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

September 7th 03, 05:59 AM

wrote in message ...

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

But you need to define Vi as the voltage
coming out of the series inductor. I don't believe
you do that here.

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.

Where does circuit theory predict
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?

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

i agree with your last statement, but do you think that the
power RC is around 64 for this circuit?

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.

Slick