RadioBanter - View Single Post - Rho = (Zload-Zo*)/(Zload+Zo), for complex Zo

September 6th 03, 07:52 PM

In article ,
(Dr. Slick) wrote:

Hello,

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

(1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A
*transmission line surge impedance* of Z_0 = 50 + 200j is impossible;
surge impedances of transmission lines must have angles between - Pi/4
radians and + Pi/4 radians.

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.
**********

(2) Not equivalent in any reasonable sense. 50 and 50 + 200j aren't equal,
nor are - 200j and 0 equal.

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

(3) You forgot the factor of 50 in the denominator. The quantity you are
calculating above is approximately a magnitude of 8.062257748 at an angle
of about - 97.12501636 degrees. Of course this is silly for a value of rho
(but not as silly as 403.1 at an angle of - 97 degrees). However see my
comment (1) above.

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.

(4) I hope most readers believe the way to calculate rho when Z_L = 0 is
rho = (Z_L - Z_0)/(Z_L + Z_0) = (0 - Z_0)/(0 + Z_0) = - 1.

Hah! the plot thickens a bit....

Yes, it does.

David, ex-W8EZE

--
David or Jo Anne Ryeburn

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