Reply
 
LinkBack Thread Tools Search this Thread Display Modes
  #1   Report Post  
Old June 6th 10, 08:01 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Apr 2010
Posts: 484
Default what happens to reflected energy ?

On Jun 6, 3:13*pm, Cecil Moore wrote:
On Jun 6, 9:22*am, "JC" wrote:

Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....


What happens to the 50 joules/second of reflected energy depends upon
the phasing between the source wave and the reflected wave at the
source impedance. *What most amateurs don't understand is there are
two mechanisms that can redistribute reflected energy back toward the
antenna. Those mechanisms are a re-reflection based on the physical
reflection coefficient (what RF engineers understand) and wave
interaction resulting in constructive/destructive interference (what
most RF engineers don't seem to understand) because, unlike optical
physicists, have not been forced to follow the energy flow.

If the reflected wave arrives 180 degrees out of phase with the source
wave, the two waves undergo destructive interference and all of the
reflected power is redistributed as constructive interference energy
back toward the antenna. This is what happens at the Z0-match
established at the input of an antenna tuner.

Thus your conditions of 100w forward power and 50w reflected power
could be accomplished with a 50w matched source.

If the reflected wave arrives 90 degrees out of phase with the source
wave, there is zero interference and the reflected power is dissipated
in the source resistor (in a source with a source resistor).

If the reflected wave arrives in phase with the source wave, all of
the reflected power and more than 1/2 of the source power can be
dissipated in the source resistor.

Such knowledge is old hat for optical physicists who don't have the
luxury of measuring voltages in light waves. They rely on a power
density (irradiance) equation.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the phase angle between the electric fields of two waves.
The last term is called the "interference term" and that value is what
most amateurs are missing in their energy analysis. If the sign of the
interference term is negative, the interference is destructive. If the
sign of the interference term is positive, the interference is
constructive.
--
73, Cecil, w5dxp.com


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. 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.
  #2   Report Post  
Old June 6th 10, 10:55 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Jun 2007
Posts: 1,336
Default what happens to reflected energy ?

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
  #3   Report Post  
Old June 6th 10, 11:10 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Apr 2010
Posts: 484
Default what happens to reflected energy ?

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


  #4   Report Post  
Old June 6th 10, 11:27 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Dec 2006
Posts: 102
Default what happens to reflected energy ?

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
  #5   Report Post  
Old June 6th 10, 11:32 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Dec 2006
Posts: 102
Default what happens to reflected energy ?

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


  #6   Report Post  
Old June 25th 10, 10:01 AM
Junior Member
 
First recorded activity by RadioBanter: Jun 2010
Posts: 1
Default

Quote:
Originally Posted by walt View Post
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.
  #7   Report Post  
Old June 7th 10, 07:00 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Aug 2006
Posts: 572
Default what happens to reflected energy ?

On Jun 6, 1:01*pm, 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.


David, you are preaching to the choir. I explained before that there
is NO interaction between forward and reflected waves because they are
traveling in different directions. The only time that coherent,
collimated waves can interact is when they are traveling in the same
direction in a transmission line.

Keith's argument requires that forward waves and reflected waves
interact. I say they cannot interact in a constant Z0 environment. You
say they cannot interact. We are on the same side of this argument.

The mechanism of the interaction of two coherent, collimated waves
traveling in the same direction is that the superposition process is
irreversible. The source photons and the reflected photons are
indistinguishable.
--
73, Cecil, w5dxp.com
  #8   Report Post  
Old June 7th 10, 11:21 PM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Apr 2010
Posts: 484
Default what happens to reflected energy ?

On Jun 7, 5:00*pm, Cecil Moore wrote:
On Jun 6, 1:01*pm, 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.


David, you are preaching to the choir. I explained before that there
is NO interaction between forward and reflected waves because they are
traveling in different directions. The only time that coherent,
collimated waves can interact is when they are traveling in the same
direction in a transmission line.

Keith's argument requires that forward waves and reflected waves
interact. I say they cannot interact in a constant Z0 environment. You
say they cannot interact. We are on the same side of this argument.

The mechanism of the interaction of two coherent, collimated waves
traveling in the same direction is that the superposition process is
irreversible. The source photons and the reflected photons are
indistinguishable.
--
73, Cecil, w5dxp.com


This is what YOU said:

What happens to the 50 joules/second of reflected energy depends upon
the phasing between the source wave and the reflected wave at the
source impedance. What most amateurs don't understand is there are
two mechanisms that can redistribute reflected energy back toward the
antenna. Those mechanisms are a re-reflection based on the physical
reflection coefficient (what RF engineers understand) and wave
interaction resulting in constructive/destructive interference (what
most RF engineers don't seem to understand) because, unlike optical
physicists, have not been forced to follow the energy flow.


There is no interaction between the waves, they may be superimposed,
and maybe their photons are indistinguishable, but there is no
'interaction'. interaction implies that one wave affects the other,
energy is transferred or fields are affected, such things occur in non-
linear media, but not in linear ones.

  #9   Report Post  
Old June 8th 10, 05:03 AM posted to rec.radio.amateur.antenna
external usenet poster
 
First recorded activity by RadioBanter: Aug 2006
Posts: 572
Default what happens to reflected energy ?

On Jun 7, 4:21*pm, K1TTT wrote:
There is no interaction between the waves, they may be superimposed,
and maybe their photons are indistinguishable, but there is no
'interaction'. *interaction implies that one wave affects the other,
energy is transferred or fields are affected, such things occur in non-
linear media, but not in linear ones.


Note that I was not talking about forward and reflected waves in a
constant Z0 transmission line. I am talking about the four wavefront
components that are generated at an impedance discontinuity. It has
been proven in experimentally that those waves indeed interact. In
fact, the transfer of destructive interference energy from wave
cancellation to the areas that permit constructive interference is
obviously interaction since the canceled waves disappear in their
original direction of travel. But, as the FSU web page says, they are
not annihilated - their energy components simply change direction.

How can you possibly argue that wave cancellation doesn't require wave
interaction? Those two waves completely disappear in the direction of
destructive interference. Dr. Best argued that those two waves don't
interact and continue propagating (completely devoid of energy)
forever in the direction of original travel. I asked him to prove that
his phantom waves exist but he could not. Is there such a proof
available?

At a 1/4WL thin-film coating on non-reflective glass, when the
internal reflected wave arrives with equal magnitude and 180 degrees
out of phase with the external reflection, wave cancellation occurs.
That is an *obvious* effect that one wave has on the other. Wave
cancellation is an obvious interaction. In the s-parameter equation:

b1 = s11*a1 + s12*a2 = 0

the s11*a1 wavefront has obviously interacted with the s12*a2
wavefront to accomplish wave cancellation.
--
73, Cecil, w5dxp.com
Reply
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules

Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Reflected Energy Cecil Moore Antenna 12 November 19th 04 10:01 PM
Reflected power ? Henry Kolesnik Antenna 328 June 9th 04 02:41 AM


All times are GMT +1. The time now is 10:48 AM.

Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
Copyright ©2004-2025 RadioBanter.
The comments are property of their posters.
 

About Us

"It's about Radio"

 

Copyright © 2017