Reply
 
LinkBack Thread Tools Search this Thread Display Modes
  #11   Report Post  
Old June 2nd 04, 05:49 PM
Walter Maxwell
 
Posts: n/a
Default

On Wed, 02 Jun 2004 08:37:30 -0500, Cecil Moore wrote:

Walter Maxwell wrote:

wrote:
Steve at first said the energy in the canceled waves continues to flow toward the
source without a voltage and current and that interference was not involved. He later
changed his mind. All that should be archived on r.r.a.a on Google for the summer
of 2001. Here's an excerpt. Steve said: "The total forward power increases as a direct
result of the vector superposition of forward voltage and current. This DOES NOT
require a corresponding destructive interference process ..." thus contradicting Hecht
in _Optics_ who says any constructive interference process must be accompanied by
an equal magnitude of destructive interference.


Superposition of forward voltage and current?


I'm sure he meant "superposition of forward voltages and superposition of
forward currents."

I don't recall Steve ever mentioning current.


I think you are right re his article. The above quote is from an
r.r.a.a. posting circa Summer 2001.

What Steve apparently doesn't understand is how the
energy direction is reversed when the rearward voltages and currents go to zero.


"How" is not explained in any of the physics references. The closest
physics reference that explains it is _Optics_, by Hecht where he says
something like, at a point some distance from a source, constructive
interference must be balanced by an equal magnitude of destructive interference.
In a matched system, there is "complete destructive interference" toward the
source side of the match point and "complete constructive interference" toward
the load side of the match point. Energy is always displaced from the "complete
destructive interference" event to the "complete constructive interference"
event. (That's what you call a "virtual short" or "virtual open" capable of
re-reflecting the reflected energy.)


Cecil, I explained the 'how', both in Reflections and in QEX. My explantion of
'how' is what Steve is continually stating is incorrect, especially in his last
3-part QEX article. Statements in that article prove he doesn't understand the
wave mechanism that reverses the direction of the reflected energy. Evidence of
this is that by simply saying the voltages cancel is insufficient description of
how the energies reverse direction. In fact, in his Oct 99 ComQuart article he
specifically states that both voltages and power cancel. This tell me that he
doesn't understand the wave action he's attempting to teach.

MIT's Slater and Harvard's Alford both explain it brilliantly, but Steve rejects
those references as 'irrelevant', and says I mistakenly used them as references
in Reflections.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!

In s-parameter terms, b1 is the reflected voltage from port 1 toward the source.
Port 1 is the input to a matched tuner (transmatch). The equation is:

rearward-traveling voltage reflected toward the source b1 = s11(a1) + s12(a2)

For b1 to be zero, i.e. zero reflections toward the source, s11(a1) must be equal
in magnitude and opposite in phase to s12(a2). That is "complete destructive
interference". Since there are only two directions, "complete constructive interference"
must occur in the direction of b2 = s21(a1) + s22(a2) toward the load which is the
opposite direction from b1.


Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this
constitutes a short circuit. Assume two generators delivering harmonically
related output voltages equal to the two 's' voltages. When the generators are
connected with their output terminals reversed, causing their voltages to be 180
degrees out of phase--this configuration is a SHORT CIRCUIT. What I've been
trying to say is that this is the same condition as when the reflected waves of
voltage and current from a mismatched termination are of equal magnitude and
opposite phase with the voltage and current waves reflected by a matching device
such as a stub, the opposing voltages in those two sets of waves constitute a
short circuit the same as the voltages delivered by the two opposing
generators.

s11 is the port 1 reflection coefficient. a1 is the port 1 incident voltage.
s21 is the port 2 to port 1 transmission coefficient. a2 is the voltage
reflected from the load that is incident upon port 2.

Match-Point
Port1 Port2
Source------Z01--------x------------Z02------------load
a1-- --a2
--b1 b2--

The only dissipative resistance in the amp is that which heats
the plate. That dissipation is the only dissipation in the source--the other
dissipation is only in the load.


Why isn't the source impedance a negative resistance, i.e. a source
of power Vs a positive resistance, a sink of power?


Cecil, the source impedance is often correctly referred to as a negative
resistance. But it must be remembered that the source resistance of Class B and
C amps is non-dissipative, and thus totally re-reflect incident reflected power.
By this I mean that the dissipative resistance that heats the plate is entirely
separate from the output resistance represented by the load line. Remember, the
DC power goes to only two places: that which is dissipated as heat, and that
which is delivered to the load. The reflected power incident on the output
terminals of the tank has no effect on the power dissipated as heat.

Here's an example. First, adjust an amp to deliver 100 watts into a 50-ohm
resistive load. Second, change the load to a reactive 50 + j50 load and readjust
the pi-network to again deliver 100 watts into the new load. The plate current
will be exactly as in the first case, and the heat dissipated will be the same.
The difference is that in the first case the output impedance of the amp was 50
+ j0, while in the second case the output impedance is 50 - j50, due to
readjusting the reactive components in the pi-network to match the 50 + j50-ohm
load. Whether one likes it or not, this constitutes a conjugate match.

As for the plate temperature remaining the same in both cases, first, the
readjustment of the pi-network returned the input resistance of the network to
the same value as in the first case. Thus the plates saw no different condition
between the two cases. And second, Eric Nichols, KL7AJ, has measured
calorimetrically the temperature of the water cooling the tubes of megawatt
transmitters with greatly differing values of reflected power incident on the
xmtr. He has shown that the water temperature remains constant whatever the
value of the reflected power.

Since you mentioned earlier that some posters would like my opinion on the
nature of the source resistance in rf amps I'll put a paragraph or two together
with measurement data to support my opinion.

Walt

  #12   Report Post  
Old June 2nd 04, 06:11 PM
Walter Maxwell
 
Posts: n/a
Default

On Wed, 02 Jun 2004 16:49:09 GMT, Walter Maxwell wrote:

On Wed, 02 Jun 2004 08:37:30 -0500, Cecil Moore wrote:

Walter Maxwell wrote:

wrote:
Steve at first said the energy in the canceled waves continues to flow toward the
source without a voltage and current and that interference was not involved. He later
changed his mind. All that should be archived on r.r.a.a on Google for the summer
of 2001. Here's an excerpt. Steve said: "The total forward power increases as a direct
result of the vector superposition of forward voltage and current. This DOES NOT
require a corresponding destructive interference process ..." thus contradicting Hecht
in _Optics_ who says any constructive interference process must be accompanied by
an equal magnitude of destructive interference.

Superposition of forward voltage and current?


I'm sure he meant "superposition of forward voltages and superposition of
forward currents."

I don't recall Steve ever mentioning current.


I think you are right re his article. The above quote is from an
r.r.a.a. posting circa Summer 2001.

What Steve apparently doesn't understand is how the
energy direction is reversed when the rearward voltages and currents go to zero.


"How" is not explained in any of the physics references. The closest
physics reference that explains it is _Optics_, by Hecht where he says
something like, at a point some distance from a source, constructive
interference must be balanced by an equal magnitude of destructive interference.
In a matched system, there is "complete destructive interference" toward the
source side of the match point and "complete constructive interference" toward
the load side of the match point. Energy is always displaced from the "complete
destructive interference" event to the "complete constructive interference"
event. (That's what you call a "virtual short" or "virtual open" capable of
re-reflecting the reflected energy.)


Cecil, I explained the 'how', both in Reflections and in QEX. My explantion of
'how' is what Steve is continually stating is incorrect, especially in his last
3-part QEX article. Statements in that article prove he doesn't understand the
wave mechanism that reverses the direction of the reflected energy. Evidence of
this is that by simply saying the voltages cancel is insufficient description of
how the energies reverse direction. In fact, in his Oct 99 ComQuart article he
specifically states that both voltages and power cancel. This tell me that he
doesn't understand the wave action he's attempting to teach.

MIT's Slater and Harvard's Alford both explain it brilliantly, but Steve rejects
those references as 'irrelevant', and says I mistakenly used them as references
in Reflections.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!


snip

Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct
1999:
"For the impedance matching network to 'work', this analysis must demonstrate
that the steady-state traveling backward power developed at the matching network
input is equal in magnitude but 180 degrees out of phase with the initial power
reflected at the matching network input. For this to occur Vback must be the
negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE
TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition."

Emphasis mine.

Walt

  #13   Report Post  
Old June 3rd 04, 08:03 PM
Cecil Moore
 
Posts: n/a
Default

Walter Maxwell wrote:

Cecil Moore wrote:
"How" is not explained in any of the physics references.


Cecil, I explained the 'how', both in Reflections and in QEX.


Yes, I know you did, Walt. By "physics references" above, I meant
books like college physics textbooks, e.g. _Optics_, by Hecht.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!


Not recognizing his power equations as classical EM physics interference
terms was a pretty huge mistake in Part 3. But alleged gurus on this
newsgroup have done the same thing. Apparently, power is simply ignored
in present-day transmission line theory.

Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this
constitutes a short circuit.


I agree it constitutes a "short circuit" for superposed rearward-
traveling voltages. But exactly the same thing happens to the current
as happens to the voltage. And an "open circuit" is what causes the
rearward-traveling currents to superpose to zero.

The two rearward-traveling superposing voltages might be:
(100v at zero degrees) superposed with (100v at 180 degrees)
The superposed sum of the two rearward-traveling voltages is zero.
This indeed acts like a short where voltages go to zero.

The two corresponding rearward-traveling superposing currents might be:
(2a at 180 degrees) superposed with (2a at zero degrees)
The superposed sum of the two rearward-traveling currents is zero.
This acts like an open where currents go to zero.

Or if you prefer, both the E-fields and the H-fields cancel to
zero when complete destructive interference occurs. In a transmission
line, it causes a surge of constructive interference energy in the
opposite direction, something you have called "re-reflection from a
virtual short".
--
73, Cecil http://www.qsl.net/w5dxp

  #14   Report Post  
Old June 3rd 04, 08:09 PM
Cecil Moore
 
Posts: n/a
Default

Walter Maxwell wrote:
Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct
1999:
"For the impedance matching network to 'work', this analysis must demonstrate
that the steady-state traveling backward power developed at the matching network
input is equal in magnitude but 180 degrees out of phase with the initial power
reflected at the matching network input. For this to occur Vback must be the
negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE
TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition."


Joules/sec possesses phase? Joules/sec can be canceled?
--
73, Cecil http://www.qsl.net/w5dxp

  #15   Report Post  
Old June 3rd 04, 08:44 PM
Jim Kelley
 
Posts: n/a
Default



Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct
1999:
"For the impedance matching network to 'work', this analysis must demonstrate
that the steady-state traveling backward power developed at the matching network
input is equal in magnitude but 180 degrees out of phase with the initial power
reflected at the matching network input. For this to occur Vback must be the
negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE
TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition."


Joules/sec possesses phase? Joules/sec can be canceled?


Therein lies part of the problem with thinking that the unit
(Joules/sec) moves along a transmission line. Energy in Joules moves.
(Joules/sec) of power does not.

73, Jim AC6XG


  #16   Report Post  
Old June 3rd 04, 08:57 PM
Walter Maxwell
 
Posts: n/a
Default

On Thu, 03 Jun 2004 14:03:26 -0500, Cecil Moore
wrote:

Walter Maxwell wrote:

Cecil Moore wrote:
"How" is not explained in any of the physics references.


Cecil, I explained the 'how', both in Reflections and in QEX.


Yes, I know you did, Walt. By "physics references" above, I meant
books like college physics textbooks, e.g. _Optics_, by Hecht.

What is really perplexing to me is that several posters on this subject said
that Steve's 3-parter is the best and most illuminating article they ever read
on the subject. How can they have missed some of the most egregious errors
appearing in that paper is unbelievable!


Not recognizing his power equations as classical EM physics interference
terms was a pretty huge mistake in Part 3. But alleged gurus on this
newsgroup have done the same thing. Apparently, power is simply ignored
in present-day transmission line theory.

Cecil, if s11(a1) is equal in magnitude but in opposite phase with s12(a2) this
constitutes a short circuit.


I agree it constitutes a "short circuit" for superposed rearward-
traveling voltages. But exactly the same thing happens to the current
as happens to the voltage. And an "open circuit" is what causes the
rearward-traveling currents to superpose to zero.

The two rearward-traveling superposing voltages might be:
(100v at zero degrees) superposed with (100v at 180 degrees)
The superposed sum of the two rearward-traveling voltages is zero.
This indeed acts like a short where voltages go to zero.


Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.

The two corresponding rearward-traveling superposing currents might be:
(2a at 180 degrees) superposed with (2a at zero degrees)
The superposed sum of the two rearward-traveling currents is zero.
This acts like an open where currents go to zero.


Of course, but the voltage doubles.

Or if you prefer, both the E-fields and the H-fields cancel to
zero when complete destructive interference occurs. In a transmission
line, it causes a surge of constructive interference energy in the
opposite direction, something you have called "re-reflection from a
virtual short".


Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously. At a short circuit the E field
collaples to zero, but its energy temporarily merges with the H field, making
the H field double it normal value. But the changing H field immediately
reestablishes the E field, both now traveling in the forward direction. And yes,
this is called re-reflection.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.

Walt
  #17   Report Post  
Old June 3rd 04, 09:11 PM
Walter Maxwell
 
Posts: n/a
Default

Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your address
appeared as . What's the ONEDOT? I now see it in your return
address on the rraa.

What's going on?

Walt
  #18   Report Post  
Old June 3rd 04, 09:24 PM
Walter Maxwell
 
Posts: n/a
Default

On Thu, 03 Jun 2004 12:44:10 -0700, Jim Kelley wrote:



Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, the following is a direct quote from Steve's Comm Quart Article, Oct
1999:
"For the impedance matching network to 'work', this analysis must demonstrate
that the steady-state traveling backward power developed at the matching network
input is equal in magnitude but 180 degrees out of phase with the initial power
reflected at the matching network input. For this to occur Vback must be the
negative of VR. In this case ALL POWER TRAVELING BACKWARD TOWARDS THE
TRANSMISTTER WILL BE CANCELED, resulting in the steady-state matched condition."


Joules/sec possesses phase? Joules/sec can be canceled?


Hi Jim,

Of course you're right, but that's not the point. The point is that reflected
energy is not canceled, nor does it disappear at the matching point, instead it
is re-reflected into the forward direction. This is the point that Steve
apparently doesn't understand. And this is the reason his power budget is
incorrect in his 3-part article, he ignored the energy appearing at the match
point, assuming that it disappeared, though his word is 'canceled'.

I thought my emphasis with capitalization would contain the necessary info, but
I can see now that I should have made the emphasis show that the 'canceled
energy' was erroneous, because energy cannot be canceled. In this case it is
re-reflected, a concept Steve ignores.

Therein lies part of the problem with thinking that the unit
(Joules/sec) moves along a transmission line. Energy in Joules moves.
(Joules/sec) of power does not.

73, Jim AC6XG


Sorry, Jim, I put my response in the wrong place.

Walt
  #19   Report Post  
Old June 3rd 04, 09:52 PM
Cecil Moore
 
Posts: n/a
Default

Jim Kelley wrote:
Therein lies part of the problem with thinking that the unit
(Joules/sec) moves along a transmission line. Energy in Joules moves.
(Joules/sec) of power does not.


What about the Poynting Vector and Power Flow Vectors?
What about the 60 Hz "power generation" and "power
distribution" system?

Are you saying that the trailing edge of an ExH wave
is not moving? Are you saying that the ExB power in
the light from Alpha Centauri didn't come from Alpha
Centauri?
--
73, Cecil http://www.qsl.net/w5dxp

  #20   Report Post  
Old June 3rd 04, 10:18 PM
Cecil Moore
 
Posts: n/a
Default

Walter Maxwell wrote:
Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.


Nope, for complete destructive interference, both the E-field and
H-field associated with the two interfering waves collapse to zero.
For the resulting complete constructive interference, the ratio of
the E-field to the H-field equals the characteristic impedance of
the medium.

The two corresponding rearward-traveling superposing currents might be:
(2a at 180 degrees) superposed with (2a at zero degrees)
The superposed sum of the two rearward-traveling currents is zero.
This acts like an open where currents go to zero.


Of course, but the voltage doubles.


Nope, again here are the two sets of reflected waves.

#1 100v at zero degrees and 2a at 180 degrees = 200W

#2 100v at 180 degrees and 2a at zero degrees = 200W

Superposing those two reflected waves yields zero volts and
zero amps.

Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously.


For "complete destructive interference" as explained in _Optics_, by
Hecht, the E-field and B-field (H-field) indeed do go to zero
simultaneously. That is what causes a completely dark ring in a
set of interference rings. Of course, a resulting corresponding
complete constructive interference causes the brightest of rings.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.


He pretty much ignored current. His power equations are exact copies of
the light irradiance interference equations from optics, but he apparently
didn't realize it until I pointed it out to him.
--
73, Cecil http://www.qsl.net/w5dxp

Reply
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules

Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On



All times are GMT +1. The time now is 11:30 PM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright ©2004-2024 RadioBanter.
The comments are property of their posters.
 

About Us

"It's about Radio"

 

Copyright © 2017