H. Adam Stevens, NQ5H wrote:
Good Lord Roy, I thought you knew better.

If the match at the load is not perfect, energy is refleced back to the
source, are you with me so far?

I can easily build a source that absorbs all the reflected power: A zero
impedance source in series with a resistor that matches the transmission
line impedance.


Let's see, I put a 100 volt zero impedance source in series with a 50
ohm resistor, connect that to a half wave transmission line terminated
with 150 ohms. The current will be 100/200 = 0.5 amp, the power in the
150 ohm load is 37.5 watts, the power in the 50 ohm source resistor is
12.5 watts. The SWR is 3:1, the forward power is 50 watts, the reverse
power is 12.5 watts. Sure enough, the power in the source resistor
equals the reverse power. Good job. That sure must be the worst case,
all right.

Just to check, I'll change the load resistor to 16.67 ohms. Now the
current is 1.5 amps, the power in the 16.67 ohm load is 37.5 watts, and
the power in the source resistor is 112.5 watts. The SWR is still 3:1,
the forward power is 50 watts just like before, and the reverse power is
12.5 watts just like before. Hm. The reverse power is 12.5 watts, but
the source resistor is now dissipating 112.5 watts. Must be worse than
the worst case.

Well, shoot, maybe the source resistor dissipates all the reverse power
*plus* some more power that comes from somewhere else. So let's try a
200 ohm load. Now the current is 0.4 amp, the power in the 200 ohm load
resistor is 32 watts, and the power in the 50 ohm source resistor is 8
watts. The SWR is 4:1, the forward power is 50 watts, and the reverse
power is 18 watts. Oops, the source resistor is only dissipating 8 watts
but the reverse power is 18 watts. Not only isn't it dissipating all the
reverse power, but it isn't even dissipating that extra power that came
from somewhere else when we connected the 16.67 ohm resistor. Wonder
where the other 10 watts of reverse power went?(*)

So using your simple criterion of a zero impedance source and resistor
equal to the transmission line impedance, and by only changing the load
resistance, we've got cases whe

-- The source resistor dissipation equals the reverse power
-- The source resistor dissipation is greater than the reverse power
-- The source resistor dissipation is less than the reverse power

And none of these will explain the loss figure you gave earlier.

Guess I don't know better after all.

Anyone who's interested can find more interesting cases in "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/. And those who aren't
interested, well, you're welcome to believe what you choose. Just don't
look too closely at the evidence.

(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

Roy Lewallen, W7EL

Roy Lewallen wrote:
(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

Here's my earlier thought example again.

100w----one second long lossless feedline----load, rho=0.707

SWR = (1+rho)/(1-rho) = 5.828:1
Source is delivering 100 watts (joules/sec)
Forward power is 200 watts (joules/sec)
Reflected power is 100 watts (joules/sec)
Load is absorbing 100 watts (joules/sec)

It can easily be shown that 300 joules of energy have been
generated that have not been delivered to the load, i.e.
those 300 joules of energy are stored in the feedline.
The 300 joules of energy are stored in RF waves which
cannot stand still and necessarily travel at the speed of
light. TV ghosting can be used to prove that the reflected
energy actually makes a round trip to the load and back.
A TDR will indicate the same thing.

Choosing to use a net energy shortcut doesn't negate the
laws of physics.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Roy Lewallen wrote:

(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

Sorry, I misquoted there. Walter Johnson doesn't say anything
about Poynting Vectors. The above is from: "Fields and Waves ..."
by Ramo, Whinnery, and Van Duzer, page 350, where they assert:

Pz-/Pz+ = |rho|^2

The reflected power Poynting Vector divided by the forward
power Poynting Vector equals the power reflection coefficient.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil,

I will presume that your reference to Walter Johnson is with regard to
his book, "Transmission Lines and Networks", published in 1950.

I have been unable to find any mention of Poynting Vectors or Power Flow
Vectors in my copy.

Would you be so kind as to identify the page number(s) describing these
concepts?

73,
Gene
W4SZ


Cecil Moore wrote:

[snip]

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

[snip]

Gene Fuller wrote:

Cecil,

I will presume that your reference to Walter Johnson is with regard to
his book, "Transmission Lines and Networks", published in 1950.

I have been unable to find any mention of Poynting Vectors or Power Flow
Vectors in my copy.

Would you be so kind as to identify the page number(s) describing these
concepts?

Gene,
My next posting admitted my senility. I was quoting Ramo & Whinnery,
not Walter Johnson. In "Fields and Waves in Communication Electronics",
page 325, an equation is given for Pz+, "The Poynting vector for the
positive traveling wave ...". It continues: "Similarly, the Poynting
vector for the negatively traveling wave is always in the negative 'z'
direction except when it is zero."

On page 350 it gives the ratio of the forward Poynting vector to the
rearward Poynting vector as the power reflection coefficient. Sorry
for my faulty memory.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

Here's my earlier thought example again.

100w----one second long lossless feedline----load, rho=0.707

SWR = (1+rho)/(1-rho) = 5.828:1
Source is delivering 100 watts (joules/sec)
Forward power is 200 watts (joules/sec)
Reflected power is 100 watts (joules/sec)
Load is absorbing 100 watts (joules/sec)

It can easily be shown that 300 joules of energy have been
generated that have not been delivered to the load, i.e.
those 300 joules of energy are stored in the feedline.

Not easy if t 2 sec. :-)

The 300 joules of energy are stored in RF waves which
cannot stand still and necessarily travel at the speed of
light.

It's ironic that the first paramater cited in the problem starts with an
'S'. :-)

TV ghosting can be used to prove that the reflected
energy actually makes a round trip to the load and back.
A TDR will indicate the same thing.

If either source were monochromatic, I bet I could come up with an
example where the surfaces reflect no energy. :-)

Choosing to use a net energy shortcut doesn't negate the
laws of physics.

Particular when characterized as a matter of opinion, it can be like
having a religious discussion.

73 ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
It can easily be shown that 300 joules of energy have been
generated that have not been delivered to the load, i.e.
those 300 joules of energy are stored in the feedline.

Not easy if t 2 sec. :-)

Of course, my statement is related to steady-state. I don't
see anything worth responding to, Jim. Where's the beef?
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:

It can easily be shown that 300 joules of energy have been
generated that have not been delivered to the load, i.e.
those 300 joules of energy are stored in the feedline.


Not easy if t 2 sec. :-)


Of course, my statement is related to steady-state. I don't
see anything worth responding to, Jim. Where's the beef?

The problem is that there should only be a 1 second lapse of time
between the beginning of gozinta at 100 Joules/sec and the beginning of
comezouta at 100 Joules/sec. At what point is the additional 2 seconds
worth of energy fed into the system?

73, AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
Of course, my statement is related to steady-state. I don't
see anything worth responding to, Jim. Where's the beef?

The problem is that there should only be a 1 second lapse of time
between the beginning of gozinta at 100 Joules/sec and the beginning of
comezouta at 100 Joules/sec. At what point is the additional 2 seconds
worth of energy fed into the system?

During the power-on transient phase. The load rejects half the incident
power. To keep things simple, assume a very smart fast tuner. After
one second, the feedline will contain 100 joules. The load will have
accepted zero joules. After two seconds, the feedline will contain the
100 joules generated plus 50 joules rejected by the load and the load
will have accepted 50 joules. Already the feedline contains 150 joules
while the source is putting out 100 joules per second. After 'n'
seconds, the line contains 300 joules, 100 from the source and 200
rejected by the load during the power-on transient stage.

seconds forward energy reflected energy load power
1 100 0 0
2 100 50 50
3 150 50 50
4 150 75 75
5 175 75 75
6 175 87.5 87.5
7 187.5 87.5 87.5
8 187.5 93.75 93.75
9 193.75 93.75 93.75
10 193.75 96.875 96.875
n 200 100 100

After 10 seconds the source has output 1000 joules. The load
has accepted 709.375 joules. 290.625 joules are already stored
in the feedline on the way to 300 joules during steady-state.
This is simple classical reflection model stuff.

If a load in rejecting half its incident power, the steady-
state reflected power will equal the steady-state load
power. The steady-state forward power will be double
either one of those.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:

Of course, my statement is related to steady-state. I don't
see anything worth responding to, Jim. Where's the beef?


The problem is that there should only be a 1 second lapse of time
between the beginning of gozinta at 100 Joules/sec and the beginning
of comezouta at 100 Joules/sec. At what point is the additional 2
seconds worth of energy fed into the system?


During the power-on transient phase. The load rejects half the incident
power. To keep things simple, assume a very smart fast tuner. After
one second, the feedline will contain 100 joules. The load will have
accepted zero joules. After two seconds, the feedline will contain the
100 joules generated plus 50 joules rejected by the load and the load
will have accepted 50 joules. Already the feedline contains 150 joules
while the source is putting out 100 joules per second. After 'n'
seconds, the line contains 300 joules, 100 from the source and 200
rejected by the load during the power-on transient stage.

seconds forward energy reflected energy load power
1 100 0 0
2 100 50 50
3 150 50 50
4 150 75 75
5 175 75 75
6 175 87.5 87.5
7 187.5 87.5 87.5
8 187.5 93.75 93.75
9 193.75 93.75 93.75
10 193.75 96.875 96.875
n 200 100 100

After 10 seconds the source has output 1000 joules. The load
has accepted 709.375 joules. 290.625 joules are already stored
in the feedline on the way to 300 joules during steady-state.
This is simple classical reflection model stuff.

If a load in rejecting half its incident power, the steady-
state reflected power will equal the steady-state load
power. The steady-state forward power will be double
either one of those.

It really is an interesting theory. And I'm willing to concede on a
certain point here. If we were to fit a curve to the data in your far
right side column, what we have is a dispersion curve. That is a
predictable phenomenon, most easily observable on long transmission
lines. However as this is not actual data, an important column is
missing. A column marked 'energy from source' is crucial to proving
your point. Without running the experiment and taking the data we can't
really know how much energy would be in any of the columns at any given
time. When we assume what that energy might be, we run the risk of
making an ass out of u and me. Well, mostly u. :-)

73, AC6XG