Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.

Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.

If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.


Now try the correct "conjugate" equation (for complex
Zo):

Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1

Which is exactly what you should get for a short,
a full reflection with a phase shift of 180 degrees, but the ratio
can never be more than one for a passive network.




And consequently, the ratio of the reflected to incident
powers can also never be more than 1 for a passive network:

If you use ratios, it doesn't matter whether you use
peak or RMS voltages.

([Vpeak.incident/Vpeak.reflected])=
([Vrms.incident/Vrms.reflected])=sqrt(Pi/Pr)=[rho]

This is because the sqrt(2) is in the numerator and denominator.
And the 2 (after squaring) is also factored out in the Power RC!

And...the Zo is also factored out for Power RC!

The Zo is not needed for the Power RC, because the
impedance of the source is identical to the load for the
reflected power! Sure, you use the Zo in relation to Zl to
get rho, but once you get rho, you have the power RC.

Hah! the plot thickens a bit....


Slick