One word answers...
 

Lets escape from mixed analysis of VSWR in transient scenarios.

Visualise a series circuit of a battery of 100V and negligible internal
resistance, a 100 ohm ideal resistor, and an open switch. This circuit is
connected to a lossless (ie ideal) transmission line with surge impedance
(or characteristic impedance, Zo) of 100 ohms which is open circuit at
the far end. It takes T seconds for a wave to travel from one end of the
line to the other.

The scenario is a source impedance matched to the transmission line, with
unmatched load.

When the switch is closed, current flows into the line. Until t=2T
seconds, the current If that flows into the line equals 100V/(100+100
ohms) = 0.5A, during which time the resistor dissipates heat at the rate
of I^2*R = 100W. The voltage of the wave Vf travelling from the source is
I*Zo = 50V. This situation is constant until t=2T seconds.

At precisely t=T seconds, the wave travelling from the source end reaches
the open end of the line, and a reflected wave is established to satisfy
the conditions that I at the end of the line must be zero. The reflected
wave must have current equal and opposite to the forward wave, so Ir=-If,
the negative sign indicates that the wave travels in the opposite
direction. Considering the reflected wave, Vr=Zo*Ir = 50V, so the total
voltage at the o/c end is Vf+Vr = 50+50 = 100V.

At precisely t=2T seconds, the reflected wave reaches the source end of
the line, and the voltage at the line terminals instantly becomes Vf+Vr
or 100V. At that instant, the current from the source falls to zero, and
the dissipation in the source resistor also falls to zero. This situation
continues indefinitely.

My questions a

How much of the energy that was contained in the reflected wave was
dissipated in the source resistor?

Does existence of a reflected wave necessarily increase dissipation in
the equivalent source resistance?

Is the reflected wave necessarily absorbed (or partly absorbed) by the
equivalent source resistance?

Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

One word answers should be sufficient.

Owen

One word answers...
 

Owen Duffy wrote:
Lets escape from mixed analysis of VSWR in transient scenarios.

Visualise a series circuit of a battery of 100V and negligible internal
resistance, a 100 ohm ideal resistor, and an open switch. This circuit is
connected to a lossless (ie ideal) transmission line with surge impedance
(or characteristic impedance, Zo) of 100 ohms which is open circuit at
the far end. It takes T seconds for a wave to travel from one end of the
line to the other.

At precisely t=2T seconds, the reflected wave reaches the source end of
the line, and the voltage at the line terminals instantly becomes Vf+Vr
or 100V. At that instant, the current from the source falls to zero, and
the dissipation in the source resistor also falls to zero. This situation
continues indefinitely.

My questions a

How much of the energy that was contained in the reflected wave was
dissipated in the source resistor?

None.

Does existence of a reflected wave necessarily increase dissipation in
the equivalent source resistance?

No.

Is the reflected wave necessarily absorbed (or partly absorbed) by the
equivalent source resistance?

No.

Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Yes.

One word answers should be sufficient.

Ok.

Owen

Best,

One word answers...
 

Dan Bloomquist wrote in
:

Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Yes.

I would have thought the answer was NO, the principles are *not*
inconsistent with the general case.

Or did I use too many negatives?

Owen

One word answers...
 

Owen Duffy wrote:

Dan Bloomquist wrote in
:


Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Yes.


I would have thought the answer was NO, the principles are *not*
inconsistent with the general case.

Hi Owen,
See the example I posted where a pulse is sent down the line. In that
case all the energy is dissipated in the source resistor.

Best, Dan.

One word answers...
 

Cecil Moore wrote in news:96hKh.483$rj1.92
@newssvr23.news.prodigy.net:

Owen Duffy wrote:
Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Of course, pure DC principles are not adequate to cover
general case distributed network problems. Else, there

The principles are those of the behaviour of transmission lines and the
theory applying to solving the source circuit. In the 'DC' case, the
analysis becomes trivial, but IMHO there were no underlying principles
that do not apply in the general case.

The example is a specific case of the general, and the outcome that
reflected power is not dissipated in the source resistance dissproves any
general statement to the contrary.

You could devise an example with an AC source that had exactly the same
outcome, more complicated mathematics to solve it, but the exactly the
same outcome. But you don't need to do that, the example is in the set
of the general, and stands against a general statement that power
reflected from a source is necessarily dissipated in the source.

One word answers...
 

Owen Duffy wrote in
:


You could devise an example with an AC source that had exactly the
same outcome, more complicated mathematics to solve it, but the
exactly the same outcome. But you don't need to do that, the example
is in the set of the general, and stands against a general statement
that power reflected from a source is necessarily dissipated in the
source.


This wasn't so well written, let me move a word around:

You could devise an example with an AC source that had exactly the
same outcome, more complicated mathematics to solve it, but the
exactly the same outcome. But you don't need to do that, the example
is in the set of the general, and stands against a general statement
that reflected power from a source is necessarily dissipated in the
source.

Owen

One word answers...
 

Owen Duffy wrote:
You could devise an example with an AC source that had exactly the same
outcome, more complicated mathematics to solve it, but the exactly the
same outcome.

But the point is that I can devise an example with an
AC source that has the opposite outcome simply by using
a Norton equivalent source. When the reflection arrives,
the dissipation doubles.
--
73, Cecil, w5dxp.com

One word answers...
 

Dan Bloomquist wrote in news:4khKh.5006$ya1.3770
@news02.roc.ny:

Owen Duffy wrote:

Dan Bloomquist wrote in
:


Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Yes.


I would have thought the answer was NO, the principles are *not*
inconsistent with the general case.

Hi Owen,
See the example I posted where a pulse is sent down the line. In that
case all the energy is dissipated in the source resistor.

Dan, I don't deny there may be cases where that may happen, but it is not
true that reflected power is necessarily dissipated (or partly
dissipated) in the equivalent source resistance.

Owen

One word answers...
 

Owen Duffy wrote:

Cecil Moore wrote in news:96hKh.483$rj1.92
@newssvr23.news.prodigy.net:


Owen Duffy wrote:

Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Of course, pure DC principles are not adequate to cover
general case distributed network problems. Else, there


The principles are those of the behaviour of transmission lines and the
theory applying to solving the source circuit. In the 'DC' case, the
analysis becomes trivial, but IMHO there were no underlying principles
that do not apply in the general case.

But your underlying principle is that the source steps and doesn't
change. So any general case where the source is at some different level
is now not covered.

Best, Dan.

One word answers...
 

Owen Duffy wrote:
Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Of course, pure DC principles are not adequate to cover
general case distributed network problems. Else, there
would have been no need to develop the distributed network
model. How would your transmission line calculator work
if you only used DC principles?

For instance, the conditions at the source are opposite
between the RF frequency where the feedline length is 1/4WL
and the RF frequency where the same feedline length is 1/2WL.

Conditions are also opposite depending upon whether a
Thevenin equivalent or a Norton equivalent is chosen.

Ramo and Whinnery, of "Fields and Waves ..." fame, warn
us not to attach any significance to the dissipation
within those two equivalent circuits. They are still
equivalent even though one is dissipating zero watts
and the other is dissipating 200 watts.

Even when the impedance is the same between the DC problem
and the 1/2WL RF problem, the DC current in the middle of
the feedline is zero while the RF current in the middle
of the 1/2WL section is at a maximum.
--
73, Cecil, w5dxp.com

One word answers...
 

Owen Duffy wrote:

Dan Bloomquist wrote in news:4khKh.5006$ya1.3770
@news02.roc.ny:


Owen Duffy wrote:


Dan Bloomquist wrote in
:



Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Yes.


I would have thought the answer was NO, the principles are *not*
inconsistent with the general case.

Hi Owen,
See the example I posted where a pulse is sent down the line. In that
case all the energy is dissipated in the source resistor.


Dan, I don't deny there may be cases where that may happen, but it is not
true that reflected power is necessarily dissipated (or partly
dissipated) in the equivalent source resistance.

And I answered accordingly when you asked the very question, I agreed
with you. And I showed just that in the step example of a previous post.

Owen

Best, Dan.

One word answers...
 

Cecil Moore wrote in news:KGhKh.7646$yW.5893
@newssvr11.news.prodigy.net:

Owen Duffy wrote:
You could devise an example with an AC source that had exactly the same
outcome, more complicated mathematics to solve it, but the exactly the
same outcome.

But the point is that I can devise an example with an
AC source that has the opposite outcome simply by using
a Norton equivalent source. When the reflection arrives,
the dissipation doubles.

Your Norton equivalent doesn't pretend to replicate my (real) source in
every way, it only pretends to supply the same voltage and current to its
load... and it does that.

Owen

One word answers...
 

Dan Bloomquist wrote in news:xYhKh.5014$ya1.2122
@news02.roc.ny:

But your underlying principle is that the source steps and doesn't
change. So any general case where the source is at some different level
is now not covered.

It semantics Dan, the excitation is part of the scenario or example, it is
not a principle of circuits or transmisssion lines.

Owen

One word answers...
 

Owen Duffy wrote:
I would have thought the answer was NO, the principles are *not*
inconsistent with the general case.

Even considering the dissipation within a Thevenin
equivalent circuit is inconsistent with any valid
significance according to Ramo and Whinnery.
--
73, Cecil, w5dxp.com

One word answers...
 

Owen Duffy wrote:

Dan Bloomquist wrote in news:xYhKh.5014$ya1.2122
@news02.roc.ny:


But your underlying principle is that the source steps and doesn't
change. So any general case where the source is at some different level
is now not covered.


It semantics Dan, the excitation is part of the scenario or example, it is
not a principle of circuits or transmisssion lines.

Like saying the impedance of a capacitor is the same no matter what the
frequency......

From here, I'll let you have the last word. I'm not into hand waving....

Owen

Best, Dan.

One word answers...
 

Dan Bloomquist wrote in
:

Owen Duffy wrote:

Dan Bloomquist wrote in
news:4khKh.5006$ya1.3770 @news02.roc.ny:


Owen Duffy wrote:


Dan Bloomquist wrote in
:



Are the principles that apply to this example inconsistent with
the general case (eg dc, ac, transient, steady state etc)?

Yes.


I would have thought the answer was NO, the principles are *not*
inconsistent with the general case.

Hi Owen,
See the example I posted where a pulse is sent down the line. In that
case all the energy is dissipated in the source resistor.


Dan, I don't deny there may be cases where that may happen, but it is
not true that reflected power is necessarily dissipated (or partly
dissipated) in the equivalent source resistance.

And I answered accordingly when you asked the very question, I agreed
with you. And I showed just that in the step example of a previous
post.


Dan we are agreed!

My conclusion is that the view put by some that reflected power is
(necessarily) fully or partly dissipated in the PA equivalent source
resistance, (possibly overheating the PA,) is a simplistic view, the
explanation doesn't apply in general and although apparently appealing,
it is wrong.

Owen

One word answers...
 

Owen Duffy wrote:
Dan, I don't deny there may be cases where that may happen, but it is not
true that reflected power is necessarily dissipated (or partly
dissipated) in the equivalent source resistance.

The question is whether reflected power is ever dissipated
in the source. I can come up with a black box source that
dissipates 100% of the reflected power. All it takes is
a circulator and a load resistor. How does your DC
principles handle a circulator?
--
73, Cecil, w5dxp.com

One word answers...
 

Owen Duffy wrote:
Your Norton equivalent doesn't pretend to replicate my (real) source in
every way, ...

I thought we were discussing general principles, not
replicating a special case example in every way.
--
73, Cecil, w5dxp.com

One word answers...
 

Owen Duffy wrote:
My conclusion is that the view put by some that reflected power is
(necessarily) fully or partly dissipated in the PA equivalent source
resistance, (possibly overheating the PA,) is a simplistic view, the
explanation doesn't apply in general and although apparently appealing,
it is wrong.

The question is: Does any reflected joules/second ever
get dissipated in the PA? To ascertain the answer, one
must calculate the interference patterns in both directions.
If the energy is not dissipated anywhere else, it is
dissipated in the PA.
--
73, Cecil, w5dxp.com

One word answers...
 

Owen Duffy wrote in
:

....
When the switch is closed, current flows into the line. Until t=2T
seconds, the current If that flows into the line equals 100V/(100+100
ohms) = 0.5A, during which time the resistor dissipates heat at the
rate of I^2*R = 100W. The voltage of the wave Vf travelling from the
source is I*Zo = 50V. This situation is constant until t=2T seconds.
....

David Ryeburn is awake, even if I wasn't at 5.38 when I wrote this.

should be ...during which time the resistor dissipates heat at the
rate of I^2*R = 25W...

It doesn't materially change anything, but I messed up!

Owen

One word answers...
 

On Thu, 15 Mar 2007 20:43:54 GMT, Owen Duffy wrote:

My conclusion is that the view put by some that reflected power is
(necessarily) fully or partly dissipated in the PA equivalent source
resistance, (possibly overheating the PA,) is a simplistic view, the
explanation doesn't apply in general and although apparently appealing,
it is wrong.

Hi Owen,

Was your example any more complex, or general? In fact it was heavily
tailored for one simplistic answer only wasn't it? The subject line
informs us it was.

You have in the past used the dictum that one counter-example
devastates a poor hypothesis. What you have here is a two degree
answer that fails for the other 358 degrees, where a one word answer
is insufficient.

Reduce this to chance for haphazard line lengths and a spectrum of
loads, then there's a 50% probability of cooling (absurd of course)
and a 50% probability of heating. Reducing that generality to one
word would give us "maybe."

The subject becomes 99.44% uninforming.

73's
Richard Clark, KB7QHC