On Sat, 2 Feb 2008 13:01:22 -0800 (PST)
Keith Dysart wrote:

On Jan 27, 10:57 am, Roger Sparks wrote:
On Sat, 26 Jan 2008 19:24:22 -0800 (PST)

Keith Dysart wrote:
On Jan 26, 12:15 pm, Roger Sparks wrote:
On Fri, 25 Jan 2008 19:13:31 -0800 (PST)

Keith Dysart wrote:
On Jan 24, 10:33 pm, Roger Sparks wrote:
[snip]
By examining this derivation, the reader can see that power and energy
is reflected when a wave encounters a discontinuity. The reader can
also see that more power is present on the transmission line than is
delivered to the load.

This is the conventional phraseology for describing the behaviour at
the impedance discontinuity.

Allow me to offer a specific example for which this phraseology is
inappropriate.

Consider a 50 V step function generator with an output impedance of
50 ohms driving a 50 ohm line that is 1 second long terminated in an
open circuit.

Turn on the generator. A 50 V step propagates down the line. The
generator is putting 50 J/s into the line. One second later it
reaches the open end and begins propagating backwards.
After two seconds it reaches the generator. The voltage at the
generator is now 100 V and no current is flowing from the
generator into the line. In the 2 seconds, the generator put
100 joules into the line which is now stored in the line.
The line is at a constant 100 V and the current is zero everywhere.

Computing Pf and Pr will yield 50 W forward and 50 W reflected.
And yet no current is flowing anywhere. The voltage on the line
is completely static.

And yet some will claim that 50 W is flowing forward and 50 W
is flowing backwards.

Does this seem like a reasonable claim for an open circuited
transmission line with constant voltage along its length and
no current anywhere?

I do not find it so.

...Keith

This is a reasonable observation for a static situation where energy is stored on a transmission line.

If the example contained an ongoing consideration, like "Where does the power move to?", then it would be reasonable to consider that the wave continued to move, simply to avoid the complication of what EXACTLY happens when a wave starts and stops.

So have you thought about "where does the power go?"

Yes, only the model I use substitutes a battery for the signal generator that you are using.

A battery is not the same since it has a very low output impedance.
A battery in series with a 50 ohm resistor would offer a reasonable
match to a 50 ohm transmission line.

The returning wave can recharge the battery, but how does the current stop? Or does it ever stop? From a practical aspect, the current must stop, but I can not explain how except by resistance that absorbs the circulating power.

That is the challenging part to understand when too much emphasis
is placed on the existance of energy being transported with the
"reflected power".

When I was learning this stuff, I did many examples with matched
generators (a battery with a 50 ohm resistor is a good example).
With step functions, it is easy to compute the final state because
you can just treat it as a DC circuit for analysis.

We must keep the limitations of our models in mind.

True, but over limiting is not good either.

When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), where does the
reflected power go?
Does it enter the generator?
Is it dissipated somewhere?

Answers to these questions will quickly lead to
doubts about the *reality* of "reflected power".

...Keith

The reflected wave that does not encounter an impedance at the generator must be on an infinitely long line, and therefore the conditions must have changed between the time of launch and return.

I don't follow the association between generator impedance and
length of line.

For a 50 ohm line, a matched generator has a 50 ohm output
impedance. The returning wave does not encounter an impedance
discontinuity so is not reflected. It disappears into the
the generator.

OK, I think we are saying the same thing. The return wave DISAPPEARS.

It makes more sense to think that the reflected voltage of (using your example) 50v meets 50v of forward voltage, and therefore finds an infinite resistance, coming to a stop.

The current does stop (since it can not flow into the
infinite resistance at the end of the line), and a 50 V step
propagates back along the line.

This 50 V reverse propagating step plus the 50 V already on
the line produces a total of 100 V on the line.

If the generator was constructed as a 100 V battery in series
with a 50 ohm resistor, then when the step arrives back at
the generator, there is 100 V on both sides of the source
resistor (battery on one side, line on the other) and the
current through the source resistor becomes zero. No further
current flows into the line.

If that happens, doesn't the same condition occur as soon as the reflected wave is first generated at the line end? The current really stops as soon as the end is reached, with the energy contained in the magnetic field converted to electric field energy visible as voltage. If this happened, the reflected wave could be better described as an electric "jump", similar to a hydraulic jump found in open channel flow of liquids, where kinetic energy is converted to potential energy.

There do seem to be some similarities, though there is likely
trouble if the analogy is carried to far.

This idea of an electric "jump" requires not a reflection occuring without a discontinuity, but a moving wave front that absorbs the traveling wave, bringing it to a stop (in this case).

The current has stoppped, but has the "wave"?

Vf, Vr, If, Ir, Pf, Pr are still computable using the normal
formulae.

...Keith

Here is a link to a web site discussing reflections and more.

http://www.ivorcatt.co.uk/em.htm

He has some drawings that pertain to this discussion. He shows reflections and crossings, but no one-way transmission lines.
--
73, Roger, W7WKB