you I make a good living, you might not have any idea
what "good living" mean, but if I told you I make 3 million dollars
salary per year, then you know what my good living mean!

Then you understand the pitfalls of unrestrained subjectivity.

We all live in a "number" world!

Not only a number world, but also a world where some numbers represent
differences that are irrelevant to human experience.

Once again, thanks for all the inputs, please keep feed me with
information and knowledge.

First off, learn how to "just listen". Free, online interactive tutorials
on that subject can be found at www.pcabx.com .

Isaac Wingfield wrote in message
...
snip
The point I think we are both making is that if the line is
short enough
(in terms of the wavelengths it is carrying), then the line has
no
particular characteristics (resistance aside) that could affect
the
signal. That is certainly the case for domestic speaker cables.

Isaac

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.

Norm Strong

Isaac Wingfield wrote in message
...
snip
The point I think we are both making is that if the line is
short enough
(in terms of the wavelengths it is carrying), then the line has
no
particular characteristics (resistance aside) that could affect
the
signal. That is certainly the case for domestic speaker cables.

Isaac

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.

Norm Strong

"normanstrong" wrote in message news:9rRqb.101076

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.


How a speaker cable behaves, whether as a transmission line or as a
discrete device, is determined by the *resistance* of the conductors,
not by it's wavelength at audio frequencies. If a transmission line
was supercooled, and had zero resistance, it would act as a "high
frequency" transmission line at *all* frequencies.

Unfortunatly, the speaker cables ( 12 gage) in my house are at room
temperature, and they don't act like "high frequency" transmission
lines at *all* audio frequencies. At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

The region beween 200 Hz and 20KHz is called the transition region. In
the transition region, lines sort of act like "high frequency"
transmission lines, and sort of act like discrete components. This
makes life very difficult. :-)

At 20 KHz, you can't really treet a speaker cable as a discrete
inductor. A 100 ft length of 12 gage, speaker cable would have 20uH of
inductance. If you calculate the loss as a simple series L, R, you
would get a 0.62 dB loss (with an 8 Ohm load).

If you calculate the speaker cable as a transmission line, (which it
mostly is at 20 KHz) you get 0.32 dB loss.

Which one is right. Dammed if I know. ;-)

Bob Stanton

"normanstrong" wrote in message news:9rRqb.101076

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.


How a speaker cable behaves, whether as a transmission line or as a
discrete device, is determined by the *resistance* of the conductors,
not by it's wavelength at audio frequencies. If a transmission line
was supercooled, and had zero resistance, it would act as a "high
frequency" transmission line at *all* frequencies.

Unfortunatly, the speaker cables ( 12 gage) in my house are at room
temperature, and they don't act like "high frequency" transmission
lines at *all* audio frequencies. At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

The region beween 200 Hz and 20KHz is called the transition region. In
the transition region, lines sort of act like "high frequency"
transmission lines, and sort of act like discrete components. This
makes life very difficult. :-)

At 20 KHz, you can't really treet a speaker cable as a discrete
inductor. A 100 ft length of 12 gage, speaker cable would have 20uH of
inductance. If you calculate the loss as a simple series L, R, you
would get a 0.62 dB loss (with an 8 Ohm load).

If you calculate the speaker cable as a transmission line, (which it
mostly is at 20 KHz) you get 0.32 dB loss.

Which one is right. Dammed if I know. ;-)

Bob Stanton

On 9 Nov 2003 03:48:40 -0800, (Bob-Stanton)
wrote:

"normanstrong" wrote in message news:9rRqb.101076

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.

At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

If you calculate the speaker cable as a transmission line, (which it
mostly is at 20 KHz) you get 0.32 dB loss.

This is plain wrong, if you ask me.

Per.

On 9 Nov 2003 03:48:40 -0800, (Bob-Stanton)
wrote:

"normanstrong" wrote in message news:9rRqb.101076

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.

At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

If you calculate the speaker cable as a transmission line, (which it
mostly is at 20 KHz) you get 0.32 dB loss.

This is plain wrong, if you ask me.

Per.

(Per Stromgren) wrote in message ...
On 9 Nov 2003 03:48:40 -0800, (Bob-Stanton)
wrote:

"normanstrong" wrote in message news:9rRqb.101076

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.

At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

Are you familiar with the Smith Chart? It is probably the most usesful
graphical tool available to the RF designer. It was originally
conceived by Phil Smith, in Bell Labs, in the 1930's and it is still
widely in use.

One of the Smith Chart's uses is to calculate the impedance change of
a transmission line. The Smith Chart is a circular graph that is
divided into wavelenths (and degrees). One full rotation around the
chart equals one wavelength. 1 deg rotation equals 0.0028 WL of a
transmission line.

The Smith chart is readable down to about 1/2 degree, which equals
0.0014 wavelength. The chart will accurately show the change of
impedance along a transmission line that is only 0.002 wavelengths
long.

I used an RF computer program to calculate the impedance(and loss) of
100 ft (of 100 Ohm) speaker cable terminated with an 8 Ohm load. Then,
I checked the calculations by hand, using a Smith Chart. The Smith
Chart results agreed with the computer program. They both showed that
fractional wavelengh transmission lines "work".

If you feel that fractional wavelengh transmission lines don't "work",
you should throw away your Smith Chart, or perhaps just whiteout the
fractional wavelength part of the Smith Chart, that you don't agree
with. :-) As for me, I'll keep my old Smith Chart, because it shows
that fractional wavelength transmission lines do "work" and it can
calculate how they work.

Bob Stanton

(Per Stromgren) wrote in message ...
On 9 Nov 2003 03:48:40 -0800, (Bob-Stanton)
wrote:

"normanstrong" wrote in message news:9rRqb.101076

I'm not sure I agree with that. Speaker cables are all short
relative to the wavelengths being carried. For your statement to be
true, there would never be a reason to be concerned with inductance,
and we all know that's not true.

At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

Are you familiar with the Smith Chart? It is probably the most usesful
graphical tool available to the RF designer. It was originally
conceived by Phil Smith, in Bell Labs, in the 1930's and it is still
widely in use.

One of the Smith Chart's uses is to calculate the impedance change of
a transmission line. The Smith Chart is a circular graph that is
divided into wavelenths (and degrees). One full rotation around the
chart equals one wavelength. 1 deg rotation equals 0.0028 WL of a
transmission line.

The Smith chart is readable down to about 1/2 degree, which equals
0.0014 wavelength. The chart will accurately show the change of
impedance along a transmission line that is only 0.002 wavelengths
long.

I used an RF computer program to calculate the impedance(and loss) of
100 ft (of 100 Ohm) speaker cable terminated with an 8 Ohm load. Then,
I checked the calculations by hand, using a Smith Chart. The Smith
Chart results agreed with the computer program. They both showed that
fractional wavelengh transmission lines "work".

If you feel that fractional wavelengh transmission lines don't "work",
you should throw away your Smith Chart, or perhaps just whiteout the
fractional wavelength part of the Smith Chart, that you don't agree
with. :-) As for me, I'll keep my old Smith Chart, because it shows
that fractional wavelength transmission lines do "work" and it can
calculate how they work.

Bob Stanton

On 9 Nov 2003 18:18:23 -0800, (Bob-Stanton)
wrote:

(Per Stromgren) wrote in message ...

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

Are you familiar with the Smith Chart? It is probably the most usesful
graphical tool available to the RF designer. It was originally
conceived by Phil Smith, in Bell Labs, in the 1930's and it is still
widely in use.

One of the Smith Chart's uses is to calculate the impedance change of
a transmission line. The Smith Chart is a circular graph that is
divided into wavelenths (and degrees). One full rotation around the
chart equals one wavelength. 1 deg rotation equals 0.0028 WL of a
transmission line.

No, I was not aware of the Smith chart, I'm afraid. But it seems to me
(after some 30 minutes of Googling I just performed) that it is just a
calculation tool, and it does not per se introduce any alternative
theory. As far as I can see, there is no mention of this tool to work
in the application you just showed us. Most sources say "RF broadband
use only" or something to that effect.

Or did I miss something?

Per.

On 9 Nov 2003 18:18:23 -0800, (Bob-Stanton)
wrote:

(Per Stromgren) wrote in message ...

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

Are you familiar with the Smith Chart? It is probably the most usesful
graphical tool available to the RF designer. It was originally
conceived by Phil Smith, in Bell Labs, in the 1930's and it is still
widely in use.

One of the Smith Chart's uses is to calculate the impedance change of
a transmission line. The Smith Chart is a circular graph that is
divided into wavelenths (and degrees). One full rotation around the
chart equals one wavelength. 1 deg rotation equals 0.0028 WL of a
transmission line.

No, I was not aware of the Smith chart, I'm afraid. But it seems to me
(after some 30 minutes of Googling I just performed) that it is just a
calculation tool, and it does not per se introduce any alternative
theory. As far as I can see, there is no mention of this tool to work
in the application you just showed us. Most sources say "RF broadband
use only" or something to that effect.

Or did I miss something?

Per.

On Mon, 10 Nov 2003 09:24:47 +0100, Per Stromgren
wrote:

No, I was not aware of the Smith chart, I'm afraid. But it seems to me
(after some 30 minutes of Googling I just performed) that it is just a
calculation tool, and it does not per se introduce any alternative
theory. As far as I can see, there is no mention of this tool to work
in the application you just showed us. Most sources say "RF broadband
use only" or something to that effect.

"RF ***narrowband*** use only", of course. Sorry about that.

Per.

On Mon, 10 Nov 2003 09:24:47 +0100, Per Stromgren
wrote:

No, I was not aware of the Smith chart, I'm afraid. But it seems to me
(after some 30 minutes of Googling I just performed) that it is just a
calculation tool, and it does not per se introduce any alternative
theory. As far as I can see, there is no mention of this tool to work
in the application you just showed us. Most sources say "RF broadband
use only" or something to that effect.

"RF ***narrowband*** use only", of course. Sorry about that.

Per.

(Bob-Stanton) wrote in message

One full rotation around the chart equals one wavelength.



Let me correct the statement (above) that I wrote in the previous
message.

One full rotation around the Smith Chart actually represents *1/2
wavelength* of transmission line.


Bob Stanton

(Bob-Stanton) wrote in message

One full rotation around the chart equals one wavelength.



Let me correct the statement (above) that I wrote in the previous
message.

One full rotation around the Smith Chart actually represents *1/2
wavelength* of transmission line.


Bob Stanton

(Bob-Stanton) wrote in message om...
At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

Are you familiar with the Smith Chart? It is probably the most usesful
graphical tool available to the RF designer. It was originally
conceived by Phil Smith, in Bell Labs, in the 1930's and it is still
widely in use.
The Smith chart is readable down to about 1/2 degree, which equals
0.0014 wavelength. The chart will accurately show the change of
impedance along a transmission line that is only 0.002 wavelengths
long.

I used an RF computer program to calculate the impedance(and loss) of
100 ft (of 100 Ohm) speaker cable terminated with an 8 Ohm load. Then,
I checked the calculations by hand, using a Smith Chart. The Smith
Chart results agreed with the computer program. They both showed that
fractional wavelengh transmission lines "work".

NO, they absolutely do NOT. They simply show that you have two
conceptual models that give similar answers. You yourself have admitted
in a previous post in this thread that you do not have any idea if
the answers you are getting are right or wrong.

If you feel that fractional wavelengh transmission lines don't "work",
you should throw away your Smith Chart, or perhaps just whiteout the
fractional wavelength part of the Smith Chart, that you don't agree
with. :-) As for me, I'll keep my old Smith Chart, because it shows
that fractional wavelength transmission lines do "work" and it can
calculate how they work.

No, it shows nothing of the kind. It shows that you have a model
of a physical phenomenon that gives an answer that satisfies you.
Until such time as you check those answers against the real physical
behavior in a way that is clear, unambiguous, consistent and predictive,
your models are worthless.

(Bob-Stanton) wrote in message om...
At 20KHz they act mostly (sort of)
like "high frequency" transmission lines, at 200 Hz they act like
discrete components.

At 20 kHz, the wavelength at speed of light is 3*E8/20*E3=1.5*E4m,
i.e. 15 km or about 9 miles. Your 100 ft of cable is 2/1000th of that,
hardly a transmission line, even if we take into consideration a
somewhat slower speed in the cable.

Or do you say that it still will act as a transmission line? It is
not, according to the books I own.

Are you familiar with the Smith Chart? It is probably the most usesful
graphical tool available to the RF designer. It was originally
conceived by Phil Smith, in Bell Labs, in the 1930's and it is still
widely in use.
The Smith chart is readable down to about 1/2 degree, which equals
0.0014 wavelength. The chart will accurately show the change of
impedance along a transmission line that is only 0.002 wavelengths
long.

I used an RF computer program to calculate the impedance(and loss) of
100 ft (of 100 Ohm) speaker cable terminated with an 8 Ohm load. Then,
I checked the calculations by hand, using a Smith Chart. The Smith
Chart results agreed with the computer program. They both showed that
fractional wavelengh transmission lines "work".

NO, they absolutely do NOT. They simply show that you have two
conceptual models that give similar answers. You yourself have admitted
in a previous post in this thread that you do not have any idea if
the answers you are getting are right or wrong.

If you feel that fractional wavelengh transmission lines don't "work",
you should throw away your Smith Chart, or perhaps just whiteout the
fractional wavelength part of the Smith Chart, that you don't agree
with. :-) As for me, I'll keep my old Smith Chart, because it shows
that fractional wavelength transmission lines do "work" and it can
calculate how they work.

No, it shows nothing of the kind. It shows that you have a model
of a physical phenomenon that gives an answer that satisfies you.
Until such time as you check those answers against the real physical
behavior in a way that is clear, unambiguous, consistent and predictive,
your models are worthless.

Per Stromgren wrote in message
line.

No, I was not aware of the Smith chart, I'm afraid. But it seems to me
(after some 30 minutes of Googling I just performed) that it is just a
calculation tool, and it does not per se introduce any alternative
theory. As far as I can see, there is no mention of this tool to work
in the application you just showed us. Most sources say "RF broadband
use only" or something to that effect.

Or did I miss something?

Per.

A Smith will show the change of impedance of a transmission line, for
a line as short as 0.002 WL. This is not a new theory, it's just the
way it is. If one rotates the cursor of a Smith chart by 0.002 WL, one
will be able to read, how the impedance of a 0.002 WL line changes.

There is no low frequency limit to Smith Charts. The only limits a
the transmission line must be at least 0.002 WL long and the line must
have very low series resistance.

The case I cited was a 100 ft line, of 12 gage wire. A line of 100 Ft
is just barely long enough to allow Smith chart calculations. A 12
gage cable has just barely low enough series resistance, to allow the
Smith chart to work accurately.

Of one thing I'm su calculating the loss of a 100 ft line, using
discrete component models, will not give correct results.

Bob Stanton