On 5/16/2011 6:54 AM, Cecil Moore wrote:
On May 15, 7:54 pm, John wrote:
100W--50ohm--+--1/4WL 100ohm--200 ohm load

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

I said: "All that matters is that the source is delivering 70.707
volts to the 50 ohm feedline." That would be at the source SO-239
terminal to PL-259 50 ohm coax junction. I did *NOT* say there is
70.7v on the Z0=100 ohm feedline. There is indeed 100 volts on the
Z0=100 ohm line.

The source is Z0 matched to 50 ohms and is
*delivering 70.707 volts and 1.414 amps to the 50 ohm coax*.
70.7/1.414=50 ohms. 70.7*1.414=100 watts.

Here is a better diagram:

http://www.w5dxp.com/enfig1.gif

Pfor1=100w, Pref1=0, Z01=50 ohms, Z02=100 ohms and 1/4WL, Load=200
ohms. Everything except the load is lossless.

I apologize if my one line diagram was confusing. It is extremely
difficult to draw ASCII schematics using a proportional font.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Okay, so your source really looks like a a 100 W generator with no
internal impedance but which feeds a 50 ohm line internally. Since the
200 ohm load appears as 50 ohms when fed with a 100 ohm 1/4WL line,
there is a match and all the power appears at the load. In that case I
agree with your 70.7 volts since 100 W into 50 ohms will result in that
voltage.

But I have no need of reflections. The fact that a 200 ohm load appears
as a 50 ohm load at the input of a 1/4WL 100 ohm line takes care of all
that for me.

73,
John