I have made a living as a professional Engineer designing transmission
equipment over
the past four decades, currently more than $4BB gross shipped to world
wide
===============================

There's an old adage "With radio any bloody thing will work".

Let's run a quick calculation as Tom has suggested several times.

Consider a transmission line with Z0 = 50 - j10 ohms connected to a 50 +
j0 ohm load. As I hope will be evident, it doesn't matter what's
connected to the source end of the line or how long it is. We'll look at
the voltage V and current I at a point within the cable, but very, very
close to the load end. Conditions are steady state. When numerical V or
I is required, assume it's RMS.

I hope we can agree of the following. If not, it's a waste of time to
read the rest of the analysis.

1. Vf / If = Z0, where Vf and If are the forward voltage and current
respectively.

2. Vr / -Ir = Z0, where Vr and Ir are the reverse voltage and current
respectively. The minus sign is due to my using the common definition of
positive Ir being toward the load. (If this is too troublesome to
anyone, let me know, and I'll rewrite the equations with positive Ir
toward the source.)

3. Gv = Vr / Vf, where Gv is the voltage reflection coefficient. This is
the usual definition. Likewise,

4. Gi = Ir / If

5. V = Vf + Vr

6. I = If + Ir

Ok so far?

Combine 1 and 2 to get

7. Vf / If = -(Vr / Ir) = Vr / Vf = -(Ir / If)

From 3, 4, and 7,

8. Gv = -Gi

From 3 and 5,

9. V = Vf(1 + Gv)

and similarly from 4 and 6,

10. I = If(1 + Gi)

From 8 and 10,

11. I = If(1 - Gv)

Dividing 9 by 10,

12. V / I = (Vf / If) * [(1 + Gv) / (1 - Gv)]

Combining 12 with 1,

13. V / I = Z0 * [(1 + Gv) / (1 - Gv)]

And finally we have to observe that, by inspection, the voltage V just
inside the line has to equal the voltage Vl just outside the line and,
by Kirchoff's current law, the current I just inside the line has to
equal the current Il just outside the line:

14. Vl = V and Il = I, where Vl and Il are the voltage and current at
the load.

From 13 and 14, and noting that Zl = Vl / Il, then,

15. Zl = Z0 * [(1 + Gv) / (1 - Gv)]

Now let's test the formulas for Gv that have been presented. We have
three to choose from:

A. The one just posted by Peter, (Zl - Z0conj) / (Zl + Z0conj)

B. Slick's, (Zl - Z0conj) / (Zl + Z0)

C. The one in all my texts and used by practicing engineers, (Zl - Z0) /
(Zl + Z0)

We'll plug the numbers into Peter's (A) first, giving us:

Gv = (50 - 50 - j10) / (50 + 50 + j10) = -j10 / (100 + j10) =
-0.009901 - j0.09901

Plugging this into the right side of 15, we get:

Zl = 46.15 -j19.23

This obviously isn't correct -- Zl is known to be 50 + j0.

Let's try Slick's (B):

Gv = (50 - 50 - j10) / (50 + 50 - j10) = -j10 / (100 - j10) =
0.009901 - j0.09901

from which we calculate, from 15,

Zl = 48.00 - j20.00

Again, obviously not right.

Finally, using the universally accepted formula (C):

Gv = (50 - 50 + j10) / (50 + 50 - j10) = j10 / (100 - j10) =
-0.009901 + j0.09901

from which, from 15, we get

Zl = 50.00 + j0

This is the value we know to be Zl. Getting the correct result for one
numerical example does *not* prove that an equation is correct. However,
getting an incorrect result for even one numerical example *does* prove
that an equation isn't correct. So we can conclude that variations A and
B aren't correct.

------------------

Actually, the accepted formula (C) can be derived directly from equation
15, so if all the steps to that point are valid, so is the accepted formula.

Why aren't Peter's or Slick's formulas correct? The real reason is that
they aren't derived from known principles by an orderly progression of
steps like the ones above. There's simply no way to get from known
voltage and current relationships to the conjugate equations. They're
plucked from thin air. That's simply not adequate or acceptable for
scientific or engineering use. (I challenge anyone convinced that either
of those equations is correct to present a similar development showing why.)

What about the seemingly sound logic that the accepted formula doesn't
work for complex Z0 because it implies that a conjugate match results in
a reflection? The formula certainly does imply that. And it's a fact --
a conjugate match guarantees a maximum transfer of power for a given
source impedance. But it doesn't guarantee that there will be no
reflection. We're used to seeing the two conditions coincide, but that's
just because we're used to dealing with a resistive Z0, or at least one
that's close enough to resistive that it's a good approximation. The
fact that the conditions for zero reflection and for maximum power
transfer are different is well known to people accustomed to dealing
with transmission lines with complex Z0.

But doesn't having a reflection mean that some power is reflected and
doesn't reach the load, reducing the load power from its maximum
possible value? As you might know from my postings, I'm very hesitant to
deal with power "waves". But what's commonly called forward power
doesn't stay constant as the load impedance is changed, nor does the
forward voltage. So it turns out that if you adjust the load for a
conjugate match, there is indeed reflected voltage, and "reflected
power". But the forward voltage and power are greater when the load is
Z0conj than when Zl = Z0 and no reflection takes place -- enough greater
that maximum power transfer occurs for the conjugate match, with a
reflection present.

I'd welcome any corrections to any statements I've made above, any of
the equations, or the calculations. The calculations are particularly
subject to possible error, so should undergo particular scrutiny. I'll
be glad to correct any errors. Anyone who disagrees with the conclusion
is invited and encouraged to present a similar development, showing the
derivation of the alternate formula and giving numerical results from an
example. That's how science, and good engineering, are done. And what it
takes to convince me.

Roy Lewallen, W7EL


Peter O. Brackett wrote:
Slick:

[snip]

[snip]

You are absolutely correct Roy, that formula given by "Slick" is just

plain

WRONG!

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

Always has been, always will be.

[snip]

After consideration, I must agree with Slick.

Slick is RIGHT and I was WRONG!

Slick please accept my apologies!!! I was wrong, and I admit it!

Indeed, the correct formula for the voltage reflection coefficient "rho"
when computed using a "reference impedance" R, which is say the, perhaps
complex, internal impedance R = r + jx of a generator/source which is loaded
by a perhaps complex load impedance Z = ro + j xo must indeed be:

rho = (Z - conj(R))/(Z + conj(R)) = (Z - r + jx)/(Z + r - jx)

For indeed as Slick pointed out elsewhere in this thread, how else will the
reflected voltage equal zero when the load is a conjugate match to the
generator.

Slick thanks for directing the attention of this "subtlety" to the
newsgroup, and again...

Slick, please accept my apologies, I was too quick to criticize!

Good work, and lots of patience... :-)

Regards,

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

Methinks someone is confused between the conditions for maximum power
transfer and no reflections.

....Keith

"Peter O. Brackett" wrote:

Slick:

[snip]
[snip]

You are absolutely correct Roy, that formula given by "Slick" is just
plain
WRONG!

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

Always has been, always will be.
[snip]

After consideration, I must agree with Slick.

Slick is RIGHT and I was WRONG!

Slick please accept my apologies!!! I was wrong, and I admit it!

Indeed, the correct formula for the voltage reflection coefficient "rho"
when computed using a "reference impedance" R, which is say the, perhaps
complex, internal impedance R = r + jx of a generator/source which is loaded
by a perhaps complex load impedance Z = ro + j xo must indeed be:

rho = (Z - conj(R))/(Z + conj(R)) = (Z - r + jx)/(Z + r - jx)

For indeed as Slick pointed out elsewhere in this thread, how else will the
reflected voltage equal zero when the load is a conjugate match to the
generator.

Slick thanks for directing the attention of this "subtlety" to the
newsgroup, and again...

Slick, please accept my apologies, I was too quick to criticize!

Good work, and lots of patience... :-)

Regards,

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

Roy Lewallen wrote:
Let's run a quick calculation as Tom has suggested several times.

Consider a transmission line with Z0 = 50 - j10 ohms connected to a 50 +
j0 ohm load. As I hope will be evident, it doesn't matter what's
connected to the source end of the line or how long it is. We'll look at
the voltage V and current I at a point within the cable, but very, very
close to the load end. Conditions are steady state. When numerical V or
I is required, assume it's RMS.

I hope we can agree of the following. If not, it's a waste of time to
read the rest of the analysis.

1. Vf / If = Z0, where Vf and If are the forward voltage and current
respectively.

2. Vr / -Ir = Z0, where Vr and Ir are the reverse voltage and current
respectively. The minus sign is due to my using the common definition of
positive Ir being toward the load. (If this is too troublesome to
anyone, let me know, and I'll rewrite the equations with positive Ir
toward the source.)

3. Gv = Vr / Vf, where Gv is the voltage reflection coefficient. This is
the usual definition. Likewise,

4. Gi = Ir / If

5. V = Vf + Vr

6. I = If + Ir

Ok so far?


I am looking at W.C. Johnson, pages 15 and 16,
Equations 1.22 through 1.25, where he shows, very
concisely and elegantly, that the reflection
coefficient is zero only when the complex
terminating impedance is identically equal to the
the complex value of Z0.

And not the complex conjugate of Z0.

In other words (read carefully), the complex
terminating impedance whose value is Z0 is
equivalent to an *infinite extension* of the coax
whose complex value is Z0. This is the "clincher".

Observe also the the reflection coefficient for
current is the negative of the reflection
coefficient for voltage.

Bill W0IYH

Roy Lewallen wrote:

But doesn't having a reflection mean that some power is reflected and
doesn't reach the load, reducing the load power from its maximum
possible value? As you might know from my postings, I'm very hesitant to
deal with power "waves". But what's commonly called forward power
doesn't stay constant as the load impedance is changed, nor does the
forward voltage. So it turns out that if you adjust the load for a
conjugate match, there is indeed reflected voltage, and "reflected
power". But the forward voltage and power are greater when the load is
Z0conj than when Zl = Z0 and no reflection takes place -- enough greater
that maximum power transfer occurs for the conjugate match, with a
reflection present.

The "primary" waves are voltage and current, and
the power value involved is
"derived" from the primary waves. Pick a point on
the line and calculate
the instantaneous power at that point. Over time,
that value of power
"travels" along the line in concert with the
voltage and current waves.

There is nothing wrong with this. All of the
hulabaloo over this subject isn't worth a hill of
beans. Everyone understands what power "flow" and
"power outage" mean.

Bill W0IYH

William E. Sabin wrote:

I am looking at W.C. Johnson, pages 15 and 16, Equations 1.22 through
1.25, where he shows, very concisely and elegantly, that the
reflection coefficient is zero only when the complex terminating
impedance is identically equal to the the complex value of Z0.

And not the complex conjugate of Z0.

In other words (read carefully), the complex terminating impedance
whose value is Z0 is equivalent to an *infinite extension* of the
coax whose complex value is Z0. This is the "clincher".

That's fine. I agree entirely, and it follows from my analysis and my
conclusion. A similar analysis can be found in many texts. My offering
to provide a large number of references has brought forth no interest
from the most vocal participants, and they've also showed a lack of
willingness to work through the simple math themselves. So I felt that
it might be a good idea to post the derivation before more converts are
made to this religion of proof-by-gut-feel-and-flawed-logic.

Observe also the the reflection coefficient for current is the
negative of the reflection coefficient for voltage.

Likewise.

Bill W0IYH


One other thing. If the reflection coefficient is zero, then all of the
real power that is dumped into the coax is being delivered to the load,
minus the losses in the coax. The system is as good as it can get.

Bill W0IYH

I agree with this only in the sense that the "system is as good as it
can get" means only that the loss in the coax is minimized for a given
delivered power. It doesn't guarantee that the maximum possible power
will be delivered to the load.

After some thought, I see I was in error in stating that terminating the
line in its complex conjugate necessarily results in the maximum power
transfer to the load. If the 50 - j0 had been a source impedance instead
of a transmission line impedance, that would be true. However, what
results in the maximum power from the source in a system like this is
that the *source* be conjugately matched to the impedance seen at the
input of the line. Although terminating the line in its characteristic
impedance minimizes the line loss, it doesn't guarantee maximum load power.

Of course, the conjugate matching theorem says that if the line is
lossless, a load which is the conjugate of the impedance looking back
toward the source from the load will result in a conjugate match at the
source and everywhere else along the line, so that will effect maximum
power transfer. But the impedance looking back toward the source from
the load isn't by any means necessarily equal to the Z0 of the line, so
the conjugate of Z0 isn't necessarily the optimum impedance for power
transfer, as I erroneously stated. And I don't believe that the
conjugate match theorem applies to a lossy line. I certainly don't want
to start up an argument about this topic, though, and will simply state
for certain that the maximum net power will be delivered to the line
when the impedance seen looking into the line is equal to the complex
conjugate of the source impedance.

It should be easy to set up a couple of simple numerical examples to
illustrate this. Unfortunately, I'm pressed for time at the moment and
have to run. I apologize for the error regarding terminating impedance
and maximum power transfer.

Thanks for spurring me to re-think the conditions for maximum power
transfer. I apologize for the error.

Roy Lewallen, W7EL

"William E. Sabin" sabinw@mwci-news wrote in message ...
....
In other words (read carefully), the complex
terminating impedance whose value is Z0 is
equivalent to an *infinite extension* of the coax
whose complex value is Z0. This is the "clincher".

Exactly, Bill! And with that, you don't need any equations. You only
need to realize that if you cut a line which had no reflection on it,
you can weld it back together without introducing reflections
(assuming you do a perfect job restoring it to its original state).
That should be obvious and easy to grasp even for a math-o-phobe.

And thanks to Roy for taking the trouble to actually go through the
algebra and post it. I honestly hoped that S. would do that, and
still cling to the belief that there may have been a lurker or two who
also did it.

Cheers,
Tom

"William E. Sabin" sabinw@mwci-news wrote in message ...
....

The usage of complex conjugate Z0* becomes
significant when calculating very large values of
VSWR, according to some authors. But for these
very large values of standing waves, the concept
of VSWR is a useless numbers game anyway. For
values of VSWR less that 10:1 the complex Z0 is
plenty good enough for good quality coax.

My working definition for SWR is (1+|rho|)/|(1-|rho|)|. (Note the
overall absolute value in the denominator, so it never goes negative.)
Rho, of course, is Vr/Vf = (Zload-Zo)/(Zload+Zo), no conjugates. In
that way, when |rho|=1, that is, when |Vr|=|Vf|, SWR becomes infinite.
When |rho|1, SWR comes back down, and corresponds _exactly_ to the
Vmax/Vmin you would observe _IF_ you could propagate that Vr and Vf
without loss to actually establish standing waves you could measure.
I agree with Bill that all this is academic, but my working definition
seems to me to be consistent with the _concept_ of SWR, and does not
require me to be changing horses in mid-stream and remembering when to
use a conjugate and when not to. I _NEVER_ use Zo* in finding rho.
To me it's not so much a matter of nitpicking an area of no real
practical importance as coming up with a firm definition I don't have
to second-guess. Except on r.r.a.a., I don't seem to ever get into
discussions about such things, but my definitions are easy to state up
front so anyone I'm talking with can understand where I'm coming from
on them.


W.C. Johnson points out on page 150 that the concept:

Pload = Pforward - Preflected

is strictly correct only when Z0 is pure
resistance. But the calculations of real power
into the coax and real power into the load are
valid and the difference between the two is the
real power loss in the coax. For these
calculations the complex value Z0 for moderately
lossy coax is useful and adequate.

The preoccupation with VSWR values is unfortunate
and excruciatingly exact answers involve more
nitpicking than is sensible.

Agreed! And thanks for the reference re powers.

Cheers,
Tom

Roy Sez -
That's fine. I agree entirely, and it follows from my analysis and my
conclusion. A similar analysis can be found in many texts. My offering
to provide a large number of references has brought forth no interest
from the most vocal participants, and they've also showed a lack of
willingness to work through the simple math themselves. So I felt that
it might be a good idea to post the derivation before more converts are
made to this religion of proof-by-gut-feel-and-flawed-logic.

===============================

YOUR analysis !

Oliver Heaviside worked it all out 120 years back.

For those antagonists who are not willing, or are unable, or are too
frightened, or are just too plain lazy to work through the simple ARITHMETIC
themselves, download in a few seconds program COAXPAIR from website below
and run immediately. It is designed to do exactly the calculations everybody
is at war about. (And many more).

Use any ordinary coax with an overal length of 30 dB or more at 1000 Hz (to
ensure a nice negative angle of Zo) and terminate the line with its own
calculated input impedance = Zo = Ro+jXo. Then terminate the line with its
conjugate. Observe what happens to the reflection coefficient, SWR and
actual overall loss. Observe what happens as line length is reduced.

Program COAXPAIR (with other transmission line progs) has been sitting there
for 2 years or more.
---
=======================
Regards from Reg, G4FGQ
For Free Radio Design Software
go to http://www.g4fgq.com
=======================

"William E. Sabin" sabinw@mwci-news wrote in message ...

I am looking at W.C. Johnson, pages 15 and 16,
Equations 1.22 through 1.25, where he shows, very
concisely and elegantly, that the reflection
coefficient is zero only when the complex
terminating impedance is identically equal to the
the complex value of Z0.

And not the complex conjugate of Z0.



This is ABSOLUTELY WRONG!

The reflection coefficient is zero only when the Zload
is the conjugate of the Zo.

Go look it up in any BASIC RF book!


Slick