Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

What is the heat dissipated in the transmitter (and why)?

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

Owen

"Owen Duffy" wrote in message
...
Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?


i think the best answer is... it depends.
it depends on what the radio looks like to the reflected power.
for instance, if the tx output network included a circulator and dummy load,
then the reflected power would be lost and there would be no way to get 100w
into the load.
however, because of the conditions given it would appear you have assumed a
perfect re-reflection of the reflected power at the tx end of the line so
everything balances perfectly.


What is the heat dissipated in the transmitter (and why)?

200w dc in, 100w rf into load, lossless line, sounds like 100w dissipated in
the tx to me.


What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

none. it is all re-reflected back toward the load.

Owen Duffy wrote:
Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

What is the heat dissipated in the transmitter (and why)?

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

IMO, you missed the most important questions.

Is the amplitude of the energy stored in the transmission
line when steady-state is reached exactly equal to the
amount it takes to support 267w+167w = 434w of total power?

What happens to the 434w of total steady-state energy stored
in the transmission line when the source is disconnected?
--
73, Cecil http://www.w5dxp.com

Owen Duffy wrote:
Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

There is a conceptual exercise that I use on such problems.

Source---1WL 70 ohm line---+---1/2WL 600 ohm line---70 ohm load
Pfor1=100w-- Pfor2=267w-- 100w
--Pref1=0 --Pref2=167w

Now the picture becomes perfectly clear. The insertion of 1WL
70 ohm line doesn't affect the steady-state condition. A 70
ohm Z0-match is achieved at '+' so none of the 167w is dissipated
in the transmitter. What happens at point '+' toward the source
is total destructive interference, i.e. no reflected energy. What
happens at point '+' toward the load is total constructive
interference. Reference "Optics", by Hecht, 4th edition, page 388.

rho at '+' is 0.791. That makes rho^2 = 0.626.

The amount of Pfor1 that makes it through the impedance discontinuity
at '+' is Pfor1(1-rho^2) = 100w(0.374) = 37.4 watts - call that P1

The amount of Pref2 that is reflected from the impedance discontinuity
at '+' is Pref2(rho^2) = 167w(0.626) = 104.5 watts - call that P2

The power equation says that Pfor2 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Since it is a Z0-match point A=0 so cos(A)=1. Therefore,

Pfor2 = 37.4w + 104.5w + 2*SQRT(37.4w*104.5w) = 267 watts

What do you know? The power equation works perfectly. Pfor2 didn't
even need to be given.
--
73, Cecil http://www.w5dxp.com

On Wed, 28 Feb 2007 20:35:27 GMT, Owen Duffy wrote:

Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

Hi Owen,

After a fashion, yes. You don't have enough information. The lesson
of past correspondence reveals the source resistance is not a benign
element that can be discarded in the examination of these problems.

Explain? I thought I had by example, so to continue by simple
analysis we have to ask: "What is the source R?"

If it is 70 Ohms, it sees a 70 Ohm load and an Impedance Match. In
effect there is no other way for a transmitter that is capable of 100W
being able to source 100W to a load, is there?

If it is 50 Ohms, it immediately sees a mismatch at the connector -
you did do your lumped equivalent of the transmission line system,
didn't you? Hence a transmitter limited to supplying 100W could not
supply 100W to a mismatched load.

If it is 600 Ohms, we again see a massive mismatch, further discussion
is unnecessary as being repetitive.

What is the heat dissipated in the transmitter (and why)?

I will skip that exercise as obvious pending discussion below.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

Presuming the 70 Ohm source as it is the only one capable of
supporting the 100W application of power to the 70 Ohm load (by your
definition and limitations) we shall proceed to the determination of
those pesky powers with absolute finality.

The load voltage is 84 volts (to sufficient accuracy). The voltage at
the source port is 84 volts (to sufficient accuracy). The line has
been defined to be lossless. The same potential at the same angle is
returned to the source through 360 degrees of travel and no loss
(analysis allows us to treat the load as a generator). That signal
applied to the transmitter finds there is no potential difference, and
thus there is no current flow. This variously describes either a
short or an open; and either way it represents an infinite mismatch -
ALL the reflected power is re-reflected to the load.

Voila, another statistical curiosity!

Tuners do this more generally and we use tuners to re-reflect the
power, and to not suffer the transmitter the additional heat burden of
reflected power arriving at the finals.

If we repeat this exercise with the other source resistances, we stand
to appreciate a paradox in the initial condition of:
the DC input power to the transmitter is 200W.
forcing us to resolve the additional heat burden (roughly 50W)
exceeding the reflected power's ability to contribute that much
(especially when all the reflected power is re-reflected); or worse,
the source going into clipping. A non-linear source is not a pleasant
thing to behold.

73's
Richard Clark, KB7QHC

On Wed, 28 Feb 2007 17:51:30 -0800, Richard Clark
wrote:

The load voltage is 84 volts (to sufficient accuracy). The voltage at
the source port is 84 volts (to sufficient accuracy). The line has
been defined to be lossless. The same potential at the same angle is
returned to the source through 360 degrees of travel and no loss
(analysis allows us to treat the load as a generator). That signal
applied to the transmitter finds there is no potential difference, and
thus there is no current flow. This variously describes either a
short or an open; and either way it represents an infinite mismatch -
ALL the reflected power is re-reflected to the load.


Hi All,

Well, in lieu of Owen providing his own solution, much less
fulminating against the BS provided by yours truly in the quote above
(how seductive is it to the rest of you?). The problem here arises
out of a mixture of models, but this topology is not my choice and it
is inclined to invite such problems.

We can look at the details provided to find a 600 Ohm line terminated
at both ends by 70 Ohm resistors (again, I am forced to the
presumption of the source resistance, but given the constraints it is
the only value available). The energy within the resonant line is
ringing (one reason why a mismatched line it is called resonant is
because of circulating energy - the allusion to ringing).

If we deposit ourselves in the line, within the fields of energy, it
would be like inhabiting a hall with two partial mirrors, one at each
end. To other descriptions, these partial mirrors are leaky
interfaces. Some of the energy is passed through, some is reflected.
The "some" can be rendered into an absolute through knowledge of the
degree of mismatch - it is already provided within the original post
as being 8.6:1 (to a sufficient accuracy). In this regard, the
resonant line has a poor Q, however "poor" is in the eye of the
beholder because a flat line has the poorest of Qs at 1 (or worse).

The topology, and the question, invited me to respond with both the
lumped equivalent (this topology is in the classic expectation of
repetition of the load, cast back identically through a lossless line
to its input) and a Thevenin analysis (200W of power applied to a
source that could only supply 100W). The topology here has the
classic expectation of 50% efficiency that only a 70 Ohm source can
provide. Given that there is special constraint of a very strongly
specified line length, that too invites an analysis (which then
corrupts the lumped equivalent analysis if they are not separated).

So, we have three avenues commingled here in pursuit of an answer as
to explaining the power. Let us return to that hall of mirrors and
traverse its length (and unfortunately it must violate another
expectation of steady state, but this topology, much less expectations
galore, was not my choice - but I wander into any discussion with so
many invitations).

The first pass of energy sees the partial mirror of the load
interface. Some energy passes, some reflects. That which passes
cannot supply the expected potential of 84 volts (to a sufficient
accuracy); that remains to be justified by further analysis. The
reflected energy approaches an IDENTICAL mirror at the source end. The
reflected energy is coherent to the source (this is enforced by the
topology) behind that mirror. However, and revealing my deception
above, the reflected energy is not the same amplitude. There are two
sources of energy now embracing the source resistance of 70 Ohms.
There is a potential difference across that source resistance that is
different from that of what would be a presumed steady state 84 volts
(to a sufficient accuracy). Others can provide that computation if
they wish as the magnitude is one of degree (pun intended).

We have COOLED the source resistance! But we have not provided the
full load (anyone want to bet we cooled it by an equal degree? - pun
intended again). And the 200W application must be accounted for
somewhere (this inclusion is, of course, a sham of moving between
topologies and expectations that are so inviting).

And so on, and so on, ad infinitum through the successive waves of
reflection and dissipation for this statistical curiosity for me to
once again conclude:
ALL the reflected power is re-reflected to the load.
which is to say, energy (through the corruption of terms that has been
generally allowed).

73's
Richard Clark, KB7QHC

I did not intend this to be difficult, or incomplete. I think it is
solveable without recourse to knowledge or speculation about source
impedance, or changing the problem by inserting additional lines, or
injecting photons, or whatever.

My answers are inline:

Owen Duffy wrote in
:

Breaking out of the previous thread to explore the "power explanation"
in a steady state situation:

The scenario for discussion is a transmitter connected to a half wave
of 600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70
ohm load, the VSWR on the transmission line is 8.6, the "forward
power" (meaning Vf^2/Zo) on the transmission line is 267W, the
"reflected power" (meaning Vr^2/Zo) on the transmission line is 167W,
the DC input power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation,
if so, identify / explain?

No, I cannot see any.


What is the heat dissipated in the transmitter (and why)?

Approximately 100W. It is reasonable to assume that all DC power into the
transmitter is converted to RF power output and heat. The heat then is
approximately DC power input minus the RF output power. Since the power
in the load is stated to be 100W, and the line is stated to be lossless,
the power out of the transmitter must be 100W, and given the stated DC
input power of 200W, the transmitter heat dissipation is approximately
100W.


What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

None. The notion of the "reflected power" is part of describing the field
setup / energy storage / energy flow on the transmission line. It follows
from trying to reconcile the restriction the Vf/If=Zo, and Vr/Ir=Zo with
the V/I ratio of the circuit attached to the transmission line.

The problem defined the standing wave effects on the line. The
transmitter sees a load of 70+j0 due to the antenna and half wave
lossless line, and must be delivering 100W to that apparent load so as to
achieve the stated antenna load power.

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this problem.

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

When I said that the reflected power explanation is not a good
explanation, I did not say it was necessarily invalid, but it is my
opinion that it provides an incomplete explanation that folk with limited
knowledge and supported by ham radio folklore like to use to explain
things that are not necessarily explained by that explanation (if you are
still with me), for example why a transmitter (sometimes or necessarily)
gets hotter when the antenna VSWR is high.

Owen

On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this problem.

Hi Owen,

Why must it flow back, when the generator itself presents a huge
mismatch to the line? The laws of reflection work at both ends of the
line and to force the presumption that initial reflected power would
flow through this discontinuity without reflection is a strained
expectation. However, I see by the alteration that follows, that we
now have a fully matching source:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

"reflected power"= 50 what? Watts? So be it. Your intent may not to
be incomplete nor difficult, but what is wrong with specifying the
load? Do we get a line length?

Let's see, I start with your termination of 50 Ohms to find 1 Amp
flowing through both source and load resistances. 50 Watts each. Now
I move to a load that will reflect 50 Watts - the former load's
complete contribution. I presume by this you mean either an open or a
short at the load end. As you left it unspecified, and it being my
choice, I can show that the source resistor will get hotter than a $5
pistol. I will, however, be complete and discuss how the source
resistance dissipates the reverse power for both short and open.

To cut to the chase, and not knowing the length of the line, we have a
spectrum of choices, but we may as well force the situation with
another resonant line any number of lossless halfwaves.

It is obvious that the two powers combine constructively or
destructively. One is with the energy reflecting a zero voltage
component and all the current (or now twice the current through the
source resistance), the other is with the energy reflecting an
identical voltage component, and none of the current flows. As I've
offered, the spectrum of possible answers yields to the same analysis,
this was simpler.

Both times the reflected power had to be absorbed by the source
resistance. It does not mean that contribution has to always resolve
to more heat (haven't I already demonstrated this in this thread?).

I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this
problem.

Hi Owen,

Why must it flow back, when the generator itself presents a huge
mismatch to the line? The laws of reflection work at both ends of the
line and to force the presumption that initial reflected power would
flow through this discontinuity without reflection is a strained
expectation. However, I see by the alteration that follows, that we
now have a fully matching source:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much
of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

"reflected power"= 50 what? Watts? So be it. Your intent may not to
be incomplete nor difficult, but what is wrong with specifying the
load? Do we get a line length?

The line is stated as a half wave of lossless line, why do you need to
know more about its length?

The load you are looking for isn't there, the line ends in an open
circuit. Sorry for the confusion, I should have been explicit that there
was no load, just o/c.

The misunderstandings will frustrate your analysis, so try again.

Owen

And, in fact, the "Reflected power" is
"Re-reflected" from the Source, Back to
the Load, if memory serves (minus loss's
accumulated from the first "Reflection"
of power, if memory serves! Jim NN7K



Owen Duffy wrote:
Richard Clark wrote in
:

On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this
problem.