On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:
On 9/28/2015 4:14 PM, rickman wrote:
On 9/28/2015 4:07 PM, Jerry Stuckle wrote:
On 9/28/2015 3:22 PM, Wayne wrote:


"Jerry Stuckle" wrote in message ...

On 9/28/2015 2:27 PM, Wayne wrote:


"Jerry Stuckle" wrote in message ...

On 9/28/2015 12:47 PM, rickman wrote:
On 9/28/2015 10:38 AM, Jerry Stuckle wrote:
On 9/28/2015 12:03 AM, rickman wrote:
On 9/27/2015 10:39 PM, Jerry Stuckle wrote:
On 9/27/2015 9:46 PM, Wayne wrote:


From LUNA web site regarding optical measurements which should
be no
different from RF...


It "shouldn't be" - but optical measurements are handled
differently
than electrical measurements. Fiber Optics have their own way of
measuring loss, reflection and refraction (which doesn't exist in
feedlines).

That's like applying electrician's color codes to electronics.
They
both have color codes - but don't hook the electrician's black
wire to
ground - or the transformer's green wires to safety ground.

I thought you would claim optical was different. That's why I
included
the VSWR vs return loss table link. You didn't comment on
that.


# I didn't because I thought it was obvious. But I guess not to you.

# Return loss is calculated with logs. Logs of values 1 are
negative.
# And -10db is smaller than -5 db.

# As the SWR approaches 1:1, the reflected power approaches 0, and
the
# returned loss approaches NEGATIVE infinity. Note that I said
NEGATIVE
# infinity. At the same point, the returned power measured in watts
is 0.

Return loss is a positive number for passive networks. The equation
has
(P out/P reflected). P out will never be less that P reflected, and
thus return loss will never be negative. (for passive networks)

As the SWR approaches 1:1, the return loss increases in a positive
direction, finally reaching infinity.


# No, return loss is calculated as P reflected / P out. P out is the
# constant with varying load; P reflected is the variable. The
ratio is
# always less than one, hence the calculation is always negative DB.

# Please point to a reliable source which agrees with you.

I have never heard return loss expressed as a negative for passive RF
networks.
In fields other than RF I suppose anything is possible.

Here are some references, searching only for RF definitions:

http://www.ab4oj.com/atu/vswr.html

http://www.mogami.com/e/cad/vswr.html

http://www.microwaves101.com/encyclo...swr-calculator

http://www.amphenolrf.com/vswr-conversion-chart/

From wikipedia: https://en.wikipedia.org/wiki/Return_loss
Return loss is the negative of the magnitude of the reflection
coefficient in dB. Since power is proportional to the square of the
voltage, return loss is given by,
(couldn't cut/paste the equation)
Thus, a large positive return loss indicates the reflected power is
small relative to the incident power, which indicates good impedance
match from source to load.

http://www.spectrum-soft.com/news/fall2009/vswr.shtm
The return loss measurement describes the ratio of the power in the
reflected wave to the power in the incident wave in units of decibels.
The standard output for the return loss is a positive value, so a
large
return loss value actually means that the power in the reflected
wave is
small compared to the power in the incident wave and indicates a
better
impedance match. The return loss can be calculated from the reflection
coefficient with the equation:





I said RELIABLE SOURCE. Wikipedia and someone's blog are not what I
would call reliable.

How about IEEE, for instance?

I provided the IEEE paper cited by wikipedia. Anyone care to pay for
it? IEEE is seldom free. Who else will you accept?


I'm not interested. I know what it says. Guess I should have kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with us?

--

Rick

On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:
On 9/28/2015 4:14 PM, rickman wrote:
On 9/28/2015 4:07 PM, Jerry Stuckle wrote:
On 9/28/2015 3:22 PM, Wayne wrote:


"Jerry Stuckle" wrote in message ...

On 9/28/2015 2:27 PM, Wayne wrote:


"Jerry Stuckle" wrote in message
...

On 9/28/2015 12:47 PM, rickman wrote:
On 9/28/2015 10:38 AM, Jerry Stuckle wrote:
On 9/28/2015 12:03 AM, rickman wrote:
On 9/27/2015 10:39 PM, Jerry Stuckle wrote:
On 9/27/2015 9:46 PM, Wayne wrote:


From LUNA web site regarding optical measurements which
should
be no
different from RF...


It "shouldn't be" - but optical measurements are handled
differently
than electrical measurements. Fiber Optics have their own way of
measuring loss, reflection and refraction (which doesn't exist in
feedlines).

That's like applying electrician's color codes to electronics.
They
both have color codes - but don't hook the electrician's black
wire to
ground - or the transformer's green wires to safety ground.

I thought you would claim optical was different. That's why I
included
the VSWR vs return loss table link. You didn't comment on
that.


# I didn't because I thought it was obvious. But I guess not to
you.

# Return loss is calculated with logs. Logs of values 1 are
negative.
# And -10db is smaller than -5 db.

# As the SWR approaches 1:1, the reflected power approaches 0, and
the
# returned loss approaches NEGATIVE infinity. Note that I said
NEGATIVE
# infinity. At the same point, the returned power measured in
watts
is 0.

Return loss is a positive number for passive networks. The
equation
has
(P out/P reflected). P out will never be less that P reflected,
and
thus return loss will never be negative. (for passive networks)

As the SWR approaches 1:1, the return loss increases in a positive
direction, finally reaching infinity.


# No, return loss is calculated as P reflected / P out. P out is
the
# constant with varying load; P reflected is the variable. The
ratio is
# always less than one, hence the calculation is always negative DB.

# Please point to a reliable source which agrees with you.

I have never heard return loss expressed as a negative for
passive RF
networks.
In fields other than RF I suppose anything is possible.

Here are some references, searching only for RF definitions:

http://www.ab4oj.com/atu/vswr.html

http://www.mogami.com/e/cad/vswr.html

http://www.microwaves101.com/encyclo...swr-calculator

http://www.amphenolrf.com/vswr-conversion-chart/

From wikipedia: https://en.wikipedia.org/wiki/Return_loss
Return loss is the negative of the magnitude of the reflection
coefficient in dB. Since power is proportional to the square of the
voltage, return loss is given by,
(couldn't cut/paste the equation)
Thus, a large positive return loss indicates the reflected power is
small relative to the incident power, which indicates good impedance
match from source to load.

http://www.spectrum-soft.com/news/fall2009/vswr.shtm
The return loss measurement describes the ratio of the power in the
reflected wave to the power in the incident wave in units of
decibels.
The standard output for the return loss is a positive value, so a
large
return loss value actually means that the power in the reflected
wave is
small compared to the power in the incident wave and indicates a
better
impedance match. The return loss can be calculated from the
reflection
coefficient with the equation:





I said RELIABLE SOURCE. Wikipedia and someone's blog are not what I
would call reliable.

How about IEEE, for instance?

I provided the IEEE paper cited by wikipedia. Anyone care to pay for
it? IEEE is seldom free. Who else will you accept?


I'm not interested. I know what it says. Guess I should have kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with us?


You want it - you pay for it. Or once again you prove you're full of it.

No, I have not read the article. But I understand the physics and math
behind it - unlike you. Someone who thinks magnitude without vector
(direction) is valid! ROFLMAO!

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:
On 9/28/2015 4:14 PM, rickman wrote:
On 9/28/2015 4:07 PM, Jerry Stuckle wrote:
On 9/28/2015 3:22 PM, Wayne wrote:


"Jerry Stuckle" wrote in message ...

On 9/28/2015 2:27 PM, Wayne wrote:


"Jerry Stuckle" wrote in message
...

On 9/28/2015 12:47 PM, rickman wrote:
On 9/28/2015 10:38 AM, Jerry Stuckle wrote:
On 9/28/2015 12:03 AM, rickman wrote:
On 9/27/2015 10:39 PM, Jerry Stuckle wrote:
On 9/27/2015 9:46 PM, Wayne wrote:


From LUNA web site regarding optical measurements which
should
be no
different from RF...


It "shouldn't be" - but optical measurements are handled
differently
than electrical measurements. Fiber Optics have their own way of
measuring loss, reflection and refraction (which doesn't exist in
feedlines).

That's like applying electrician's color codes to electronics.
They
both have color codes - but don't hook the electrician's black
wire to
ground - or the transformer's green wires to safety ground.

I thought you would claim optical was different. That's why I
included
the VSWR vs return loss table link. You didn't comment on
that.


# I didn't because I thought it was obvious. But I guess not to
you.

# Return loss is calculated with logs. Logs of values 1 are
negative.
# And -10db is smaller than -5 db.

# As the SWR approaches 1:1, the reflected power approaches 0, and
the
# returned loss approaches NEGATIVE infinity. Note that I said
NEGATIVE
# infinity. At the same point, the returned power measured in
watts
is 0.

Return loss is a positive number for passive networks. The
equation
has
(P out/P reflected). P out will never be less that P reflected,
and
thus return loss will never be negative. (for passive networks)

As the SWR approaches 1:1, the return loss increases in a positive
direction, finally reaching infinity.


# No, return loss is calculated as P reflected / P out. P out is
the
# constant with varying load; P reflected is the variable. The
ratio is
# always less than one, hence the calculation is always negative DB.

# Please point to a reliable source which agrees with you.

I have never heard return loss expressed as a negative for
passive RF
networks.
In fields other than RF I suppose anything is possible.

Here are some references, searching only for RF definitions:

http://www.ab4oj.com/atu/vswr.html

http://www.mogami.com/e/cad/vswr.html

http://www.microwaves101.com/encyclo...swr-calculator

http://www.amphenolrf.com/vswr-conversion-chart/

From wikipedia: https://en.wikipedia.org/wiki/Return_loss
Return loss is the negative of the magnitude of the reflection
coefficient in dB. Since power is proportional to the square of the
voltage, return loss is given by,
(couldn't cut/paste the equation)
Thus, a large positive return loss indicates the reflected power is
small relative to the incident power, which indicates good impedance
match from source to load.

http://www.spectrum-soft.com/news/fall2009/vswr.shtm
The return loss measurement describes the ratio of the power in the
reflected wave to the power in the incident wave in units of
decibels.
The standard output for the return loss is a positive value, so a
large
return loss value actually means that the power in the reflected
wave is
small compared to the power in the incident wave and indicates a
better
impedance match. The return loss can be calculated from the
reflection
coefficient with the equation:





I said RELIABLE SOURCE. Wikipedia and someone's blog are not what I
would call reliable.

How about IEEE, for instance?

I provided the IEEE paper cited by wikipedia. Anyone care to pay for
it? IEEE is seldom free. Who else will you accept?


I'm not interested. I know what it says. Guess I should have kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with us?


You want it - you pay for it. Or once again you prove you're full of it.

No, I have not read the article. But I understand the physics and math
behind it - unlike you. Someone who thinks magnitude without vector
(direction) is valid! ROFLMAO!

Ok, so you mispoke when you said, "I know what it says."

You have said repeatedly that the return loss should be calculated by
using the power in as the reference and the reflected power as the thing
being measured which results in a negative log. I am pretty sure the
paper says this is not the correct way to calculate it and many people
are making a mistake doing it this way.

I'll see if I can get my hands on the paper. I'm not going to pay for
it. If I thought it would get you to admit you were mistaken, I'd pay
the $100. But I'm sure you will find a way to berate the authors or
twist their logic and I'm not will to pay $100 for that.

So stand by. Someone may be getting it for me.

--

Rick

On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:
On 9/28/2015 4:14 PM, rickman wrote:
On 9/28/2015 4:07 PM, Jerry Stuckle wrote:
On 9/28/2015 3:22 PM, Wayne wrote:


"Jerry Stuckle" wrote in message
...

On 9/28/2015 2:27 PM, Wayne wrote:


"Jerry Stuckle" wrote in message
...

On 9/28/2015 12:47 PM, rickman wrote:
On 9/28/2015 10:38 AM, Jerry Stuckle wrote:
On 9/28/2015 12:03 AM, rickman wrote:
On 9/27/2015 10:39 PM, Jerry Stuckle wrote:
On 9/27/2015 9:46 PM, Wayne wrote:


From LUNA web site regarding optical measurements which
should
be no
different from RF...


It "shouldn't be" - but optical measurements are handled
differently
than electrical measurements. Fiber Optics have their own
way of
measuring loss, reflection and refraction (which doesn't
exist in
feedlines).

That's like applying electrician's color codes to electronics.
They
both have color codes - but don't hook the electrician's black
wire to
ground - or the transformer's green wires to safety ground.

I thought you would claim optical was different. That's why I
included
the VSWR vs return loss table link. You didn't comment on
that.


# I didn't because I thought it was obvious. But I guess not to
you.

# Return loss is calculated with logs. Logs of values 1 are
negative.
# And -10db is smaller than -5 db.

# As the SWR approaches 1:1, the reflected power approaches 0,
and
the
# returned loss approaches NEGATIVE infinity. Note that I said
NEGATIVE
# infinity. At the same point, the returned power measured in
watts
is 0.

Return loss is a positive number for passive networks. The
equation
has
(P out/P reflected). P out will never be less that P reflected,
and
thus return loss will never be negative. (for passive networks)

As the SWR approaches 1:1, the return loss increases in a
positive
direction, finally reaching infinity.


# No, return loss is calculated as P reflected / P out. P out is
the
# constant with varying load; P reflected is the variable. The
ratio is
# always less than one, hence the calculation is always
negative DB.

# Please point to a reliable source which agrees with you.

I have never heard return loss expressed as a negative for
passive RF
networks.
In fields other than RF I suppose anything is possible.

Here are some references, searching only for RF definitions:

http://www.ab4oj.com/atu/vswr.html

http://www.mogami.com/e/cad/vswr.html

http://www.microwaves101.com/encyclo...swr-calculator

http://www.amphenolrf.com/vswr-conversion-chart/

From wikipedia: https://en.wikipedia.org/wiki/Return_loss
Return loss is the negative of the magnitude of the reflection
coefficient in dB. Since power is proportional to the square of
the
voltage, return loss is given by,
(couldn't cut/paste the equation)
Thus, a large positive return loss indicates the reflected
power is
small relative to the incident power, which indicates good
impedance
match from source to load.

http://www.spectrum-soft.com/news/fall2009/vswr.shtm
The return loss measurement describes the ratio of the power in
the
reflected wave to the power in the incident wave in units of
decibels.
The standard output for the return loss is a positive value, so a
large
return loss value actually means that the power in the reflected
wave is
small compared to the power in the incident wave and indicates a
better
impedance match. The return loss can be calculated from the
reflection
coefficient with the equation:





I said RELIABLE SOURCE. Wikipedia and someone's blog are not
what I
would call reliable.

How about IEEE, for instance?

I provided the IEEE paper cited by wikipedia. Anyone care to pay
for
it? IEEE is seldom free. Who else will you accept?


I'm not interested. I know what it says. Guess I should have
kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not
me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with us?


You want it - you pay for it. Or once again you prove you're full of it.

No, I have not read the article. But I understand the physics and math
behind it - unlike you. Someone who thinks magnitude without vector
(direction) is valid! ROFLMAO!

Ok, so you mispoke when you said, "I know what it says."



No, I didn't. I didn't say I read it. I said I know what it says. And
I do from other IEEE peer-reviewed articles. I don't need to read it to
find out it agrees with other documentation. And if it didn't, it
wouldn't have gotten published.

You have said repeatedly that the return loss should be calculated by
using the power in as the reference and the reflected power as the thing
being measured which results in a negative log. I am pretty sure the
paper says this is not the correct way to calculate it and many people
are making a mistake doing it this way.


So you've read other IEEE documentation which supports what you say?

I'll see if I can get my hands on the paper. I'm not going to pay for
it. If I thought it would get you to admit you were mistaken, I'd pay
the $100. But I'm sure you will find a way to berate the authors or
twist their logic and I'm not will to pay $100 for that.

So stand by. Someone may be getting it for me.


I really don't give a damn. You would argue the sun rises in the west.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 9/29/2015 9:22 AM, Jerry Stuckle wrote:
On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:

I'm not interested. I know what it says. Guess I should have
kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not
me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with us?


You want it - you pay for it. Or once again you prove you're full of it.

No, I have not read the article. But I understand the physics and math
behind it - unlike you. Someone who thinks magnitude without vector
(direction) is valid! ROFLMAO!

Ok, so you mispoke when you said, "I know what it says."



No, I didn't. I didn't say I read it. I said I know what it says. And
I do from other IEEE peer-reviewed articles. I don't need to read it to
find out it agrees with other documentation. And if it didn't, it
wouldn't have gotten published.

You have said repeatedly that the return loss should be calculated by
using the power in as the reference and the reflected power as the thing
being measured which results in a negative log. I am pretty sure the
paper says this is not the correct way to calculate it and many people
are making a mistake doing it this way.


So you've read other IEEE documentation which supports what you say?

I'll see if I can get my hands on the paper. I'm not going to pay for
it. If I thought it would get you to admit you were mistaken, I'd pay
the $100. But I'm sure you will find a way to berate the authors or
twist their logic and I'm not will to pay $100 for that.

So stand by. Someone may be getting it for me.

I really don't give a damn. You would argue the sun rises in the west.

I have a copy of the paper.

Trevor Bird
Editor-in-ehief,Engineering
IEEE Transactions on Antennas and Propagation
CSIRO leT Centre, PO Box 76
Epping, NSW 1710, Australia
Tel: +61 2 9372 4289
Fax: +61 2 9372 4446
E-mail:

Definition and Misuse of Return Loss
Trevor S. Bird

Here is the equation from the article

Pin
RL = 10 log,10 ( ---- ) dB, (1)
Pref

The author explicitly states the resulting value will be positive when
Pin is greater than Pref. He goes on to say, "That is, return loss is
the negative of the reflection coefficient expressed in decibels."

He goes on to quote from the "IEEE Standard Dictionary of Electrical
and Electronic Terms, Fourth Edition".

(1 ) (data transmission) (A) At a discontinuity in a
transmission system the difference between the power
incident upon the discontinuity. (B) The ratio in deci-
bels of the power incident upon the discontinuity to the
power reflected from the discontinuity. Note: This ratio
is also the square of the reciprocal to the magnitude of
the reflection coefficient. (C) More broadly, the return
loss is a measure of the dissimilarity between two
impedances, being equal to the number of decibels that
corresponds to the scalar value of the reciprocal of the
reflection coefficient, and hence being expressed by the
following formula:

|Z1 + Z2|
20 log,10 |-------| decibel
|Z1 - Z2|

where Z1 and Z2 = the two impedances.

(2) (or gain) (waveguide). The ratio of incident to
reflected power at a reference plane of a network.


So is this what you "knew" the paper said? Seems to be the opposite of
what you have been promoting.

Anyone feel this paper is incorrect?

--

Rick

On 9/29/2015 10:32 AM, rickman wrote:
On 9/29/2015 9:22 AM, Jerry Stuckle wrote:
On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:

I'm not interested. I know what it says. Guess I should have
kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not
me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with
us?


You want it - you pay for it. Or once again you prove you're full
of it.

No, I have not read the article. But I understand the physics and math
behind it - unlike you. Someone who thinks magnitude without vector
(direction) is valid! ROFLMAO!

Ok, so you mispoke when you said, "I know what it says."



No, I didn't. I didn't say I read it. I said I know what it says. And
I do from other IEEE peer-reviewed articles. I don't need to read it to
find out it agrees with other documentation. And if it didn't, it
wouldn't have gotten published.

You have said repeatedly that the return loss should be calculated by
using the power in as the reference and the reflected power as the thing
being measured which results in a negative log. I am pretty sure the
paper says this is not the correct way to calculate it and many people
are making a mistake doing it this way.


So you've read other IEEE documentation which supports what you say?

I'll see if I can get my hands on the paper. I'm not going to pay for
it. If I thought it would get you to admit you were mistaken, I'd pay
the $100. But I'm sure you will find a way to berate the authors or
twist their logic and I'm not will to pay $100 for that.

So stand by. Someone may be getting it for me.

I really don't give a damn. You would argue the sun rises in the west.

I have a copy of the paper.

Trevor Bird
Editor-in-ehief,Engineering
IEEE Transactions on Antennas and Propagation
CSIRO leT Centre, PO Box 76
Epping, NSW 1710, Australia
Tel: +61 2 9372 4289
Fax: +61 2 9372 4446
E-mail:

Definition and Misuse of Return Loss
Trevor S. Bird

Here is the equation from the article

Pin
RL = 10 log,10 ( ---- ) dB, (1)
Pref

The author explicitly states the resulting value will be positive when
Pin is greater than Pref. He goes on to say, "That is, return loss is
the negative of the reflection coefficient expressed in decibels."

Since the reflection coefficient is never greater than 1, its value is
negative. It follows that the return loss is positive.

He goes on to quote from the "IEEE Standard Dictionary of Electrical
and Electronic Terms, Fourth Edition".

(1 ) (data transmission) (A) At a discontinuity in a
transmission system the difference between the power
incident upon the discontinuity. (B) The ratio in deci-
bels of the power incident upon the discontinuity to the
power reflected from the discontinuity. Note: This ratio
is also the square of the reciprocal to the magnitude of
the reflection coefficient. (C) More broadly, the return
loss is a measure of the dissimilarity between two
impedances, being equal to the number of decibels that
corresponds to the scalar value of the reciprocal of the
reflection coefficient, and hence being expressed by the
following formula:

|Z1 + Z2|
20 log,10 |-------| decibel
|Z1 - Z2|

where Z1 and Z2 = the two impedances.

(2) (or gain) (waveguide). The ratio of incident to
reflected power at a reference plane of a network.


So is this what you "knew" the paper said? Seems to be the opposite of
what you have been promoting.

Anyone feel this paper is incorrect?


Good find Rick. That should put it to bed.

"John S" wrote in message ...

On 9/29/2015 10:32 AM, rickman wrote:
On 9/29/2015 9:22 AM, Jerry Stuckle wrote:
On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:

I'm not interested. I know what it says. Guess I should have
kept up
my IEEE membership, but it just wasn't worth it.

So share with the rest of us. What does it say?


Exactly what your table showed. But you mentioned the resource, not
me.
You pay for it or you've just once again you're full of it.

You said you *know* what the IEEE article says. Why not share with
us?


You want it - you pay for it. Or once again you prove you're full
of it.

No, I have not read the article. But I understand the physics and math
behind it - unlike you. Someone who thinks magnitude without vector
(direction) is valid! ROFLMAO!

Ok, so you mispoke when you said, "I know what it says."



No, I didn't. I didn't say I read it. I said I know what it says. And
I do from other IEEE peer-reviewed articles. I don't need to read it to
find out it agrees with other documentation. And if it didn't, it
wouldn't have gotten published.

You have said repeatedly that the return loss should be calculated by
using the power in as the reference and the reflected power as the thing
being measured which results in a negative log. I am pretty sure the
paper says this is not the correct way to calculate it and many people
are making a mistake doing it this way.


So you've read other IEEE documentation which supports what you say?

I'll see if I can get my hands on the paper. I'm not going to pay for
it. If I thought it would get you to admit you were mistaken, I'd pay
the $100. But I'm sure you will find a way to berate the authors or
twist their logic and I'm not will to pay $100 for that.

So stand by. Someone may be getting it for me.

I really don't give a damn. You would argue the sun rises in the west.

I have a copy of the paper.

Trevor Bird
Editor-in-ehief,Engineering
IEEE Transactions on Antennas and Propagation
CSIRO leT Centre, PO Box 76
Epping, NSW 1710, Australia
Tel: +61 2 9372 4289
Fax: +61 2 9372 4446
E-mail:

Definition and Misuse of Return Loss
Trevor S. Bird

Here is the equation from the article

Pin
RL = 10 log,10 ( ---- ) dB, (1)
Pref

The author explicitly states the resulting value will be positive when
Pin is greater than Pref. He goes on to say, "That is, return loss is
the negative of the reflection coefficient expressed in decibels."

Since the reflection coefficient is never greater than 1, its value is
negative. It follows that the return loss is positive.

He goes on to quote from the "IEEE Standard Dictionary of Electrical
and Electronic Terms, Fourth Edition".

(1 ) (data transmission) (A) At a discontinuity in a
transmission system the difference between the power
incident upon the discontinuity. (B) The ratio in deci-
bels of the power incident upon the discontinuity to the
power reflected from the discontinuity. Note: This ratio
is also the square of the reciprocal to the magnitude of
the reflection coefficient. (C) More broadly, the return
loss is a measure of the dissimilarity between two
impedances, being equal to the number of decibels that
corresponds to the scalar value of the reciprocal of the
reflection coefficient, and hence being expressed by the
following formula:

|Z1 + Z2|
20 log,10 |-------| decibel
|Z1 - Z2|

where Z1 and Z2 = the two impedances.

(2) (or gain) (waveguide). The ratio of incident to
reflected power at a reference plane of a network.


So is this what you "knew" the paper said? Seems to be the opposite of
what you have been promoting.

Anyone feel this paper is incorrect?


# Good find Rick. That should put it to bed.

.....LOL

"rickman" wrote in message ...

On 9/29/2015 9:22 AM, Jerry Stuckle wrote:
On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:

snip
So stand by. Someone may be getting it for me.

I really don't give a damn. You would argue the sun rises in the west.

I have a copy of the paper.

Trevor Bird
Editor-in-ehief,Engineering
IEEE Transactions on Antennas and Propagation
CSIRO leT Centre, PO Box 76
Epping, NSW 1710, Australia
Tel: +61 2 9372 4289
Fax: +61 2 9372 4446
E-mail:

Definition and Misuse of Return Loss
Trevor S. Bird

Here is the equation from the article

Pin
RL = 10 log,10 ( ---- ) dB, (1)
Pref

The author explicitly states the resulting value will be positive when
Pin is greater than Pref. He goes on to say, "That is, return loss is
the negative of the reflection coefficient expressed in decibels."

He goes on to quote from the "IEEE Standard Dictionary of Electrical
and Electronic Terms, Fourth Edition".

(1 ) (data transmission) (A) At a discontinuity in a
transmission system the difference between the power
incident upon the discontinuity. (B) The ratio in deci-
bels of the power incident upon the discontinuity to the
power reflected from the discontinuity. Note: This ratio
is also the square of the reciprocal to the magnitude of
the reflection coefficient. (C) More broadly, the return
loss is a measure of the dissimilarity between two
impedances, being equal to the number of decibels that
corresponds to the scalar value of the reciprocal of the
reflection coefficient, and hence being expressed by the
following formula:

|Z1 + Z2|
20 log,10 |-------| decibel
|Z1 - Z2|

where Z1 and Z2 = the two impedances.

(2) (or gain) (waveguide). The ratio of incident to
reflected power at a reference plane of a network.


So is this what you "knew" the paper said? Seems to be the opposite of
what you have been promoting.

Anyone feel this paper is incorrect?

Well, it is exactly what I thought I learned on the subject.

On 9/29/2015 2:10 PM, Wayne wrote:


"rickman" wrote in message ...
On 9/29/2015 9:22 AM, Jerry Stuckle wrote:
On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote:
On 9/28/2015 8:09 PM, rickman wrote:
On 9/28/2015 7:55 PM, Jerry Stuckle wrote:
On 9/28/2015 5:18 PM, rickman wrote:
On 9/28/2015 4:34 PM, Jerry Stuckle wrote:

snip
So stand by. Someone may be getting it for me.

I really don't give a damn. You would argue the sun rises in the west.

I have a copy of the paper.

Trevor Bird
Editor-in-ehief,Engineering
IEEE Transactions on Antennas and Propagation
CSIRO leT Centre, PO Box 76
Epping, NSW 1710, Australia
Tel: +61 2 9372 4289
Fax: +61 2 9372 4446
E-mail:

Definition and Misuse of Return Loss
Trevor S. Bird

Here is the equation from the article

Pin
RL = 10 log,10 ( ---- ) dB, (1)
Pref

The author explicitly states the resulting value will be positive when
Pin is greater than Pref. He goes on to say, "That is, return loss is
the negative of the reflection coefficient expressed in decibels."

He goes on to quote from the "IEEE Standard Dictionary of Electrical
and Electronic Terms, Fourth Edition".

(1 ) (data transmission) (A) At a discontinuity in a
transmission system the difference between the power
incident upon the discontinuity. (B) The ratio in deci-
bels of the power incident upon the discontinuity to the
power reflected from the discontinuity. Note: This ratio
is also the square of the reciprocal to the magnitude of
the reflection coefficient. (C) More broadly, the return
loss is a measure of the dissimilarity between two
impedances, being equal to the number of decibels that
corresponds to the scalar value of the reciprocal of the
reflection coefficient, and hence being expressed by the
following formula:

|Z1 + Z2|
20 log,10 |-------| decibel
|Z1 - Z2|

where Z1 and Z2 = the two impedances.

(2) (or gain) (waveguide). The ratio of incident to
reflected power at a reference plane of a network.


So is this what you "knew" the paper said? Seems to be the opposite
of what you have been promoting.

Anyone feel this paper is incorrect?

Well, it is exactly what I thought I learned on the subject.

To be honest, it seems logical that the power in should be the reference
and the reflected power should be the property being measured which is
what Jerry is saying. But clearly for this particular term "return
loss" this is not the case.

Does it seem intuitively correct that "return loss" should be a higher
number when the reflection is smaller?

--

Rick