On Wed, 20 Aug 2003 09:36:52 +0100, "Ian White, G3SEK"
wrote:

The fact is that is if this were true[, then] a thread would never exceed
five postings after a particular expert had posted.
bracketed [ ] inclusions made by me.

Hi Ian,

This posting (#5) is simply to confirm suspicions. :-)

73's
Richard Clark, KB7QHC

wrote:

W5DXP wrote:

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.

Let me see if I have this straight. You cut off 1/4WL of feedline and
current and voltage still exist in the cut off parts? Have you patented
that perpetual motion machine? All I have to do to prove you wrong is
measure the current and voltage in the part that was isolated from the
source. In the previous example:
1/4WL coax
source---------coax--------------------+---wattmeter-----short

When you cut the line at '+', the wattmeter will go to zero. It's
a no brainer. What do you mean I can't tell the difference?

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."

But I can tell the difference because I choose to measure the current and
voltage in the parts of the feedline that have been cut off and isolated
from the source. It's a no brainer.
--
73, Cecil http://www.qsl.net/w5dxp



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My mistake. I forgot to make clear that the transmission line was
ideal for this thought experiment; that is, R=G=0; no loss. The signal
does indeed continue to circulate for ever.

Please re-evaluate your answer after this clarification.

....Keith

W5DXP wrote:

wrote:

W5DXP wrote:

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.

Let me see if I have this straight. You cut off 1/4WL of feedline and
current and voltage still exist in the cut off parts? Have you patented
that perpetual motion machine? All I have to do to prove you wrong is
measure the current and voltage in the part that was isolated from the
source. In the previous example:
1/4WL coax
source---------coax--------------------+---wattmeter-----short

When you cut the line at '+', the wattmeter will go to zero. It's
a no brainer. What do you mean I can't tell the difference?

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."

But I can tell the difference because I choose to measure the current and
voltage in the parts of the feedline that have been cut off and isolated
from the source. It's a no brainer.
--
73, Cecil http://www.qsl.net/w5dxp

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(Tom Bruhns) wrote in message om...

That's impossible, unless you have a free source of energy, in
which case you should send me the schematic!


Excuse me? As a matter of fact, the way I discovered that you can
have a reflection coefficient greater than unity was to set up the
boundary conditions for a real transmission line feeding a moderately
high Q inductor and discover that |rho|1. Up to that point, nobody
had told me it was possible. At the time, it was a somewhat
surprising result to me. Try it yourself. On a line, Vf/If = -Vr/Ir
= Zo. At the end of the line connected to an load whose impedance is
Zl, the current is V/Zl, but V = Vf+Vr, and the current there must
also be If+Ir. You should be able to reduce those to the well-known
equation for reflection coefficient, and if you plug in numbers such
as Zo=50-j5 and Zl=1+j100, you'll see that |Vr/Vf|1. It only takes
three or four lines of simple linear algebra and perhaps plugging the
complex numbers into a calculator to find the result for that example.
Again, TRY it for yourself. DO NOT just believe what I tell you. If
you have trouble with the algebra, I'll be happy to help you out.

Cheers,
Tom


Actually, my first posting was right all along, if Zo is always
real.

From Les Besser's Applied RF Techniques:

For passive circuits, 0=[rho]=1,

And strictly speaking: Reflection Coefficient =
(Zl-Zo*)/(Zl-Zo)

Where * indicates
conjugate.

But most of the literature assumes that Zo is real, therefore
Zo*=Zo.


And then i looked at the trusty ARRL handbook, 1993, page 16-2,
and lo and behold, the reflection coefficient equation doesn't have a
term for line reactance, so both this book and Pozar have indeed
assumed that the Zo will be purely real.

That doesn't mean Zl cannot have reactance.

Try your calculation again, and you will see that you can never
have a [rho] (magnitude of R.C.)greater than 1 for a passive network.

How could you get more power reflected than what you put in? If
you guys can tell us, we could fix our power problems in CA!

But thanks for checking my work, and this is a subtle detail that
is good to know.


Slick

"Reg Edwards" wrote in message ...

Dear Dr Slick, it's very easy.

Take a real, long telephone line with Zo = 300 - j250 ohms at 1000 Hz.

Load it with a real resistor of 10 ohms in series with a real inductance of
40 millihenrys.

The inductance has a reactance of 250 ohms at 1000 Hz.

If you agree with the following formula,

Magnitude of Reflection Coefficient of the load, ZL, relative to line
impedance

= ( ZL - Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity,

and has an angle of -59.9 degrees.

The resulting standing waves may also be calculated.

Are you happy now ?
---
Reg, G4FGQ


Nope. Re-do your calculations, and you will get something like
0.935:

Actually, my first posting was right all along, if Zo is always
real.

From Les Besser's Applied RF Techniques:

"For passive circuits, 0=[rho]=1,

And strictly speaking: Reflection Coefficient =
(Zl-Zo*)/(Zl-Zo)

Where * indicates
conjugate.

But most of the literature assumes that Zo is real, therefore
Zo*=Zo."


And then i looked at the trusty ARRL handbook, 1993, page 16-2,
and lo and behold, the reflection coefficient equation doesn't have a
term for line reactance, so both this book and Pozar have indeed
assumed that the Zo will be purely real.

That doesn't mean Zl cannot have reactance (be complex).

Try your calculation again, and you will see that you can never
have a [rho] (magnitude of R.C.)greater than 1 for a passive network.

How could you get more power reflected than what you put in? If
you guys can tell us, we could fix our power problems in CA!

But thanks for checking my work, and this is a subtle detail that
is good to know.


Slick

"Reg Edwards" wrote in message ...

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.


Actually, my first posting was right all along, if Zo is always real.

From Les Besser's Applied RF Techniques:

"For passive circuits, 0=[rho]=1,

And strictly speaking: Reflection Coefficient =
(Zl-Zo*)/(Zl-Zo)

Where * indicates
conjugate.

But most of the literature assumes that Zo is real, therefore
Zo*=Zo."


And then i looked at the trusty ARRL handbook, 1993, page 16-2,
and lo and behold, the reflection coefficient equation doesn't have a
term for line reactance, so both this book and Pozar have indeed
assumed that the Zo will be purely real.

That doesn't mean Zl cannot have reactance (be complex).

Try your calculation again, and you will see that you can never
have a [rho] (magnitude of R.C.)greater than 1 for a passive network.

How could you get more power reflected than what you put in? If
you guys can tell us, we could fix our power problems in CA!

But thanks for checking my work, and this is a subtle detail that
is good to know.


Slick

wrote:
My mistake. I forgot to make clear that the transmission line was
ideal for this thought experiment; that is, R=G=0; no loss. The signal
does indeed continue to circulate for ever.

So exactly how many angels can dance on the head of a pin?
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
Given an ideal transmission line (necessary to achieve 0 volts), there
are no measurements you can make which will indicate whether the short
is present or not.

Given a Transmission Line God, there are no measurements you can make
which will indicate whether the short is present or not. How about we
limit these technical discussions to the real world?
--
73, Cecil http://www.qsl.net/w5dxp



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"J. McLaughlin" wrote -
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
===================================

Mac, you didn't make an arithmetical error. I think you used the wrong value
of terminating reactance Xt. Just a Sherlock Holmes deduction. ;o)

The value of terminating reactance which gives greatest possible magitutude
of reflection coefficient is equal to the magnitude of Zo which is equal to
|Zo| = Sqrt( Sqr( Ro) + Sqr( Xo) ).

As you have already deduced, maximum value occurs when the angle of Zo is
equal to -45 degrees at which Ro = -Xo.

You incorrectly set the terminating inductive reactance XL equal to -Xo
whereas XL should have been Sqrt( 2 ) times greater.

Now if XL is incorrectly made -Xo then the reflection coefficient calculates
to Sqrt( 5). Which is where your 5 comes from.

Somewhere in these threads I made the exactly same error myself in a
non-calculating context where it was not likely to be noticed. The pair of
errors, yours and mine, were probably just coincidental.

I do hope I have not just introduced another. ;o)

You appear to be in an educational establishment. There are two programs
available from the website below which you may find useful. Download in a
few seconds and run immediately programs COAXPAIR and RJELINE3. They have
been written according to classical transmission line formulae, generally
accurate to all figures displayed and may be used to check other work which
uses only engineering approximations. There are others also of educational
value.
----
=======================
Regards from Reg, G4FGQ
For Free Radio Design Software
go to http://www.g4fgq.com
=======================

Ahhhhh, finally. The not unexpected descent into non-sequitors.

Much safer than actually thinking about questions which might cause
you to change your world view.

....Keith

W5DXP wrote:

wrote:
My mistake. I forgot to make clear that the transmission line was
ideal for this thought experiment; that is, R=G=0; no loss. The signal
does indeed continue to circulate for ever.

So exactly how many angels can dance on the head of a pin?
--
73, Cecil http://www.qsl.net/w5dxp