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Old August 26th 03, 12:25 AM
Roy Lewallen
 
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Hey, I've got it. It's sort of a "virtual reflection coefficient". Now I
understand fully. Thanks!

Roy Lewallen, W7EL

W5DXP wrote:
. . .
The energy analysis on my web page deals only with physical reflection
coefficients. If 'rho' is not a physical reflection coefficient, then
it is the END RESULT of a mathematical calculation and is not the
CAUSE of anything. If a source doesn't "see" a physical impedance
discontinuity, it doesn't "see" anything except forward and reflected
waves. Coherent waves traveling in opposite directions are "unaware" of
each other. Coherent waves traveling in the same direction merge, lose
their separate identies, and become indistinguishable from one another.



  #92   Report Post  
Old August 26th 03, 02:48 AM
W5DXP
 
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Roy Lewallen wrote:
If people want to argue about the reflection of power waves, I'll gladly
bow out and let Cecil and his colleagues resume their interminable
arguments without me.


I don't think there's anything to argue about. From _Optics_, by Hecht:
"We define the reflectance R to be the ratio of the reflected power to
the incident power." "When we talk about the 'amount' of light illuminating
a surface, we are referring to something called the irradiance, denoted by
I - the average energy per unit area per unit time." Light and RF are both
EM waves. These scientific facts concerning light have only been known to
physics for about 300 years.
--
73, Cecil http://www.qsl.net/w5dxp



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  #93   Report Post  
Old August 26th 03, 02:55 AM
W5DXP
 
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Peter O. Brackett wrote:
But I maintain that is "unatural" and "Mother Nature" naturally likes the
classical

rho = (Z - Zo)/(Z + Zo)

Thoughts, comments?


I agree with you, Peter. That's what I have been calling the physical
reflection coefficient. At the junction of two transmission lines of
different characteristic impedances, it becomes s11=(Z02-Z01)/(Z02+Z01)
and is usually not equal to rho=Sqrt(Pref/Pfwd).
--
73, Cecil http://www.qsl.net/w5dxp



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  #94   Report Post  
Old August 26th 03, 03:20 AM
Roy Lewallen
 
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I'm trying to follow this, but have gone astray on the first couple of
steps.

Peter O. Brackett wrote:

The definition of the reflection coefficient is dependent upon what you
define the "reflected voltage" to be.

Consider the classical bridge circuit for measuring reflection coefficients.

If Z is an unknown load presented to a generator of internal impedance R and
that internal impedance R is used as the "reference" impedance to
observe/measure the reflected voltage b and that reflected voltage b is
calculated/measured by observation of the voltage v across the load Z and
the current i through the load Z then the classical definition of a
reflected voltage would be calculated as:

b = v - Ri = Zi - Ri = (Z - R)i Volts.


First of all, you're speaking of a circuit with a source impedance R and
load impedance Z, rather than a terminated transmission line. Forward
and reflected wave terminology is widely used in S parameter analysis,
which also uses this model, so I'll be glad to follow along to see if
and how S parameter terminology differs from the transmission line
terminology we've been discussing so far. Please correct me where my
assumptions diverge from yours.

Your "classical definition" of b isn't one familiar to me. v + Ri would
of course be the source voltage (which I'll call Vs). So v - Ri is Vs -
2*Ri. Where does this come from and what does it mean?

From your equation, and given source voltage Vs, i = Vs/(R+Z).
Therefore, your "classical definition" of reflected voltage b is, in
terms of Vs, Vs*((Z-R)/(Z+R)).

and the incident voltage a would be the Thevinins equivalent voltage across
the sum of Z and R, i.e.

a = (Z + R)i


Since i = Vs(Z+R), you're saying that a = the source voltage Vs (from
your two equations). So what you're calling the "incident voltage" is
simply the source voltage Vs.

Let's do a consistency check. The voltage at the load should be a + b =
Vs + Vs*((Z-R)/(Z+R)) = Vs*2Z/(Z+R). Inspection of the circuit as I
understand it shows that the voltage at the load should be half this
value. So, we already diverge. Which is true:

1. I've goofed up my algebra (a definite possibility)
2. I've misinterpreted your circuit, or
3. The voltage at the load is not equal to the sum of the forward and
reflected voltages a and b, as you use the terms "forward voltage" and
"reflected voltage". If v isn't equal to a + b, then what is the
relationship between v, a, and b, and what are the physical meanings of
the forward and reflected voltages?

I'd like to continue with the remainder of the analysis, but can't
proceed until this problem is cleared up.

. . .


Roy Lewallen, W7EL

  #95   Report Post  
Old August 26th 03, 03:23 AM
W5DXP
 
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Roy Lewallen wrote:
As soon as anyone starts arguing about average power waves, I'm outta
here. I'd just as soon argue about the temperature of ghosts.


You probably don't discuss light waves all that often, eh? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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  #96   Report Post  
Old August 26th 03, 03:24 AM
Roy Lewallen
 
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Thanks for the suggestion, Reg, and it's absolutely true. But even more
information is thrown away when you take the average of the power.
Remember the statistician who drowned when crossing the creek whose
average depth is only three feet.

As soon as anyone starts arguing about average power waves, I'm outta
here. I'd just as soon argue about the temperature of ghosts.

Roy Lewallen, W7EL

Reg Edwards wrote:
Dear Roy,

To reduce the amount of bafflegab to an absolute minimum why don't you just
say that immediately the value of the reflection coefficient is squared (to
allow 'power' to be introduced) half of the information it contains is
tossed into the nearest garbage heap.

Any conclusions drawn from following calculations are inevitably ambiguous
and highly suspicious to say the least.

To mention one well-known example, that's why it is impossible to deduce the
value of the line-terminating impedance from the calculated SWR.

Ignorance is the root cause of these silly, time-wasting arguments.

Get back to basics, erase Smith Charts, mis-quoted worshipped idols, and
ill-conceived inventions from your minds and start afresh from R, L, C, G, F
and little t.

Fortunately, the success of proposed missions to Mars does not depend on the
deliberations of this newsgroup.

Cec, by the way, are there any vinyards in Texas? It *is* as big as France.
----
Yours, Reg, G4FGQ



  #97   Report Post  
Old August 26th 03, 04:18 AM
Dr. Slick
 
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"David Robbins" wrote in message ...

Incorrect. You need the conjugate in the numerator if the Zo is
complex. If it is purely real, WHICH MOST TEXTS ASSUME, then you can
use the normal equation.


sorry, the derivation for the table in the book i sent before is for the
general case of a complex Zo. they then go on to simplify for an ideal line
and for a nearly ideal line... nowhere does a conjugate show up.


Please post this derivation again.

When they say "ideal line" do they mean purely real?


and that reference you give is not for a load on a transmission line, it is
talking about a generator supplying power to a load... a completely
different animal.


Not at all really. The impedance seen by the load can be from
either a source or a source hooked up with a transmission line. It
doesn't matter with this equation.


As Reg points out about the normal equation:


"Dear Dr Slick, it's very easy.

Take a real, long telephone line with Zo = 300 - j250 ohms at 1000 Hz.

Load it with a real resistor of 10 ohms in series with a real
inductance of
40 millihenrys.

The inductance has a reactance of 250 ohms at 1000 Hz.

If you agree with the following formula,

Magnitude of Reflection Coefficient of the load, ZL, relative to line
impedance

= ( ZL - Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity,

and has an angle of -59.9 degrees.

The resulting standing waves may also be calculated.

Are you happy now ?"
---
Reg, G4FGQ



Well, i wasn't happy, because how can you have a R.C. greater than
one into a passive network??? Quite impossible.

But, if you use the conjugate formula, the R.C. will indeed be
less than one.

Convince yourself.


Slick
  #98   Report Post  
Old August 26th 03, 05:19 AM
Tdonaly
 
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(Tdonaly) wrote in message
...

Very impressive. You've designed 5/8s vertical ground planes,
1/4 wavelength [something or others, I guess] and dipoles.
Where are you working now? Did you go to Lowell?
73,
Tom Donaly, KA6RUH



Did you skip the part about U. C. Davis?

I'm working part-time in the RF field, after being laid off among
what seems like everyone else. Gives me time to paint my next masterpiece!

Tit for Tat, maybe you can tell us something about you, Tom.

last school attended? Job responsibilities?


Slick



My pitifully inadequate education could be of no interest to you, Garvin; I'm
just a humble ham. (This is an _amateur_ newsgroup after all.) It is
interesting to me, though, that a person of your age and attainments
would pose as a potty-mouth little black-faced god whenever someone
disagreed with you about something as abstruse as the reflection
coefficient on a transmission line. I can only suppose that your social
education was deficient, or that you really do want your name to be the
most popular in the group's collective killfile. Anyway, you're wasting your
time with the infantile behavior. Most of the fellows on this group are old
men
who gave up that form of discourse when they learned to talk.
By the way, some of your art isn't half bad and shows the influence of some
training. Did you have an art minor in college?
73,
Tom Donaly, KA6RUH
  #99   Report Post  
Old August 26th 03, 07:54 AM
Ian White, G3SEK
 
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W5DXP wrote:

Nope, they're not Reg. For Z0-matched systems (which most ham systems are),
rho^2 is all you need to know along with the forward and reflected powers
to completely solve the voltage, current, and phase conditions at the
impedance discontinuity.


There in one sentence is the whole problem with the "power" approach.
For a complete solution including phase conditions, you have to assume a
Z0-match, and even Cecil acknowledges that is only true for "most ham
systems".

As Reg says, this is because the power approach throws away the phase
information at the start, and if you want it back again, you have to
make assumptions.

So the problem is not that the "power" approach cannot give a complete
solution, but that it cannot do it for all cases. In other words, it
isn't completely general - and that flaw is fatal.

The conventional approach based on voltage (or current) waves doesn't
discard the phase information, but uses it to give a complete solution
for every case. It is completely general, and that's precisely why
engineers use it.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek
  #100   Report Post  
Old August 26th 03, 08:03 AM
Peter O. Brackett
 
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Roy:

[snip]
b = v - Ri = Zi - Ri = (Z - R)i Volts.


First of all, you're speaking of a circuit with a source impedance R and
load impedance Z, rather than a terminated transmission line. Forward
and reflected wave terminology is widely used in S parameter analysis,
which also uses this model, so I'll be glad to follow along to see if
and how S parameter terminology differs from the transmission line
terminology we've been discussing so far. Please correct me where my
assumptions diverge from yours.

Your "classical definition" of b isn't one familiar to me. v + Ri would
of course be the source voltage (which I'll call Vs). So v - Ri is Vs -
2*Ri. Where does this come from and what does it mean?

[snip]

Yes it should be familiar to you because it is the most common definition
and one you seem to agree with. I presume that you are not used to seeing
the use
of the symbols "a" and "b" for those quantities. The use of "a" and "b" is
widely
used in Scattering Formalism and is less confusing to many than using
subscripts.
In your terminology above the "Vs" symbol is nothing more than the incident
voltage
usually given the symbol "a" in the Scattering Formalism, or the symbol V
with a "+"
sign subscript in many developments. Often you will find authors use a V
with a plus
sign "+" as a subscript to indicate the "a" voltage and a V with a minus
sign "-" subscript
to idicate the "b" voltage. Personally I find the use of math symbols "-"
and "+" or other
subscripts to variables to be confusing, I much prefer the use of "a" and
"b" for forward
or incident and reflected voltages.

Simply put, if a generator with "open circuit" voltage "a" and "internal
impedance" R
is driving a load Z [Z could be a transmission line driving point impedance,
for instance
Z would be the characteristic or surge impedance Zo of a transmission line
if
the generator was driving a semi-infinite line.] then v is the voltage drop
across Z
and I is the current through Z, and so...

a = v + Ri = Zi - Ri = (Z - R)i

is simply the [usual] forward voltage or incident voltage applied by the
generator to the
to the load Z, which may be a lumped element load or if you prefer to talk
about transmission
lines, Z can be just the driving point impedance of a transmsision line,
whatever you wish.

Then its'just a simple application of Ohms Law tosee that b = v - Ri = Zi -
Ri = (Z - R)i is
the [usual] reflected voltage. b is just the difference between the voltage
across Z which is
calculated as Zi and the voltage that would be across Z if Z was actually
equal to R. i.e. the
reflected voltage b is just the voltage that would exist across Z if there
was an "image match"
between Z and R. [If Z is the Zo of a semi-infinite transmsision line you
could call this a Zo match].

Taking the ratio of "b" to "a" just yeilds the [usual] reflection
coefficient as
b/a = (Z - R)i/(Z + R)i = (Z - R)/(Z + R). A well known result. Simple?

[snip]
From your equation, and given source voltage Vs, i = Vs/(R+Z).
Therefore, your "classical definition" of reflected voltage b is, in
terms of Vs, Vs*((Z-R)/(Z+R)).

and the incident voltage a would be the Thevinins equivalent voltage

across
the sum of Z and R, i.e.

a = (Z + R)i

[snip]

Yep you got it all right!

[snip]
Since i = Vs(Z+R), you're saying that a = the source voltage Vs (from
your two equations). So what you're calling the "incident voltage" is
simply the source voltage Vs.

[snip]

Yes, mathematically "a" = "Vs", what else would it be? Nothing mysterious
about
that. The incident voltage is always simply the open circuit voltage of the
source. In
words a is not the source voltage because the source is a Thevinin
equivalent made up
of the ideal voltage generator Vs = a behind the "internal" source impedance
R. A better
way to describe Vs = a in words would be the incident voltage a is the "open
circuit source
voltage".

[snip]
Let's do a consistency check. The voltage at the load should be a + b =
Vs + Vs*((Z-R)/(Z+R)) = Vs*2Z/(Z+R). Inspection of the circuit as I
understand it shows that the voltage at the load should be half this
value. So, we already diverge. Which is true:

[snip]

No, the voltage at the load is not (a + b) rather it is [the quite obvious
by Ohms Law] v = Zi.

and the sum of the incident and reflected voltage is simply

a + b = (v + Ri) + (v - Ri) = 2v = 2Zi

Now if there is an "image match" and the "unknown" Z is actually equal to R,
i.e. let Z = R
in all of the above, then...

a = Vs

b = 0

a + b = 2Ri

and i = Vs/2R = a/2R.

[snip]
1. I've goofed up my algebra (a definite possibility)

[snip]

Only a little :-)

[snip]
2. I've misinterpreted your circuit, or

[snip]

No you have it correct!

[snip]
3. The voltage at the load is not equal to the sum of the forward and
reflected voltages a and b, as you use the terms "forward voltage" and
"reflected voltage". If v isn't equal to a + b, then what is the
relationship between v, a, and b, and what are the physical meanings of
the forward and reflected voltages?

[snip]

I showed those relationships above. There is nothing new here... these are
the [most] widely accepted definitions of incident and reflected voltages.

[snip]
I'd like to continue with the remainder of the analysis, but can't
proceed until this problem is cleared up.

Roy Lewallen, W7EL

[snip]

OK, let's carry on.

--
Peter K1PO
Indialantic By-the-Sea, FL.






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