Thread: Parallel coax
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Old September 29th 15, 08:58 AM posted to rec.radio.amateur.antenna
Ian Jackson[_2_] Ian Jackson[_2_] is offline
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First recorded activity by RadioBanter: Nov 2007
Posts: 568
Default Parallel coax

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