Home |
Search |
Today's Posts |
#2
![]() |
|||
|
|||
![]()
On Dec 20, 8:06*am, Cecil Moore wrote:
Here's a request to anyone capable of performing the necessary math. Can anyone verify or disprove my two voltage values in the following example? Shortened Stub Example: * * * * * * * * * * * rho = -0.7143 --43.4 deg 600 ohm line--+--10 deg 100 ohm line--open * * * * * * * * * Vfor1--|--Vfor2 * * * * *Vi--| * * * * * * * * * Vref1--|--Vref2 Assume the RMS voltage, Vi, incident upon the open end of the stub is: Vi = 100 volts at 0 degrees At point '+' 10 degrees back from the open, Vfor2 = 100 volts at -10 degrees Vref2 = 100 volts at +10 degrees It appears to me that you have the signs wrong above. The voltage at the joint will be a wave later in time than the one at the open end so it should have a greater angle, i.e. plus. Except for that, the computation seems correct, though you have not explained your work. Vfor1 = 143.33 volts at -46.6 degrees Vref1 = 143.33 volts at +46.6 degrees This step has no explanation. Can you provide a justification from first principles? Since the reference is still the open end, you need to show that jumping across the joint results in a 36.6 (I use your numbers) degree phase change. Then you can compute 46.6, not the other way around. First principles would be the rules for voltages at an impedance discontinuity and the expressions for summing sine waves. Please state any other assumptions you make. It would also help if you made exactly clear what information is needed to compute the phase shift. Just rho? What else? Next, what are the useful properties of this computed phase shift. Does it facilitate the solving of problems? Which ones? Is using the computed phase shift an improvement over the conventional ways of solving the problem? Or is the computed phase shift just an alternate way of viewing what is happening? If the latter, does it have any downsides? Does it mislead in any way? Will it cause the student to assume something that will have to be unlearned later? The phase shift between Vfor1 and Vfor2 is 36.6 degrees. Silly math error? 36.6 + 46.6 +10 - 93.2, which does not align with your previous posts. The phase shift between Vref2 and Vref1 is 36.6 degrees. The total RMS voltage at all points up and down the line is 200 volts at 0 deg. I don't follow this statement. Which line? From the original experiment, is not the total voltage at the driven end zero? ...Keith |