50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
73,
Tom Donaly, KA6RUH

Tom, KA6RUH wrote:
"Will somebody explain how this works to Cecil?"

Cecil is capable of solving practically any transmission line problem
without help from anyone. Tom could have been more descriptive by saying
if his line were 50-ohm coax with a velocity factor of about 2/3.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Tom, KA6RUH wrote:
"Will somebody explain how this works to Cecil?"

Cecil is capable of solving practically any transmission line problem
without help from anyone. Tom could have been more descriptive by saying
if his line were 50-ohm coax with a velocity factor of about 2/3.

Best regards, Richard Harrison, KB5WZI


No he isn't, Richard, or he wouldn't have made the ignorant statement
that there is no phase information in a standing wave. As to the
velocity factor, the line is one of Cecil's ideal, lossless lines. The
velocity factor is 1. Nice try.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
No he isn't, Richard, or he wouldn't have made the ignorant statement
that there is no phase information in a standing wave.

Please get it right, Tom. I said there is no phase
information in the standing-wave phase. The phase
information is certainly there but it is in the
amplitude, not in the phase. The following graph
proves it:

http://www.w5dxp.com/travstnd.gif

The phase of the standing-wave is absolutely flat.
--
73, Cecil http://www.w5dxp.com

Richard Harrison wrote:
Tom, KA6RUH wrote:
"Will somebody explain how this works to Cecil?"

Cecil is capable of solving practically any transmission line problem
without help from anyone. Tom could have been more descriptive by saying
if his line were 50-ohm coax with a velocity factor of about 2/3.

Tom apparently doesn't realize that he needs to be dealing
with more than one Z0 to observe the difference between
electrical degrees and physical degrees. I invite everyone
to solve the dual-Z0 stub problem that I earlier presented
and repeated tonight.
--
73, Cecil http://www.w5dxp.com

On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

....Keith

Keith Dysart wrote:
On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

...Keith
Right. I lied.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Keith Dysart wrote:
On Dec 4, 9:06 pm, "Tom Donaly" wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode)

arc cosine, perhaps?

of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

Cecil either knows this, in which case it is unnecessary to explain
it,
or he does not, in which case it will be impossible.

...Keith
Right. I lied.
73,
Tom Donaly, KA6RUH
To amplify:
Arc Cosine is correct. And the comment on Cecil is right on the mark.
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.

If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.
73,
Tom Donaly, KA6RUH