Parallel coax
 

On 9/29/2015 3:47 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 9/28/2015 7:12 PM, John S wrote:
On 9/28/2015 1:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote:





You said return loss increases with lower SWR. It does not.


It does.

Sorry, a lower SWR does not increase the amount of loss.

Of course it doesn't. No one said it did. It does the opposite, ie a
lower SWR gives less loss on the feeder.

Please cite a reliable reference that says it does. Even the table Rick
cited shows a negative value for return SWR.

What is this 'Return SWR'? I'm not familiar with it.


Sorry, writing too quickly. I meant return loss.

Do you mean Return Loss Ratio (RLR)? This is a simple, easily
measurable, and meaningful statement of how strong the returning
reflected signal is compared with the outgoing forward signal.

The reflected signal is a weaker version of the forward signal. It's
expressed as a loss, an attenuation, or relatively how much down the
level of the reflection is. You can express this as a numerical ratio -
the reflection coefficient (rho) - or (as often more convenient) rho in dB.

As others have suggested, what is apparently a negative sign in the
chart is presumably more artistic licence than scientific accuracy. If
you lose $10, you don't say that you lost 'minus $10'. Similarly, when
you lose 10dB of signal, you don't say you lost 'minus 10dB'.

Which is greater - 10db or -30db?

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

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

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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

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

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On 9/29/2015 4:36 AM, Jeff wrote:
On 29/09/2015 00:57, Jerry Stuckle wrote:
On 9/28/2015 6:18 PM, Ian Jackson wrote:
In message , Roger Hayter
writes



It would probably make more sense to call return loss "return gain",
but, since it is always less than one, that would merely cause a
different set of ambiguities.

The reflected signal is an attenuated version of the forward (source)
signal. It is a 'loss' - in exactly the same way as the signal at the
output end of an attenuator is an attenuated version of the source
signal at the input end. The RLR (in dB) is the ratio of the what you
put in to what you get out. It cannot be less than 1, so the RLR is in
positive dB. There is absolutely no ambiguity. No one in RF engineering
quotes or uses negative values for RLR (or for attenuators). The greater
the RLR, the less signal is reflected.




The ratio of output to input can never be greater than one - so the log
of that can never be positive.

I.E. 100W in and 50W out is -3db, not +3db.


Yes, that is a GAIN of -3db. When considering RL you are looking for a
reduction (LOSS) in the reflected power so the sign changes.

It is just common sense.

Jeff

OK, so I have a system with gains and losses of +3, +1, +5, +2. What it
the total gain of the system?

In college we learned to consider everything a "black box". You don't
know whether it is active or passive, an amplifier or an attenuator.
All you know are the characteristics. If the output has +db over the
input, it is an amplifier. -db indicates a loss.

The same is true when modeling antennas. The antenna connection is a
black box. Returned power is always less than power going into it.

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

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

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On 9/29/2015 3:58 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 9/28/2015 6:21 PM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 9/28/2015 3:10 PM, rickman wrote:
On 9/28/2015 2:55 PM, Jerry Stuckle wrote:
On 9/28/2015 2:19 PM, rickman wrote:
On 9/28/2015 2:01 PM, Wayne wrote:


"rickman" wrote in message ...

On 9/28/2015 11:59 AM, Wayne wrote:


"Ian Jackson" wrote in message
...
In message , rickman

writes



Definition of Return Loss

In technical terms, RL is the ratio of the light reflected back
from
a device under test, Pout, to the light launched into that
device,
Pin, usually expressed as a negative number in dB.

RL = 10 log10(Pout/Pin)

Here is a link for a table of return loss and VSWR....

http://www.jampro.com/uploads/tech_d.../VSWRChart.pdf

It shows a higher return loss (assuming you mean magnitude
since the
values are all negative) for lower VSWR.


I'm surprised to see negative quantities. For 50 years, I've
always
understood the Return Loss Ratio (RLR) to be exactly what it
says on
the tin, ie the ratio (in dB) of the LOSS (the attenuation) of
the
reflected signal wrt the incident signal. This is a +ve quantity.
Things are already sufficiently confusing without having to start
thinking in unnecessary -ve figures!

I think the table headings are using a dash, not a negative sign.
Return loss- dB

# Look at the equation and you will understand. When the ratio is
less
# than one, the log is negative.

But the ratio is never less than one for passive devices.

If all the power forward is reflected, then the power ratio is 1
to 1.
That's 0 dB return loss from the equation.

Return loss is a positive number.

I'm not so sure. It depends on how you define it. What if half the
power is reflected? The equation above and *many* other sources say
that is 10 log(0.5/1) = -3 dB. A few sources take exception to
this and
say it is 10 log(1/0.5) = 3 dB.

At this point I dunno.


Rickman, you are correct. The return loss is calculated as return
value
(the variable) divided by the output value (the constant).

I haven't seen any reliable sources which say otherwise.

I haven't seen any reliable sources that say either. Have you?



Not for 40 years or so. But then I haven't looked for one since
college. I've just dealt with RF engineers, who have used the same
terminology.

They use negative numbers for loss to figure the gain of the entire
system. It doesn't matter which direction the loss is in; loss is a
negative number (and gain is a positive number).

A loss is only negative if it is being thought of as gain. If it's a
loss, its value is positive.

Physicists and engineers do not mix gain and loss. Gain is always shown
as a positive number and loss as a negative number.

Physicists and engineers don't get themselves into situations where
'gain' and 'loss' are used ambiguously. Unfortunately, the same cannot
be said of certain radio amateurs.

Yes, there are some here who are trying to use the terms ambiguously -
saying +10db can be both gain and loss.


For instance - a system shows a gain and loss of +3, +5, +2, +1. What
is the total gain or loss of the system?

A physicists or engineer would never ask such a meaningless question.

No, because the physicists and engineers use +db for gain and -db for
loss. And they understand -100db is less than -10db. However, they are
also math literate, unlike many non-scientists nowadays.

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

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

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On 9/29/2015 4:40 AM, Jeff wrote:

Physicists and engineers do not mix gain and loss. Gain is always shown
as a positive number and loss as a negative number.

For instance - a system shows a gain and loss of +3, +5, +2, +1. What
is the total gain or loss of the system?


Of course they do, particularly when dealing with a quantity that is
defined as a LOSS.

I have never heard any engineer when asked the question 'what is that
attenuator' reply minus 3 dB. It is always 3dB. It is always called a
3dB attenuator, not a minus 3dB attenuator.

Jeff

Yes, and the power out is never +3db relative to the input. It is
always -3db.

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

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

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In message , Jerry Stuckle
writes
On 9/29/2015 3:47 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 9/28/2015 7:12 PM, John S wrote:
On 9/28/2015 1:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote:





You said return loss increases with lower SWR. It does not.


It does.

Sorry, a lower SWR does not increase the amount of loss.

Of course it doesn't. No one said it did. It does the opposite, ie a
lower SWR gives less loss on the feeder.

Please cite a reliable reference that says it does. Even the table Rick
cited shows a negative value for return SWR.

What is this 'Return SWR'? I'm not familiar with it.


Sorry, writing too quickly. I meant return loss.

Do you mean Return Loss Ratio (RLR)? This is a simple, easily
measurable, and meaningful statement of how strong the returning
reflected signal is compared with the outgoing forward signal.

The reflected signal is a weaker version of the forward signal. It's
expressed as a loss, an attenuation, or relatively how much down the
level of the reflection is. You can express this as a numerical ratio -
the reflection coefficient (rho) - or (as often more convenient) rho in dB.

As others have suggested, what is apparently a negative sign in the
chart is presumably more artistic licence than scientific accuracy. If
you lose $10, you don't say that you lost 'minus $10'. Similarly, when
you lose 10dB of signal, you don't say you lost 'minus 10dB'.

Which is greater - 10db or -30db?

In voltage ratios, they are 1/3 and 1/30 respectively, and in power
ratios, 1/10 and 1/000 respectively. But if you lost 30dB down your
coax, you'd be losing 20dB MORE than if you were only losing 10dB. But
surely even you wouldn't say "My coax has a loss of minus 30dB"? [Or
would you?!]
--
Ian

Parallel coax
 

In message , Jerry Stuckle
writes
On 9/29/2015 4:40 AM, Jeff wrote:

Physicists and engineers do not mix gain and loss. Gain is always shown
as a positive number and loss as a negative number.

For instance - a system shows a gain and loss of +3, +5, +2, +1. What
is the total gain or loss of the system?


Of course they do, particularly when dealing with a quantity that is
defined as a LOSS.

I have never heard any engineer when asked the question 'what is that
attenuator' reply minus 3 dB. It is always 3dB. It is always called a
3dB attenuator, not a minus 3dB attenuator.

Jeff

Yes, and the power out is never +3db relative to the input. It is
always -3db.

And that's because it has suffered a positive LOSS of 3dB.

As I have already asked you, if you lose $10, do you say "I've lost
MINUS $10?
--
Ian

Parallel coax
 

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

Parallel coax
 

On 9/29/2015 8:14 AM, Jerry Stuckle wrote:
On 9/29/2015 3:27 AM, John S wrote:
On 9/28/2015 6:54 PM, Jerry Stuckle wrote:

You said return loss increases with lower SWR. It does not.


It does.

Sorry, a lower SWR does not increase the amount of loss.

It increases the amount of loss that is reflected. Hence, return loss.

Please cite a reliable reference that says it does. Even the table Rick
cited shows a negative value for return SWR.

No, the table is correct and does not show negative values for return
loss. What is return SWR?



No, a 1:1 SWR has no reflection, therefore no reflective loss.

And yes, it does show negative values. Don't you see the '-' sign?


No. I see a hyphen, not a minus sign. It indicates that the data column
is in units of dB.

Parallel coax
 

"Jerry Stuckle" wrote in message ...

On 9/29/2015 3:58 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 9/28/2015 6:21 PM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 9/28/2015 3:10 PM, rickman wrote:
On 9/28/2015 2:55 PM, Jerry Stuckle wrote:
On 9/28/2015 2:19 PM, rickman wrote:
On 9/28/2015 2:01 PM, Wayne wrote:


"rickman" wrote in message ...

On 9/28/2015 11:59 AM, Wayne wrote:


"Ian Jackson" wrote in message
...
In message , rickman

writes



Definition of Return Loss

In technical terms, RL is the ratio of the light reflected back
from
a device under test, Pout, to the light launched into that
device,
Pin, usually expressed as a negative number in dB.

RL = 10 log10(Pout/Pin)

Here is a link for a table of return loss and VSWR....

http://www.jampro.com/uploads/tech_d.../VSWRChart.pdf

It shows a higher return loss (assuming you mean magnitude
since the
values are all negative) for lower VSWR.


I'm surprised to see negative quantities. For 50 years, I've
always
understood the Return Loss Ratio (RLR) to be exactly what it
says on
the tin, ie the ratio (in dB) of the LOSS (the attenuation) of
the
reflected signal wrt the incident signal. This is a +ve quantity.
Things are already sufficiently confusing without having to start
thinking in unnecessary -ve figures!

I think the table headings are using a dash, not a negative sign.
Return loss- dB

# Look at the equation and you will understand. When the ratio is
less
# than one, the log is negative.

But the ratio is never less than one for passive devices.

If all the power forward is reflected, then the power ratio is 1
to 1.
That's 0 dB return loss from the equation.

Return loss is a positive number.

I'm not so sure. It depends on how you define it. What if half the
power is reflected? The equation above and *many* other sources say
that is 10 log(0.5/1) = -3 dB. A few sources take exception to
this and
say it is 10 log(1/0.5) = 3 dB.

At this point I dunno.


Rickman, you are correct. The return loss is calculated as return
value
(the variable) divided by the output value (the constant).

I haven't seen any reliable sources which say otherwise.

I haven't seen any reliable sources that say either. Have you?



Not for 40 years or so. But then I haven't looked for one since
college. I've just dealt with RF engineers, who have used the same
terminology.

They use negative numbers for loss to figure the gain of the entire
system. It doesn't matter which direction the loss is in; loss is a
negative number (and gain is a positive number).

A loss is only negative if it is being thought of as gain. If it's a
loss, its value is positive.

Physicists and engineers do not mix gain and loss. Gain is always shown
as a positive number and loss as a negative number.

Physicists and engineers don't get themselves into situations where
'gain' and 'loss' are used ambiguously. Unfortunately, the same cannot
be said of certain radio amateurs.

# Yes, there are some here who are trying to use the terms ambiguously -
# saying +10db can be both gain and loss.

I don't have many engineering books left, and my 1950s version of Terman
doesn't cover return loss.
My perspective on return loss happened late in my career using a network
analyzer to evaluate reentry vehicle antennas.
The network analyzer we used had RL expressed as a positive number.

There is another equation to consider:

RL=(Power incident in dBm) minus (Power reflected in dBm)

Thus the bigger the RL number, the bigger the difference between forward and
reflected power.