Owen Duffy wrote:
So what is the answer to your example, the load Z, with and without
consideration of the losses?

Sorry, I just got a new Dell and don't have my software loaded.
A quick pencil whipping of a Smith Chart graph yields a
load around 1140 - j1900 ohms which would be about right
for a 145 ft. dipole on 40m.
--
73, Cecil http://www.qsl.net/w5dxp

Owen Duffy wrote:
Cecil, you have a single solution, and you are inclined to transform
every problem to require that single solution (read your posts).

No, my solution allows me to pull Reg's leg in arguments about
SWR meters since I don't use a tuner. :-) My solution also allows
me to get by without a 500 watt tuner for my SGC-500. The other
advantages are just frosting on the cake.
--
73, Cecil http://www.qsl.net/w5dxp

On Tue, 29 Nov 2005 21:35:33 GMT, Cecil Moore wrote:

Owen Duffy wrote:
So what is the answer to your example, the load Z, with and without
consideration of the losses?

Sorry, I just got a new Dell and don't have my software loaded.
A quick pencil whipping of a Smith Chart graph yields a
load around 1140 - j1900 ohms which would be about right
for a 145 ft. dipole on 40m.

You didn't differentiate between the lossy and lossless solutions. The
difference isn't very much, I figure about 1100-j2000 vs 1000-j2170
which is real hard to resolve on the Smith chart. Smith chart programs
solve the same tranmission line problems as underlies the Smith chart,
but using a program provides much higher resolution, and the
convenience and accuracy benefit of not having to find the solution in
a normalised domain.

The Smith chart does not directly indicate the mismatch loss in the
general case.

My online calculator at http://www.vk1od.net/tl/tllce.php produces an
answer (considering loss) of 422.87-j1441.91 if the transmission line
was 90' Wireman 552 and using Wes's characterisation of 552 for
derivation of the fundamental line parameters (slightly different Zo
to yours). Indicated line loss is 0.93 dB. (SWR is not used to arrive
at the solution.)

Dan's TLD gives 432.7-j1459.0 using a slightly different transmission
line approximation, and independently deriving the model from Wes's
characterisation. Indicated line loss is 0.93 dB.

This demonstrates the sensitivity of the result to very slight
variations in line parameters, consideration of loss, and the
approximations used in modelling.

Owen
--

"Owen Duffy" wrote in message
...
On Tue, 29 Nov 2005 15:02:15 GMT, Cecil Moore wrote:

Reg Edwards wrote:
You then include in the calculation the measurement or assumption of
the Zo of the 50-ohm coax, and the measurement or assumption of Zo of
the twin-line, and the forward and reverse powers, and the SWR on the
twin line can be deduced or assumed.

Actually, nowadays I use my MFJ-259B to read the resistance at
the choke-balun where I have adjusted the ladder-line length to
guarantee the existence of a current maximum point. It's actually
easier to do than to write about it. An assumption that Z0=50 ohms
is not necessary.

But if you think you are measuring SWR on anything you are cheating
and fooling yourself.

I actually have an SWR meter calibrated for balanced 380 ohms but it's
in a box somewhere in my garage. I found my indirect measurements to
be entirely accurate enough. In general, if one can isolate the problem
to 10% of the Smith Chart, one can solve any problem by tweaking.

Speaking of indirect measurements - let's say the feedline Z0 is 380
ohms with a VF of 0.9 and a length of 90 ft. The measured resistance
at the current maximum point is 30 ohms on 7.15 MHz. The SWR on the
ladder-line is 380/30 = 12.7:1. The feedline is 0.727 wavelengths
long. Plot the point 30/380 = 0.079 + j0 on a Smith Chart. Draw an
SWR circle through that point. Backtrack from that point around the
circle for 0.727 wavelengths and there's your antenna feedpoint
impedance (neglecting losses). Losses can be taken into account by
using SWR spirals instead of SWR circles. And of course, all of this
is done by a computer program after just a few seconds of data entry.

So what is the answer to your example, the load Z, with and without
consideration of the losses?

Hi Owen

Did you mean to write the load impedance as 0.079 + j0? Or did you mean
0.079 + j1?

Jerry

Jerry Martes wrote:
Did you mean to write the load impedance as 0.079 + j0? Or did you mean
0.079 + j1?

0.079 + j0 is the normalized impedance at a current
maximum point at the transmitter. The load impedance
is not given.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
. net...
Jerry Martes wrote:
Did you mean to write the load impedance as 0.079 + j0? Or did you
mean 0.079 + j1?

0.079 + j0 is the normalized impedance at a current
maximum point at the transmitter. The load impedance
is not given.
--
73, Cecil http://www.qsl.net/w5dxp

I have *no* excuse for my not being able to read and think. i did
re-read the post and see whats being done.

Jerry

On Tue, 29 Nov 2005 22:30:16 GMT, "Jerry Martes"
wrote:


Did you mean to write the load impedance as 0.079 + j0? Or did you mean
0.079 + j1?

I didn't write that, it was Cecil, and j0 would be correct.

Owen
--

"Reg Edwards" wrote in news:dmf4fr$4cu$1
@nwrdmz01.dmz.ncs.ea.ibs-infra.bt.com:


"Saandy wrote

you can't measure SWR.
=========================================

I am pleased you agree with me.

=========================================
You can CALCULATE the SWR using the formula.
=========================================

But of what use is the SWR it after you have calculated it?

To what transmission line does it apply? Where is it? What are the
locations of max-volts and min-volts? It does NOT apply to the line
between transmitter and antenna. I suggest it exists only in your
imagination. ;o)

Heh heh. I remember one time a friend of mine was wailing that his fancy
new outdoor 2m antenna wasn't working as good as the indoor mag-mount on
a pie plate that he had been using.

I asked about his installation and he informed me that the antenna was
properly installed and that the SWR was 1 to 1. So I inquired further.
He was using a bridge to measure reflected power at the transmitter. And
there really wasn't any. Then I asked him what he was using for
transmission line. Turns out it was about 75 feet of cheap RG58. So I
told him to take it off the antenna and see what the bridge said with no
antenna. It climbed all the way to 1.2 to 1. In short, the coax was
simply eating the power. Changing it out to better quality line proved
to be the answer there.

SWR is so overrated. I'm in the process of putting together a modest
balcony-based HF station. I'm much more interested in the efficiency of
loading coils than in actual SWR on the coax. A 3-to-1 SWR on coax is
meaningless at 80 meters. It adds only a fraction of a decibel to
losses, even in RG58. But your transmitter might not like the complex
load it is seeing at the end of that coax, hence the utility of a tuning
device. Back in the day, I used to just ignore such issues because my
transmitters had pretty good output tuning networks. But with the advent
of broadbanded solid state finals, it is necessary to match the radio to
the transmission line's complex input impedance.

--
Dave Oldridge+
ICQ 1800667

"Saandy , wrote
Guys, leave it alone! Just make sure that the SWR is a reasonable
value, something that the transmitter can handle and leave it at
that.
========================================
Saandy,

You are a man after my own heart.

I would further simplify it. No need to mention SWR. It is
meaningless. Just make sure the transmitter is loaded with about 50
ohms and leave it at that.

To make things nice and tidy, just change the name of the meter to TLI
= Transmitter Loading Indicator.
----
Reg, G4FGQ

Saandy , 4Z5KS wrote:
...if already going into it, a little bit of history.
when the cows had bigger heads and the air was greener, there was no
such thing as coax. what we used was the ubiquitous ladder wire, with
an unknown impedance and with a frequency response depending on what
the Gods ate at lunch! the VSWR story was not invented yet.
What we did was one of two things: either tune the system for maximum
current in the line or else used a light bulb in conjunction with a
small light bulb and tuned for maximum brilliance. in neither case was
SWR involved in the mess.
the whole SWR uproar began after WW2 with the advent of coax and the
new fangled theories. that was also the time when all kind of
directional couplers came up.in due time a few wise guys developed all
kinds of theories on the subject, and manged to convey the impression
that SWR is king! nothing further from truth. what's really true is
that reflections can cause the apparent impedance at the network's
input to differ from Zo. SO WHAT? if you can adjust your matching
network between the transmitter and the line for a match what do you
care?
actually the hitch is that, with a high SWR on the line, the losses go
up. if the cable can take it, without melting no harm's done: whatever
remains will get radiated. this was the good pint of open feeders: the
losses were very low. an SWR fo 10 and more was insignificant from the
losses' point ov view.
Guys, leave it alone! Just make sure that the SWR is a reasonable
value, something that the transmitter can handle and leave it at that.
Saandy 4Z5KS

Some of what you say is very true. Especially in the world of HAM
radio. How ever, this being an open forum, truth is of great
importance. Your thesis on the "whys" of the importance of vswr
measurements are incorrect in some areas.
Yes, importance grew with the advent of coaxial lines simply because of
the relatively small distances between inner and outer conductors. That
part is true.

However, until you've seen a 6" universal coaxial transmission line with
no insulators or inner conductor remaining over a 350" run, I guess you
can't really appreciate the need for monitoring and maintaining good
"system" vswr characteristics.

Now to the measuring of said vswr. It can be done. In the broadcast
world it's accomplished through the measurement of "Return Loss". By
measuring the system return loss at the line input (generator end) and
deducting twice the line attenuation, we get an indication of the load
return loss thesis value is easily converted to vswr.

Return Loss:
This is the dB value of absolute reflection coefficient.
It is rather curious concept of transmission engineering.
This loss value becomes 0 for 100% reflection and becomes infinite for
an ideal connection.

RL = 20log((VSWR+1) / (VSWR-1))


Voltage Standing Wave Ratio (VSWR):
This is the ratio of maximum voltage to minimum voltage
in standing wave pattern.
It varies from +1 to infinite.

VSWR = (1+(10^RL/20)) / ((10^RL/20)-1)

These are good and valid measurements which should be performed at
initial installation of the system and periodically verified throughout
the system life.

Return Loss/VSWR is only one of many measurements that should be
periodically done. DC measurements such as megger. LO-Ohms are also
very important.




--
Over The Hill
__________________________________________________ ___________________________

The question of whether computers can think is like the question of
whether submarines can swim.

***Edsgar Dijkstra***