Jim Kelley wrote:
Yes, Cecil. I have considered it, and I agree. It just bothers me when
people forget what the minus sign means, and try using it to make
unrealistic claims.

At least for real Z0's, it should be consistent. Wouldn't
a rho of 0.5 at 20 degrees would be the same as a rho of
-0.5 at 200 degrees?
--
73, Cecil, W5DXP

On Fri, 26 Sep 2003 12:53:50 -0500, Cecil Moore
wrote:
At least for real Z0's, it should be consistent. Wouldn't
a rho of 0.5 at 20 degrees would be the same as a rho of
-0.5 at 200 degrees?

Hi Cecil,

The Reflection Coefficient is a characteristic of the Load or Source,
not a value projected all along the line. This is the teaching of
Chipman that you undoubtedly speed-read past on your way to the
cut-and-paste opportunities you sought.

When are you going to ride your bike back to the library to fill all
these shortfalls of his teachings you so liberally sprinkle your
missives with?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
The Reflection Coefficient is a characteristic of the Load or Source,
not a value projected all along the line. This is the teaching of
Chipman that you undoubtedly speed-read past on your way to the
cut-and-paste opportunities you sought.

For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't
even need to know the load and/or source impedances.

When are you going to ride your bike back to the library to fill all
these shortfalls of his teachings you so liberally sprinkle your
missives with?

Just as soon as I am over my sinusitis and have a day off.
--
73, Cecil, W5DXP

On Fri, 26 Sep 2003 14:35:33 -0500, Cecil Moore
wrote:

For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't
even need to know the load and/or source impedances.

Hi Cecil,

How did you get a -1 out of your |rho|? Take a box of kleenex on
your bike ride to the library - your logic is dribbling.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

wrote:
For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't
even need to know the load and/or source impedances.

How did you get a -1 out of your |rho|?

I probably should have said rho^2 = Pref/Pfwd. When Pref = Pfwd,
rho can be plus or minus one. I used |rho| to indicate a magnitude,
sans phase angle, not an absolute value.
--
73, Cecil http://www.qsl.net/w5dxp



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On Fri, 26 Sep 2003 23:15:44 -0500, Cecil Moore
wrote:

Richard Clark wrote:

wrote:
For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't
even need to know the load and/or source impedances.

How did you get a -1 out of your |rho|?

I probably should have said rho^2 = Pref/Pfwd. When Pref = Pfwd,
rho can be plus or minus one. I used |rho| to indicate a magnitude,
sans phase angle, not an absolute value.

Hi Cecil,

You obviously don't respect/know the difference between a dependant
variable (rho) and independent variables (P).

Rho is a dependency of the interface, not a translatable value you are
forcing illogic to perform. You really need to ride your bike to the
library more and offer these poor examples less. Since Rho is the
dependant variable, even squared (for you to force a -1 into this
charade) requires a concurrent observance of a negative in the right
hand side (negative power - perhaps if you were in a black hole).

The long and short of it is that this confirms Jim's observance of
your forced math serving your canards rather than logic. Oh, and
please stop offering and polluting Chipman as a resource when you've
only copied one page.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Since Rho is the
dependant variable, even squared (for you to force a -1 into this
charade) requires a concurrent observance of a negative in the right
hand side (negative power - perhaps if you were in a black hole).

You seem to have forgotten some junior high math, Richard. There is
no requirement for a negative anywhere in order for the square root
of a number to be negative. The square root of 100W/100W has two
values, plus or minus one, and sure enough, an open or a short will
cause 100% reflection.

BTW, I copied that page in Chipman with which you are having such a
problem and I don't see the problem you described. Absolutely nothing
said about reflections from the source. In fact, the source has the
same impedance as the transmission line so there are no reflections
from the source.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote in message ...
....
Wouldn't
a rho of 0.5 at 20 degrees would be the same as a rho of
-0.5 at 200 degrees?

Yes, and any other complex quantity would likewise be the same
expressed either way. But it would certainly be confusing to the
readers. It would be much better to stick with rectangular or with
polar and not mix them in the same quantity. Of course, sometimes one
is easier to work with, or offers more insight, than the other and
you're welcome to convert between them at any time.

Let's see if we can keep it more along the lines of 0.5 at 20 degrees
being (very nearly) the same as .4698+j.1710

Cheers,
Tom