Not QED at all. You claimed to have proved that maximum power is
delivered to a load when a transmission line is terminated such that the
reflected voltage on the line is zero. Here, you're agreeing that the
reflected voltage is zero when the line is terminated in its
characteristic impedance. So where's the proof that this condition leads
to maximum power to the load?

Roy Lewallen, W7EL

Peter O. Brackett wrote:
Roy:

[snip]

Oops. No it's not. Consider circuit "A", a voltage source in series with
a lumped impedance equal to r + jx. And circuit "B", a voltage source in
series with a transmission line of length, say, one wavelength, with
characteristic impedance Z0 = r + jx. The two are not at all equivalent.
To show that they're not, connect a 50 ohm resistor to the output of
each. The resistor current in circuit "A" is Vi/(50 + r + jx). The
current in circuit "B" is Vi/50.

[snip]

Y'all are missing my point.

First off... you, and Dave have just changed situation I proposed. I did
not say
that there was an unterminated line of one wavelength! You and Dave said
that. Forget it. That has nothing to do with the discussion/proof!

Read my "typing"! I said you must have *either* a semi-infinite line, whose
driving point impedance is known by all to be exactly Zo, *or*, but it is
simply beside the point, if you must insist on using a finite length
line then that is OK as well as long as it is terminated at it's far end
in an impedance equal to Zo.

In either case of the semi-infinite Zo line or the finite line terminated in
Zo the
driving point impedance is identical, i.e. it is Zo and you can't tell the
difference
between the two at the driving point.

[snip]

David Robbins recently made several postings explaining that these two
circuits are not equivalent. Hopefully the above example illustrates that.

[snip]

I disagree with Dave. A semi-infinite Zo line and a finite Zo line
terminated
in Zo are indistinguisable at the driving point and at all points along
the lines, until you come to the end of the finite line, but even there the
voltage and current across then terminating Zo is identical to the voltage
and current at the same point on the semi-infinite line.

All of this discussion about there being a difference is moot. There is no
difference. That's exactly why I set it up that way, simply because there
is no difference. You guys are missing the point.

Check out any book on transmission lines...

When you look into the end of a transmission line of semi-infinite extent
you
see a driving point impedance of Zo. That is the very definition of Zo!

If you cut the line off to a finite length and terminate it in it's
characteristic
impedance Zo the driving point remains the same, i.e. Zo. That again is
part of the defining charateristic of characteristic or surge impedance!

Take a chunk of 50 Ohm lossless coax and terminate it in 50 Ohms. What's
the impedance you see looking into the end? 50 Ohms. Now change the length
to something else and again terminate it in 50 Ohms, what 's the impedance?
50 Ohms, it doesn't matter how long the line is if it is terminated in it's
characteristic impedance it's driving point impedance is Zo! End of story.

You simply can't tell the difference between a line terminated in it's
characteristic
impedance and the characteristic impedance of an semi-infinite line as
found by
observiations at the driving point, they are one and the same. Zo!

[snip]

Can't you do your "proof" with, say, a one wavelength transmission line
of characteristic impedance Z0?

Roy Lewallen, W7EL

[snip]

Yep, sure can... sigh, why is this soooo hard.

Use a one wavelength line if you insist, but terminate it at the far end
with an
impedance equal to Zo!,

The length of the line has nothing to do with the proof.

As a matter of fact that is one of the neat things about the so-called
"proof" I cooked up.

The "proof" is just about whether there are reflections at a conjugate
match or not.

The proof, mathematical or experimental, just proves that there are no
reflections [as defined
by the classical rho] when the generator sees it's image Zo, but there are
reflections when the
generator sees it's conjugate! And the same in the reverse direction. i.e.
there are no reflections
when the line sees it's image Zo but there are reflections when it sees it's
conjugate.

And if Zo is a real resistance say Zo = R, then both situations are
identical.

That's all, nothing complicated, simple straightforward stuff, when Zo is
complex, there *will* always
be reflections at a conjugate match.

This means that when there is a conjugate match on a complex line a
"classical" reflectometer
will not indiacate zero reflected voltage. And if the scale on the
reflectometer is actually calibrated
in "Watts" even though it measures Volts, it will indicate the presence of
[[a "false"] reflected power
at a conjugate match.

But then everybody already knew that!

QED!

--
Peter K1PO
Indialantic By-the-Sea, FL.