On Jun 15, 8:22*am, "J.B. Wood" wrote:
On 06/12/2011 03:05 PM, walt wrote:



Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient Vñ equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or
57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.

Walt, *I can only get your 1.0 - j1.1547 normalized impedance value
(looking into the line towards the normalized load of 3.0) if I move 60
degrees, not 30, from the load towards the generator. *I'm also assuming
a lossless line. *Sincerely, and 73s from N4GGO,

--
J. B. Wood * * * * * * * * *e-mail:

Hello JB, thank you for the response. I'm sure I understand what's
going on. When you view the Smith Chart at the unity resistance circle
with a 3:1 mismatch the normalized impedance there is 1.0 - j1.1547,
as you stated. However, a radius through that point yields a voltage
reflection coefficient rho = 0.5 @ -60°. In other words, the radial
line intersects the periphery of the Smith Chart at -60°.

As I'm sure you know, reflection degrees equals two electrical
degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs
at 30° rearward of the load. That OK?

In the absence of a Smith Chart here's an easy way to determine the
reactance appearing at the unity resistance circle for any given
degree of mismatch:

1/(sqrt SWR/
SWR-1)

Walt, W2DU