Keith Dysart wrote:
clip text...........


After considerable thought, I think the math you presented above is for one of two cases of reflective waves, the reflection from a higher impedance load. When the load is less than the Zo of the line, the currents add but voltages subtract. Right?



I don't think so. Vt = Vf + Vr, It = If - Ir, Vf = If * Z0 and Vr = Ir * Z0 are the fundamental equations defining forward and reverse waves. Perhaps you arrive at two choices because sometimes Vr and Ir are negative, which after simplification appears to give an alternate form?



The end result is the same for both cases.



This is good. If you chase the signs, though, I think you will find that there is only one case. We probably should not toss Power into the mix until agreement is reached on this. Power is fraught with issues which seriously confuse some. ....Keith

I can see that I need to further explain. 

My analysis always begins with the source because the first formation of the wave comes from the source, then travels through the transmission line system.  The source defines the wave only until the wave reaches any discontinuity(s) or the line end.   Thereafter, discontinuities and end conditions define the system,.

Why might I say that?   Initiation of the wave at the source results in a sine wave with the impedance of the transmission line, and the power and frequency of the  source.   This is a steady state condition until the first discontinuity or reflection point is reached by the traveling wave.   Each successive reflection point (discontinuity) reflects power which travels back to the source and changes the feed point impedance conditions.   The most distant possible reflection point is the end of the transmission line (ignoring reflections which might occur on the antenna) and might be an open circuit, a reactive resistance, or a short circuit.   Any power reflected from the end will change the measured impedance found at any point on the transmission line all the way back to the source, and will define the steady state conditions of the system.

If we accept that the steady state conditions are defined by the load, then we should examine the conditions on the source side of the load, assuming it is the end of the transmission line.  The forward wave spawns the reflective wave in one of two ways, one way of  load resistance higher than line impedance, and a second way of  load resistance lower than line impedance.   In both cases the power of both forward and reflective wave add, but the voltages and currents both add and subtract.   (Cecil explained it very well in his follow up postings.   Thanks Cecil.)  I presented the power equations to illustrate the two conditions.

It is convenient that both cases result in the same math for the directional watt meter.  

73, Roger, W7WKB

On Wed, 12 Dec 2007 21:52:48 -0800, Roger wrote:

!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
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Hi Roger,

99.9999% of posters here use unformatted text which makes responses
very readable.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Wed, 12 Dec 2007 21:52:48 -0800, Roger wrote:


!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
html
head
meta content="text/html;charset=ISO-8859-1" http-equiv="Content-Type"
title/title
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Hi Roger,

99.9999% of posters here use unformatted text which makes responses
very readable.

73's
Richard Clark, KB7QHC


Thanks Richard, I will use text from now on. Sorry for the inconvenience.

Roger

Roger wrote:

...
Thanks Richard, I will use text from now on. Sorry for the
inconvenience.

Roger

HTML is not a problem for many of us, any decade old/decent newsreader
handles it fine--if you have .html enabled ...

Regards,
JS

Roger wrote:
Thanks Richard, I will use text from now on. Sorry for the
inconvenience.

I didn't even notice with Thunderbird since I had the
display HTML as plain text option selected.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

...
I didn't even notice with Thunderbird since I had the
display HTML as plain text option selected.

That's because you have good taste and Thunderbird ROCKS! (the
newsreader, NOT the wine ;-) )

Regards,
JS

On Dec 13, 12:52 am, Roger wrote:
Keith Dysart wrote:
clip text...........After considerable thought, I think the math you presented above is for one of two cases of reflective waves, the reflection from a higher impedance load. When the load is less than the Zo of the line, the currents add but voltages subtract. Right?I don't think so. Vt = Vf + Vr, It = If - Ir, Vf = If * Z0 and Vr = Ir * Z0 are the fundamental equations defining forward and reverse waves. Perhaps you arrive at two choices because sometimes Vr and Ir are negative, which after simplification appears to give an alternate form?The end result is the same for both cases.This is good. If you chase the signs, though, I think you will find that there is only one case. We probably should not toss Power into the mix until agreement is reached on this. Power is fraught with issues which seriously confuse some. ...KeithI can see that I need to further explain.
My analysis always begins with the source because the first formation of the wave comes from the source, then travels through the transmission line system. The source defines the waveonlyuntil the wave reaches any discontinuity(s) or the line end. Thereafter, discontinuities and end conditions define the system,.
Why might I say that? Initiation of the wave at the source results in a sine wave with the impedance of the transmission line, and the power and frequency of the source. This is a steady state condition until the first discontinuity or reflection point is reached by the traveling wave. Each successive reflection point (discontinuity) reflects power which travels back to the source and changes the feed point impedance conditions. The most distant possible reflection point is the end of the transmission line (ignoring reflections which might occur on the antenna) and might be an open circuit, a reactive resistance, or a short circuit. Any power reflected from the end will change themeasuredimpedance found atany pointon the transmission line all the way back to the source, and will define the steady state conditions of the system.
If we accept that the steady state conditions are defined by the load, then we should examine the conditions on the source side of the load, assuming it is the end of the transmission line. The forward wavespawnsthe reflective wave in one of two ways, one way of load resistance higher than line impedance, and a second way of load resistance lower than line impedance. In both cases the power of both forward and reflective wave add, but the voltages and currents both add and subtract. (Cecil explained it very well in his follow up postings. Thanks Cecil.) I presented the power equations to illustrate the two conditions.
It is convenient that both cases result in the same math for the directional watt meter.

I'd still suggest that you have the cart and the horse backwards.

The math came first and that is why all the example cases turn out
to be consistent with the math.

And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

....Keith

....Keith

Keith Dysart wrote:
On Dec 13, 12:52 am, Roger wrote:

Keith Dysart wrote:


Clipping text.............
I'd still suggest that you have the cart and the horse backwards.

The math came first and that is why all the example cases turn out
to be consistent with the math.

And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

...Keith

...Keith

Interesting! The important thing is to get answers that agree with
our experiments.

I have done some computations for DC voltage applied to transmission
lines. The real surprise for me came when I realized that transmission
line impedance could be expressed as a function of capacitance and the
wave velocity. Z0 = 1/cC where c is the velocity of the wave and C is
the capacitance of the transmission line per unit length.

73, Roger , W7WKB

On Thu, 13 Dec 2007 08:40:53 -0800, Roger wrote:

And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

...Keith

...Keith

Interesting! The important thing is to get answers that agree with
our experiments.

I have done some computations for DC voltage applied to transmission
lines. The real surprise for me came when I realized that transmission
line impedance could be expressed as a function of capacitance and the
wave velocity. Z0 = 1/cC where c is the velocity of the wave and C is
the capacitance of the transmission line per unit length.

Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Thu, 13 Dec 2007 08:40:53 -0800, Roger wrote:


And just for completeness...
The fundamental equations also work when:
- the signal is not sinusoidal, e.g. pulse, step, square, ...
- rather than a load at one end, there is a source at each end
- the sources at each end produce different arbitrary functions
- the arbitrary functions at each end are DC sources
It is highly instructive to compute the forward and reverse
voltage and current (and then power) for a line with the same
DC voltage applied to each end.

...Keith

...Keith


Interesting! The important thing is to get answers that agree with
our experiments.

I have done some computations for DC voltage applied to transmission
lines. The real surprise for me came when I realized that transmission
line impedance could be expressed as a function of capacitance and the
wave velocity. Z0 = 1/cC where c is the velocity of the wave and C is
the capacitance of the transmission line per unit length.


Hi Roger,

This last round has piqued my interest when we dipped into DC. Those
"formulas" would lead us to a DC wave velocity?

73's
Richard Clark, KB7QHC

Hi Richard,

Here are two links to pages that cover the derivation of the formula Zo
= 1/cC and much more.

http://www.speedingedge.com/PDF-File..._Impedance.pdf
http://www.ece.uci.edu/docs/hspice/h...001_2-269.html

Here is the way I proposed to Kevin Schmidt nearly seven years ago after
seeing him use the formula on a web page:

*ASSUME*:
1) An electrical wave travels at the speed of light, c
2) A 'perfect' voltage source without impedance, V
3) A 'perfect' transmission line having no resistance but uniform
capacitance per unit length, C

*CONDITIONS AND SOLUTION*
The perfect voltage source has one terminal connected to the
transmission line prior to beginning the experiment. The experiment
begins by connecting the second terminal to the transmission line. The
voltage source drives an electrical wave down the transmission line at
the speed of light. Because of the limitation of speed, the wave
travels in the shape of a square wave containing all frequencies
required to create a square wave.

The square wave travels down the transmission line at the speed of light
(c). After time (T), the wave has traveled distance cT down the
transmission line, and has charged the distributed capacity CcT of the
line to voltage V over that distance. The total charge Q on the
distributed capacitor is VCcT.

Current (I) is expressed as charge Q per unit time. Therefore the
current into the transmission line can be expressed as

I = Q/T =
VCcT / T = VCc

Impedance (Zo) is the ratio of voltage (V) to current (I). Therefore
the impedance can be expressed as

Zo = V / I =
V / VCc = 1/Cc

We can generalize this by using the velocity of the electrical wave
rather than the speed of light, which allows the formula to be applied
to transmission line with velocities slower than the speed of light.

Of course, only the wave front and wave end of a DC wave can be
measured to have a velocity.

73, Roger, W7WKB