Roy Lewallen wrote:
There are lots of problems with analyzing waves of average power
bouncing around in a transmission line. I'm sure that every one of them
has been pointed out many, many times in postings directed at Cecil.

And none of them are a problem to explain. If there's energy in a
system and it's not presently coming from the source, then it
exists as forward and reflected wave energy that previously came
from the source (in a single source system). All you need to do
to figure everything out is keep track of all the energy.
--
73, Cecil http://www.qsl.net/w5dxp



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Dilon Earl wrote:
If you have a 100 watt transmitter, the watt meter shows 3 watts
reflected. I deliver 103 watts to the antenna. I now know where the
reflected power go's. But where did it come from? If I could find a
way to have 100 watts reflected I could put 200 watts to the antenna
from a 100 watt transmitter.

The key word is "to", not "accepted by". You can indeed get 200 watts
to (incident upon) the antenna with a 100 watt transmitter. Trouble
is, the antenna only accepts half of that power.

For some reason I need a circulator on my SB-401.

Only if you allow reflected energy to reach your SB-401.

To get max power out my 6146's I need to turn them upside down in a
glass of water? :-)

Only if they are metal.
--
73, Cecil http://www.qsl.net/w5dxp



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"W5DXP" wrote in message
...
Jim Kelley wrote:
You're sourcing and sinking an additional 33.33 watts, and yet the
wattmeter can't discern the difference between this scenario and the 100
watt, single source scenario.

But that sourcing and sinking is occurring *INSIDE* the SGCL(150). The net
power inside SGCL(150) is 100W dissipated, *exactly* like the other
scenario.

Exactly like the other scenario, yeah, like the extra .033 Kw makes no
difference. That's a good one, Cecil :-)

The example illustrates perfectly the shortcomings of the idea of power
flow, as well as some of the faulty conclusions that can be drawn from
measurements made by a directional power meter.

The shortcoming I notice is your sidestepping of the question:

More notable are the shortcomings you _don't_ care to notice.

What happened
to Pref1=25W? It just seems to have disappeared when we turned on
SGCL(150)
and the 33.33W wavefront arrived at the impedance discontinuity. What
could
have possibly made Pref1 disappear?

Hmmmm, never thought about that before. Could wave cancellation be
involved?

73, Jim AC6XG

William E. Sabin wrote:
Ian White, G3SEK wrote:
* The output impedance of the transistor doesn't come into the
story at all - not when characterizing RF power devices that are not
operating in class A. Even the device manufacturer doesn't know or
care what it is. Neither need we.


Tubes and transistor power amplifiers quite oftem use negative feedback
to improve SSB linearity. Improvements of 5 to 10 dB are common. The
negative feedback reduces the internal impedance of the tube and
transistor amplifiers. The tube/transistor data sheets do not consider
this factor.

Again, we usually don't really know or care much about the values of
the internal impedances.

Agreed.

But there is a special case. Voice/music/data tube transmitters
operating at low frequencies have a problem called "sideband clipping"
where the plate tank selectivity may be too sharp and reduces the
modulation bandwidth. The internal impedance tends to broaden the
response at resonance. When designing the tank circuit this effect may
have to be included.

Thanks for that information. A related topic would be the effect of
tank circuit Q on the bandwidth of HF amplifiers; I seem to remember
reading something about, but don't recall what it implied about the
magnitude of the tube internal impedance, as compared with the load
impedance.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian,

Thanks for the lowdown. I am saving the message. I looked through the Moto
RF book last night and noticed some specs spell out by what they mean by
output impedance (conjugate of load), while others don't. Newer ones seem to
qualify it. Since the efficiency of these devices is usually around 50%, you
can't tell which is which by looking at the numbers.

I have designed several 6m amplifiers by choosing the parallel load
resistance based solely on VCC and desired PO, and it always worked.

Tam/WB2TT

Roy,

I think I have the complete solution to your 70.7V generator and 1/2 wave
line, but with a finite load. Part of it implies that you are correct. Part
of it seem weird, even though the numbers add up. I analyzed it both as a
circuit element and as a transmission line problem. Kind of long, but I'll
post it if you want to see it.

Tam/WB2TT
"Roy Lewallen" wrote in message
...
There are lots of problems with analyzing waves of average power
bouncing around in a transmission line. I'm sure that every one of them
has been pointed out many, many times in postings directed at Cecil.

Jim Kelley wrote:
"W5DXP" wrote in message
Exactly like the other scenario, yeah, like the extra .033 Kw makes no
difference. That's a good one, Cecil :-)

In the first scenario, the 0.033 Kw was loaded into the system before
steady-state conditions were established. The s-parameter equations
consider the power reflected from a mismatched load as another source
of power in the system. The two scenarios are very similar.

But forget the first scenario. What happens to Pref1 in this scenario?
It takes a 25 joules/sec wave to cancel a 25 joules/sec wave.

Vfwd1(rho) = 35.36V at zero degrees, Ifwd1(rho) = 0.707A at 180 degrees.

Vref2(tau) = 35.36V at 180 degrees, Iref2(tau) = 0.707A at zero degrees.

Each of these rearward-traveling waves contains |35.36V*0.707A| = 25 joules/sec

They superpose to zero. What happens to that 50 joules/sec of rearward-
traveling energy? We already know it winds up in a forward-traveling wave
toward the load. You have already admitted that wave cancellation is
responsible for Pref1 being zero. Waves simply cannot exist without energy.
When waves cease to exist, they are forced to give up their intrinsic energy.
We know that Vref2(tau) is traveling rearward. That's all we need to know.
Vref2(tau) is the same voltage term as s12*a2 in the s-parameter equation.

When b1=0=s11*a1+s12*a2, what happens to |s11*a1|^2 and |s12*a2|^2? We
know that s11*a1 and s12*a2 are rearward-traveling voltages.
--
73, Cecil http://www.qsl.net/w5dxp



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No need, I've done both transient and steady state sinusoidal solutions
for the general case many times. The source impedance dissipation and
power supplied by the source can be entirely determined simply by
replacing the transmission line and terminating impedance with an
impedance equal to that seen looking into the input end of the line. It
becomes a simple, three-component electrical circuit. You can replace
the transmission line and load with any combination of length, Z0, and
load impedance you'd like, as long as the input impedance is the same,
creating just about any amount of "reflected power" you want, without
changing the source dissipation.

The "reflected power" is not dissipated in the source, matched or not.
It's trivial to show that this is true. But people still want to believe.

Roy Lewallen, W7EL

Tarmo Tammaru wrote:
Roy,

I think I have the complete solution to your 70.7V generator and 1/2 wave
line, but with a finite load. Part of it implies that you are correct. Part
of it seem weird, even though the numbers add up. I analyzed it both as a
circuit element and as a transmission line problem. Kind of long, but I'll
post it if you want to see it.

Tam/WB2TT
"Roy Lewallen" wrote in message
...

There are lots of problems with analyzing waves of average power
bouncing around in a transmission line. I'm sure that every one of them
has been pointed out many, many times in postings directed at Cecil.

Roy Lewallen wrote in message ...

What needs to be thrown away is the belief that all impedances are the
ratio of a voltage to a current, along with the notion that only
resistors can have resistance.

Roy Lewallen, W7EL



You have convinced me that you are correct about both of these
points.

Good. Then the effort was worthwhile.


Absolutely.



But i don't think that an antennas impedance will not be affected
by the permeability of the medium that surrounds it. An antennas
input impedance will be different in free space as opposed to being
immersed in water, for example.

Indeed it will.


This indicates to me that the antenna is indeed "matching" 50
Ohms to the impedance of free space, even if it is a different type of
impedance.

That's a leap I'm unable to make or to follow.



Clearly, neither of us are PhDs in EM wave propagation, but water
certainly has a different E versus H impedance than the 377 Ohms of
free-space, which is why the input impedance of the antenna will
change. This is not the same, but similar to how the load on the
secondary will affect the primary impedance of a transformer.



Do you think that the characteristics of a transformer of a
specific turns ratio, gauge wire, and core geometry, will NOT depend
on the core material? I would say definitely it WILL depend on the
material.

Actually, an adequate core shouldn't appear as a significant factor in
transformer performance. Naturally, an inadequate core will adversely
affect it. But I just don't accept that as evidence, let alone "proof"
that an antenna is fundamentally an impedance matching device.



Well, you've already agreed that an antenna/transducer can be
considered one half of a transformer, but what i'm saying is that the
permeability of the core or medium will certainly affect the impedance
of the transducer.

What do you mean by "adequate core"? One that suits your purpose
i suppose. But a material of the wrong permeability will definitely
affect your transformer performance. So the impedance of the core
definitely affects the transformer characteristics, as does the
impedance of the air (or water) between two antennas.



I see that you won't be swayed from your visualization. But hopefully
some of the other readers can see the fallacy of the concept. I think
I've done all I can, so I'll leave this topic now.



You've convinced me that antennas are transducers, which are one
half of a transformer, by giving me logical statements.

But you have not come up with anything to convince me otherwise
on this point, which i don't believe is a fallacy at this time.



*Chuckle* I was just reminded of something that happened years ago, when
my son was a small boy. He learned that I was an engineer, so he
couldn't wait to see the train I drove. After a great deal of repeated,
patient, explanation, I finally got across (I thought) a description of
what I did, and that it had nothing to do with trains. Well, he had
occasion to visit me at work quite a long time later. He kept wandering
off. When I asked why, he explained that he was trying to find where the
train was kept. Yeah, I might not drive trains, but I must have
*something* to do with trains.


I'm not a small boy Roy, and I'm an engineer too. Your NG
inspired sarcasm doesn't change my opinion at all, and cannot even be
compared to logical reasoning.

Roy, thanks for your insight, and you have definitely helped me
out with your strict semantics (sometimes needed, especially in the
engineering world!).

But your need to be always right closes your mind to new ideas
and new learning. This is the sign of someone who claims to know
everything about a subject, which i personally believe to be
impossible, even for such a specialized topic as antennas (actually,
it's quite broad, isn't it?), and even for someone as bright and
knowledgable as you are.


Again, much thanks for your input.


Dr. Slick

On Sat, 19 Jul 2003 10:11:36 -0500, W5DXP
wrote:

Dilon Earl wrote:
If you have a 100 watt transmitter, the watt meter shows 3 watts
reflected. I deliver 103 watts to the antenna. I now know where the
reflected power go's. But where did it come from? If I could find a
way to have 100 watts reflected I could put 200 watts to the antenna
from a 100 watt transmitter.

The key word is "to", not "accepted by". You can indeed get 200 watts
to (incident upon) the antenna with a 100 watt transmitter. Trouble
is, the antenna only accepts half of that power.


Where does the other 100 watts go?

For some reason I need a circulator on my SB-401.

Only if you allow reflected energy to reach your SB-401.

How can I stop it from reaching my SB-401?

Then all ham transmitters should have a circulator?