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....
JC

On 6 jun, 16:22, "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....
JC

Hello,

If the source has a linear 50 Ohms output impedance (assuming 50 Ohm
cable, no loss), all power will go back into the source. Partly in the
form of heat, partly in the form of power saving. However a PA is not
a 50 Ohms source, so what happens may vary.

It is very likely that some reverse power reflects back to the antenna
(as the transmitter is not a source with 50 Ohms output impedance). So
the reading on an instrument that measures true forward power is the
sum of the real incident power plus the reflected power from the
transmitter towards the antenna. Changing the load may result in a
reduction or increase of total forward power. In case of a real 50
Ohms source, the forward power will not change, no matter the load.

The bad match seen from the transmitter may result in a reduction of
DC input power (input current reduces), but may also result in an
increase of DC input power.

So when you have reverse power reading of 50W, you cannot just say
that the PA has to dissipate an extra 50W, it can be more, less and
even negative. The negative case is when the input power reduces
significantly and the active element has to dissipate less (with
respect to the best match condition). In particular high efficiency
amplifiers (where the active devices are used as switches), show
strong variation in supply current versus load change. Under certain
load conditions such amplifiers may show a strong decrease in
efficiency resulting in high relative increase of device dissipation.

Best regards,

Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me

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

On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote:

However a PA is not a 50 Ohms source

Hi Wimpie,

You say this like others, with the air of "knowing." However, when I
ask in response of those who "know" what the PA is NOT, what IS it?

Give me the Z value of your transmitter. Specify all initial
conditions.

We have had lengthy correspondence with Walt Maxwell's very rigorously
measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly
that as is practicable (say +/- 20%); and yet your voice was missing
from this discussion with evidence to the contrary.

73's
Richard Clark, KB7QHC

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.

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, 10:22*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.....
JC

I assume that your numbers were obtained using a directional
wattmeter,
that is, forward power is 100W, 50W are reflected and by subtraction,
it is computed that 50W enter the antenna, some of which is dissipated
in the antenna losses and some of which is radiated.

The 100W forward and 50W reflected have no relation to actual powers
but
are simply values that are constructed so that when they are
subtracted
the result is the average power flowing towards the load.

From these numbers alone it is impossible to decide whether the system
is operating as it should or to compute where the losses might be, or
whether there are any losses at all. With an appropriate antenna tuner
it is entirely possible that the transmitter is delivering its design
output of 50W to the line and all of this energy is reaching the
antenna and being radiated.

Thus the whole question of "where does the reflected power go?", is
rather misleading.

....Keith

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