Jim Kelley wrote:
You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of the
discussion that you've been pretty closed minded about.

Here's a brainteaser for you, Jim. Assume the following example
under steady-state conditions:

XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms
Ps=100W Pfor=200W-- --100W=Pref PL=100W

The lossless tuner is tuned for a Z0-match so there is zero
reflected power at point 'X'. The Z0 of the feedline is 50
ohms. The voltage reflection coefficient at the load is
0.707 making the power reflection coefficient 0.5

We suddenly disconnect the source at point 'Y'. After the
source is disconnected, how many joules of energy are delivered
to the load? _____

How many seconds does it take to dissipate all the energy in
the feedline? _____

On a second by second basis, how much power is delivered to the
load after the disconnect? _____ _____ _____ _____ _____
--
73, Cecil, http://www.qsl.net/w5dxp

Cecil Moore wrote:
Jim Kelley wrote:

You should think about it because amateur radios don't usually have
circulators on their output, and the result illuminates a corner of
the discussion that you've been pretty closed minded about.


Here's a brainteaser for you, Jim. Assume the following example
under steady-state conditions:



XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms
Ps=100W Pfor=200W-- --100W=Pref PL=100W

If you aren't going to solve the problem I posed, then I don't see why
I should feel obliged to bother with any more of yours. Only seems
fair. There is a significant point to my question that you still
have not addressed. Would you have us believe that some new
revelation is being presented by this latest problem of yours? Or
perhaps it's simply intended to divert attention away, once again,
from my question.

Or maybe it was just too hard for you. How about a simpler problem then:

XMTR--X--one second long lossless feedline---infinite ohms
PS=100W

How many Joules are stored in the transmission line? Why?

73, ac6xg

On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote:

How many Joules are stored in the transmission line?

100

Why?

Spillage. The cup is two halves full (or no halves empty for the
pessimists in the crowd)

Richard Clark wrote:

On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote:


How many Joules are stored in the transmission line?


100

That may be the correct solution to a somewhat different problem.

Why?


Spillage. The cup is two halves full (or no halves empty for the
pessimists in the crowd)

Pessimists would report that "there are a disasterous and
insurmountable number of full halves", when in fact there are just two.

73, ac6xg

On Fri, 01 Sep 2006 12:49:43 -0700, Jim Kelley
wrote:

Richard Clark wrote:

On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote:


How many Joules are stored in the transmission line?


100

That may be the correct solution to a somewhat different problem.


Hi Jim,

So, with a wrong answer (and a direct answer at that, imagine!). Do I
warrant the "correct" answer?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
So, with a wrong answer (and a direct answer at that, imagine!). Do I
warrant the "correct" answer?

With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer. He didn't even
tell you what the forward and reflected power readings
were. He also doesn't seem to realize that a 100W XMTR
cannot force any power into an infinite load. I am trying
to apply some boundary conditions that will remedy that
problem.

So Richard, what do you think happens when a constant
power output XMTR is facing an infinite load? What
happens when an irresistible force meets an immovable
object?
--
73, Cecil http://www.qsl.net/w5dxp

On Fri, 01 Sep 2006 22:13:32 GMT, Cecil Moore
wrote:
With Jim's lackadaisical approach to conservation of
energy, there is no "correct" answer.

A long winded answer for you don't know. Talk about lackadaisical
with all the fluff cut off the end of this too.

Jim Kelley wrote:
Or maybe it was just too hard for you.

It was impossible for anyone to solve without having
a math model for the sources.

How about a simpler problem then:

This is not simpler - it is contradictory. It provides a
constant power source with nowhere for the source power
to go during steady-state. If you change the source to
a Thevenin equivalent then it can be solved.

XMTR--X--one second long lossless feedline---infinite ohms
PS=100W

Please explain how the XMTR can supply 100W during steady-
state? Where does the 100 joules/sec go? As stated, this
problem, like your other one, is impossible to resolve.

If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable. The answer is
that 200 joules exist in the feedline and, contrary to what
W7EL states, will dissipate in the source resistance after
the source voltage goes to zero. Note the source resistance
is the only resistance in the entire circuit.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
If you change the source to a Thevenin equivalent 141.42V
and 50 ohm source impedance and specify Z=50 ohms feedline
impedance then the problem becomes solvable.

Going with this configuration which is similar to the 1/2
WL stub suggested by W7EL in his Food For Thought #1:

SW
Source nc c Pfor=100W-- --Pref=100W
141.4V---o---o---one second long lossless 50 ohm stub--open
50 ohm
o---/\/\/\/\/\--Gnd
no 50 ohms

For the first two seconds after power up, the Thevenin
source delivers 200 joules into the stub while dissipating
200 joules as heat in the source resistance. Those 200
joules of energy in the stub must be conserved.

During steady-state, assuming the stub is an exact integral
number of wavelengths, the source will see an open circuit
and be delivering zero power. A wattmeter calibrated for
50 ohms will read 100 watts forward power and 100 watts
reflected power during steady-state.

After steady-state is reached, we throw the switch and connect
the stub to a 50 ohm dummy load. 100 watts will be supplied
to the dummy load for two seconds for a total of 200 joules.

Who was it who said that reflected power cannot be
recovered? (Food For Thought #1 on www.eznec.com).
--
73, Cecil http://www.qsl.net/w5dxp