On Fri, 26 Nov 2004 00:24:13 -0000, "Dave" wrote:

the 1/4 wave open end coax looks like a short circuit at the feed point. so
your reading makes perfect sense.


Dear Dave,

Yes, I believe it does - that is, it makes perfect sense to have a low
resistance and to have a near zero reactive component. What does not
make sense is that the high SWR is supposed to produce outrageous
losses. I don't see values that I can interpret as high losses - quite
the opposite. Maybe I just don't interpret it correctly, but I would
expect it to be several ohms - not 0.57 ohms.

In fact, and this is where it gets ridiculous, the examples in the
ARRL Antenna Book would lead me to believe that the above quarter wave
line would exhibit 20 dB of total losses. In order to get those
numbers the SWR at the load of say 8000 would have to decrease to
1.01:1 at the source end in order to account for 20 dB in losses. (See
the example on page 24-9 of the 17th Edition.)


Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

On Fri, 26 Nov 2004 04:19:06 GMT, (Robert Lay
W9DMK) wrote:

What does not
make sense is that the high SWR is supposed to produce outrageous
losses. I don't see values that I can interpret as high losses - quite
the opposite. Maybe I just don't interpret it correctly, but I would
expect it to be several ohms - not 0.57 ohms.

Hi Bob,

Thanx for the explicit results. Now what is needed is the explicit
expectation. I note that you circumspectly describe it as being
ARRL Antenna Book would lead me to believe that the above quarter wave
line would exhibit 20 dB of total losses. In order to get those
numbers the SWR at the load of say 8000 would have to decrease to
1.01:1 at the source end in order to account for 20 dB in losses. (See
the example on page 24-9 of the 17th Edition.)
which is an inference which being an interpretation is open to errors
of mis-interpretation.

Reference Data for Radio Engineers, "Mismatch and Transducer Loss,"
"One End Mismatched," pg. 22-12:
Transducer Loss = A0 + 10 · log (Pm/P) decibels
where
A0 = normal attenuation of the line
Pm = power that would be delivered were system matched
P = power delivered to the load

Of particular note is that this is one of my references as to the
nature of Source Z which is often neglected in academic treatments
with the presumption that the engineer has already been schooled in
the nature of Real sources (this may shock some complaisant readers
here). However, this citation offers that explicit lesson in figure
10 and makes use of this commonplace characteristic in illustrations
of Mismatch Uncertainty. They go as far as to explicitly offer a
section entitled "Generator and Load Mismatched." You may wish to
review this treatment as it offers the math that would present the
most loss available in a line, above and beyond the typical charts
offered for line loss (which are confined to both ends being matched).

73's
Richard Clark, KB7QHC

On Fri, 26 Nov 2004 07:33:04 GMT, Richard Clark
wrote:

.....snip
Reference Data for Radio Engineers, "Mismatch and Transducer Loss,"
"One End Mismatched," pg. 22-12:
Transducer Loss = A0 + 10 · log (Pm/P) decibels
where
A0 = normal attenuation of the line
Pm = power that would be delivered were system matched
P = power delivered to the load

Of particular note is that this is one of my references as to the
nature of Source Z which is often neglected in academic treatments
with the presumption that the engineer has already been schooled in
the nature of Real sources (this may shock some complaisant readers
here). However, this citation offers that explicit lesson in figure
10 and makes use of this commonplace characteristic in illustrations
of Mismatch Uncertainty. They go as far as to explicitly offer a
section entitled "Generator and Load Mismatched." You may wish to
review this treatment as it offers the math that would present the
most loss available in a line, above and beyond the typical charts
offered for line loss (which are confined to both ends being matched).

Dear Richard,

I'm finally ready to comment on the above - it is my great fortune to
be blessed with copies of both the Fourth and Fifth Editions of the
ITT Handbook.

I studied over the first 13 pages of Chapter 22 and found that, just
as Wes said, it's entirely the work of MacAlpine as published in 1953.

I went over Equations (1) through (4) in the Mismatch section very
carefully and found no heartburn with anything in that section. This
is NOT to say that I LIKE it, but I do understand it and have no
problem with the math model and the figures. My problems with the two
mismatch topics is simply that I just don't like to call it a loss
when energy that COULD have been delivered to the load does NOT get
delivered to the load as a result of mismatch. For me, lost energy in
a transmission line problem is energy actually lost in the
transmission line, not energy that is being lost elsewhere as a result
of the transmission line not being matched properly. I realize that
I'm probably alone in that thinking, but I like to feel that such
terms as efficiency and losses should be associated strongly with the
item under evaluation, namely the transmission line, and not the
ancillary equipment which feed it or terminate it. Those items get
their own hearings relative to efficiency and losses and those
evaluations do not require the presence of the transmission line. In
fact, those items are usually evaluated as to their performance in
ways that do not in any way relate to how well some transmission line
is or is not working.

However, this is not the nub of the problem that I was encountering -
a problem which has now been partly resolved. At least I think I have
a far, far better understanding of the problem now than I had a few
days ago. The problem centers on the Additional Losses Due to SWR and
how to model them. Since it is, perhaps, more appropriate to continue
that topic under the responses from Wes, I will not go into it here.

I want to thank you and Wes, both, for leading me to Chapter 22 - it
is much more readable than MacAlpine's original paper.

Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

"Robert Lay W9DMK" wrote in message
...
On Fri, 26 Nov 2004 00:24:13 -0000, "Dave" wrote:

the 1/4 wave open end coax looks like a short circuit at the feed point.
so
your reading makes perfect sense.


Dear Dave,

Yes, I believe it does - that is, it makes perfect sense to have a low
resistance and to have a near zero reactive component. What does not
make sense is that the high SWR is supposed to produce outrageous
losses. I don't see values that I can interpret as high losses - quite
the opposite. Maybe I just don't interpret it correctly, but I would
expect it to be several ohms - not 0.57 ohms.

In fact, and this is where it gets ridiculous, the examples in the
ARRL Antenna Book would lead me to believe that the above quarter wave
line would exhibit 20 dB of total losses. In order to get those
numbers the SWR at the load of say 8000 would have to decrease to
1.01:1 at the source end in order to account for 20 dB in losses. (See
the example on page 24-9 of the 17th Edition.)


the cases they talk about in there are figuring the loss in power that you
would be supplying to a load. in your case the load is an infinite
resistance so it receives zero power which is what the arrl book says... in
this case all the power that is sent down the line is reflected back minus a
little bit of heating so the swr at the feedpoint should be near infinite,
but not quite. the actual loss in the wave going down and coming back is
very small hence the very low impedance. this is an effect that is used to
make coaxial stub filters and transformers.

(Robert Lay W9DMK) wrote:
Yes, I believe it does - that is, it makes perfect sense to have a low
resistance and to have a near zero reactive component. What does not
make sense is that the high SWR is supposed to produce outrageous
losses. I don't see values that I can interpret as high losses - quite
the opposite. Maybe I just don't interpret it correctly, but I would
expect it to be several ohms - not 0.57 ohms.

Hi Bob, I'm still at my relatives' house posting through Google. I'll
expand on this when I get back to my computer.

Those equations in The ARRL Antenna Book (15th Edition) make an assumption
that may or may not be true - I don't know. What they are assuming is that
the losses are due to the additional power associated with high SWR. In
general, if the forward power is 100w for the matched case and the sum
of the forward and reflected power is 300w for the unmatched case, the
losses will be three times higher for the unmatched case. That seems a
reasonable assumption. However, rho at the shorted or open end of a stub
is equal to |1| so rho^2 will be equal to 1. That puts (1-rho^2) = 0 in
the denominator of the equation and makes the addditional losses undefined.

In fact, and this is where it gets ridiculous, the examples in the
ARRL Antenna Book would lead me to believe that the above quarter wave
line would exhibit 20 dB of total losses. In order to get those
numbers the SWR at the load of say 8000 would have to decrease to
1.01:1 at the source end in order to account for 20 dB in losses. (See
the example on page 24-9 of the 17th Edition.)

Here's an example. Assume 100w is delivered to the load for both the
matched and unmatched conditions. Assume 3dB matched line loss in the
transmission line. Assume an SWR of 5.83:1 (rho=0.707) at the load for
the mismatched condition.

Forward 200w------------3dB loss-------------Matched Load 100w


Forward 400w------------3dB loss-------------Mismatched Load 100w
Reflected 50w both Forward 200w
directions Reflected 100w

The equations gives an additional loss of 5.44dB. This is based on an
assumption that the losses are directly proportional to forward power
plus reflected power.

Remember that at the mouth of the stub, the impedance is equal to
(Vf + Vr)/(If + Ir) so, knowing the Z0, that should allow you to
calculate those four values existing at the mouth of the stub. From
that, you can calculate the total losses. More when I get back.
--
73, Cecil, W5DXP