Re: The Rest of the Story

Cecil Moore wrote:
Roger Sparks wrote:
Vrs(t) = Vs(t) - Vg(t) + Vref(t)

Good job, Roger. Let's simplify the equation through
substitution.

Let V1(t) = [Vs(t) - Vg(t)]

Let V2(t) = Vref(t)

Vrs(t) = V1(t) + V2(t)

This works. It certainly helps to link my work with yours.

The power to Rs would be (Vrs^2)/50 = (70.7^2)/50 = 100w, all coming from the reflection.

Yes, because when Vs(t)=0, there is zero instantaneous
interference.

In summary, power to the resistor Rs comes via two paths, one longer than the other by 90 degrees (in this example). The short path is the series path of two resistors composed of the source resistor Rs and the input to the 50 ohm transmission line measured by Vg. The long path is the series path of one resistor Rs and one capacitor composed of the shorted transmission line. Both paths are available at all times. Power flows through both paths to Rs at all times, but because of the time differential in arrival timing, at some point Rs will receive power only from Vs, and at another point, receive power only from Vref.

What Keith is missing is that:

Vrs(t)^2 = [V1(t) + V2(t)]^2 NOT= V1(t)^2 + V2(t)^2

In his math, Keith is asserting that since

Prs(t) NOT= P1(t) + P2(t), then the reflected power is not
being dissipated in the source resistor. But every sophomore
EE knows NOT to try to superpose powers like that. Roger,
I'll bet you know not to try to superpose powers?

Using circuit theory, at the peak under steady conditions, we have 141.4v applied to 70.7 ohms which gives a current of 2a. The 70.7 ohms is sqrt(50^2 + 50^2). Two amps flowing through Rs = 50 ohms, the power to Rs is (2^2) * 50 = 4 * 50 = 200w.

Using your simplified equation for the voltage across Rsj

Vrs(t) = V1(t) + V2(t)

Vrs(t) = V1*sin(wt + 90) + V2*sin(wt)

In our case, V1 = V2 because both voltages are developed over a 50 ohm load. As a result, in our case, 100w will be delivered to Rs both at wt = 0 and wt = -90, two points 90 degrees apart. If we are looking for the total power delivered to Rs, then it seems to me like we SHOULD add the two powers in this case. This recognizes that power is delivered to Rs via two paths, each carrying 100w. Alternatively, we could add the two voltages together to find the peak voltage, and then square that number and divide by the resistance of Rs. Both methods should give the same result.

So it looks to me like Keith is right in his method, at least in this case.


Since Keith doesn't listen to me, would you pass that
technical fact on to him?

When Keith uses the correct equation:

Prs(t) = P1(t) + P2(t) +/- 2*SQRT[P1(t)*P2(t)]

he will see that the equation balances and therefore
100% of the reflected energy is dissipated in the
source resistor since the interference term averages
out to zero over each cycle.

I am still uncomfortable with my summary statement that both paths, the long and short, are available and active at all times. On the other hand, it is consistent with your advice that the waves never reflect except at a discontinuity. The conclusion was that 100w is delivered to Rs via each path, with the path peaks occurring 90 degrees apart in time. Surprisingly, 100 percent of the reflected energy is ALWAYS absorbed by Rs. There is no further reflection from Vs or Rs.

This is counter intuitive to me. I like to resolve the circuit into one path, one wave form. We do that with circuit analysis. With traveling waves, we frequently have two or more paths the exist independently, so we must add the powers carried on each path, just like we add the voltages or currents. (But watch out and avoid using standing wave voltages or currents to calculate power!)

Most surprising to me is the observation that my beginning statement (from my previous post) about the circuit evolving from a circuit with two resistive loads, into a circuit with a resistive load and capacitive load, is really incorrect. Once steady state is reached, BOTH circuits are active at the same time, forming the two paths bringing power to RS. Using that assumption, we can use the traveling waves to analyze the circuit.

73, Roger, W7WKB
.

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