View Single Post
  #2   Report Post  
Old December 21st 07, 01:59 PM posted to rec.radio.amateur.antenna
Keith Dysart[_2_] Keith Dysart[_2_] is offline
external usenet poster
 
First recorded activity by RadioBanter: May 2007
Posts: 492
Default Please verify or disprove

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