Dear Reg:
Lab equipment was not used. The lab was just a convenient gathering
place with a blackboard. (No cigarettes in our lab or anywhere inside
buildings.)
I suspect that we made an arithmetic error. Need to revisit what we
did - much too late at night right now.
Thanks for the poke. 73 Mac N8TT

--
J. Mc Laughlin - Michigan USA
Home:

"Reg Edwards" wrote in message
...
"J. McLaughlin" wrote
The issue of the max. value of the magnitude of the reflection
coefficient came up in our student mess. It was shown in lab that
it
must approach the square root of 5, which is close to 2.41.

================================

You don't need a lab. All you need is a pencil and the back of a
cigarette
packet.

Theoretical Max possible value = 1+Sqrt(2) exactly = 2.4142136 . .
.. . .

Can't imagine where you get 5 from.

It occurs when line Zo = Ro - jXo has an angle of -45 degrees, ie.,
when Xo
= -Ro, and when the line is terminated in an inductive reactance of
+jXo.

----
Reg.

Roy Lewallen wrote:
p(t) = v(t) * i(t).

What is 100 volts at zero degrees multiplied by 2 amps at 90 degrees?
You really want us to believe there is not a cos(90deg) in there
somewhere? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
It does go back to 'double think' unless you can explain how energy
can flow when the power is zero.

Already explained. The forward power flow vector is equal to the
reflected power flow vector so the *NET* power is zero just as
explained in Ramo & Whinnery.

So I will repeat the challenge:
"It appears that you are prepared to ignore Pinst = Vinst * Iinst. Could
you expand on when Pinst IS equal to Vinst * Iinst and when it isn't?

Sorry, this appears to be just a diversion from a subject you don't
want to discuss. According to Hecht, Pinst brings nothing of value
to the discussion.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
Why the resistance to examining this case?

I don't chase irrelevant logical diversions.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
You can now remeasure any current or voltage at any
points you choose to tell me whether the shorts or cuts are in or out.

That's easy. The line is shorted or cut where the signals disappear.

Since you can not tell the difference, the circuit with cuts and opens
is identical to the one without.

I can't tell the difference when someone cuts my transmission line?
Can I have a hit off of whatever you are smokin'?
--
73, Cecil http://www.qsl.net/w5dxp



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Goodness, didn't you do this at Texas A & M? Or did they just give you
the formula for average power and tell you to forget the rest? But the
calculation that follows requires only high school level trigonometry,
not engineering mathematics.

v = 100 cos(wt)
i = 2 cos(wt + 90)

v * i = 200 cos(wt)cos(wt + 90)
= 100 cos(-90) + 100 cos(2wt + 90)
= 100 cos(2wt + 90).

What you have here is a sinusoidal waveform of radian frequency 2w,
centered about zero. Cos(90) is zero. The result above, the power, or
rate of energy transfer, is not zero. It shows that energy moves back
and forth at twice the rate of v and i, and that the same amount that
moves one way moves back on the alternating cycle.

Roy Lewallen, W7EL

W5DXP wrote:
Roy Lewallen wrote:

p(t) = v(t) * i(t).


What is 100 volts at zero degrees multiplied by 2 amps at 90 degrees?
You really want us to believe there is not a cos(90deg) in there
somewhere? :-)

Roy Lewallen wrote:
Goodness, didn't you do this at Texas A & M?
= 100 cos(2wt + 90).

I thought you said there wasn't a cos(90) in there anywhere.
If t=0, there's a cos(90) term.

What you have here is a sinusoidal waveform of radian frequency 2w,
centered about zero. Cos(90) is zero. The result above, the power, or
rate of energy transfer, is not zero. It shows that energy moves back
and forth at twice the rate of v and i, and that the same amount that
moves one way moves back on the alternating cycle.

And exactly why is that information important?
--
73, Cecil http://www.qsl.net/w5dxp



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Art Unwin KB9MZ wrote:
I don't see from what what Reg said that this should have been unloaded
on him in such a way. I sure hope the " establishment" or "cabal" as I
call them
don't follow you and pile it on . Seems to me that Reg is a giver not
a taker as far as ham radio is concerned, and just as much as a asset
to this group as those wish to fire off a shot at the first oportunity
to give dutch courage to all those that wil follow.
Now you can stick the label on me also, audacity was it? After all I
did say "cabal" ? What really surprises me is that you have not noticed
over the years how many people have been attacked so wickedly on this
newsgroup by people that infer that "all is known" to which you
whimsically referred to.
The fact is that is if this were true a thread would never exceed
five postings after a particular expert had posted. I can only imagine
that Reg has not swallowed all you have said and thus has raised your
ire such that you have lost your cool

If you still think there is a cabal, then you missed the whole point.

If you thought it was a personal attack against Reg, then you completely
misunderstood my motives in writing all that. (And anyway, Reg probably
regards "audacity" as the highest of compliments :-)

And if you thought any of it was written in anger, then you never
understood a single word.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

W5DXP wrote:

wrote:
You can now remeasure any current or voltage at any
points you choose to tell me whether the shorts or cuts are in or out.

That's easy. The line is shorted or cut where the signals disappear.

You must have misread the question. Shorts are only applied at points
with zero volts and only wires with zero current are cut. Since there
is no 'signal' present at these points, the 'signal' will not disappear.

For your convenience, I have provided the experiment again, below.

"Just to recap, the experiment was: a matched source connected to an
open
transmission line. After the source is turned on (and a brief wait for
settling), measure any voltage or current you desire at any point on
the line. I short the points of zero voltage and cut the points with
zero current. You can now remeasure any current or voltage at any
points you choose to tell me whether the shorts or cuts are in or out.

Since you can not tell the difference, the circuit with cuts and opens
is identical to the one without."

....Keith

Keith wrote:
"Since you can not tell the difference, the circuit with cuts and opens
is identical to one without."

There he goes again!

If you short an r-f transmission line where SWR has produced zero volts,
you make a difference. You change the reflection point to the short you
impose from its former location.

Best regards, Richard Harrison, KB5WZI