Dr. Slick wrote:
"Therefore, the reflected voltage can never be greater than the input
voltage for a passive network, and the reflection coeficient can never
be greater than 1 for such a case."

A reflection coefficient implies that transmission lines long in terms
of wavelength may qualify as passive networks.

At the open-circuit end of a long transmission line, current is
interrupted by the open circuit. Having nowhere else to go, energy in
the current wave must be accepted by the voltage wave which doubles on
the spot. 2X the incident voltage is the sum of the voltage in the
incident wave and the voltage in the new reflected wave as they are
in-phase. The reflected voltage is no greater than the incident voltage.
It is equal to the incident voltage. It`s the sum that doubles.

Best regards, Richard Harrison, KB5WZI

Dr. Slick wrote:
"Therefore, the reflected voltage can never be greater than the input
voltage for a passive network, and the reflection coefficient can never
be greater than 1 for such a case."

Terman, despite Reg`s disdain for experts, seems to agree with Dr.
Slick. Here is Terman`s gist:
1. The reflected wave is identical to the incident wave except it
travels toward the generator.
2. Ereflected / I reflected = -Zo. Just as Eforward / I forward = Zo
3. Line loss causes the reflected wave to decline as it travels toward
the generator.
4. Phase of the reflected wave drops back as distance back from the load
increases.
5. Volts at the load are the sum of the incident and reflected wave
volts. Likewise for currents.
6. E/I at the load equals Zload.
7. The vector ratio Ereflected / E incident erquals rho, the reflection
coefficient..
8. In a lossless line, rho is the same everywhere on the line.
9. The effect of a reactive load is merely to displace the SWR pattern
on a transmission line.
There is no opportunity in the stated conditions on a transmission line
for a reflected voltage to exceed the incident voltage or for the
reflection coefficient to exceed one (1).

Best regards, Richard Harrison, KB5WZI

(Richard Harrison) wrote in message ...
Dr. Slick wrote:
"Therefore, the reflected voltage can never be greater than the input
voltage for a passive network, and the reflection coefficient can never
be greater than 1 for such a case."

Terman, despite Reg`s disdain for experts, seems to agree with Dr.
Slick. Here is Terman`s gist:
1. The reflected wave is identical to the incident wave except it
travels toward the generator.
2. Ereflected / I reflected = -Zo. Just as Eforward / I forward = Zo


Thank you Richard! You and Richard Clark should be named twin
pillars of reason in a sea of irrationality spurred by the Lust of
Always Being Right.

This is a very important point, that the reflected voltage will
be received by the same impedance that launched it to begin with.



3. Line loss causes the reflected wave to decline as it travels toward
the generator.
4. Phase of the reflected wave drops back as distance back from the load
increases.
5. Volts at the load are the sum of the incident and reflected wave
volts. Likewise for currents.
6. E/I at the load equals Zload.
7. The vector ratio Ereflected / E incident erquals rho, the reflection
coefficient..
8. In a lossless line, rho is the same everywhere on the line.
9. The effect of a reactive load is merely to displace the SWR pattern
on a transmission line.
There is no opportunity in the stated conditions on a transmission line
for a reflected voltage to exceed the incident voltage or for the
reflection coefficient to exceed one (1).

Best regards, Richard Harrison, KB5WZI



It's really an elegant and simple equation:

[s11]**2 + [s21]**2 = 1

For a lossless passive two port network, where the brackets
indicate magnitude only. If the sum of the squares is less than one,
the network has losses.


Slick

Everything's fine down to point 8, but it's for a line with loss, and
therefore with a characteristic impedance which is reactive, that you
get magnitude of rho greater than unity. You have all the info you
need to show that to yourself with the first few points and additional
fact that load current equals the sum of Ifwd and Irefl at the load.
Just work through the simple algebra to verify the well-known equation
for Vr/Vf, and then plug in Zo = 50-j5 and Zl = 1+j100 and evaluate
the magnitude of Vr/Vf at the load.

Cheers,
Tom

(Richard Harrison) wrote in message ...
Dr. Slick wrote:
"Therefore, the reflected voltage can never be greater than the input
voltage for a passive network, and the reflection coefficient can never
be greater than 1 for such a case."

Terman, despite Reg`s disdain for experts, seems to agree with Dr.
Slick. Here is Terman`s gist:
1. The reflected wave is identical to the incident wave except it
travels toward the generator.
2. Ereflected / I reflected = -Zo. Just as Eforward / I forward = Zo
3. Line loss causes the reflected wave to decline as it travels toward
the generator.
4. Phase of the reflected wave drops back as distance back from the load
increases.
5. Volts at the load are the sum of the incident and reflected wave
volts. Likewise for currents.
6. E/I at the load equals Zload.
7. The vector ratio Ereflected / E incident erquals rho, the reflection
coefficient..
8. In a lossless line, rho is the same everywhere on the line.
9. The effect of a reactive load is merely to displace the SWR pattern
on a transmission line.
There is no opportunity in the stated conditions on a transmission line
for a reflected voltage to exceed the incident voltage or for the
reflection coefficient to exceed one (1).

Best regards, Richard Harrison, KB5WZI

Deacon Dave wrote:
"Is it possible for a resonant condition to exist " (with a short
cable)?

If the cable is too short to enforce Zo, the source for Zload is not
necessarily a resistance and the load for the source can be resonant.

A short-circuited transmission line, shorter than 1/4-wavelength, is an
inductive reactance. An open-circuited transmission line, shorter than
1/4-wavelength, is a capacitive reactance.

A transmission line is not required for mismatch between source and
load. If a source can only supply a certain voltage to current ratio,
and it`s different from the immutable Zload, you have a mismatch and a
surplus of volts or amps which are rejected by the load.

The simple analysis of a series-resonant situation involving line
inductance and a load capacitance is that volts across either is (I)(Z).
The Z`s are about equal at resonanvce.

Trivia: 1/8-wavelength of line has a reactance which is equal to Zo.

Best regards, Richard Harrison, KB5WZI

Keith wrote:
"Not an "of course" at all."

Terman and Bird Electronic Corporation say:

Usually, power delivered by the transmitter and to the load equals
Forward Power less Reflected Power.

Best regards, Richard Harrison, KB5WZI

Reading Keith`s posting, somebody wrote:
"The power is Vavg"Iavg"cos(theta)"

This is mistaken. The average value of a sine wave is 0.637 times the
peak value. We use 0.707 toimes the peak value of a sine wave, the rms
value, which is the effective value, that is, it is as effective as d-c
in making power calculations.

The average power must be the same as d-c. This is peak volts times peak
amps divided by two, or 1/2 the peak power. 0,707 x 0.707 = 0.5, so we
use rms, not avg. volts and amps to calculate average power, which is
the same in capability when producing light and heat from resistive
devices.

Best regards, Richard Harrison, KB5WZI

Roy Lewallen wrote:
"But the avreage value of a sine wave, assuming no DC offset, is zero,
not 0.637 times the peak value."

Yes. The value of any undistorted sine wave which is symmetrical about
the zero axis is zero over one complete cycle.

0.637 and 0.707 are half-cycle values which are quoted as the average
and rms fractions of the peak value of a sine wave in references
containing such trivia. I just remember tthat the average in this case
is about 0.9x the rms value.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Keith wrote:
"It does go back to "double think" unless you can explain how energy can
flow when power is zero."

Think! The only way the voltage can go to zero on a good transmission
line which is energized is by interference between an incident wave and
a reflected wave. Zero volts on the line is merely a manifestation of
VSWR.

No disagreement from me as long as you replace 'incident wave' and
'reflected wave' with the more precise 'incident voltage wave' and
'reflected voltage wave'. This aligns with the precise use of V in
VSWR.

Adopting Roy's more precise notation for power
P(t) = V(t) * I(t) (1)
The question is: Do you accept this expression as describing power?

a) If not, then there is no "double think", but a lot of electrical
engineering will have to be tossed as well.

b) If you do accept (1), then in a case where V(t) is zero for all time,
power must be zero as well. V(t) is zero for all time at a current
maximum in a shorted transmission line, so the power (energy flowing)
must be zero as well.

c) If you accept (1), but also claim that there is energy flowing when
V(t) is zero, then "double think" is an appropriate description.

I think these are the only 3 options.

If you choose b), then I think we are in agreement.
If you choose c), then ... I'm not sure what the 'then' is.
If you choose a), we can explore all the difficulties that will arise
when (1) is not true and with some effort you may arrive at b).

Forward (incident) power and reverse (reflected) power are both on the
line. A zero voltage on the line requires a complete load reflection so
that the reflection volts are as strong as the forward volts.

The sentence with 'volts' is correct. The sentence with 'power' leads
to a great deal of difficulty as described above.

A directional coupler, at the very spot where a slotted line probe would
sense zero volts, would show you have full power (with its volts and
amps) coming and going.

Many people use directional couplers as a reason to stay out of
camp b), but this necessarily means they are in a) or c). The first
step to enlightenment is to briefly set aside directional couplers
and 'Bird watt' meters, and realize that in a choice between a), b)
or c), b) is the only place it makes sense to be. Then go back and
figure out how directional couplers are not inconsistent with b).

In reality there is not zero volts in the incident wave or in the
reflected wave. There`s full voltage coming and going. The volts just
happen to be out-of-phase at this point.

Yes, indeed. But there is no power.

....Keith

wrote:
In reality there is not zero volts in the incident wave or in the
reflected wave. There`s full voltage coming and going. The volts just
happen to be out-of-phase at this point.

Yes, indeed. But there is no power.

Power is the same as irradiance in optics. When total V=0, it is simply
the result of destructive interference. Perhaps this quote from _Optics_,
by Hecht, will enlighten you. "The principle of conservation of energy
makes it clear that if there is constructive interference at one point,
the "extra" energy at that location must have come from elsewhere. There
must therefore be destructive interference somewhere else."

The voltage goes to zero because two voltage waves are engaged in destructive
interference. The current goes to maximum because two current waves are engaged
in constructive interference. The momentum in the voltage waves simply transfers
to the current waves and they just keep on rolling along. There is no mechanism
of physics existing at that point to change the momentum of the waves. Believing
that no energy crosses a superposed V=0 boundary is just a wet dream. The current
is at an absolute maximum point so plenty of charge carriers are crossing that
boundary.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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