If you have a resonant antenna, supposedly all of the power is radiated.

(SWR 1:1)

If you have a nonresonant antenna, some of the power is reflected back to
the transmitter.

(SWR 1:1)

If you connect a tuner and it is "tuned", none of the power is reflected
back to the transmitter.

(SWR at transmitter 1:1, at antenna still 1:1)

Obviously it has to go somewhere.

Where?

If you are designing a tuner, where would you design it to go?

Thanks in advance

Geoff.

--
Geoffrey S. Mendelson, Jerusalem, Israel N3OWJ/4X1GM
I do multitasking. If that bothers you, file a complaint and I will start
ignoring it immediately.

"Geoffrey S. Mendelson" wrote in
:

Geoffrey,

I have bad news for you, your post is based heavily on misconceptions,
archetypal ham myths even.

I wrote a short article entitled "A simple VSWR analysis without mirrors"
at http://vk1od.net/blog/?p=1259 , which analyses a mismatched line example
without resorting to dodgy concepts often (and almost exlusively) employed
by hams. The article does deal with why there is a reflected wave, and the
result of that reflected wave on line loss and transmitter loading.

IMHO, you would do best by starting with a reputable text book, and
studying the topic. Certainly, the S/N ratio in ham forums is low, 'facts'
seem to be determined by vote, and your wrong concepts are more likely to
be reinforced than seriously challenged.

Owen

Geoffrey S. Mendelson Inscribed thus:

If you have a resonant antenna, supposedly all of the power is
radiated.

Yes

(SWR 1:1)

With respect to what !

If you have a nonresonant antenna, some of the power is reflected back
to the transmitter.

(SWR 1:1)

If you connect a tuner and it is "tuned", none of the power is
reflected back to the transmitter.

(SWR at transmitter 1:1, at antenna still 1:1)

Obviously it has to go somewhere.

Where?

Its radiated, less losses.

If you are designing a tuner, where would you design it to go?

That would depend upon desired system characteristics.

Thanks in advance

Geoff.


A antenna at resonance does not necessarily match the feeder impedance !
The whole idea of a tuner is to make one end match the other.
Stating a VSWR of unity implies a 50 ohm measurement system, which an
antenna at resonance rarely is.

73's
--
Best Regards:
Baron.

Geoffrey S. Mendelson wrote:
If you have a resonant antenna, supposedly all of the power is radiated.

(SWR 1:1)

If you have a nonresonant antenna, some of the power is reflected back to
the transmitter.

(SWR 1:1)

If you connect a tuner and it is "tuned", none of the power is reflected
back to the transmitter.

(SWR at transmitter 1:1, at antenna still 1:1)

Obviously it has to go somewhere.

Where?

If you are designing a tuner, where would you design it to go?

Thanks in advance

Geoff.

All of the power produced by a transmitter is either radiated or
dissipated as heat, period. But you'll never understand where the
"reflected power" goes until you understand where it comes from.

Consider a 100 watt transmitter connected to a 50 ohm dummy load or
antenna via a half wavelength of 50 ohm lossless transmission line.

The transmitter sees an impedance of 50 ohms resistive.
The transmitter delivers 100 watts to the transmission line.
The transmission line delivers 100 watts to the load.
The load dissipates 100 watts.
The VSWR on the line is 1:1
The "forward power" in the transmission line is 100 watts.
The "reverse power" in the transmission line is zero.

Ok so far? Now replace the transmission line with one having 200 ohm
impedance.

The transmitter sees an impedance of 50 ohms resistive.
The transmitter delivers 100 watts to the transmission line.
The transmission line delivers 100 watts to the load.
The load dissipates 100 watts.
The VSWR on the line is 4:1.
The "forward power" in the transmission line is 156.25 watts.
The "reverse power" in the transmission line is 56.25 watts.

By changing the line impedance, we somehow created an extra 56.25 watts
of "forward power". Neither the transmitter nor the load was aware of
this wonderful event, since the transmitter sees the same load impedance
as before, and the load sees the same source impedance. The transmitter
is delivering exactly the same amount of power as before, and the load
is dissipating the same amount. (If the transmission line had loss, it
would have increased very slightly, but probably not enough to measure,
and certainly much, much less than 56 watts.)

Likewise, neither the transmitter nor the load knows anything about the
new "reverse power" which was created at the same time.

So the answer to your question is that the "reverse power" goes to
wherever the new excess "forward power" comes from. Once you figure out
where that is, you'll have a better understanding of the topic than
nearly everyone who has been arguing on this forum about it for years.

As for a tuner, it has the magical property of having different amounts
of "forward power" and "reverse power" at its input and output when the
amount of actual power flowing is the same on both sides (neglecting
tuner loss). So it can create or destroy "forward power" as well as
"reverse power" by a simple twist of its knobs. Little does the tuner,
transmitter, or load know that as soon as "reverse power" is created,
interminable arguments will take place debating about where it goes.

I now return you to your regularly scheduled programming. . .

Roy Lewallen, W7EL

On Jun 9, 2:44*pm, Roy Lewallen wrote:
So the answer to your question is that the "reverse power" goes to
wherever the new excess "forward power" comes from. Once you figure out
where that is, you'll have a better understanding of the topic than
nearly everyone who has been arguing on this forum about it for years.

Let's be careful to point out that "forward power" is a measurement
*at a fixed point* on a transmission line of the rate of flow of the
forward energy which is traveling at the speed of light in the medium.
Same concept for "reflected power" - reflected energy flowing in the
reverse direction at the speed of light in the medium. EM waves cannot
stand still.

It's easy to figure out where the excess forward energy comes from
especially in an ideal lossless system which I will assume for the
rest of this posting. Consider the following thought experiment. An
ideal lossless system with a one-second long Z0=300 ohm feedline. Why
one-second long? Because the watts (joules/second) and the joules will
be the same magnitude, i.e. If it's a matched line with 100 watts of
forward power, it is easy to show that a one-second feedline has 100
joules of energy in it that has been sourced but not yet delivered to
the load - like a delay line.

Here's an example of a Z0-matched system with an SWR of 6:1 on the
Z0=300 ohm feedline. *Steady-state* conditions a forward power =
204 watts and reflected power = 104 watts. Power delivered to the load
= 100 watts. There is a steady-state Z0-match at '+'. Zero reflected
energy reaches the source, i.e. the SWR on the 50 ohm line is 1:1.

100w source---50 ohm line---+---1-sec-long 300 ohm line---50 ohm load

During the transient key-down state, the source supplies a number of
joules to the transmission line that do not reach the load. At the
beginning of steady-state that value is 204 joules plus 104 joules
equals 308 joules.

When steady-state is reached there are 308 joules of energy "stored"
in the one second long transmission line that have not been delivered
to the load. Only after the transmission line is charged with steady-
state energy does the load start to accept the same amount of power as
is being sourced. There is no magic here - just a simple accounting
for all the energy. Why otherwise intelligent, educated RF engineers
cannot perform that simple second-grade math is a subject for another
thread.

During steady-state, there are 308 joules of energy "stored" in the
transmission line that are performing the Z0-match function. There are
204 joules in the forward wave and 104 joules in the reflected wave.
Since RF cannot actually be stored as in a battery, this energy is
moving at the speed of EM waves - the speed of light in the medium.
104 joules/second is continuously being reflected at the load. 104
joules/second is continuously being redistributed back toward the load
at the Z0-match.

At key-up, steady-state ends and the key-up transient state occurs.
During this transient state, all of the 308 joules in the forward wave
and reflected wave during steady-state will be dissipated. During the
key-down transient state, 308 joules are stored in the one second long
transmission line. During the key-up transient state, the 308 joules
are dissipated. For those who feel overwhelmed, it is just simple
second-grade math

In a Z0-matched system, the entire analysis can be performed using an
energy analysis instead of a voltage analysis. The common way to
analyze the above system is to assume that 70.7 volts is present on
the 50 ohm side of the Z0-match with 1.414 amps flowing into the Z0-
match. But with only an energy analysis, we can reverse-engineer the
voltage and current conditions at any point in a Z0-matched system.

Fact of Physics: There are exactly enough joules of energy in the
mismatched transmission line to support the forward wave power and the
reflected wave power. It's not magic nor rocket science. Anyone
willing to count the joules that have not reached the load will
realize that the energy in the forward and reflected waves consists of
real-world joules that cannot be swept under the rug just because some
alleged RF guru is ignorant of the laws of physics.

Here's a simple math problem for all of the readers having trouble
with this concept. On 10 MHz, we have a 50 ohm antenna being fed with
one wavelength of lossless 300 ohm line from a 100w source. During
steady-state, how many microjoules of energy are "stored" in the
transmission line that have not yet reached the load? How are those
microjoules split between the forward wave and reflected wave?
--
73, Cecil, w5dxp.com

On Jun 9, 2:44*pm, Roy Lewallen wrote:

Consider a 100 watt transmitter connected to a 50 ohm dummy load or
antenna via a half wavelength of 50 ohm lossless transmission line.
...
(clipped)
...
Ok so far? Now replace the transmission line with one having
200 ohm impedance.

The transmitter sees an impedance of 50 ohms resistive.
The transmitter delivers 100 watts to the transmission line.
The transmission line delivers 100 watts to the load.
The load dissipates 100 watts.
The VSWR on the line is 4:1.
The "forward power" in the transmission line is 156.25 watts.
The "reverse power" in the transmission line is 56.25 watts.

Could you or someone please post the same analysis when the electrical
length of the 200 ohm transmission line is 1/4-wave rather than 1/2-
wave, including the physical locations/components where r-f output
power is dissipated?

RF

Richard Fry wrote in news:32a66fcc-a742-45c6-bdf1-
:
....

Could you or someone please post the same analysis when the electrical
length of the 200 ohm transmission line is 1/4-wave rather than 1/2-
wave, including the physical locations/components where r-f output
power is dissipated?

Well Richard, since the example discussion implies lossless line, you
know that zero power is dissipated in line losses.

Assuming that you are talking steady state...

If literally, you changed the line length in Roy's example to an
electrical quarter wave, then the transmitter will see a load impedance
of 800+j0, and it will develop exactly the same power in the actual load
(quarter wave 200 ohm line+50+j0 load) as it would in a 800 ohm resistor
connected to its output terminals. Since the line is lossless, 100% of
the actual transmitter output power is absorbed by the load, dissipated
if the load was a resistor. If you were to plot the instantaneous power
at the transmitter output terminals vs time, it would always be positive,
ie there would not be a reactive component.

The transmitter output power is probably different to that developed in a
50+j0 load for several reasons, but there is not enough information to
determine that output power.

Of course, a practical line has line loss, and the heat generated under
standing waves varies along the line (ie W/m is not constant), and the
impedance seen by the transmitter will be a little different (including
being reactive) due to the fact that Zo is not real, the VSWR is not 4,
and effects of line attenuation, so actual transmitter power may be a
little different.

Owen

On Jun 11, 6:16*am, Owen Duffy wrote:

... The transmitter output power is probably different ...

Thank you, Owen.

Do your comments apply to a transmitter designed/adjusted for, and
expecting a 50 + j 0 ohm load?

IOW, if the net output power of such a transmitter (which equals that
dissipated in the load) probably is different with such a mismatch, do
you expect the reason for that to be related to "reflected power?"

RF

On Jun 11, 11:32*am, Richard Fry wrote:
On Jun 11, 6:16*am, Owen Duffy wrote:

*... The transmitter output power is probably different ...

Thank you, Owen.

Do your comments apply to a transmitter designed/adjusted for, and
expecting a 50 + j 0 ohm load?

IOW, if the net output power of such a transmitter (which equals that
dissipated in the load) probably is different with such a mismatch, do
you expect the reason for that to be related to "reflected power?"

RF

NO! NO! NO! the difference is not due to 'reflected power'... any
difference is due to the impedance change seen by the transmitter at
its terminals.

A VERY important method of analyzing sinusoidal steady state response
of a system like a transmission line with a load on the end takes the
load impedance and transforms it using the transmission line equations
(with or without loss) back to the source terminals and then replacing
it with an equivalent impedance. as long as the load is constant and
linear this is a well known and easily proved substitution. you can
then solve for conditions seen by the source ignoring the length of
the line and load and using only a lumped impedance, therefore any
reflected wave is irrelevant when analyzing the response of the
source.

Richard Fry wrote:
On Jun 11, 6:16 am, Owen Duffy wrote:

... The transmitter output power is probably different ...

Thank you, Owen.

Do your comments apply to a transmitter designed/adjusted for, and
expecting a 50 + j 0 ohm load?

IOW, if the net output power of such a transmitter (which equals that
dissipated in the load) probably is different with such a mismatch, do
you expect the reason for that to be related to "reflected power?"

RF

We've had this discussion many times before, but with no apparent change
in your perceptions of the nature of "reflected power".

But why would you expect the "reflected power" to have a different
effect when the transmission line is a quarter wavelength than when it's
a half wavelength? Can you write an equation giving the supposed
dissipation caused by "reflected power" as a function of transmission
line length?

Roy Lewallen, W7EL