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.

However, I guess it's just too difficult for you to understand negative
numbers and how to relate db to watts.

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

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

On 9/28/2015 11:47 AM, Ian Jackson wrote:
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!



Ian,

Yes, you are correct. But the return signal can never be greater than
the incident signal, so the actual ratio of the reflected signal loss to
the incident signal must always be 1. And the log of a value 1 is a
negative number.

And yes, it is specified as a negative quantity because the result of
the calculation is a negative quantity. Physicists and engineers
specify loss as a negative number rather than confuse issues by changing
signs to suit them. Changing signs only leads to problems in later
calculations.

Non-professionals do change signs because it's often easier to
understand that a 10db loss is greater than a 5 db loss. Not so easy to
grasp quickly is that a -10db gain is more lossy (less gain) than a -5db
gain.

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

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

On 9/28/2015 12:05 PM, John S wrote:
On 9/28/2015 10:47 AM, Ian Jackson wrote:
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 you are correct. I think the confusion is the word *loss*. If
you have a positive *loss* number, the return signal is reduced. To have
a negative return loss number, you need to refer to *gain*.

For example, a return *loss* of 20dB is the same as a return *gain* of
-20dB.

In my other replies on this I mistakenly said the return loss in dB is
negative for all but zero reflection. That should have been for all but
100% reflection.

However, I did a bit more research on the topic and I noticed that
Wikipedia says for the general case a loss ratio in decibels should be a
positive number. But they have a paragraph explaining that by
convention the return loss ratio is a negative number which makes it
identical to the reflection coefficient.

I could not check the reference wikipedia gives for this as it is a
subscription document. I did find in the Electronic Engineers' Handbook
a formula for "system" losses starting at the antenna terminal which are
expressed as

Ls = 10 log (pt/pa) = Pt - Pa, dB

where Ls is the loss in decibels, pt is the power delivered to the
antenna terminals, pa is the power at the receiving antenna. The
capital form of these power levels are in dB.

I think the reason we see a different formula for return loss is because
in the above equation the ratio is the power sent to the power received.
The formula for the return loss is the ratio of the power provided to
the power reflected which is not at all the same ratio. To make the
loss equations measure the same thing return loss would be

Lr = 10 log (pt/(pt-pr))

where pr is the power reflected yielding the same ratio as pt/pa. This
would still be a positive value in dB. So I'm not at all sure why the
convention of negative dB for return loss got started. Maybe it is one
of those issues where there is a lot of misinformation that never gets
corrected properly.

So I withdraw my statements since the issue is very confused in the
literature. Clearly there is no universal convention.

--

Rick

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?

--

Rick

"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.

In message , Wayne
writes


"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
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

Phew - what a relief!


--
Ian

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.

--

Rick

"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.

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.

Where do you get the equation? Do you have a reference?

--

Rick

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.

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

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