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!

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