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

September 7th 03, 12:32 AM

(David or Jo Anne Ryeburn) wrote in message .. .
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.

Ok, a source impedance then.

I don't fully understand why your last statement needs to be so.

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.

I understand your point, but the reactances WILL cancel. And if
you are feeding from a lossless 50 ohm transmission line, the circuit
won't know the difference.

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.

My mistake. Wrote too quickly. A gain of about 8 is STILL insane
for a passive network!

(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.

rho = (Z_L - Z_0*)/(Z_L + Z_0)

I agree with you. But the incident voltage in this case will be
coming
out of a series inductor of +j200 reactance at the test frequency.

It will be charging up a capacitor, but the reflected voltage will
not be
8 times the incident.

Again, the reactances will cancel at the series resonance, so in
effect, if you are feeding a lossless 50 ohm tranmission line, you
will not be able to tell the difference. It will appear exactly like
a 50 ohm line shorted at the end.

Where do you stand David?

Slick