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.

On Jun 11, 9:13*am, K1TTT wrote:

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.

If the "forward" power that can be generated by a transmitter is
dependent on the degree of match of the output load to the load
specified for that transmitter, would that not mean that there would
be little/no risk of damage to the transmitter if it was operated into
a load with ~infinite SWR relative to the load it expects, such as a
short or open directly at its r-f output connector?

RF

On Jun 11, 10:45*am, K1TTT wrote:
On Jun 9, 10:38*pm, Jim Lux wrote:



In fact, the ration between that stored energy and the amount flowing
"through" (i.e. radiated away) is related to the directivity of the
antenna: high directivity antennas have high stored energy (large
magnetic and electric fields): *the ratio of stored to radiated energy
is "antenna Q" (analogous to the stored energy in a LC circuit leading
to resonant rise).

So, high directivity = high stored energy = high circulating energy =
high I2R losses.

this is a relationship i haven't heard of before... and would be very
wary of stating so simply. *it may be true for a specific type of
antenna, MAYBE Yagi's, MAYBE rhombics or or close coupled wire arrays,
but some of the most directive antennas are parabolic dishes which i
would expect to have very low Q and extremely low losses. *you could
also have an antenna with very high Q, very high i^2r losses, but very
low directivity, so i would be careful about drawing a direct link
between the two.

and yes, this does work for complex loads and multiple stubs and
connections to the line.

this is a reasonable description of the derivation of these
techniques, study especially the Thevenin equivalent impedance
representation on page 2-13 and how it is applied:
http://ee.sharif.edu/~comcir/readings/tran%20line.pdf

On Jun 11, 9:13*am, K1TTT wrote:
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.

Please let's be careful to give all the details. The *virtual*
impedance, the ratio of Vtotal/Itotal, seen by the transmitter at its
terminals is:

Z = Vtotal/Itotal = (Vfor + Vref)/(Ifor + Iref)

where all math is phasor (vector) math. Vref and Iref are components
of the reflected wave. So the mismatch is certainly related to the
magnitude and phase of the reflected wave. If the Z0 of the
transmission line is the impedance for which the transmitter was
designed, we can go as far as to say that the reflected wave causes
the mismatch, the virtual impedance that deviates from the Z0 of the
line. Before the reflected wave arrives back at the transmitter, the
transmitter sees the Z0 of the transmission line. The mismatch
develops when the reflected wave arrives.
--
73, Cecil, w5dxp.com

On Jun 11, 3:03*pm, Cecil Moore wrote:
On Jun 11, 9:13*am, K1TTT wrote:

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.

Please let's be careful to give all the details. The *virtual*
impedance, the ratio of Vtotal/Itotal, seen by the transmitter at its
terminals is:

Z = Vtotal/Itotal = (Vfor + Vref)/(Ifor + Iref)

where all math is phasor (vector) math. Vref and Iref are components
of the reflected wave. So the mismatch is certainly related to the
magnitude and phase of the reflected wave. If the Z0 of the
transmission line is the impedance for which the transmitter was
designed, we can go as far as to say that the reflected wave causes
the mismatch, the virtual impedance that deviates from the Z0 of the
line. Before the reflected wave arrives back at the transmitter, the
transmitter sees the Z0 of the transmission line. The mismatch
develops when the reflected wave arrives.
--
73, Cecil, w5dxp.com

i wouldn't call it a 'virtual' impedance, it is a very real impedance.
it is the steady state impedance seen by the transmitter at its output
terminals. once the transient response rings down the waves in the
line beyond the terminals are irrelevant and the constant impedance
can be used to calculate the voltage, current, and power in the
source. in the steady state (which except in special cases is
entirely adequate for amateur use) the mechanism of generating that
impedance is irrelevant, waves or not it can be represented by a
constant value and the source won't know the difference.

On Jun 11, 10:36*am, K1TTT wrote:
i wouldn't call it a 'virtual' impedance, it is a very real impedance.

Yes, it meets the (B) non-dissipative definition of impedance from
"The IEEE Dictionary". It doesn't meet the (A) dissipative definition
and it is not an impedor as it only exists as a V/I ratio. That's why
I call it a virtual impedance. There is no resistor, there is no
inductor, and there is no capacitor. There is only a V/I ratio caused
by something else.

it is the steady state impedance seen by the transmitter at its output
terminals. *

And it is a V/I ratio caused by the superposition of the forward wave
and reflected wave, i.e. it would NOT be the same impedance without
the reflected wave. That fact of physics is undeniable.

When we have an actual impedor, i.e. a resistor plus an inductor or a
capacitor, the voltage/current ratio is caused by the impedor. When we
have a virtual impedance, the cause/effect procedure is reversed and
the impedance is caused by the voltage/current ratio which may (or may
not) contain both forward and reflected values.

For the two definitions of impedance, (A) and (B), given in "The IEEE
Dictionary", cause and effect are reversed.

Let's take an example. Assume a 50 ohm load resistor fed with 1/2WL of
300 ohm lossless line. If we assume a 100 watt (50 ohm) source, the
forward power will be 204 watts and the reflected power will be 104
watts.

100w Source----1/2WL 300 ohm feedline----50 ohm load

In the feedline, a forward power of 204 watts is a forward voltage of
247.3 volts and a forward current of 0.825 amps.

In the feedline, a reflected power of 104 watts is a reflected voltage
of 176.6 volts and a reflected current of 0.589 amps.

The reflected voltage is 180 degrees out of phase with the forward
voltage at the transmitter, so the superposed voltage at the
transmitter is 70.7 volts. The reflected current is in phase with the
forward current at the transmitter so the superposed current is 1.414
amps. Since everything is either in phase or 180 degrees out of phase,
phasor addition is not needed. So I repeat, the impedance seen by the
transmitter is:

Z = (Vfor - Vref)/(Ifor + Iref)

Z = (247.3 - 176.6)/(0.825 + 0.589)

Z = 70.7/1.414 = 50 ohms, NON-DISSIPATIVE!

It is clear that the superposition of the forward wave with the
reflected wave is the CAUSE of the 50 ohm impedance. That particular
impedance cannot exist without the effects of the reflected wave.

Using a math model for answers to problems for so long that one comes
to believe that the model dictates reality is a major contributor to
the myths and old wives' tales that exist in amateur radio today. It's
time to get back to the basics even if it causes a few old rusty
brains to be put in gear for the first time in decades.
--
73, Cecil, w5dxp.com

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

On Jun 11, 4:18*pm, Cecil Moore wrote:
On Jun 11, 10:36*am, K1TTT wrote:

i wouldn't call it a 'virtual' impedance, it is a very real impedance.

Yes, it meets the (B) non-dissipative definition of impedance from
"The IEEE Dictionary". It doesn't meet the (A) dissipative definition
and it is not an impedor as it only exists as a V/I ratio. That's why
I call it a virtual impedance. There is no resistor, there is no

and that is part of the problem in here... 'you call it'. what is the
definition of 'virtual impedance' in the ieee dictionary?