On May 10, 5:24*am, Richard Fry wrote:
RF

Ignore my earlier post in this thread. I used the wrong multiplier.

On further study it appears that the sum of the power contained in all
perfect, coherent re-reflections from the source that will be absorbed
by the load is asymptotic to the value of the power absorbed from the
original incident wave (lossless transmission line).

On 5/9/2011 6:13 PM, walt wrote:


Wim, why is it that you can't seem to answer my question concerning
the amount of power reflected from a 2:1 mismatch? The following is
what you said in reply, totally ignoring the details in my question:

The problem really is that a simple small signal linear circuit theory
approach (which is what Thevenin equivalents and maximum power transfer
theorem are all about) breaks down, especially when you consider that
actual amplifiers have operating area limits that are some combination
of current, power, and voltage, and that amplifiers and power supplies
have efficiencies that vary a lot depending on the operating point,

Your response above is totally evasive, and I can't understand why you
would do this. There is a specific correlation between source power,



You're beating up on Wim, who didn't write that, I did.

I don't have time right now, but I will respond in due course.

73
Jim, W6RMK

On May 10, 10:01*am, Richard Fry wrote:
On further study it appears that the sum of the power contained in all
perfect, coherent re-reflections from the source that will be absorbed
by the load is asymptotic to the value of the power absorbed from the
original incident wave (lossless transmission line).

There are two mechanisms responsible for the redistribution of
incident reflected power on/at the source back toward the load.

1. The incident reflected EM wave energy can be partially re-reflected
from a physical impedance discontinuity at the source and merge with
the forward EM wave energy. Note that "reflection" is something that
happens to a single wave.

2. The components of the reflected EM wave energy that are not
directly re-reflected can undergo superposition with other coherent EM
waves and, associated with destructive/constructive interference,
their energies can be redistributed back toward the load. Some
consider this to be a re-reflection but it is NOT since it involves
superposition of two (or more) EM waves.

Given that b1 = s11*a1 + s12*a2 = 0, what happens to the powers,
(s11*a1)^2 and (s12*a2)^2? If that energy is not moving toward the
source, it is moving toward the load and contained in (s21*a1)^2 and
(s22*a2)^2. That power is NOT a reflection. That's the conservation of
energy principle in action associated with destructive/constructive
interference.

When one calculates the source impedance based on the total
redistributed energy, one makes a large error when one assumes that
all of the redistributed energy is "reflected". It is possible for
there to exist zero re-reflected power while all of the reflected
power is redistributed back toward the load through superposition
associated with destructive/constructive interference. That's exactly
what happens in one of W7EL's food-for-thought#1 examples.

When W7EL's source sees an infinite impedance, there is zero power
dissipated in the 50 ohm source resistor. However, since we are using
50 ohm lossless coax, the Z0 of the coax is equal to the Zg of the
source so there are no re-reflections at the source, according to the
reflection model. Is that a contradiction as W7EL seems to suggest?
Absolutely not! He is simply missing the redistribution of reflected
energy back toward the load as constructive interference in balance
with the destructive interference at the source resistor.

In s-parameter terms, if b1 = s11*a1 + s12*a2 = 0, then all of the
incident reflected power must be redistributed back toward the load -
but not through re-reflection - just a necessary result of destructive
interference at the source resistor. I'm sure they exist but I
personally do not know any hams who understand that fact of physics
well recognized from the field of optics.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 10 mayo, 03:13, walt wrote:
On May 9, 3:42*pm, Wimpie wrote:



On 7 mayo, 01:51, Cecil Moore wrote:

On May 6, 5:24*pm, walt wrote:

You have a 100-ohm source terminated with a 50-ohm load that provides
100w. But you say with the VSWR = 2 the net power will increase to
112w. How does this happen?

Walt, here is probably what Wim means by that.

Source------1/2WL 50 ohm lossless------RLoad

Vsource = 212v, Rsource=100 ohms

If Rload = 50 ohms, PLoad = 100w

If Rload = 100 ohms, PLoad=112.5w

Wim must be assuming a 50 ohm SWR in both cases.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

Your assumption is correct! *To avoid confusion, I mentioned: "that is
VSWR = 2 for a 50 Ohms reference", but maybe it is because of English
isn't my mother tongue.

I think it is not unusual to use 50 Ohms as reference when the output
specification says: "100W into 50 Ohms".

With kind regards,

Wim
PA3DJSwww.tetech.nl

Wim, why is it that you can't seem to answer my question concerning
the amount of power reflected from a 2:1 mismatch? The following is
what you said in reply, *totally ignoring the details in my question:

The problem really is that a simple small signal linear circuit theory
approach (which is what Thevenin equivalents and maximum power transfer
theorem are all about) breaks down, especially when you consider that
actual amplifiers have operating area limits that are some combination
of current, power, and voltage, and that amplifiers and power supplies
have efficiencies that vary a lot depending on the operating point,

Your response above is totally evasive, and I can't understand why you
would do this. There is a specific correlation between source power,
forward power and reflected power, such that with a 100w source in a
50-ohm system, a 2:1 mismatch, when the reflected power is re-
reflected the forward power is 111.11w, reflected power is 11.111w,
and power absorbed in the load is 100w.

Now please explain with rigorous physics and math, how with a 100w
source and a 2:1 mismatch you can justify a forward power of 130w and
a reflected power of 15w. Is the fact that there is no way these
values can exist is the reason you're evading answering the question?
If someone who is unfamiliar with the truth in this instance came up
with these absurd numbers, and who is willing to admit a mistake, I'm
willing to forgive a mistake. But if you are stonewalling, is what it
appears to be, then I'm not amused. And you should be ashamed for
doing it.

Walt, W2DU

Hello Walt,

I thought you would be able to do this yourself, so I only stated the
net delivered power to a 100 Ohms load (that is the 111W), though the
delivered power into the stated 50 ohms is 100W. The example was to
show you with numbers what I experienced in practice when I was 16
years old and repeated it when I had access to good equipment
(mentioned before).

In addition, it is not relevant to the output impedance issue of a PA,
as you can calculate the increase in power due to mismatch (referenced
to 50 Ohms) by just using Ohm's Law.

But here it comes:

We have 111W dissipated into a 100 Ohms load (from the EMF
calculation, assuming 100 Ohms Zout for the PA and EMF = 212.1 Vrms).
Reference for waves is 50 Ohms (according to the text on the back of
the PA: "100 Watts into 50 Ohms"). We assume an electrically very
small piece of 50 Ohms coaxial cable between the socket and the 100
Ohms load. This could be your reflectometer. If you like you may
insert 0.5 lambda so that the PA sees 100 Ohms.

100 Ohms equals |rc| = 0.3333 (50 Ohms reference), I assume you know
how to calculate this.

Hence "reflected power" = 0.333^2*"incident power" = 0.111*"incident
power".

"net delivered power" = "incident power" – "reflected power" = 111W.

"net delivered power" = (1-0.11111)*"incident power" = 111W.

"incident power"= 125W, "reflected power" = 14 W

Do you realize that when using transmission line resonators (or
filters) the forward and reflected power INSIDE the resonator can be
very high, while the net input power can be very low?



Wim
PA3DJS
www.tetech.nl

On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie
wrote:

Do you realize that when using transmission line resonators (or
filters) the forward and reflected power INSIDE the resonator can be
very high, while the net input power can be very low?

And this asked of the designer of the Telstar antenna.

73's
Richard Clark, KB7QHC

On 10 mayo, 21:32, Richard Clark wrote:
On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie
wrote:

Do you realize that when using transmission line resonators (or
filters) the forward and reflected power INSIDE the resonator can be
very high, while the net input power can be very low?

And this asked of the designer of the Telstar antenna.

73's
Richard Clark, KB7QHC

Hi Richard,

It doesn't matter what somebody designed in the past. I think my
example regarding net power increase due to mismatch (referenced to 50
ohms) was fully clear.

Stressing on forward and reflected power (several times) may give rise
to the thought that somebody can't solve such a transmission line
problem himself. Note that I wasn't the one that brought up this issue
(as Jim noted also), but I know the stated figures aren't strange.

I use a similar example in my antenna measurement classes to show that
just calculating additional loss due to mismatch may go wrong
completely when Rsource or Rload don't equal the reference impedance
(mostly 50 Ohms).


Wim
PA3DJS
www.tetech.nl

On Tue, 10 May 2011 13:36:13 -0700 (PDT), Wimpie
wrote:

On 10 mayo, 21:32, Richard Clark wrote:
On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie
wrote:

Do you realize that when using transmission line resonators (or
filters) the forward and reflected power INSIDE the resonator can be
very high, while the net input power can be very low?

And this asked of the designer of the Telstar antenna.

73's
Richard Clark, KB7QHC

Hi Richard,

It doesn't matter what somebody designed in the past.

This is a very curious reply given that everyone of us here has
responded to the topic with past accomplishments. So it MUST matter,
otherwise you have just impeached your own credentials citing
experience at age 16, in work for a thesis, and work of the (recent)
past.

I think my
example regarding net power increase due to mismatch (referenced to 50
ohms) was fully clear.

So clear that it has to presume Walt doesn't know transmission line
fundamentals?

Stressing on forward and reflected power (several times) may give rise
to the thought that somebody can't solve such a transmission line
problem himself. Note that I wasn't the one that brought up this issue
(as Jim noted also), but I know the stated figures aren't strange.

So, by your emphasis on forward and reflected power (several times) I
presume you can expand upon your application of transmission line
mechanics to offer something other than an oblique comment that
nothing is strange about:
Now please explain with rigorous physics and math, how with a 100w
source and a 2:1 mismatch you can justify a forward power of 130w and
a reflected power of 15w. Is the fact that there is no way these
values can exist is the reason you're evading answering the question?

If, by any chance of miscommunication, you do find this last quoted
statement strange; it would be clarity itself to simply say so and we
will wait for Jim to sort this out.

73's
Richard Clark, KB7QHC

On May 10, 3:32*pm, Richard Clark wrote:
On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie
wrote:

Do you realize that when using transmission line resonators (or
filters) the forward and reflected power INSIDE the resonator can be
very high, while the net input power can be very low?

And this asked of the designer of the Telstar antenna.

73's
Richard Clark, KB7QHC


This is for Richard Fry,

Richard, because the output source impedance of a p-network following
a tube RF power amp is non-dissipative, no reflected wave enters it to
be dissipated. Instead, the reflected power is totally re-reflected
back into the line. The only loss from there on is due to line
attenuation. The same is true when using an antenna tuner, total re-
reflection. Thus, when ignoring the line loss, with a source power of
100w and a load mismatch of 2:1, the reflected power after reaching
the steady state is 11.111w, and when added to the source power, the
forward power is 111.11w.

Here is what I believe you missed: When the 111.11w arrives at the
mismatch, 11.111w is reflected, leaving 100w dissipated in the load.

Correction for Richard Clark: My designs were for the dish antennas
that went on the Moon buggies, and the antennas that flew on the 1st
nine weather satellites. I also designed one of the quadrifilar helix
antennas now flying on the polar-orbiting NOAA weather satellites. Not
Telstar.

Now Wim, your math is very challenging: You state 100w delivered by
the source, but at one point you also state 111w is delivered to the
100-load--at another point you state that 100w is delivered to the 100-
ohm load. Which is it?

You also state that voltage out of the source is 212.1v--sum ting wong
here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm
yields 449.86w. These power values are nowhere near the values
appearing in you statements.

In my calculations, with 100w the voltage across a 50-ohm line is
70.71v, and across a 100-ohm line the voltage is 100v.

With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you
stated, thus the power-reflection coefficient is 0.1111. This means,
as I've stated continually, that with 100w delivered by the source,
the power reflected at the 2:1 mismatch is 11.111w, which when added
to the 100w supplied by the source, makes the forward power 111.11w.
Now with 11.111w of power reflected at the mismatch, this leaves 100w
delivered to the 100-ohm load, not 111w. Please tell me where the 111w
came from. You also haven't told us why you calculate 130w forward and
15w reflected. Someone else may have made these calculations, but it
was you I asked for an explanation, because you repeated those
calculations.

When using the correct physics and math in this example, how can the
results be so different?

To close, let me present the procedure I use to calculate the total
forward power--I guess it is related to Ohm's Law:

With the power-reflection coefficient as prc, the forward power PF = 1/
(1 - prc).

So Wim, can you clarify the confusion you appear to have made?

Walt

On Tue, 10 May 2011 14:37:35 -0700 (PDT), walt wrote:

Correction for Richard Clark: My designs were for the dish antennas
that went on the Moon buggies, and the antennas that flew on the 1st
nine weather satellites. I also designed one of the quadrifilar helix
antennas now flying on the polar-orbiting NOAA weather satellites. Not
Telstar.

Hi Walt,

Your paygrade is still intact.

Now Wim, your math is very challenging: You state 100w delivered by
the source, but at one point you also state 111w is delivered to the
100-load--at another point you state that 100w is delivered to the 100-
ohm load. Which is it?

There seems to be a contagion of problems infecting this thread. Quite
typical for this class of discussion, unfortunately, with so many
"substitutions" offered in place of sticking with the subject.

You also haven't told us why you calculate 130w forward and
15w reflected. Someone else may have made these calculations, but it
was you I asked for an explanation, because you repeated those
calculations.

However, it is still Jim's cross to bear on this slow path to Calvary.

73's
Richard Clark, KB7QHC

On May 10, 4:37*pm, walt wrote:
This is for Richard Fry,

Richard, because the output source impedance of a p-network following
a tube RF power amp is non-dissipative, no reflected wave enters it to

Here is what I believe you missed: When the 111.11w arrives at the
mismatch, 11.111w is reflected, leaving 100w dissipated in the load.

Walt,

Let us start at T=0. The transmitter is keyed, and instantaneously
delivers 100W to the input of a lossless transmission line. That 100W
travels to the end of the transmission line and encounters a load
connected there having a 2:1 mismatch to the Zo of the transmission
line.

That load will dissipate about 88.889W, and reflect about 11.111W back
to the source.

Likewise, the load will dissipate only 88.889% of the RE-reflected
power from the source for the same reason it dissipated only 88.889%
of the power in the original incident wave.

So even if the RE-reflection from the source is lossless, and phase-
coherent with the original incident power at the load, how does this
yield the 100W of load dissipation that you posit?

RF