"Henry Kolesnik" wrote in message
. ..
The previous thread got too big and convoluted for me so I started another
if for no other reason than my own clarity and convenience. It's still the
same question "I know that any power not dissipated by an antenna is
reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open or
a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt".

Before going to bed I got out the book REFLECTIONS II by Walt Maxwell,
W2DU. I'm typing verbatim from page 2-2 and 23-1 for those who don't
have
the book.

Page 2-2 : "Contrary to what many believe, it is not true that when a
transmitter delivers power into a line with reflections, a returning wave
sees an internal generator resistance as a dissipative load. Nor is the
reflected wave converted to heat and, while at the same time damaging the
final amplifier....the reflected power is entirely conserved...."

From page 23-1 "One of the most serious misconceptions concerned
reflected
power reaching the tubes in the RF amplifier of the transmitter. The
prevalent, but erroneous thinking was that the reflected power enters the
amplifier, causing tube overheating and destruction. However, I dispelled
this misconception in the above mentioned publications, using
wave-mechanics
treatment, discussed here in greater detail, by showing that when the
pi-network tank is tuned to resonance, a virtual short circuit to rearward
traveling waves is created at the input of the network. Consequently,
instead of the reflected power reaching the tubes of the amplifier, it is
totally re-reflected toward the load by the virtual short circuit
appearing
only to waves at the network input".

I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR



Henry,

I had posted the following on the other thread, about the time you started
the new one.

Tam
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line the reflected wave will be the same amplitude as the generator, there
will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know" what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.