billcalley wrote:

snip

I've been reading the posts on this. One poster said this has been going
on for twenty years! (For the other groups, this thread has life on
rec.radio.amateur.antenna) It doesn't need to be so.

First, there should be no doubt that reflected power on a transmission
line is real. Sure, you can replace the line with a lump but that
doesn't clear up the question for others.

For the next two examples, see page 179:
http://cp.literature.agilent.com/lit...4753-97015.pdf
All examples assume the same impedance for source and line.

First example, step into an open line with a Thevenin source. The energy
is divided between the source and the line. Half the energy is moving
down the line and when it returns changes the impedance the source sees
to an open circuit. The energy does not flow back into the source, so,
where did it go? It is stored in the capacitance of the line.

Second example, step into a shorted line. When the energy returns the
source now sees a short. The energy does not flow back into the source,
so, where did it go? It is stored in the inductance of the line.

So here are two examples where the energy sent down the line do not
return to the source.

Third example. Send a pulse down the line. The Thevenin voltage source
will go to short, as it should, when the pulse falls. The pulse is
reflected from either an open or a short at the end of the line. All the
energy is dissipated in the source impedance when this pulse returns.
That is where the energy goes. And it is obviously the _same_ energy
created at the source.

Sure, non of the cases above represent steady state AC. But they do show
that energy may or may not be returned to the real component of the source.

With the above in mind, it can be shown, (in some part II), that a real
accounting of energy from source to load and back is possible.
Equivalent circuits are just that, the trading of line for lump. But,
and this is really important, the only reason the effective impedance at
the input of a 50 ohm line is not 50 ohms is because of reflected energy.

Best, Dan.

Dan Bloomquist wrote:

billcalley wrote:

snip

I've been reading the posts on this. One poster said this has been going
on for twenty years! (For the other groups, this thread has life on
rec.radio.amateur.antenna) It doesn't need to be so.

First, there should be no doubt that reflected power on a transmission
line is real. Sure, you can replace the line with a lump but that
doesn't clear up the question for others.

For the next two examples, see page 179:
http://cp.literature.agilent.com/lit...4753-97015.pdf
All examples assume the same impedance for source and line.

First example, step into an open line with a Thevenin source. The energy
is divided between the source and the line. Half the energy is moving
down the line and when it returns changes the impedance the source sees
to an open circuit. The energy does not flow back into the source, so,
where did it go? It is stored in the capacitance of the line.

Second example, step into a shorted line. When the energy returns the
source now sees a short. The energy does not flow back into the source,
so, where did it go? It is stored in the inductance of the line.

So here are two examples where the energy sent down the line do not
return to the source.

Wrong, wrong, wrong. Energy is not created or destroyed, but it can be
converted back and forth to mass. See mass defect and hydrogen fusion.


Third example. Send a pulse down the line. The Thevenin voltage source
will go to short, as it should, when the pulse falls. The pulse is
reflected from either an open or a short at the end of the line. All the
energy is dissipated in the source impedance when this pulse returns.
That is where the energy goes. And it is obviously the _same_ energy
created at the source.

Less wrong, line length makes no significant difference (unless it is really
lossy).


Sure, non of the cases above represent steady state AC. But they do show
that energy may or may not be returned to the real component of the
source.


Yes they do. Same model, same reflected power, same heating effect on the
last amplifier stage.

With the above in mind, it can be shown, (in some part II), that a real
accounting of energy from source to load and back is possible.
Equivalent circuits are just that, the trading of line for lump. But,
and this is really important, the only reason the effective impedance at
the input of a 50 ohm line is not 50 ohms is because of reflected energy.


Wrong, wrong, wrong again. Transmission line impedance is strictly a matter
of physical dimensions, and surrounding materials permitivity and
permeablity.

Best, Dan.

--
JosephKK
Gegen dummheit kampfen die Gotter Selbst, vergebens.Â*Â*
--Schiller