"Dr. Slick" wrote:

Hello,

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

An excellent example.

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

Corrected arithmetic error - -1-j8 = 8.062/_ -97.125

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.

So for this example using the 'revised' rho Vr = -Vi
so the voltage across the capacitor would be Vi + Vr = 0 .

Let us do some circuit analysis.

As you say above, the equivalent circuit is 3 elements in
series: a 50 ohm resister, a +j200 ohm inductor and -j200
ohm capacitor.

Let us apply 1 volt to this circuit...
Total impedance
50+j200+0-j200 = 50 ohms
Total current (volts/impedance)
1/50 = .02 A
Voltage across resistor
.02 * 50 = 1 V
Voltage across inductor
.02 * (0+j200) = 4/_ 90 Volts
Voltage across capacitor (the load)
.02 * (0-j200) = 4/_ -90

Now for the check...
Vi = 0.5 V
With classic rho
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125
Vload = Vi+Vr = 0.5 + 4.031/_ -97.125 = 4/_ -90
The same as computed using circuit theory.

With revised rho
Vr = -.5
Vload = 0.5 -0.5 = 0

Which is more useful?
classic rho which properly predicts the voltage across the load
or
revised rho which just provides some number

I know which one I'd pick.

The thing to remember is that in circuits with inductors it is
very easy to achieve voltages greater than any of the supplies.

....Keith