"Roy Lewallen" wrote

Yep, you don't have to mess with those pesky reflections when the
line
is terminated in a load equal to its characteristic impedance. Then
the
problem reduces to the trivial one you've posed. Simple problem,
simple
solution.
=======================================

As I say, I've lost the original question.

But if a steady-state answer is needed for the original question then
the answer can be found by the same simple means without the
unnecessary complication of reflections.

It's the Devil which makes Cecil pose such loaded questions.
----
Reg.

Reg Edwards wrote:
If you ask silly questions you can expect silly answers.

Are you looking for a reason to introduce your irrelevant reflections
again.

No, just trying to correlate your example to mine. In
your example, how long does it take from power up for
the load to accept 100 joules/second?
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:
Yep, you don't have to mess with those pesky reflections when the line
is terminated in a load equal to its characteristic impedance.

Which begs the question of what is to be done when "those
pesky reflections" exist in a mismatched system.
--
73, Cecil http://www.qsl.net/w5dxp

Sigh. Yes, I believe you HAVE made a mistake in your calculations,
right at the beginning. C = tau/Z0 = 1/100 farad; L = tau*Z0 = 100
henries. But of course, the concept is perfect. 100V, 1A, 100W,
E^2*C/2 = 50 joules stored in the capacitance. I^2*L/2 = 50 joules
stored in the inductance. The two are equal only for a termination
equal to the line Z0 of course.

Too bad Cec doesn't understand that lossless line has unvarying Z0 as a
function of frequency from DC up, as long as TEM mode is supported.

Cheers,
Tom

Reg Edwards wrote:
I've lost the original question.

But there is a lossless transmission line 1 second long.
With a velocity factor of 1.0 it is 30,000 kilometres long.

Let Zo = 100 ohms.
Let applied volts = 100 VDC.
Let terminating resistance be 100 ohms.

Line Inductance = 10,000 Henrys.
Line Capacitance = 1 Farad.

Under steady state conditions -
Line current = 1 amp.
Energy stored in inductance = Sqr(I)*L/2 = 5000 Joules.

Volts across capacitance = 100 volts.
Energy stored in capacitance = SqrV)*C/2 = 5000 Joules.

Total energy stored = 10, 000 Joules = 10,000 watt.seconds.

Which has nothing whatsoever to do with all this nonsense about
reflections. Note that energy in inductance equals energy in
capacitance.

(I trust I have not made a mistake with the arithmetic.)
----
Reg.

"Cecil Moore" wrote in message
. net...
"Buck" wrote:
I can't tell you how many joules are in my coax, but, if you ask,
I
can tell you how many jewels are in my logbook.

This is not aimed at people who don't care about such things.
This is aimed at people who do care about such things especially
those who are disseminating false technical information about
such things.

Some say that since there's 100 watts in and 100 watts out
during steady-state, there's nothing left over for the reflected
waves. They apparently forgot about the time immediately
following power up when there was 100 watts in and less
than 100 watts out.
--
73, Cecil, W5DXP

Reg Edwards wrote:
It's the Devil which makes Cecil pose such loaded questions.

It's only loaded with the conservation of energy
principle, Reg. Here's a simple example. The SG-CR
is a 50 ohm signal generator equipped with a
circulator resistor to dissipate all reflected
power incident upon the source.

200W SG-CR---one second long lossless 50 ohm line---291.42 ohm

SWR is 5.83:1 rho^2 = 0.5 Steady-state forward power
is 200 watts. Steady-state reflected power is 100 watts.

The question is: After steady-state has been reached, how
many joules of energy have been generated but have not been
dissipated in the load resistor or circulator resistor?
--
73, Cecil http://www.qsl.net/w5dxp

K7ITM wrote:

Sigh. Yes, I believe you HAVE made a mistake in your calculations,
right at the beginning. C = tau/Z0 = 1/100 farad; L = tau*Z0 = 100
henries. But of course, the concept is perfect. 100V, 1A, 100W,
E^2*C/2 = 50 joules stored in the capacitance. I^2*L/2 = 50 joules
stored in the inductance. The two are equal only for a termination
equal to the line Z0 of course.

Too bad Cec doesn't understand that lossless line has unvarying Z0 as a
function of frequency from DC up, as long as TEM mode is supported.

I do understand that. That's why Reg's 10,000 joules don't make
any sense to me. Now if the Z0 was one ohm at DC, it would make
sense for 100 volts to be able to supply 10,000 joules in one
second.
--
73, Cecil http://www.qsl.net/w5dxp

Reg Edwards wrote:
I've lost the original question.
....

Actually, it was "200 watts forward, 100 watts reflected" -- but it
works the same way: 100 ohm line, 1 second long: source is DC
200*sqrt(2) volts in series with 100 ohms. Load is such that it
absorbs 100 watts from that source: 100*(3 +/- sqrt(8)) ohms. About
17.16 ohms works OK. Then the line voltage is 41.42V and the energy
stored in the 1/100F line capacitance is 8.579 joules, and the line
current is 2.414A and the energy stored in the 100H line inductance is
291.421 joules. Total energy stored is 300 joules.

It's somewhat comforting that it's just as you'd expect from power
delivered versus time. It's an illustration of something I believe Ian
White was trying to get across in another thread: there are commonly
multiple ways to analyze a problem, and if they agree, you MAY have the
right answer. If they disagree, you have at least as many wrong
answers as the number of disagreements (or perhaps the disagreements
only seem to be disagreements).

Now, I have no idea who it is that might be saying that there's no
energy stored in the fields in a line, so I really don't know what the
to-do is all about. Maybe it's just another misinterpretation of what
people have actually said...perhaps about lines that are very far
indeed from being lossless.

Cheers,
Tom

"K7ITM" wrote in message
ups.com...
Sigh. Yes, I believe you HAVE made a mistake in your calculations,
right at the beginning. C = tau/Z0 = 1/100 farad; L = tau*Z0 = 100
henries. But of course, the concept is perfect. 100V, 1A, 100W,
E^2*C/2 = 50 joules stored in the capacitance. I^2*L/2 = 50 joules
stored in the inductance. The two are equal only for a termination
equal to the line Z0 of course.

=================================
Tom,

Furthermore, the energy dissipated each second in the 100-ohm
termination (100 watts) is equal to the energy stored in the line.

Damned decimal points. I just KNEW I had made a mistake somewhere
along the line. Thank you.

By the way - what is tau ?
----
Reg.

K7ITM wrote:
Actually, it was "200 watts forward, 100 watts reflected" -- but it
works the same way: 100 ohm line, 1 second long: source is DC
200*sqrt(2) volts in series with 100 ohms. Load is such that it
absorbs 100 watts from that source: 100*(3 +/- sqrt(8)) ohms. About
17.16 ohms works OK. Then the line voltage is 41.42V and the energy
stored in the 1/100F line capacitance is 8.579 joules, and the line
current is 2.414A and the energy stored in the 100H line inductance is
291.421 joules. Total energy stored is 300 joules.

As can be seen from http://www.qsl.net/w5dxp/1secsgcr.gif

Do you think it's just a coincidence that 200 watts of forward
power in a one second long line would require 200 joules and
100 watts of reflected power in a one second long line would
require 100 joules for a total of 300 joules?

Now, I have no idea who it is that might be saying that there's no
energy stored in the fields in a line, so I really don't know what the
to-do is all about.

Please read what W7EL says about forward and reflected power
in his "food for thought" writings. In particular: "Some inventive
people have supposed they can separate forward and reverse power with
a circulator. That really sounds attractive, particularly with an open
or short circuited load. In that condition, the forward and reverse
powers are each 100 watts, yet the transmitter (or the transmitter
voltage source) doesn't have to produce any power at all."

What W7EL doesn't seem to realize is that the source produced the
power in the forward and reverse power waves while the feedline
was charging up during the transient period before the reverse power
waves reached the source so there is bona fide energy in those waves.
He also says there is no model for handling such yet one exists in
detail in the magazine article on my web page.
--
73, Cecil http://www.qsl.net/w5dxp

Reg Edwards wrote:
Damned decimal points. I just KNEW I had made a mistake somewhere
along the line.

Now do you see why my question wasn't simple or loaded?
I just wondered how you shoved 10,000 joules down a
Z0=100 ohm feedline in one second using a fixed 100
volt source.
--
73, Cecil http://www.qsl.net/w5dxp