So do it right and show us how it really should be done.

Roy Lewallen, W7EL

Cecil Moore wrote:

You obviously did a something-parameter analysis (maybe a z-parameter
analysis?). Whatever you did results in four power terms, not two plus
a third. When you introduced 'x' you introduced an analysis that produces
a reflected wave on each side of 'x' and a forward wave on each side of
'x'. That's four waves. You went too far when you combined two of those
waves into one especially since one is a forward wave and one is a
reflected wave.

Roy Lewallen wrote:
So do it right and show us how it really should be done.

Sorry, that diversion won't work. Correct your error first
and effort on my part will be avoided.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

What all you experts have forgotten is that SWR on a
lossless line is the ratio of two voltages, max and
min, SPACED APART BY 1/4-WAVELENGTH. That is if the
line is long enough to contain both a max and a min.

When the line is not lossless, ie., it has appreciable
attenuation in dB per 1/4-wavelength, then the ratio is
'distorted' and has a phase angle. So negative values
of indicated SWR can be expected at some values of |
Vmax | / | Vmin |

SWR is calculated from the square of | rho |. As I've
said before, immediately | rho | is squared, half the
information it contains is junked. Any
discussion/argument about power waves following
rho-squared on a lossy (a real ) line is meaningless
piffle.

Anybody who writes books about power waves, selling
them to make a living, is obtaining money under false
pretences.

On the other hand we should be kind to otherwise
unemployed Ph.D's. They too have wive's and kid's to
clothe, feed and provide a roof over their heads.
That's life!
---
Reg.

"Reg Edwards" wrote in message ...

When the line is not lossless, ie., it has appreciable
attenuation in dB per 1/4-wavelength, then the ratio is
'distorted' and has a phase angle. So negative values
of indicated SWR can be expected at some values of |
Vmax | / | Vmin |


What are you talking about? If it have losses, and they
are dissipative losses, the amplitude of the voltage will
decrease due to voltage drops. That would be moving AWAY
from having a greater reflected voltage than an incident one.

But, that's impossible anyways with a passive network.

The concept of Negative SWRs is rubbish.

SWR is calculated from the square of | rho |. As I've
said before, immediately | rho | is squared, half the
information it contains is junked. Any
discussion/argument about power waves following
rho-squared on a lossy (a real ) line is meaningless
piffle.

Anybody who writes books about power waves, selling
them to make a living, is obtaining money under false
pretences.

On the other hand we should be kind to otherwise
unemployed Ph.D's. They too have wive's and kid's to
clothe, feed and provide a roof over their heads.
That's life!
---
Reg.

Remind me not to be YOUR book when it comes out!



The ratio Pref/Pfwd is directly related to the ratio [rho].

Pref/Pfwd = [rho]**2 Absolute value brackets are a must!


Consider that after the absolute value brackets, the phase information
is gone. But since we are going to a ratio of average (RMS)
values OR peak values of power, it doesn't matter.

In other words, if you use V**2/R, the "V" can be either peak or
RMS, it doesn't matter, because it is a ratio. And of course, the "R"
doesn't matter either. And of course, the phase information is gone
with
the absolute value brackets.

If you agree that the Pref/Pfwd ratio cannot be greater than 1
for a passive network, then neither can the [Vref/Vfwd]= rho be
greater
than 1 either.

Some people wanna rewrite some books here.


Slick

If the loss per unit wavelength is large enough, and you produced a plot of
voltage vs. distance x. The voltage maximum would be at the source, and the
voltage minimum at the load. Try a thousand miles or so of RG58 at 60 Hz.
I suspect that to see anything that looks like a standing wave you would
have to look at dV/dx. Remember, I can always define a lossier line.

Tam/WB2TT

"Reg Edwards" wrote in message
...
What all you experts have forgotten is that SWR on a
lossless line is the ratio of two voltages, max and
min, SPACED APART BY 1/4-WAVELENGTH. That is if the
line is long enough to contain both a max and a min.

When the line is not lossless, ie., it has appreciable
attenuation in dB per 1/4-wavelength, then the ratio is
'distorted' and has a phase angle. So negative values
of indicated SWR can be expected at some values of |
Vmax | / | Vmin |

SWR is calculated from the square of | rho |. As

VSWR is defined as |Vmax|/|Vmin| and so can never be negative. in lossless
lines this expression can be reduced to a function of rho, but that method
is not valid in lossy lines. VSWR is not a constant in lossy lines and
probably doesn't really mean much of anything as each voltage maximum and
minimum is a different value, so which ones do you use???

VSWR is not a constant in lossy lines and
probably doesn't really mean much of anything as each
voltage maximum and
minimum is a different value, so which ones do you
use???

-------------------------------------------------------
---------

Dear David,

You have expressed my sentiments exactly. I have never
used either or any of them. What does anybody do with
value of SWR when they imagine they know it? I'm
pleased to make your acquaintance!

For some years I have mildly advertised the idea of
changing the name the name of the common-or-garden, so
called SWR meter / combined forward-and-reflected power
meter, to the TLI (Transmitter Loading Indicator) which
is all it does. Although I must admit, at the present
state of the art, it is a very useful instrument when
changing antennas.

Is the transmitter loaded with a resistance of 50 ohms
or is it not?

{ Actually, the meter on my top-band transmitter
indicates relative to 75 ohms }

And there HAS to be SOMETHING more than the weather to
talk about in QSO's and, of course, on this newsgroup.
;o)
----
Reg, G4FGQ