The previous thread got too big and convoluted for me so I started another
if for no other reason than my own clarity and convenience. It's still the
same question "I know that any power not dissipated by an antenna is
reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt".

Before going to bed I got out the book REFLECTIONS II by Walt Maxwell,
W2DU. I'm typing verbatim from page 2-2 and 23-1 for those who don't have
the book.

Page 2-2 : "Contrary to what many believe, it is not true that when a
transmitter delivers power into a line with reflections, a returning wave
sees an internal generator resistance as a dissipative load. Nor is the
reflected wave converted to heat and, while at the same time damaging the
final amplifier....the reflected power is entirely conserved...."

From page 23-1 "One of the most serious misconceptions concerned reflected
power reaching the tubes in the RF amplifier of the transmitter. The
prevalent, but erroneous thinking was that the reflected power enters the
amplifier, causing tube overheating and destruction. However, I dispelled
this misconception in the above mentioned publications, using wave-mechanics
treatment, discussed here in greater detail, by showing that when the
pi-network tank is tuned to resonance, a virtual short circuit to rearward
traveling waves is created at the input of the network. Consequently,
instead of the reflected power reaching the tubes of the amplifier, it is
totally re-reflected toward the load by the virtual short circuit appearing
only to waves at the network input".

I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR

"Henry Kolesnik" wrote in message
. ..
The previous thread got too big and convoluted for me so I started another
if for no other reason than my own clarity and convenience. It's still the
same question "I know that any power not dissipated by an antenna is
reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open or
a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt".

Before going to bed I got out the book REFLECTIONS II by Walt Maxwell,
W2DU. I'm typing verbatim from page 2-2 and 23-1 for those who don't
have
the book.

Page 2-2 : "Contrary to what many believe, it is not true that when a
transmitter delivers power into a line with reflections, a returning wave
sees an internal generator resistance as a dissipative load. Nor is the
reflected wave converted to heat and, while at the same time damaging the
final amplifier....the reflected power is entirely conserved...."

From page 23-1 "One of the most serious misconceptions concerned
reflected
power reaching the tubes in the RF amplifier of the transmitter. The
prevalent, but erroneous thinking was that the reflected power enters the
amplifier, causing tube overheating and destruction. However, I dispelled
this misconception in the above mentioned publications, using
wave-mechanics
treatment, discussed here in greater detail, by showing that when the
pi-network tank is tuned to resonance, a virtual short circuit to rearward
traveling waves is created at the input of the network. Consequently,
instead of the reflected power reaching the tubes of the amplifier, it is
totally re-reflected toward the load by the virtual short circuit
appearing
only to waves at the network input".

I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR



Henry,

I had posted the following on the other thread, about the time you started
the new one.

Tam
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line the reflected wave will be the same amplitude as the generator, there
will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know" what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Henry Kolesnik wrote:

The previous thread got too big and convoluted for me so I started another
if for no other reason than my own clarity and convenience. It's still the
same question "I know that any power not dissipated by an antenna is
reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt".

Hank, Here's what happens at the match point in an amateur radio antenna
system. Walter Maxwell himself references J. C. Slater's book.

From _Microwave_Transmission_, by J. C. Slater: "We can get a better
understanding of (matching) devices by considering reflections at
discontinuities. This method is useful in considering any impedance-
matching device, and we shall think of it first in connection with the
ordinary quarter-wave transformer. The object of this transformer is
to eliminate the reflection that would be present if the impedance
Z1 were connected directly to Zt."

The diagram is: Z1----1/4WL Z0----Zt where Z0 is a 1/4WL transformer
section that matches Z1 to Zt.

"The method of eliminating reflections is based on the interference
between waves. Two waves half a wavelength apart are in opposite
phases, and the sum of them, if their amplitudes are numerically
equal, is zero. The fundamental principle behind the elimination of
reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase. In order that this second
wave may have traveled half a wavelength farther than the first, it is
obvious that it must have gone a quarter wavelength farther up the line,
and correspondingly a quarter of a wavelength back, before it meets the
original reflected wave. In other words, two discontinuities in characteristic
impedance, of such magnitude as to give equal amplitudes of reflected waves
and spaced a quarter of a wavelength apart, will give no net reflection
and hence will not introduce reflections into the line."

Here's how to perform an experiment to answer your questions. The source
(SGCL) is a Signal Generator equipped with a Circulator and Load. It puts
out a constant 100 watts. All reflected power is dissipated in the circulator
load.
very short
100W SGCL---50 ohm line---tuner----------450 ohm line------------Load

Initially, the system is tuned for a match so there are no reflections
on the 50 ohm line. Then the load is changed to 450 ohms. Assuming the
load was not 450 ohms to start with, this will cause reflections to
appear on the 50 ohm line. We can use an oscilloscope to measure the
magnitude and phase of the voltage across the 50 ohm circulator load
resistor. That's the reflected voltage. We can calculate the reflected
current through the 50 ohm circulator load resistor. So now we have the
component voltage and current reflected from the input to the tuner.
We can reference the phase of the reflected voltage and current to
the phase of the source voltage. Let's say the source voltage has a
reference phase of zero degrees.

Assume we measured a reflected voltage of 30 volts at 10 degrees and
calculated a reflected current of 0.6 amps at 190 degrees. (Note that
the reflected current is always 180 degrees out of phase with the
reflected voltage.)

Now switch the load from 450 ohms back to the original load. What
happens at the tuner input to achieve a match is the superposition
of two reflected waves. We know what the first one looks like so
we deduce what the second one looks like.

The first reflected wave, reflected from match point at the
tuner input is:

30 volts at 10 degrees and 0.6 amps at 190 degrees

The second reflected wave, returning from the load and reaching
the match point at the tuner input is:

30 volts at 190 degrees and 0.6 amps at 10 degrees

Adding those two voltages yields zero volts. Adding those two
currents yields zero amps. Thus the match is accomplished.

What the tuner does is shift the magnitude and phase of the voltage
and current reflected from the load to cancel the reflection of the
source wave that occurs at the input of the tuner.

The reflected voltage flowing toward the source is canceled to zero
through superposition of the two reflected voltage waves.

The reflected current flowing toward the source is canceled to zero
through superposition of the two reflected current waves.

That's how all matching devices work according to J. C. Slater.
--
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! =-----

I hate this reverse posting, so I'm changing.

First Henry (Thinks for the swell phone conversation) has the idea in basic
form:

Henry Kolesnik wrote:

[...snip...] "I know that any power not dissipated by an antenna is
reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone.
....I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt".

Hank uses the term "virtual open" a rather loose term but a good model,
however, keep in mind that he is talking about a system with NO "matching
device" as we call it, just a transmitter. He is also talking about the
REVERSE DIRECTION where all the power gets reflected back to the load and we
all live happily after. (the same concept applies for a matching device, as
well)

Then in a longer version, Cecil quotes (see below), but he is referring to a
matching device. He is also directing his attention to the forward
direction, unlike Hank above, and he moves to the matching device's input.
What the concept here is, concerns what the forward wave sees at the
matching input so that there is no net reflected power. The concept is
this:
One wave from the transmitter (call it wave "A") hits the matching device.
Some makes it through toward the load and some is reflected back toward the
Tx due to the matching device's weird impedance...HOWEVER, the wave coming
from the load has some component which makes it through the matching section
to the input (call this wave "B"). It is wave "A" abd wave B"B which must
be equal in amplitude and 180 degrees out - therefore cencelling for a net
wave back toward the Tx of null, naught, zip..all is happy in the world of
ham radio

--
Steve N, K,9;d, c. i My email has no u's.

P.S. Get this MPT blockage out of your minds... The "maximum power therom"
(ZL=Zs) ONLY applies to ONE special case, NOT all cases. That case is where
the source's output power (or if you like current) capability is limited
ONLY by the two resistances. That is, the case is when the source can put
out all the power needed by these resistors and no other internal limit
dominates. A common circuit can be shown to give maximum power at other
than Zs=ZL (aparently violating the above referred-to therom). There are
things other than these resistances that limit the output power of a
practical source.... (see how long this thread goes.....



"Cecil Moore" wrote in message
...

Hank, Here's what happens at the match point in an amateur radio antenna
system. Walter Maxwell himself references J. C. Slater's book.

From _Microwave_Transmission_, by J. C. Slater: "We can get a better
understanding of (matching) devices by considering reflections at
discontinuities. This method is useful in considering any impedance-
matching device, and we shall think of it first in connection with the
ordinary quarter-wave transformer. The object of this transformer is
to eliminate the reflection that would be present if the impedance
Z1 were connected directly to Zt."

The diagram is: Z1----1/4WL Z0----Zt where Z0 is a 1/4WL transformer
section that matches Z1 to Zt.

"The method of eliminating reflections is based on the interference
between waves. Two waves half a wavelength apart are in opposite
phases, and the sum of them, if their amplitudes are numerically
equal, is zero. The fundamental principle behind the elimination of
reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase. In order that this second
wave may have traveled half a wavelength farther than the first, it is
obvious that it must have gone a quarter wavelength farther up the line,
and correspondingly a quarter of a wavelength back, before it meets the
original reflected wave. In other words, two discontinuities in
characteristic
impedance, of such magnitude as to give equal amplitudes of reflected
waves
and spaced a quarter of a wavelength apart, will give no net reflection
and hence will not introduce reflections into the line."

Here's how to perform an experiment to answer your questions. The source
(SGCL) is a Signal Generator equipped with a Circulator and Load. It puts
out a constant 100 watts. All reflected power is dissipated in the
circulator
load.
very short
100W SGCL---50 ohm line---tuner----------450 ohm line------------Load

Initially, the system is tuned for a match so there are no reflections
on the 50 ohm line. Then the load is changed to 450 ohms. Assuming the
load was not 450 ohms to start with, this will cause reflections to
appear on the 50 ohm line. We can use an oscilloscope to measure the
magnitude and phase of the voltage across the 50 ohm circulator load
resistor. That's the reflected voltage. We can calculate the reflected
current through the 50 ohm circulator load resistor. So now we have the
component voltage and current reflected from the input to the tuner.
We can reference the phase of the reflected voltage and current to
the phase of the source voltage. Let's say the source voltage has a
reference phase of zero degrees.

Assume we measured a reflected voltage of 30 volts at 10 degrees and
calculated a reflected current of 0.6 amps at 190 degrees. (Note that
the reflected current is always 180 degrees out of phase with the
reflected voltage.)

Now switch the load from 450 ohms back to the original load. What
happens at the tuner input to achieve a match is the superposition
of two reflected waves. We know what the first one looks like so
we deduce what the second one looks like.

The first reflected wave, reflected from match point at the
tuner input is:

30 volts at 10 degrees and 0.6 amps at 190 degrees

The second reflected wave, returning from the load and reaching
the match point at the tuner input is:

30 volts at 190 degrees and 0.6 amps at 10 degrees

Adding those two voltages yields zero volts. Adding those two
currents yields zero amps. Thus the match is accomplished.

What the tuner does is shift the magnitude and phase of the voltage
and current reflected from the load to cancel the reflection of the
source wave that occurs at the input of the tuner.

The reflected voltage flowing toward the source is canceled to zero
through superposition of the two reflected voltage waves.

The reflected current flowing toward the source is canceled to zero
through superposition of the two reflected current waves.

That's how all matching devices work according to J. C. Slater.
--
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! =-----

Richard Clark has some good points. My 30S-1 has load and tune still in it
and the final a 3CX1500B is ceramic and can't be seen but it never lost
it's Eimac stencil. However my Collins 30L-1 was a different story. Once
or twice I had 811As with holes melted in their plate structure. They
continued to work well and I always thought this was from overdriving but
now I'm not so sure because it ran at the same high swr as the 30S-1. I
used the 30L-1 in summer because the air conditioner couldn't keep up with
the heat generated by the 30S-1 so it was my winter amp. I've seen 3-500Z
and 4-1000 plates running red hot but I think it was from overdrive and not
SWR but really wan't paying enough attention, I wish I had.
I notice that Walt Maxwell's Reflections II was published by World Radio
whereas Reflections was by the ARRL. There is some disagreement on
"conjugate match'" and probably other things between ARRL and Walt. So now
I have to wonder if the final tube is dissipative or non-dissipative for my
reflections and not Walts. Pun unavoidable.

Looks like my new thread isn't working but I'm posting this to it just in
case since I don't really know the customary protocol on a long thread that
sometime digresses.

--
73
Hank WD5JFR
"Steve Nosko" wrote in message
...
I hate this reverse posting, so I'm changing.

First Henry (Thinks for the swell phone conversation) has the idea in
basic
form:

Henry Kolesnik wrote:

[...snip...] "I know that any power not dissipated by an antenna is
reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone.
....I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt".

Hank uses the term "virtual open" a rather loose term but a good model,
however, keep in mind that he is talking about a system with NO "matching
device" as we call it, just a transmitter. He is also talking about the
REVERSE DIRECTION where all the power gets reflected back to the load and
we
all live happily after. (the same concept applies for a matching device,
as
well)

Then in a longer version, Cecil quotes (see below), but he is referring to
a
matching device. He is also directing his attention to the forward
direction, unlike Hank above, and he moves to the matching device's input.
What the concept here is, concerns what the forward wave sees at the
matching input so that there is no net reflected power. The concept is
this:
One wave from the transmitter (call it wave "A") hits the matching device.
Some makes it through toward the load and some is reflected back toward
the
Tx due to the matching device's weird impedance...HOWEVER, the wave coming
from the load has some component which makes it through the matching
section
to the input (call this wave "B"). It is wave "A" abd wave B"B which must
be equal in amplitude and 180 degrees out - therefore cencelling for a net
wave back toward the Tx of null, naught, zip..all is happy in the world of
ham radio

--
Steve N, K,9;d, c. i My email has no u's.

P.S. Get this MPT blockage out of your minds... The "maximum power therom"
(ZL=Zs) ONLY applies to ONE special case, NOT all cases. That case is
where
the source's output power (or if you like current) capability is limited
ONLY by the two resistances. That is, the case is when the source can put
out all the power needed by these resistors and no other internal limit
dominates. A common circuit can be shown to give maximum power at other
than Zs=ZL (aparently violating the above referred-to therom). There are
things other than these resistances that limit the output power of a
practical source.... (see how long this thread goes.....



"Cecil Moore" wrote in message
...

Hank, Here's what happens at the match point in an amateur radio antenna
system. Walter Maxwell himself references J. C. Slater's book.

From _Microwave_Transmission_, by J. C. Slater: "We can get a better
understanding of (matching) devices by considering reflections at
discontinuities. This method is useful in considering any impedance-
matching device, and we shall think of it first in connection with the
ordinary quarter-wave transformer. The object of this transformer is
to eliminate the reflection that would be present if the impedance
Z1 were connected directly to Zt."

The diagram is: Z1----1/4WL Z0----Zt where Z0 is a 1/4WL transformer
section that matches Z1 to Zt.

"The method of eliminating reflections is based on the interference
between waves. Two waves half a wavelength apart are in opposite
phases, and the sum of them, if their amplitudes are numerically
equal, is zero. The fundamental principle behind the elimination of
reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase. In order that this second
wave may have traveled half a wavelength farther than the first, it is
obvious that it must have gone a quarter wavelength farther up the line,
and correspondingly a quarter of a wavelength back, before it meets the
original reflected wave. In other words, two discontinuities in
characteristic
impedance, of such magnitude as to give equal amplitudes of reflected
waves
and spaced a quarter of a wavelength apart, will give no net reflection
and hence will not introduce reflections into the line."

Here's how to perform an experiment to answer your questions. The source
(SGCL) is a Signal Generator equipped with a Circulator and Load. It
puts
out a constant 100 watts. All reflected power is dissipated in the
circulator
load.
very short
100W SGCL---50 ohm line---tuner----------450 ohm line------------Load

Initially, the system is tuned for a match so there are no reflections
on the 50 ohm line. Then the load is changed to 450 ohms. Assuming the
load was not 450 ohms to start with, this will cause reflections to
appear on the 50 ohm line. We can use an oscilloscope to measure the
magnitude and phase of the voltage across the 50 ohm circulator load
resistor. That's the reflected voltage. We can calculate the reflected
current through the 50 ohm circulator load resistor. So now we have the
component voltage and current reflected from the input to the tuner.
We can reference the phase of the reflected voltage and current to
the phase of the source voltage. Let's say the source voltage has a
reference phase of zero degrees.

Assume we measured a reflected voltage of 30 volts at 10 degrees and
calculated a reflected current of 0.6 amps at 190 degrees. (Note that
the reflected current is always 180 degrees out of phase with the
reflected voltage.)

Now switch the load from 450 ohms back to the original load. What
happens at the tuner input to achieve a match is the superposition
of two reflected waves. We know what the first one looks like so
we deduce what the second one looks like.

The first reflected wave, reflected from match point at the
tuner input is:

30 volts at 10 degrees and 0.6 amps at 190 degrees

The second reflected wave, returning from the load and reaching
the match point at the tuner input is:

30 volts at 190 degrees and 0.6 amps at 10 degrees

Adding those two voltages yields zero volts. Adding those two
currents yields zero amps. Thus the match is accomplished.

What the tuner does is shift the magnitude and phase of the voltage
and current reflected from the load to cancel the reflection of the
source wave that occurs at the input of the tuner.

The reflected voltage flowing toward the source is canceled to zero
through superposition of the two reflected voltage waves.

The reflected current flowing toward the source is canceled to zero
through superposition of the two reflected current waves.

That's how all matching devices work according to J. C. Slater.
--
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! =-----

Steve Nosko wrote:
One wave from the transmitter (call it wave "A") hits the matching device.
Some makes it through toward the load and some is reflected back toward the
Tx due to the matching device's weird impedance...HOWEVER, the wave coming
from the load has some component which makes it through the matching section
to the input (call this wave "B"). It is wave "A" abd wave B"B which must
be equal in amplitude and 180 degrees out - therefore cencelling for a net
wave back toward the Tx of null, naught, zip..all is happy in the world of
ham radio

Yep, you got it! J. C. Slater stated such in _Microwave_Transmissions_
before most present day hams were born. I'm glad you understand.
--
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! =-----

And I didn't read any of the stuff any of you quote from. Just using a
solid understanding of fundamentals and applying them correctly.

We'll see whom else (dis)agrees...

Thanks for the stimulating subject. Gets the remainder of the gray matter
really warm...
73, Steve

--
Steve N, K,9;d, c. i My email has no u's.

No single raindrop believes it is responsible for the flood.


"Cecil Moore" wrote in message
...
Steve Nosko wrote:
One wave from the transmitter (call it wave "A") hits the matching
device.
Some makes it through toward the load and some is reflected back toward
the
Tx due to the matching device's weird impedance...HOWEVER, the wave
coming
from the load has some component which makes it through the matching
section
to the input (call this wave "B"). It is wave "A" abd wave B"B which
must
be equal in amplitude and 180 degrees out - therefore cencelling for a
net
wave back toward the Tx of null, naught, zip..all is happy in the world
of
ham radio

Yep, you got it! J. C. Slater stated such in _Microwave_Transmissions_
before most present day hams were born. I'm glad you understand.
--
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! =-----