H. Adam Stevens, NQ5H wrote:

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the source.
This is not necesssarily the case, but gives us the least signal strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as well.

That calculation completely baffles me.

The ALC in my Icom rig keeps the forward power constant up to the point
where it reduces power, at around 3:1 SWR. This is typical for
commercial rigs. The rig delivers 100 watts to a 1:1 SWR load.
(Techically, this really means a load which, if terminating a 50 ohm
line, will produce 1:1 SWR on that line.) 1.7:1 SWR is a reverse/forward
power ratio of 0.067. The ALC keeps the forward power at 100 watts, so
with 1.7:1 SWR, the reverse power is 6.7 watts. The net power delivered
to the load is 100 - 6.7 = 93.3 watts, or 0.3 dB, not 2.3 dB, less than
the power delivered to a 1:1 load.

Oh, and the "reverse power" isn't "absorbed in the source". Anyone
interested in learning more about this might take a look at "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/.

Roy Lewallen, W7EL

Roy
Worst case.
Get it?
H.

"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal
strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db
is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

That calculation completely baffles me.

The ALC in my Icom rig keeps the forward power constant up to the point
where it reduces power, at around 3:1 SWR. This is typical for
commercial rigs. The rig delivers 100 watts to a 1:1 SWR load.
(Techically, this really means a load which, if terminating a 50 ohm
line, will produce 1:1 SWR on that line.) 1.7:1 SWR is a reverse/forward
power ratio of 0.067. The ALC keeps the forward power at 100 watts, so
with 1.7:1 SWR, the reverse power is 6.7 watts. The net power delivered
to the load is 100 - 6.7 = 93.3 watts, or 0.3 dB, not 2.3 dB, less than
the power delivered to a 1:1 load.

Oh, and the "reverse power" isn't "absorbed in the source". Anyone
interested in learning more about this might take a look at "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/.

Roy Lewallen, W7EL

H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

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.

73
H.


"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

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]

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

In my third example, where does the other 10 watts of reflected power
go? If it goes to the load and back, why does it reflect off the source
resistor?

Roy Lewallen, W7EL

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.
. . .