On Sat, 04 Dec 2004 21:51:09 GMT, Richard Clark
wrote:

On Sat, 04 Dec 2004 18:40:20 GMT, Richard Clark
wrote:

At 30 MHz, this 1.5pF capacitance represents a reactance of 3.5K Ohm.
This rather sweeps aside your specification for the load and replaces
it with 4700 -j3537 Ohms.

Actually 1700 -j2258 Ohms

As you can see, this is the reason why open stubs are avoided.
"Knowing" the parasitic capacitance is problematic. The fringing
effect at the end is difficult to manage whereas a short is much
simpler to define and implement.

73's
Richard Clark, KB7QHC

Dear Richard,
Each and every one of your concerns is deeply appreciated. I had
seriously about measuring that 4700 ohm resistor by itself for just
the reasons that you mention. The reason that I did not is quite
simple. The Resistance dial on the General Radio Model 1606B RF Bridge
only goes to 1000 ohms. C'est la Vie!

I could have chosen a much lower resistance to start with, but then I
would not have had the desired high SWR, which is critical to the
experiment. You might well ask why I made the switch from a 1/4 wave
open ended stub (at 10.6 Mhz) to the 30 MHz test using the same piece
of line. That was to get a high SWR and still have reasonable
confidence in the value without actually measuring it. Perhaps I
should use a capacitor instead. That's still a possibility. I am
thinking of doing it over with a measured 1 k resistor, and just
accepting the SWR of only 20:1. I don't think I have any low ohm
resistors that would give me something around 1 to 3 ohms and still be
low inductance - not from my parts box. Actually, I doubt seriously
that the capacitance effects are as bad as you are suggesting, but I
won't argue that without some additional measurements.

My reason for not going with a short circuit was even simpler. I
wanted a concrete value for a non-zero load power.

Regarding the artificial adjustment of the attenuation loss from
around 0.9 up to about 1.72 did not bother me at the time that I did
it, but I can see why it bothers an independant observer. I will have
to do something about that, and perhaps doing it over with a 1 K
resistor is the answer to all of that - so, here goes, and I will get
back to you in a few hours.

BTW - this is becoming more and more academic, since I think the real
mystery has been resolved, but there's nothing like getting it wrapped
up with a proper ribbon.

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

On Sun, 05 Dec 2004 02:53:35 GMT, (Robert Lay
W9DMK) wrote:

Each and every one of your concerns is deeply appreciated. I had
seriously about measuring that 4700 ohm resistor by itself for just
the reasons that you mention. The reason that I did not is quite
simple. The Resistance dial on the General Radio Model 1606B RF Bridge
only goes to 1000 ohms. C'est la Vie!

Say la-la!

Hi Bob,

Good, you have a great piece of equipment, but the implicit question
remains, what do you use to measure the voltage? That is going to be
a serious load across even a 4700 +j0 resistor. Even X10 Scope probes
would be a poor method.

I could have chosen a much lower resistance to start with, but then I
would not have had the desired high SWR, which is critical to the
experiment. You might well ask why I made the switch from a 1/4 wave
open ended stub (at 10.6 Mhz) to the 30 MHz test using the same piece
of line. That was to get a high SWR and still have reasonable
confidence in the value without actually measuring it. Perhaps I
should use a capacitor instead. That's still a possibility.

Hardly worth the thought actually. The 1.5pF stray capacitance would
be a coup-de-grace for the resistor, do you think you could find less
capacitance (with or without the resistor) when there is probably
already that much in fringe capacitance?

I am
thinking of doing it over with a measured 1 k resistor, and just
accepting the SWR of only 20:1. I don't think I have any low ohm
resistors that would give me something around 1 to 3 ohms and still be
low inductance - not from my parts box. Actually, I doubt seriously
that the capacitance effects are as bad as you are suggesting,

They are, life is cruel.

but I
won't argue that without some additional measurements.

It's better than sitting on the park bench with a box of chocolates
and spouting the philosophy of optical interference. For that you get
my Kudos.

My reason for not going with a short circuit was even simpler. I
wanted a concrete value for a non-zero load power.

Regarding the artificial adjustment of the attenuation loss from
around 0.9 up to about 1.72 did not bother me at the time that I did
it, but I can see why it bothers an independant observer. I will have
to do something about that, and perhaps doing it over with a 1 K
resistor is the answer to all of that - so, here goes, and I will get
back to you in a few hours.

Well, one way is to take your measurements and stick with the original
attenuation loss and transform from the input to this 4700 Ohm load.
Does the data suggest this load is closer to my hip shot guess? You
have a reactance built into your results that is unaccounted for, but
comes with this parasitic inclusion I offer.

BTW - this is becoming more and more academic, since I think the real
mystery has been resolved, but there's nothing like getting it wrapped
up with a proper ribbon.

It's all academic, Bob. When you have to take more than one exam
(with questions that go beyond parallel parking) to pursue a hobby
it's part of the turf. Owning a GR-1606 caps it.

73's
Richard Clark, KB7QHC

Cecil Moore wrote:
Reg Edwards wrote:

Instead of messing about calculating the additional loss due to SWR
and then
adding it to the matched loss, I've just had a wonderful idea.

Why not calculate the actual line loss directly and solve all your
problems
at one fell swoop.

What is the formula for the total dB loss?

I don't know of one, but here's a procedure which will get you the answer:

1) Determine the transmission line characteristics R, G, L & C at the
frequency of interest.

2) Determine the transmission line termination impedance (or admittance)
at the frequency of interest.

3) For convenience, define the excitation at the input to the
transmission line. e.g. defined voltage, current, or generator with
defined source impedance.

4) Solve the transmission line equation for both the voltage and
current, as a function of time and position, subject to the imposed
boundary conditions. Use any method you wish, though the usual approach
is the so called method of reflections where the complete solution is
taken as the summation of damped forward and reverse traveling waves.
Calculation of Zo and Gamma require knowledge of R,G,L & C.

The solution can be expressed algebraically in terms of hyperbolic
functions and will be found in any good transmission line reference.

Now known are v(x,t) and i(x,t).

5) Separately calculate the dissipative losses due to series resistance
and shunt conductance (or dielectric power factor) as follows:

a) The loss due to series resistance at any instant in time for some
section of transmission line, is i(x,t)^2 * R * dx, integrated over the
subject transmission line section.

b) The loss due to shunt conductance at any instant in time for some
section of transmission line, is v(x,t)^2 * G * dx, integrated over the
subject transmission line section.

c) The total dissipative power loss is the the sum of the time averaged
series and shunt loss components, integrated over the entire
transmission line length.

Consider:

The SWR (reflection coefficient at the termination) alone is not
sufficient to determine the loss in general. This is an important
concept to understand.

For illustration, assume the limiting case where the shunt conductance
can be disregarded. This is a useful approximation for many
transmission line installations used at HF where the SWR is reasonably
low (50). In such practical cases the losses will be dominated by
series resistance.

Consider a short (45 electrical degrees) transmission line terminated
in a pure resistance which is much less than Ro. The standing wave
pattern will be such that the current will be maximum, and the voltage
minimum, at the termination. The losses can be calculated using the
procedure given above.

Now consider the same transmission line terminated in a pure resistance
larger than Ro such that the SWR is the same as previously. The current
will now be minimum at the termination and the total dissipative loss
will be considerably less than when the termination resistance is low,
even though the SWR is identical.

Transmission line losses depend on both the actual load impedance and
the transmission line parameters. Knowing matched attenuation
(dB/length) and load SWR is not sufficient to calculate losses exactly.
Any formulas using only SWR and matched transmission line loss are
approximations and must be used carefully!

Feeding electrically short antennas with 50 ohm coax as is commonly
attempted by amateurs on 80 & 160m can result in transmission line
losses much larger than predicted using the simple approximations given
in the amateur literature.

bart
wb6hqk

You are all floundering about like fish out of water.

The ONLY way to predict the performance of a small resistor at UHF is to
treat it as a transmission line - starting with the simple dimensions of
length and diameter.

L, C and R, and short-circuit input resistance versus frequency
automatically drop into your laps with adequate ball-park accuracy.

As so-called engineers, if you can't measure it, such obvious arithmetical
techniques should already have been included in your elementary analytical
educations.

Spice (of which I have heard about only by reputation on newsgroups) is
useless unless one already knows what it's all about beforehand.

My criticism is not personal - only of the western world's educational
systems and insistent dependence on weapons of mass-destruction.
----
Reg.

Cecil Moore wrote:
Reg Edwards wrote:

Cecil, when there are several different power levels at different
places in
a circuit, it is entirely up to you how you reference one to another
in dB.

Now Reg, that just cannot be true.

It is true.

Otherwise, there would exist no
conventions.

Multiple conventions exist in all the technical disciplines. A well
versed practioneer is familiar with the various conventions commonly in
use and will choose the one most appropriate to the problem at hand.
General terms such as 'gain' and 'loss' always require the context be
carefully established before the term can be used with any precision.
For example, dissipative loss in a transmission line is different from
transducer loss, but both are often expressed in decibels. One relates
to an actual ratio of two physically meaningful power levels, and the
other is notional.

It's a simple question: When you tell me that the losses
in a transmission line with reflections are 2 dB, exactly what power
are you referencing those losses against?

Reasonable question. Without knowing the context the statement is
essentially meaningless.

Reflected volts - yes!
Reflected current - yes!
Reflected power - NO!

Reflected volts and reflected current existing without any associated
joules/sec????

Reg didn't say any such thing. The wave equation is usually expressed
in terms related to the two complementary field quantities associated
with energy storage. In the case of TEM transmission lines, the two
variables are voltage, and current. The method of reflections (images)
used to facilitate solution of the wave equation, utilizes the concept
of voltage and current reflection coefficient respectively. These
concepts are used to find the v(x,t) and i(x,t) without any reference to
to power or energy. Once v(x,t) and i(x,t) are known, the power, v(x,t)
* i(x,t) can easily be calculated at any point x and time t. The
energy storage density can be calculated as i(x,t)^2 * L/2 + v(x,t)^2 *
C/2. As you might expect, the v(x,t)*i(x,t) is the derivative of the
energy storage density.

I've heard of waves without any trace of energy before,
Reg, but I certainly didn't expect to hear miracle metaphysics from you.

'Solving' the wave equation for TEM transmission lines usually means
determining the values of the v(x,t) and i(x,t) as a function of place
and time. Energy distribution, and it's movement, is easily calculated
from the knowledge of v(x,t) and i(x,t).

Whatever happened to V*I*cos(theta) being power?

Hasn't changed; works for DC out to upper microwave frequencies.

The power companies
would be surprised to learn that they are not transferring any joules/sec
to their customers.

Indeed they would. Who do you know who believes they are not
transferring energy to their customers?

bart
wb6hqk

wrote:

"Reg Edwards" wrote in message
...

"Cecil Moore" wrote

Reg Edwards wrote:

SNIP
I wonder why I ever bothered to introduce Chipman to this newsgroup.

Try him.

----
Reg.



Reg,
This thread is purely a platform for snide remarks or for the pursuit of
appearing clever
to readers. It is not for technical education but instead it is a duking out
of smarmy comments so as to produce a suedo pecking order for onlookers
to assuage who has the most accumen with respect antenna education.

Art, I think most of us "onlookers" are just chuckling, at least those
of us who are left reading the thread.

Have you ever seen the videos of the American Prairie Chicken during
its mating ritual? That is the mental picture I keep geting from this
kind of stuff.

- Mike KB3EIA -

On Sun, 5 Dec 2004 17:30:23 +0000 (UTC), "Reg Edwards"
wrote:
You are all floundering about like fish out of water.
....
My criticism is not personal

Of course not Reggie,

But neither is it original, and if you had been following the
technical correspondence instead of the fluff offerings, you would
have observed neither of us is in danger of falling prey to all the
ills you offer cautions against. Preach to those who need redemption
and confine your comments along our technical interchanges to coin in
kind.

73's
Richard Clark, KB7QHC

Bart Rowlett wrote:
A well
versed practioneer is familiar with the various conventions commonly in
use and will choose the one most appropriate to the problem at hand.

My point exactly, Bart. It is not up to me to define a new standard.

Indeed they would. Who do you know who believes they are not
transferring energy to their customers?

There are people who believe that energy being transferred past
a point in the transmission line is not power. However, the IEEE
dictionary says that power is the rate of generation, transfer,
or consumption of energy.
--
73, Cecil http://www.qsl.net/w5dxp

Bart, I don't doubt that your description of how to calculate line loss is
correct.

But it's more simple than that.

The input and output currents (or voltages) of any line are very rigid and
well defined. Regardless of generator internal impedance. Just as a
transformer's input and output currents and voltages are defined.

Their ratio is even more simply defined, albeit in complex terms.

So when the line input RESISTANCE, regardless of input reactance, and the
terminating RESISTANCE, regardless of termiating reactance, are known then
the power input to power output ratio is easily directly calculated from I
squared R.

The hardest but still straightforward part of the calculation is for the
input impedance, Rin+jXin, for given line attenuation and phase shift, and
for given terminating impedance, Rt+jXt.

The power dissipated in the line, although distributed in a highly complex
manner along its length due to standing waves, is very simply Pin - Pout.

Throughout the calculations, things like reflection coefficients and SWR do
not appear, You cant use them as spin-off benefits. What practical use could
you make of them anyway? The calculations have already been done, finished,
ended, caput!
----
Reg.

Preach to those who need redemption
and confine your comments along our technical interchanges to coin in
kind.

73's
Richard Clark, KB7QHC

==============================
I am preaching to anybody who needs redemption.

It is YOU who are taking it personally despite me explicitly letting you off
the hook.

But Rich, it's very unlike you to take cover in "off topic". ;o)
---
Reg.