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...

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.


But I assumed that Rick was talking about the reflected power used in
the return loss calculation. That part goes to zero for a perfect
match, hence the infinite return loss.


You cannot have a return loss when there is no returned signal. 0
divided by anything is still 0.

Yes, that is why the value of return loss goes to minus infinity, log of
zero is not technically defined, but in the limit, it goes to negative
infinity.


Since my ATU is closer to the transmitter than the antenna, I tune for
lowest SWR from the transmitter to the ATU and don't worry about the ATU
to antenna SWR.


But that is where the loss occurs. The loss will be dependent on the
SWR and the length of the coax. If your coax is short, you won't have a
significant loss with a reasonable SWR.

Return loss doesn't refer to the loss of signal in the cable. It refers
to the loss of signal due to the reflection from the antenna rather than
being transferred to the antenna. It is true that some of that signal
may be reflected again from the transmitter or other irregularities, but
that is not relevant to the return loss measurement.


I believe we are all talking about the same thing.

I don't think so.

I agree that we are not all on the same page.

Then some of us much prefer to argue rather than discuss. Is there
anything about Waynes post you like? Are the facts more clear now at
least?

--

Rick

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.

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

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

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.

--

Rick

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 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

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

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

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."

--

Rick

In message , Jerry Stuckle
writes


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

It does if you are used to +ve values of RLR!

--
Ian

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.

"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.