There has been some discussion in the past months about conjugate matching,
VSWR, and power transfer from source to load.
I've come across a puzzle while noodling on this.
My main issue here is how the heck my VSWR meter is measuring the way it is.

My elementary hook-up is an RF power amp feeding directly into a VSWR meter,
and from there into a load consisting of a carbon resistor and a variable
capacitor rigged in series. The meter connects to the load via about a foot of
50 ohm coax. The frequency is between 1 and 10 MHz.
Model the source impedance as Zs = R + jX, and the load impedance as Zl = r +
jx (or use phasors if you prefer .
The following two statements are true:
1) The power dissipated in the load (r) is maximised when x = -X (so-called
"conjugate matching"), whatever the value of (r).
2) The classical VSWR is minimised (zero "reflected power") when x = +X,
whatever the value of (r).

However, my VSWR meter, whch is a conventional 2-diode bridge and short
transmission line, indicates that minimum indicated VSWR
corresponds to max power dissipated in (r).!! (i.e. at conjugate match, and
NOT when reflected power is zero).

The equation normally used for VSWR is
VSWR = ABS( (1 + |p|) / (1 - |p|) )
where
p = (Zl - Zs) / (Zl + Zs)
and p is a measure of the amount of power reflected back to the source, called
the "voltage reflection coefficient"

I plotted something I call "conjugate VSWR" or VSWR*. which is the same
expression as above, but with p defined as
p = (Zl - Zs*) / (Zl + Zs)
where Zs* indicates the complex conjugate of Zs.
and the behaviour of this VSWR* thingie absolutely matches what I see on my
meter. Aye, there's the rub.

Some points to note
a) Classical VSWR shows NO minimum for all r, when x has the opposite sign to
X
b) VSWR* always has a minimum at the same r-value which causes maximum power
to be dissipated in r, whatever the value of x.

Again, I flat don't understand how my VSWR meter can indicate VSWR* when I
know it should indicate VSWR.

Here are a couple of links to flesh out the theory.

1. Wade through this at your peril - it's you lot fighting abou this issue and
is VERY long
http://www.ibiblio.org/pub/academic/...S/20030831.ant

2. This is much more succint - cut to the chase on p47
http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf

Best,
Andrew

ok, bonzo, i'll bite on the troll bait. but only because its early in the
morning and the normal endless discussion of this stuff hasn't taken over
yet.

"Lord Snooty" wrote in message
nk.net...
There has been some discussion in the past months about conjugate
matching,
VSWR, and power transfer from source to load.
I've come across a puzzle while noodling on this.
My main issue here is how the heck my VSWR meter is measuring the way it
is.

My elementary hook-up is an RF power amp feeding directly into a VSWR
meter,
and from there into a load consisting of a carbon resistor and a variable
capacitor rigged in series. The meter connects to the load via about a
foot of
50 ohm coax. The frequency is between 1 and 10 MHz.
Model the source impedance as Zs = R + jX, and the load impedance as Zl =
r +
jx (or use phasors if you prefer .
The following two statements are true:
1) The power dissipated in the load (r) is maximised when x = -X
(so-called
"conjugate matching"), whatever the value of (r).

wrong. you must transform the R+jX along the transmission line to get back
to the load seen by the source. you stipulate a low frequency and short
line, so you are close anyway.

2) The classical VSWR is minimised (zero "reflected power") when x = +X,
whatever the value of (r).

doubly wrong. vswr is on a cable and is independent of the source. it
knows nothing of R+jX only the characteristic impedance of the cable. all
following calculations are wrong for this reason alone.


However, my VSWR meter, whch is a conventional 2-diode bridge and short
transmission line, indicates that minimum indicated VSWR
corresponds to max power dissipated in (r).!! (i.e. at conjugate match,
and
NOT when reflected power is zero).

The equation normally used for VSWR is
VSWR = ABS( (1 + |p|) / (1 - |p|) )
where
p = (Zl - Zs) / (Zl + Zs)

wrong again, the impedance used must be that of the cable not of the source.
its not worth commenting further until you understand this.

and p is a measure of the amount of power reflected back to the source,
called
the "voltage reflection coefficient"

I plotted something I call "conjugate VSWR" or VSWR*. which is the same
expression as above, but with p defined as
p = (Zl - Zs*) / (Zl + Zs)
where Zs* indicates the complex conjugate of Zs.
and the behaviour of this VSWR* thingie absolutely matches what I see on
my
meter. Aye, there's the rub.

Some points to note
a) Classical VSWR shows NO minimum for all r, when x has the opposite sign
to
X
b) VSWR* always has a minimum at the same r-value which causes maximum
power
to be dissipated in r, whatever the value of x.

Again, I flat don't understand how my VSWR meter can indicate VSWR* when I
know it should indicate VSWR.

Here are a couple of links to flesh out the theory.

1. Wade through this at your peril - it's you lot fighting abou this issue
and
is VERY long

http://www.ibiblio.org/pub/academic/...S/20030831.ant

2. This is much more succint - cut to the chase on p47

http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf

Best,
Andrew

Andrew,

Conjugate matching may be a very interesting subject and is indeed a much
discussed topic on these walls.

However, amongst the amateur fraternity, a "conjugate match" should be
classified as being off topic. It does not exist. Not even when the tuner
has been adjusted exactly for a VSWR equal to 1 to 1.

Why?

Because the internal impedance of the transmitter is not 50 ohms. It is not
relevant.

What is the internal impedance of YOUR transmitter? You will not find it
mentioned in the manufacturer's operating or maintenance handbooks. In all
likelihood you will not know what it is even if you designed and built the
transmitter yourself!
---
Reg, G4FGQ

From the previous posting, I can guess who is going to jump all over this.
Keep up the good work.

Tam
"Dave" wrote in message
...
ok, bonzo, i'll bite on the troll bait. but only because its early in the
morning and the normal endless discussion of this stuff hasn't taken over
yet.

"Lord Snooty" wrote in message
nk.net...
There has been some discussion in the past months about conjugate
matching,
VSWR, and power transfer from source to load.
I've come across a puzzle while noodling on this.
My main issue here is how the heck my VSWR meter is measuring the way it
is.

My elementary hook-up is an RF power amp feeding directly into a VSWR
meter,
and from there into a load consisting of a carbon resistor and a
variable
capacitor rigged in series. The meter connects to the load via about a
foot of
50 ohm coax. The frequency is between 1 and 10 MHz.
Model the source impedance as Zs = R + jX, and the load impedance as Zl
=
r +
jx (or use phasors if you prefer .
The following two statements are true:
1) The power dissipated in the load (r) is maximised when x = -X
(so-called
"conjugate matching"), whatever the value of (r).

wrong. you must transform the R+jX along the transmission line to get
back
to the load seen by the source. you stipulate a low frequency and short
line, so you are close anyway.

2) The classical VSWR is minimised (zero "reflected power") when x = +X,
whatever the value of (r).

doubly wrong. vswr is on a cable and is independent of the source. it
knows nothing of R+jX only the characteristic impedance of the cable. all
following calculations are wrong for this reason alone.


However, my VSWR meter, whch is a conventional 2-diode bridge and short
transmission line, indicates that minimum indicated VSWR
corresponds to max power dissipated in (r).!! (i.e. at conjugate match,
and
NOT when reflected power is zero).

The equation normally used for VSWR is
VSWR = ABS( (1 + |p|) / (1 - |p|) )
where
p = (Zl - Zs) / (Zl + Zs)

wrong again, the impedance used must be that of the cable not of the
source.
its not worth commenting further until you understand this.

and p is a measure of the amount of power reflected back to the source,
called
the "voltage reflection coefficient"

I plotted something I call "conjugate VSWR" or VSWR*. which is the same
expression as above, but with p defined as
p = (Zl - Zs*) / (Zl + Zs)
where Zs* indicates the complex conjugate of Zs.
and the behaviour of this VSWR* thingie absolutely matches what I see on
my
meter. Aye, there's the rub.

Some points to note
a) Classical VSWR shows NO minimum for all r, when x has the opposite
sign
to
X
b) VSWR* always has a minimum at the same r-value which causes maximum
power
to be dissipated in r, whatever the value of x.

Again, I flat don't understand how my VSWR meter can indicate VSWR* when
I
know it should indicate VSWR.

Here are a couple of links to flesh out the theory.

1. Wade through this at your peril - it's you lot fighting abou this
issue
and
is VERY long


http://www.ibiblio.org/pub/academic/...S/20030831.ant

2. This is much more succint - cut to the chase on p47


http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf

Best,
Andrew

nope, all done. reg is here and cecil can't be far behind. i've had my
fun, time to do other more productive things than watch them re-hash
conjugal matches for the next month or two.

"Tam/WB2TT" wrote in message
...
From the previous posting, I can guess who is going to jump all over this.
Keep up the good work.

Tam
"Dave" wrote in message
...
ok, bonzo, i'll bite on the troll bait. but only because its early in
the
morning and the normal endless discussion of this stuff hasn't taken
over
yet.

"Lord Snooty" wrote in message
nk.net...
There has been some discussion in the past months about conjugate
matching,
VSWR, and power transfer from source to load.
I've come across a puzzle while noodling on this.
My main issue here is how the heck my VSWR meter is measuring the way
it
is.

My elementary hook-up is an RF power amp feeding directly into a VSWR
meter,
and from there into a load consisting of a carbon resistor and a
variable
capacitor rigged in series. The meter connects to the load via about a
foot of
50 ohm coax. The frequency is between 1 and 10 MHz.
Model the source impedance as Zs = R + jX, and the load impedance as
Zl
=
r +
jx (or use phasors if you prefer .
The following two statements are true:
1) The power dissipated in the load (r) is maximised when x = -X
(so-called
"conjugate matching"), whatever the value of (r).

wrong. you must transform the R+jX along the transmission line to get
back
to the load seen by the source. you stipulate a low frequency and short
line, so you are close anyway.

2) The classical VSWR is minimised (zero "reflected power") when x =
+X,
whatever the value of (r).

doubly wrong. vswr is on a cable and is independent of the source. it
knows nothing of R+jX only the characteristic impedance of the cable.
all
following calculations are wrong for this reason alone.


However, my VSWR meter, whch is a conventional 2-diode bridge and
short
transmission line, indicates that minimum indicated VSWR
corresponds to max power dissipated in (r).!! (i.e. at conjugate
match,
and
NOT when reflected power is zero).

The equation normally used for VSWR is
VSWR = ABS( (1 + |p|) / (1 - |p|) )
where
p = (Zl - Zs) / (Zl + Zs)

wrong again, the impedance used must be that of the cable not of the
source.
its not worth commenting further until you understand this.

and p is a measure of the amount of power reflected back to the
source,
called
the "voltage reflection coefficient"

I plotted something I call "conjugate VSWR" or VSWR*. which is the
same
expression as above, but with p defined as
p = (Zl - Zs*) / (Zl + Zs)
where Zs* indicates the complex conjugate of Zs.
and the behaviour of this VSWR* thingie absolutely matches what I see
on
my
meter. Aye, there's the rub.

Some points to note
a) Classical VSWR shows NO minimum for all r, when x has the opposite
sign
to
X
b) VSWR* always has a minimum at the same r-value which causes maximum
power
to be dissipated in r, whatever the value of x.

Again, I flat don't understand how my VSWR meter can indicate VSWR*
when
I
know it should indicate VSWR.

Here are a couple of links to flesh out the theory.

1. Wade through this at your peril - it's you lot fighting abou this
issue
and
is VERY long



http://www.ibiblio.org/pub/academic/...S/20030831.ant

2. This is much more succint - cut to the chase on p47



http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf

Best,
Andrew

Dave wrote:
nope, all done. reg is here and cecil can't be far behind. i've had my
fun, time to do other more productive things than watch them re-hash
conjugal matches for the next month or two.

I guess I need to say this again. My take on discussions of conjugate
matching in ham antenna systems is that it is a waste of time. If reflected
energy is not allowed to reach the source, e.g. typical ham Z0-matched
systems, the source impedance is irrelevant and doesn't affect anything
in the system except for efficiency.
--
73, Cecil http://www.qsl.net/w5dxp



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Lord Snooty wrote:
Indeed so, because the whole idea of characterising my SWR meter is towards
the goal of measuring the output impedance of my RF amp!
I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof
procedure to find R and X. Got one?

R + jX is a linear function. As your amp is probably push-pull
AB1 class, you could only ever obtain the averaged impedance of
two non-linear devices. Is that what you want?
--
73, Cecil http://www.qsl.net/w5dxp



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-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

On Sun, 23 May 2004 14:17:51 +0000 (UTC), "Reg Edwards"
wrote:

Because the internal impedance of the transmitter is not 50 ohms. It is not
relevant.

Old Son,

Lord Kelvinator would be saddened by such a dismissal of actually
delving into "knowing" by your meagre understanding that fails to
offer an actual value that it IS. Such superstition that dominates
this paucity would lead us to believe the Z is as slight as an angel
wafting past in a dream (would that be in English Units, or metric?).

Definitions by negatives is an amusing troll however. I do enjoy
participation in kind:

Let's see, the Z of a transmitter is NOT
a dead parrot, nor
five farthings, nor
10 inches, nor
a polka dot dress, nor
the gross national income of Lithuania, nor
the combined weight of all your dead white scientists.

I'm sure I missed a few.... ;-)

73's
Richard Clark, KB7QHC

On Sun, 23 May 2004 16:39:19 GMT, "Lord Snooty" wrote:

Indeed so, because the whole idea of characterising my SWR meter is towards
the goal of measuring the output impedance of my RF amp!
I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof
procedure to find R and X. Got one?

Cheers,
Andrew
G3UHD

Hi Andrew,

You have asked an inescapable question that will lead to a deluge of
scribbling commemorating the best attempts of Houdini.

There are several many methods to determine exactly what you want to
know. The simplest and certainly the one that contains as much
information necessary is called "load pulling." To even mention this
time and bench proven method will result in hoots from those who would
be the last to offer you a fixed answer; however, we shall proceed.

This requires that you have access to known, but non-standard value
loads capable of sustaining the power you will perform your
measurement at. This is not a trivial requirement. It also requires
that you can in some way defeat your ALC which will attempt to offset
the pull of the non-standard load.

It is simplicity itself that only demands you consider the elements of
a Thevenin model and how to determine the model's source Z (or
likewise, the Norton model's source Z). You will need a means to
measure the voltage across the load, or the current through it. Even
here, proportionality is all that is required as long as the Load is
characterized and thus the tools can be rather spartan.

In the long run, this will mean you have to construct and verify your
own non-standard loads. Take care that through your verification you
confirm their value across all power applications (resistors are very
susceptible to drift with temperature). You should also take care to
insure that all paths and leads are as short as possible. Loading
directly at the terminals will save grief of complex compensation math
(and reduce introducing other errors). However, you can choose to
employ remote loads if you take care to characterize the lines through
which they are attached (this means you should be adept at the Smith
Chart).

There is more to be said, but this enough to offer you a significant
lead to find that, yes, the source exhibits nearly 50 Ohms (the common
Ham transmitter running at rated power will fall between 30 and 70
Ohms) - as specified and designed.

73's
Richard Clark, KB7QHC

Wise words. The true load-pulling technique advocated professionally uses a
known model of the output stage using S parameters. Yuk. Not for this soldier.
I'd rather sniff underwear in a geriatric clinic thank you. Whatever one did
in this direction, as an amateur, would doubtless be either wrong or so
inaccurate as to be not even wrong as Dr. Pauli was wont to say.

The amp is single-ended out of an MRF136, so I presume it's Class A. The amp's
designation is H-10 (I bought it surplus). It's rated at around 15W, 0.1 - 30
MHz. The circuit diagram shows no hint of current limiting circuitry.

If one is serious about proper design of a matching network - a network, I
might add, which attaches *directly* (near as dammit) to the Tx output - then
one is all at sea without a proper knowledge of source impedance. See my
comments in the other thread about this.

I tried to use a technique advised by K7ITM to measure source impedance (R +
jX), and it produces a negative value of R by calculation. This is the
technique, based on a load setting of (r + jx) ohms. And once again, there is
no cable, folks. And it wouldn't matter if I did have a few inches of it
either.

1. Determine X.
a) Set r ~= R based on a best guess (which would be R=50 ohms nominally in
many systems).
b) Monitor the voltage, current, or power in the load (r).
c) Adjust (x) to maximise the monitored value.
This setting corresponds to x = -X, and we have determined X.

2. Determine R
a) Leave x = -X set as in step 1,so now the circuit is pure resistive.
b) Monitor the voltage across the resistor.
c) Set r to R(nominal) plus and minus a small percentage, and measure the
monitored voltage at both values of r.
The voltage across r is given by
V/V0 = r / (R + r)
Solving the 2 simultaneous equations to eliminate V0 shows that R is
determined by the equation
R = r1*r2*(V2 - V1) / (V1*r2 - V2*r1)
-------------------------------------

My problems
-----------
1. My measurements of V1,V2 lead to the inescapable conclusion that the above
model fails, because the calculated value of R comes out negative.
Let us assume that we set (r2 r1) and we obtain (V2 V1), which is
predicted from the model, and is also the case for my measurements. Under
these conditions, a negative value of R can only be obtained from the
equation if V2/V1 r2/r1, which in my case is true.
I undertook a full (and rather exhaustive and tedious!) calculation of the
expression for R when the value of (x) was not set correctly to -X, thinking
that perhaps this was the cause of the discrepancy. It turns out in this case
that, if the calculated value of R is negative, it has nothing to do with the
setting of (x), and depends ONLY on the condition V2/V1 r2/r1. Since we know
that the actual value of R cannot be negative, this implies a failure of the
model.
How then can the model fail? Since we are maintaining frequency constant, any
collection of resistances and reactances, however complicated, can be modelled
as (R + jX), so it cannot be that. The only assumption left to question is the
constancy of V0, and this is what the failure must be.
This leaves me with more questions than answers, because the way forward is
now completely unclear.

2. I should also mention another, less serious, problem I had, and that is
with the determination of X. The value of (x) I determine from measurement
would be expected to be constant at a given frequency. It is in my case not
so. The derived value of X appears to depend on
a) the power level setting of my amplifier
b) the value of (r) I set in the circuit when determining X.
Quite probably this second problem relates to the first problem's
identification of the failure of the model, and can probably be subsumed under
that category.

Best,
Andrew



"Richard Clark" wrote in message
...
On Sun, 23 May 2004 16:39:19 GMT, "Lord Snooty" wrote:

Indeed so, because the whole idea of characterising my SWR meter is towards
the goal of measuring the output impedance of my RF amp!
I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof
procedure to find R and X. Got one?

Cheers,
Andrew
G3UHD

Hi Andrew,

You have asked an inescapable question that will lead to a deluge of
scribbling commemorating the best attempts of Houdini.

There are several many methods to determine exactly what you want to
know. The simplest and certainly the one that contains as much
information necessary is called "load pulling." To even mention this
time and bench proven method will result in hoots from those who would
be the last to offer you a fixed answer; however, we shall proceed.

This requires that you have access to known, but non-standard value
loads capable of sustaining the power you will perform your
measurement at. This is not a trivial requirement. It also requires
that you can in some way defeat your ALC which will attempt to offset
the pull of the non-standard load.

It is simplicity itself that only demands you consider the elements of
a Thevenin model and how to determine the model's source Z (or
likewise, the Norton model's source Z). You will need a means to
measure the voltage across the load, or the current through it. Even
here, proportionality is all that is required as long as the Load is
characterized and thus the tools can be rather spartan.

In the long run, this will mean you have to construct and verify your
own non-standard loads. Take care that through your verification you
confirm their value across all power applications (resistors are very
susceptible to drift with temperature). You should also take care to
insure that all paths and leads are as short as possible. Loading
directly at the terminals will save grief of complex compensation math
(and reduce introducing other errors). However, you can choose to
employ remote loads if you take care to characterize the lines through
which they are attached (this means you should be adept at the Smith
Chart).

There is more to be said, but this enough to offer you a significant
lead to find that, yes, the source exhibits nearly 50 Ohms (the common
Ham transmitter running at rated power will fall between 30 and 70
Ohms) - as specified and designed.

73's
Richard Clark, KB7QHC