On Feb 5, 2:41*pm, Cecil Moore wrote:
Keith Dysart wrote:
How does this reflected wave of 6.25 W affect the already
present forward wave of 25 W when it arrives back at the
generator. It affects it not one iota. The forward power
remains the same, the forward voltage remains the same
and the forward current remains the same. This is the
meaning of no reflection.

Maybe a more drastic example will help.

* * * * * * Rs * * * Vg * * * * * * * * * * Vl
* * * *+----/\/\/-----+----------------------+
* * * *| * *50 ohm * * * * * * * * * * * * * |
* * * *| * * * * * * * * * * * * * * * * * * /
* * * Vs * * * * * * * * 90 degrees * * * * *\ * Rl
* *100 cos(wt) * * * * * 50 ohm line * * * * / *1 ohm
* * * *| * * * * * * * * * * * * * * * * * * \ *load
* * * *| * * * * * * * * * * * * * * * * * * |
* * * *+--------------+----------------------+
* * * gnd

Current through the source: Is = 0.0392 cos(wt).
The source is supplying 1.96 watts. That is all the
power available to the entire system.

You say the forward power is 25 watts and that none
of the reflected power is reflected at the source.

How can there be 25 watts of forward power when
the source is supplying only 1.96 watts and none
of the reflected energy is joining the forward
wave?

The answer would be clear to you if you were to
complete the analysis.

Please compute:
Pfor
Pref
Vfor
Vref
at the point Vg, both before and after the
reflected wave makes it back to the point Vg.

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

...Keith

Keith Dysart wrote:
Cecil Moore wrote:
How can there be 25 watts of forward power when
the source is supplying only 1.96 watts and none
of the reflected energy is joining the forward
wave?

The answer would be clear to you if you were to
complete the analysis.

Please don't ignore the question. How does a 1.96
watt source support a forward power of 25 watts
without reflected energy joining the forward energy?
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Cecil Moore wrote:

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm ~23w--Pref |
| 0.04w / 1.92w
Vs 90 degrees \ Rl
100 cos(wt) 1.96w 50 ohm line / 1 ohm
| \ load
| |
+--------------+----------------------+
gnd

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection. The redistribution of
energy due to interference is a well known and
well understood phenomenon in optical physics but has
been virtually ignored in the field of RF engineering.

It is not a conventional reflection. It is wave
cancellation in action. It is obeying the following
equation:

Ptot = P1 + P2 - 2*SQRT(P1*P2) where
-2*SQRT(P1*P2) is the destructive interference term.

If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

Nope, not true. The phase angle
between the voltages determines which direction the
energy flows. If the above example is changed to
1/2WL instead of 1/4WL, the forward and reflected
voltages and powers remain the same but the interference
at the source resistor changes from destructive to
constructive and the source resistor heats up.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm ~23w--Pref |
| 0.04w / 1.92w
Vs 90 degrees \ Rl
100 cos(wt) 1.96w 50 ohm line / 1 ohm
| \ load
| |
+--------------+----------------------+
gnd

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection. The redistribution of
energy due to interference is a well known and
well understood phenomenon in optical physics but has
been virtually ignored in the field of RF engineering.

It is not a conventional reflection. It is wave
cancellation in action. It is obeying the following
equation:

Ptot = P1 + P2 - 2*SQRT(P1*P2) where
-2*SQRT(P1*P2) is the destructive interference term.

Is this the total applied power or the remaining power (Prem)(which is
really the reflected power), where

Prem = Z0(If - Ir)^2
= Z0((If^2) +(Ir^2) - 2(If*Ir))

I think these (yours and mine) are both the same equation except that I
changed the terms so that we could use it on a transmission line. Z0 is
transmission line impedance, If is forward current, Ir is reflected current

Just to clarify,
2*SQRT(P1*P2)
= 2*SQRT((Z*I1^2)*(Z*I2^2)
= 2*Z(I1*I2)

The equation should apply at all locations on the transmission line.


If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

Nope, not true. The phase angle
between the voltages determines which direction the
energy flows. If the above example is changed to
1/2WL instead of 1/4WL, the forward and reflected
voltages and powers remain the same but the interference
at the source resistor changes from destructive to
constructive and the source resistor heats up.

73, Roger, W7WKB

Roger Sparks wrote:
The equation should apply at all locations on the transmission line.

What you say is true but there is a caveat. Remember
that I have previously said that interference can exist
without wave cancellation. As long as Z0 remains constant
up and down a transmission line, there is no forward and
reflected wave interaction. They pass each other like
"ships in the night".

The resistor is a physical object. When the forward and
reflected waves superpose at the *resistor*, there is
an obvious interaction, i.e., a permanent interference.
One only has to observe the power dissipated in the resistor
to ascertain that it is not the same as the forward power
plus the reflected power unless the E-fields of the two
waves are 90 degrees apart, a condition for which zero
interference exists.

What is interesting is a procedure for determining the
current through Rs using only powers. We have to be
pretty accurate with the powers to do that. We know
the forward power is 25w. The power reflection coefficient
at the load is [(50-1)/(50+1)]^2 = 0.92311. That makes
the reflected power from the load equal to 23.077 watts.
When the forward wave and reflected wave superpose at
the 50 ohm source resistor, they are 180 degrees out of
phase which makes the interference term minus and
therefore destructive.

25w + 23.077w - 2*SQRT(25w*23.077w) = 0.03845w

That's the dissipation in the 50 ohm source resistor.
So the RMS current is SQRT(0.03845w/50ohm)= 0.02773a

Ig(t) = 1.414(0.02773)cos(wt) = 0.039216 cos(wt)

And that is, indeed, the current through the 50 ohm
source resistor, Rs, using only an energy analysis.

All energy, except 0.03845w, is "redistributed", i.e.
reflected back toward the load. This is all explained
in my energy analysis article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
. net...
Roger Sparks wrote:
The equation should apply at all locations on the transmission line.

What you say is true but there is a caveat. Remember
that I have previously said that interference can exist
without wave cancellation. As long as Z0 remains constant
up and down a transmission line, there is no forward and
reflected wave interaction. They pass each other like
"ships in the night".

The resistor is a physical object. When the forward and
reflected waves superpose at the *resistor*, there is
an obvious interaction, i.e., a permanent interference.
One only has to observe the power dissipated in the resistor
to ascertain that it is not the same as the forward power
plus the reflected power unless the E-fields of the two
waves are 90 degrees apart, a condition for which zero
interference exists.

What is interesting is a procedure for determining the
current through Rs using only powers. We have to be
pretty accurate with the powers to do that. We know
the forward power is 25w. The power reflection coefficient
at the load is [(50-1)/(50+1)]^2 = 0.92311. That makes
the reflected power from the load equal to 23.077 watts.
When the forward wave and reflected wave superpose at
the 50 ohm source resistor, they are 180 degrees out of
phase which makes the interference term minus and
therefore destructive.

25w + 23.077w - 2*SQRT(25w*23.077w) = 0.03845w

That's the dissipation in the 50 ohm source resistor.
So the RMS current is SQRT(0.03845w/50ohm)= 0.02773a

Ig(t) = 1.414(0.02773)cos(wt) = 0.039216 cos(wt)

And that is, indeed, the current through the 50 ohm
source resistor, Rs, using only an energy analysis.

All energy, except 0.03845w, is "redistributed", i.e.
reflected back toward the load. This is all explained
in my energy analysis article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com


I disagree. If the current through the 50 ohm source resistor is 0.039216,
then the power in the 50 ohm source resistor cannot be 0.03845w. Namely
because 0.039216 * 0.039216 * 50 = 0.076895w (eye-squared-are, matey).
Either your power is wrong or your current is wrong. My own calculations
show that your current is correct. Therefore, your energy analysis appears
to be in error.

John

John KD5YI wrote:
I disagree. If the current through the 50 ohm source resistor is 0.039216,
then the power in the 50 ohm source resistor cannot be 0.03845w. Namely
because 0.039216 * 0.039216 * 50 = 0.076895w (eye-squared-are, matey).
Either your power is wrong or your current is wrong. My own calculations
show that your current is correct. Therefore, your energy analysis appears
to be in error.

You seem to have forgotten that since 0.039216 is the maximum
value of current you need to divide the 0.076895w by two.

0.076895w/2 = 0.03845 watts

If we used the RMS value of the current, i.e. 0.039216/1.414
equals 0.027734a, then 50(0.027734)^2 = 0.03845 watts and
all is well.
--
73, Cecil http://www.w5dxp.com

On Feb 6, 8:02*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:

* * * * * * Rs * * * Vg * Pfor--25w * * * * Vl
* * * *+----/\/\/-----+----------------------+
* * * *| * *50 ohm * * * ~23w--Pref * * * * |
* * * *| * *0.04w * * * * * * * * * * * * * */ *1.92w
* * * Vs * * * * * * * * 90 degrees * * * * *\ * Rl
* *100 cos(wt) 1.96w * * 50 ohm line * * * * / *1 ohm
* * * *| * * * * * * * * * * * * * * * * * * \ *load
* * * *| * * * * * * * * * * * * * * * * * * |
* * * *+--------------+----------------------+
* * * gnd
When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection.

When we analyze this circuit we find that there is
no voltage re-reflection when the wave gets back
to the generator. This is clear because Vf does not
change, which it would have to do if any of Vr was
re-reflected.

And yet it appears that you are claiming that power
is reflected at the generator. How can power be
reflected if voltage is not?

You may find the Excel spreadsheet on this page
http://keith.dysart.googlepages.com/...oad,reflection
interesting.

It computes the various results for the circuit we
have been discussing. Input Vs, Rs, Rl and line length
and it will compute the various voltages, currents and
powers.

After entering the circuit parameters, click "First".
The circuit conditions at the end of the first round
trip will be computed. Click "Next" to have any
re-reflection included into the forward conditions
for the next round trip.

When Rs is 50 ohms (i.e. it matches the line impedance),
the circuit conditions settle to their final state
immediately, as would be expected when there is no
reflection at the generator. Change Rs to a value other
than 50 and it takes many clicks of "Next" before the
conditions settle. At each click, Vf changes as some of
Vr is re-reflected and added to Vf.

...Keith

"Cecil Moore" wrote in message
...
John KD5YI wrote:
I disagree. If the current through the 50 ohm source resistor is
0.039216, then the power in the 50 ohm source resistor cannot be
0.03845w. Namely because 0.039216 * 0.039216 * 50 = 0.076895w
(eye-squared-are, matey). Either your power is wrong or your current is
wrong. My own calculations show that your current is correct. Therefore,
your energy analysis appears to be in error.

You seem to have forgotten that since 0.039216 is the maximum
value of current you need to divide the 0.076895w by two.


Yes, I forgot to use the RMS value of the source.

John

Keith Dysart wrote:
When we analyze this circuit we find that there is
no voltage re-reflection when the wave gets back
to the generator. This is clear because Vf does not
change, which it would have to do if any of Vr was
re-reflected.

Nope, Vf doesn't have to change for there to be a
reflection. Looking at what happens to the power
dissipation in the source resistor, Rs, when the
reflected wave arrives, we see that the energy being
supplied by the source drops by 99.85%. Most of
the energy in the forward wave is supplied by the
reflected wave, starting at the time the reflected
wave arrives and causes destructive interference.

And yet it appears that you are claiming that power
is reflected at the generator. How can power be
reflected if voltage is not?

To be technically correct, reflected energy is redistributed
back toward the load during the process of destructive
interference.

The conditions before the reflected wave arrives and
after the reflected wave arrives are extremely different.

Before the reflected wave arrives, the source resistor,
Rs, is dissipating 25 watts.

After the reflected wave arrives, the source resistor,
Rs, is dissipating 0.03845 watts.

Clearly, something drastic has happened and that is:
99.85% of the forward energy originally supplied by the
source has been replaced with reflected energy being
redistributed back toward the load.

Because of destructive interference, reflected energy
*never* flows through the source resistor, Rs, and is
instead redistributed back toward the load.

Nothing else is possible since the source is supplying
only 1.9608 watts during steady-state. 92.3% of the forward
power is not being supplied by the source during steady-state.

The energy incident upon a point must equal the energy
exiting the point. The energy incident upon the
generator terminals is 1.92234 joules/sec from the
source and 23.0777 joules/sec from the reflected wave.
The energy exiting that point is 25 joules/sec. The
reflected wave energy obviously reverses direction and
joins the forward wave and that is what we call a
"reflection".
--
73, Cecil http://www.w5dxp.com