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.
--
Ian

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.

I have seen a number of references that give the formula as Pout - Pin.
Clearly both forms are in use. I can offer a book that shows a graph
of negative RL values.

https://books.google.com/books?id=nH...0ratio&f=false

--

Rick

On 9/28/2015 1:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote:
On 9/28/2015 12:54 PM, Jerry Stuckle wrote:
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:


"John S" wrote in message ...

On 9/27/2015 1:20 PM, Wayne wrote:


"rickman" wrote in message ...

On 9/27/2015 10:41 AM, kg7fu wrote:

Matching the antenna won't make the Return Loss go away but it
will
make
the transmitter happy.

Can you explain this? I thought matching the antenna would
*exactly*
make the return loss go away because it would eliminate the
mismatch.

Not wanting to put words in his mouth....
I read that to mean that the high SWR between the ATU and the
antenna
would remain, but the transmitter would be happy with the SWR on
the
transmitter/ATU coax.


# Rick is correct. If the antenna (load) is matched to the line,
there is
# no return loss, hence no SWR. The ATU will be adjusted
(hopefully) to
# make the transmitter operate properly with the impedance as
seen at
the
# transmitter end of the line.

# Yes, the SWR due to mismatch of the antenna (load) and line will
remain.
# Even if the real part of your load impedance is matched to the
line, you
# will still have a high SWR if the reactance remains.

# Does this make sense?

Yes. That's what I was trying to say using SWR instead of return
loss.
Return loss numbers get bigger with lower SWR.
For example: SWR 1:1 = infinite return loss.


Incorrect. Return loss increases with an increased SWR. An SWR
of 1:1
has no return loss because there is no returned signal to lose.
100% of
the signal is radiated.

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.

I believe that is exactly what I said in the portions of my post which
you trimmed. These values for RF return loss match exactly the equation
which you are saying is not used in RF. So which is it, the return loss
table is correct with negative values of return loss or the equation I
posted is incorrect even though it gives the values in the table?


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


It does.

On 9/28/2015 5:17 PM, rickman wrote:
On 9/28/2015 4:33 PM, Jerry Stuckle wrote:
On 9/28/2015 4:15 PM, rickman wrote:
On 9/28/2015 4:02 PM, Jerry Stuckle wrote:
On 9/28/2015 3:02 PM, rickman wrote:
On 9/28/2015 2:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote:
On 9/28/2015 12:54 PM, Jerry Stuckle wrote:
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:


"John S" wrote in message ...

On 9/27/2015 1:20 PM, Wayne wrote:


"rickman" wrote in message
...

On 9/27/2015 10:41 AM, kg7fu wrote:

Matching the antenna won't make the Return Loss go away
but it
will
make
the transmitter happy.

Can you explain this? I thought matching the antenna would
*exactly*
make the return loss go away because it would eliminate the
mismatch.

Not wanting to put words in his mouth....
I read that to mean that the high SWR between the ATU and the
antenna
would remain, but the transmitter would be happy with the
SWR on
the
transmitter/ATU coax.


# Rick is correct. If the antenna (load) is matched to the
line,
there is
# no return loss, hence no SWR. The ATU will be adjusted
(hopefully) to
# make the transmitter operate properly with the impedance as
seen at
the
# transmitter end of the line.

# Yes, the SWR due to mismatch of the antenna (load) and line
will
remain.
# Even if the real part of your load impedance is matched
to the
line, you
# will still have a high SWR if the reactance remains.

# Does this make sense?

Yes. That's what I was trying to say using SWR instead of
return
loss.
Return loss numbers get bigger with lower SWR.
For example: SWR 1:1 = infinite return loss.


Incorrect. Return loss increases with an increased SWR. An
SWR
of 1:1
has no return loss because there is no returned signal to lose.
100% of
the signal is radiated.

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.

I believe that is exactly what I said in the portions of my post
which
you trimmed. These values for RF return loss match exactly the
equation
which you are saying is not used in RF. So which is it, the return
loss
table is correct with negative values of return loss or the
equation I
posted is incorrect even though it gives the values in the table?


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

Are you being pedantic that -1 dB is not lower than -10 dB? It is not
numerically lower in value, but is lower in magnitude and it is a
lower
loss. I even referred to the magnitude in my post. But then that was
the same part you snipped which I referred to earlier.

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


No, I am not being pedantic. -1 is greater than -10. The fact it is a
negative number makes all the difference - as any engineer who knows
what he's talking about will tell you.

Except that I clearly stated I was referring to the magnitude.


Magnitude without direction is meaningless.

Lol. There is only one direction for return loss.


Yes - the loss is a negative gain - hence the use of negative db values.

You can't just give a magnitude. You need to specify the direction,
also. You seem to have a very hard time understanding such a simple
concept - one that was taught in freshman physics.
--
==================
Remove the "x" from my email address
Jerry, AI0K

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

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:
On 9/28/2015 12:54 PM, Jerry Stuckle wrote:
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:


"John S" wrote in message ...

On 9/27/2015 1:20 PM, Wayne wrote:


"rickman" wrote in message ...

On 9/27/2015 10:41 AM, kg7fu wrote:

Matching the antenna won't make the Return Loss go away but it
will
make
the transmitter happy.

Can you explain this? I thought matching the antenna would
*exactly*
make the return loss go away because it would eliminate the
mismatch.

Not wanting to put words in his mouth....
I read that to mean that the high SWR between the ATU and the
antenna
would remain, but the transmitter would be happy with the SWR on
the
transmitter/ATU coax.


# Rick is correct. If the antenna (load) is matched to the line,
there is
# no return loss, hence no SWR. The ATU will be adjusted
(hopefully) to
# make the transmitter operate properly with the impedance as
seen at
the
# transmitter end of the line.

# Yes, the SWR due to mismatch of the antenna (load) and line will
remain.
# Even if the real part of your load impedance is matched to the
line, you
# will still have a high SWR if the reactance remains.

# Does this make sense?

Yes. That's what I was trying to say using SWR instead of return
loss.
Return loss numbers get bigger with lower SWR.
For example: SWR 1:1 = infinite return loss.


Incorrect. Return loss increases with an increased SWR. An SWR
of 1:1
has no return loss because there is no returned signal to lose.
100% of
the signal is radiated.

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.

I believe that is exactly what I said in the portions of my post which
you trimmed. These values for RF return loss match exactly the equation
which you are saying is not used in RF. So which is it, the return loss
table is correct with negative values of return loss or the equation I
posted is incorrect even though it gives the values in the table?


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.

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

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

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.


--
==================
Remove the "x" from my email address
Jerry Stuckle

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

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.

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

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

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.

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

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

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

On 9/28/2015 7:52 PM, Jerry Stuckle wrote:
On 9/28/2015 5:17 PM, rickman wrote:
On 9/28/2015 4:33 PM, Jerry Stuckle wrote:
On 9/28/2015 4:15 PM, rickman wrote:
On 9/28/2015 4:02 PM, Jerry Stuckle wrote:
On 9/28/2015 3:02 PM, rickman wrote:
On 9/28/2015 2:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote:
On 9/28/2015 12:54 PM, Jerry Stuckle wrote:
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:


"John S" wrote in message ...

On 9/27/2015 1:20 PM, Wayne wrote:


"rickman" wrote in message
...

On 9/27/2015 10:41 AM, kg7fu wrote:

Matching the antenna won't make the Return Loss go away
but it
will
make
the transmitter happy.

Can you explain this? I thought matching the antenna would
*exactly*
make the return loss go away because it would eliminate the
mismatch.

Not wanting to put words in his mouth....
I read that to mean that the high SWR between the ATU and the
antenna
would remain, but the transmitter would be happy with the
SWR on
the
transmitter/ATU coax.


# Rick is correct. If the antenna (load) is matched to the
line,
there is
# no return loss, hence no SWR. The ATU will be adjusted
(hopefully) to
# make the transmitter operate properly with the impedance as
seen at
the
# transmitter end of the line.

# Yes, the SWR due to mismatch of the antenna (load) and line
will
remain.
# Even if the real part of your load impedance is matched
to the
line, you
# will still have a high SWR if the reactance remains.

# Does this make sense?

Yes. That's what I was trying to say using SWR instead of
return
loss.
Return loss numbers get bigger with lower SWR.
For example: SWR 1:1 = infinite return loss.


Incorrect. Return loss increases with an increased SWR. An
SWR
of 1:1
has no return loss because there is no returned signal to lose.
100% of
the signal is radiated.

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.

I believe that is exactly what I said in the portions of my post
which
you trimmed. These values for RF return loss match exactly the
equation
which you are saying is not used in RF. So which is it, the return
loss
table is correct with negative values of return loss or the
equation I
posted is incorrect even though it gives the values in the table?


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

Are you being pedantic that -1 dB is not lower than -10 dB? It is not
numerically lower in value, but is lower in magnitude and it is a
lower
loss. I even referred to the magnitude in my post. But then that was
the same part you snipped which I referred to earlier.

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


No, I am not being pedantic. -1 is greater than -10. The fact it is a
negative number makes all the difference - as any engineer who knows
what he's talking about will tell you.

Except that I clearly stated I was referring to the magnitude.


Magnitude without direction is meaningless.

Lol. There is only one direction for return loss.


Yes - the loss is a negative gain - hence the use of negative db values.

You can't just give a magnitude. You need to specify the direction,
also. You seem to have a very hard time understanding such a simple
concept - one that was taught in freshman physics.

Ok Jerry. You just said the value can only be negative, so if the
magnitude is specified it is a complete specification.

I'll toss in the towel as you will never stop arguing such a silly point.

--

Rick

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