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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Jim Kelley wrote in message ...
Ur right, thanks Dave. I meant to say voltage rather than power. Let
me ask the question properly.

Tom,
Since rho represents the fraction of forward voltage that is reflected,
what does a negative value for rho indicate?

Jim,

I suppose your question has been answered sufficiently (Thanks, Roy,
Cecil and Tam), but I'd like to offer a bit different viewpoint than
is implied by your "fraction that is reflected."

I prefer to think of it not as "a fraction that's reflected" but
rather as a resolution of a particular voltage and current into two
modes. There are two modes of propagation supported by TEM line, one
in each direction along the line. If you excite a line to
steady-state at one frequency, there will be some sinusoidal current
at each point along the line, and some sinusoidal voltage across the
line at each point along its length. (You can have a load at one end
and a source at the other, or two sources each with its own internal
impedance, one at each end, so long as they are on the same
frequency.) That set of voltages and currents can be resolved
mathematically into two components, one corresponding to the mode of
propagation in one direction and one corresponding to the mode in the
other direction. Rho is simply a number representing that resolution
at the point on the line where that rho is measured (or calculated).
It's a complex number because it represents both phase and amplitude.

(Note that our resolution of measured voltage and current into the two
modes generally assumes that we know the line's Zo, and the degree to
which we don't know that will introduce an error in our determination
of rho. But that's a whole 'nuther topic...)

Cheers,
Tom

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

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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On Sat, 27 Sep 2003 16:34:43 -0500, Cecil Moore
wrote:

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.

Hi Cecil,

If neither powers are negative, the square root of them cannot
possibly enclose a negative. There is no possibility of Rho being
negative by your description. You should start biking to junior high.


BTW, I copied that page in Chipman with which you are having such a
problem and I don't see the problem you described.

My problem? Quote me rather than give me your tarted up remembrance
of what I said. Clearly your head cold cannot answer for such
consistently unreliable correspondence.

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.

Duh!

Cecil, You are going to run your bike's mileage warrantee out by
pedalling to the library for one page copies at a time. Why don't you
spend a couple hours there and read it instead?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
If neither powers are negative, the square root of them cannot
possibly enclose a negative.

Huh?????? The square root of +100W/+100W cannot be negative?????
WOW! Sounds like you are letting your personal feelings get in
the way of accepted math principles. Hint: If one of those powers
is negative, the square root will be imaginary.

Cecil, You are going to run your bike's mileage warrantee out by
pedalling to the library for one page copies at a time. Why don't you
spend a couple hours there and read it instead?

I've got page 139, the one you referenced, in front of me. It says
absolutely nothing about reflections from the source. All it seems
to say is that conjugately matched loads accept more power than
non-conjugately matched loads but we knew that already.

Incidentally, pages 140-143 discusses "Transmission line sections
as two-port networks" using the h-parameter analysis. Who said
transmission lines didn't have ports?
--
73, Cecil http://www.qsl.net/w5dxp



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