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 9/28/2015 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?

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.

Please cite a reliable reference that says it does.


From the ARRL:

http://www.arrl.org/news/amateur-radio-quiz-a-log-of-dbs

"9) D -- Higher positive values of Return Loss (RL) in dB indicate less
power returning from a load, indicating a lower SWR."

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.

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

OH! I see your problem. You think the column heading dashes are minus
signs. They are not. Loss - dB indicates that the column data have the
units of dB, not that they are negative. I usually use Loss (dB) for my
column headings.

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.

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'.
--
Ian

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.

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.
--
Ian

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.

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


Create your own chart that shows you are incorrect:

http://www.jampro.com/uploads/tech_calc/vswr.htm

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?

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

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

On 9/29/2015 3:36 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.

Please cite a reliable reference that says it does.


From the ARRL:

http://www.arrl.org/news/amateur-radio-quiz-a-log-of-dbs

"9) D -- Higher positive values of Return Loss (RL) in dB indicate less
power returning from a load, indicating a lower SWR."



OK, I'll have to take that up with N0AX. My university professors and
the IEEE disagree with him from an engineering view.

However, hams often try to make things easier - and not always correctly.

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

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