On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:

as in the other thread, what is the mechanism of that 'interaction'
between waves? i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source.

There is no mixing as in multiplication of waveforms. Perhaps I can
offer a simple analogy. Instead of two AC waveforms (forward and
reflected), use a DC equivalent. Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

Similarly, the coax cable acts much in the same way. The two
batteries are replaced by the incident and reflected signals. At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated.

if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).



--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On Jun 6, 8:55*pm, Jeff Liebermann wrote:
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. *so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. *

There is no mixing as in multiplication of waveforms. *Perhaps I can
offer a simple analogy. *Instead of two AC waveforms (forward and
reflected), use a DC equivalent. *Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. *The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. *Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

it is very misleading to try to make a lumped circuit analogy out of a
transmission line problem. the first step in any circuits 101 course
would be to simplify the two batteries into one then solve for the
single simple current across the resistor... no waves, no back and
forth, no interaction between batteries, just a single current and
voltage.


Similarly, the coax cable acts much in the same way. *The two
batteries are replaced by the incident and reflected signals. *At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated. *

they can be, but what does it prove? i want to freeze it just as the
first wave gets to the load, there is no reflected wave yet, so what
good is that? in order to use our common equations there are many
unstated but necessary assumptions. the most restrictive of which is
that we normally only solve for the sinusoidal steady state waves...
this requires that a long time(in terms of the length of the line) has
passed since the source was energized, that it is a single frequency
pure sine wave, and that nothing is changing in time in the load or
line characteristics. given all that there is no need for
instantaneous values, sure they can be calculated or measured, but
they are of no use in most cases.


if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. *This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).

and obviously if either one is non-linear then all the simple
equations can be abandoned and a more complete analysis must be done.


--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

On Jun 6, 4:55*pm, Jeff Liebermann wrote:
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. *so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. *

There is no mixing as in multiplication of waveforms. *Perhaps I can
offer a simple analogy. *Instead of two AC waveforms (forward and
reflected), use a DC equivalent. *Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. *The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. *Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

Similarly, the coax cable acts much in the same way. *The two
batteries are replaced by the incident and reflected signals. *At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated. *

if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. *This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).

--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

Looks like I should step in here, as the answer to this question is
the main theme in the book Reflections--Transmission Lines and
Antennas, the first edition published in 1990, the second in 2001, and
the third in just this past month of May, released at Dayton.

The notion that ANY reflected power enters the source, such as an RF
power amp using tubes and a pi-network, is FALSE!!! The output source
resistance of these amps is non-dissipative, and totally re-reflects
all reflected power from a mismatched antenna. The same is true when
using an antenna tuner. When correctly adjusted the antenna tuner
totally reflects all reflected power, resulting in a conjugate match
at the antenna-coax mismatch, canceling all reactances in the system
to zero, thus tuning the non-resonant antenna to resonance. This
action if fundamental, and has been a misunderstood myth for
centuries.

For proof of the above statements I invite you to read Chapter 23 of
Reflections, which you can find on my web page at www.w2du.com. Click
on 'Read Chapters from Reflections 2' and then click on Chapter 23.

In addition, Chapter 19 gives more insight, and the addition to
Chapter 19 can be found by clicking on 'Preview Chapters from
Reflections 3'. The addition shows measured data proving that the
output source impedance of the RF amp is the conjugate of the complex
load impedance when the pi-network is adjusted to deliver all the
available power at a given level of grid drive.

Furthermore, a completely revised edition of Chapter 23 and the total
Chapter 19 appear in Reflections 3, which is now available from CQ
Magazine.

Walt Maxwell, W2DU

On Jun 6, 5:27*pm, walt wrote:
On Jun 6, 4:55*pm, Jeff Liebermann wrote:



On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. *so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. *

There is no mixing as in multiplication of waveforms. *Perhaps I can
offer a simple analogy. *Instead of two AC waveforms (forward and
reflected), use a DC equivalent. *Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. *The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. *Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

Similarly, the coax cable acts much in the same way. *The two
batteries are replaced by the incident and reflected signals. *At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated. *

if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. *This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).

--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

Looks like I should step in here, as the answer to this question is
the main theme in the book Reflections--Transmission Lines and
Antennas, the first edition published in 1990, the second in 2001, and
the third in just this past month of May, released at Dayton.

The notion that ANY reflected power enters the source, such as an RF
power amp using tubes and a pi-network, is FALSE!!! The output source
resistance of these amps is non-dissipative, and totally re-reflects
all reflected power from a mismatched antenna. The same is true when
using an antenna tuner. When correctly adjusted the antenna tuner
totally reflects all reflected power, resulting in a conjugate match
at the antenna-coax mismatch, canceling all reactances in the system
to zero, thus tuning the non-resonant antenna to resonance. This
action if fundamental, and has been a misunderstood myth for
centuries.

For proof of the above statements I invite you to read Chapter 23 of
Reflections, which you can find on my web page atwww.w2du.com. Click
on 'Read Chapters from Reflections 2' and then click on Chapter 23.

In addition, Chapter 19 gives more insight, and the addition to
Chapter 19 can be found by clicking on 'Preview Chapters from
Reflections 3'. The addition shows measured data proving that the
output source impedance of the RF amp is the conjugate of the complex
load impedance when the pi-network is adjusted to deliver all the
available power at a given level of grid drive.

Furthermore, a completely revised edition of Chapter 23 and the total
Chapter 19 appear in Reflections 3, which is now available from CQ
Magazine.

Walt Maxwell, W2DU

Forgot to mention that the output of the RF power amp is LINEAR, even
though the input is non-linear. The reason is that the the pi-network
tank circuit is not only an impedance transformer, it's an energy-
storage device that isolates the output from the input. The linearity
of the output is indicated by the sinusoidal shape of the output wave,
and that the voltage and current are in phase when the load impedance
is resistive.

Walt, W2DU

I also think so.