Owen Duffy wrote:

The myth: Measurements with a Bird 43 of the conditions on the
Thruline section are invalid unless it has some minimum length of 50
ohm line on both sides of itself.

I have performed a test using components at hand where the Bird 43 has
75 ohm line on both sides of itself, and the test configuration is
designed to present a 50+j0 ohm load at the point where the Bird 43
sampling element is located. The question is, how does the Bird
respond?

The test in detail. Each component is in a list from the source to the
load:

IC706IIG
1m RG58 with UHF connectors
MFJ949E ATU
3m RG6 (Zo=75 ohm) with BNC connectors
Bird 43
5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors
50 ohm dummy load

Short adapters were used to connect to the Bird's Type N connectors,
the dummy load's Type N connectors and the MFJ949 UHF connector.

The half wave resonance of the 5.27m length of RG6 was determined by
s/c one end and connecting the other end via an adapter to a MFJ259B
and finding the impedance dip at 23.05MHz. The calculated vf from this
test is 0.81, which reconciles reasonably with the specs. Free space
wavelenght at the test frequency is 13m.

The transmitter was set to 23.05MHz, and the ATU tuned to develop
rated power output. The ATU is only used to present the rated load to
the transmitter so as to obtain 100W for the test, to suit the Bird 43
element. It is inconsequential to the DUT (the Bird 43).

With this configuration, it is expected that the impedance at the Bird
43 is approximately 50+j0, and that there would be almost zero
reflected power.

The Bird 43 indicated 100W forward power and a quarter of a needle
width detection on reflected power.

The Bird 43 would appear to provide valid readings for the conditions
on the Bird 43 Thruline section in this case, notwithstanding that
there is not any 50 ohm transmission line attached to the Bird 43 +
N-BNC adapters.

The myth that measurements with a Bird 43 of the conditions on the
Thruline section are invalid unless it has some minimum length of 50
ohm line on both sides of itself is BUSTED.

Actually, the results of your experiment proves the myth to be true
and not to be a myth at all. There's 104.17 watts of forward power
through the Bird and 4.17 watts of reflected power back through the
Bird. Why does the Bird ignore those actual power values?

Has anyone experimental evidence to the contrary?

Yes, your experiment. Assuming 100 watts delivered to the load, the
forward power on the 75 ohm coax is actually about 104.17 watts so
the Bird's forward power reading is in error by 4.17 watts.

The reflected power on the 75 ohm coax is about 4.17 watts so the
Bird's reflected power reading is in error by close to an infinite
percentage. The Bird 43 is reading neither of the actual power values
correctly. All you have just proven is that the Bird 43 gives invalid
readings when it is in a 75 ohm environment.

THERE ARE ABOUT 4.17 WATTS OF REFLECTED ENERGY FLOWING BACK THROUGH THE
BIRD AND THE BIRD COMPLETELY IGNORES IT. So the Bird is not even yielding
valid readings for forward and reflected power through itself. That's
exactly what I have been saying all along. If it were calibrated for
75 ohms, it would indicate the correct values.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
news
Owen Duffy wrote:

The myth: Measurements with a Bird 43 of the conditions on the
Thruline section are invalid unless it has some minimum length of 50
ohm line on both sides of itself.

I have performed a test using components at hand where the Bird 43 has
75 ohm line on both sides of itself, and the test configuration is
designed to present a 50+j0 ohm load at the point where the Bird 43
sampling element is located. The question is, how does the Bird
respond?

The test in detail. Each component is in a list from the source to the
load:

IC706IIG
1m RG58 with UHF connectors
MFJ949E ATU
3m RG6 (Zo=75 ohm) with BNC connectors
Bird 43
5.27m Belden 9275 (Zo=75 ohm, vf=0.83) with BNC connectors
50 ohm dummy load

Short adapters were used to connect to the Bird's Type N connectors,
the dummy load's Type N connectors and the MFJ949 UHF connector.

The half wave resonance of the 5.27m length of RG6 was determined by
s/c one end and connecting the other end via an adapter to a MFJ259B
and finding the impedance dip at 23.05MHz. The calculated vf from this
test is 0.81, which reconciles reasonably with the specs. Free space
wavelenght at the test frequency is 13m.

The transmitter was set to 23.05MHz, and the ATU tuned to develop
rated power output. The ATU is only used to present the rated load to
the transmitter so as to obtain 100W for the test, to suit the Bird 43
element. It is inconsequential to the DUT (the Bird 43). With this
configuration, it is expected that the impedance at the Bird
43 is approximately 50+j0, and that there would be almost zero
reflected power.

The Bird 43 indicated 100W forward power and a quarter of a needle
width detection on reflected power.

The Bird 43 would appear to provide valid readings for the conditions
on the Bird 43 Thruline section in this case, notwithstanding that
there is not any 50 ohm transmission line attached to the Bird 43 +
N-BNC adapters.

The myth that measurements with a Bird 43 of the conditions on the
Thruline section are invalid unless it has some minimum length of 50
ohm line on both sides of itself is BUSTED.

Actually, the results of your experiment proves the myth to be true
and not to be a myth at all. There's 104.17 watts of forward power
through the Bird and 4.17 watts of reflected power back through the
Bird. Why does the Bird ignore those actual power values?

Has anyone experimental evidence to the contrary?

Yes, your experiment. Assuming 100 watts delivered to the load, the
forward power on the 75 ohm coax is actually about 104.17 watts so
the Bird's forward power reading is in error by 4.17 watts.

The reflected power on the 75 ohm coax is about 4.17 watts so the
Bird's reflected power reading is in error by close to an infinite
percentage. The Bird 43 is reading neither of the actual power values
correctly. All you have just proven is that the Bird 43 gives invalid
readings when it is in a 75 ohm environment.

THERE ARE ABOUT 4.17 WATTS OF REFLECTED ENERGY FLOWING BACK THROUGH THE
BIRD AND THE BIRD COMPLETELY IGNORES IT. So the Bird is not even yielding
valid readings for forward and reflected power through itself. That's
exactly what I have been saying all along. If it were calibrated for
75 ohms, it would indicate the correct values.
--
73, Cecil http://www.qsl.net/w5dxp

math please??? where do you get 4.17watts?? to me it looks like a 50 ohm
load on a 50 ohm meter so zero reflected power.

Dave wrote:
math please??? where do you get 4.17watts?? to me it looks like a 50 ohm
load on a 50 ohm meter so zero reflected power.

Here's the setup.

100W--tuner--75 ohm coax--Bird--1/2WL 75 ohm coax--50 ohm load

Assumptions: Losses are negligible. The tuner provides a Z0-match
so 100 watts is delivered to the load, i.e. all reflected power
is re-reflected. I think this was actually the case for the experiment.
Also assume that the impedance bump caused by the insertion of the
Bird is negligible, i.e. the same net voltage and net current exists
whether the Bird is in or out of the circuit.

The voltage reflection coefficient at the load is (50-75)/(50+75)
equals -0.2. That makes the power reflection coefficient (-0.2)^2
equals 0.04, i.e. 4% of the power incident upon the load is reflected.
If 100 watts is delivered to the load while 4% is being reflected
then the power incident upon the load is 100w/(1-0.04)=100w/0.96=
104.1667 watts. Incident power minus delivered power = reflected
power so the reflected power on the 75 ohm coax is 4.1667 watts.

The SWR on the 75 ohm coax is (1+|rho|)/(1-|rho|)= 1.2/0.8 = 1.5
THE SWR ON THE 75 OHM COAX IS 1.5:1. The Bird is in error when
it reports the SWR to be 1:1. The SWR is *NOT* 1:1 anywhere on
the load side of the system.

The Bird reports a forward power of 100w. The actual forward power
is 104.1667w. The Bird reports a reflected power of near zero. The
actual reflected power is 4.1667w. The Bird reports an SWR of 1:1.
The actual SWR is 1.5:1. The Bird is reporting false values because
it is embedded in a 75 ohm environment and is calibrated for 50 ohms.

If the ratio of voltage to current equals 50 within a 75 ohm system,
there exists an SWR of 1.5:1 and an SWR meter calibrated for 75 ohms
will verify that fact.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
The SWR is *NOT* 1:1 anywhere on the load side of the system.

Sorry, this should have read: The SWR is *NOT* 1:1 anywhere on
the load side of the tuner.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
.. .
Dave wrote:
math please??? where do you get 4.17watts?? to me it looks like a 50 ohm
load on a 50 ohm meter so zero reflected power.

Here's the setup.

100W--tuner--75 ohm coax--Bird--1/2WL 75 ohm coax--50 ohm load

Assumptions: Losses are negligible. The tuner provides a Z0-match
so 100 watts is delivered to the load, i.e. all reflected power
is re-reflected. I think this was actually the case for the experiment.
Also assume that the impedance bump caused by the insertion of the
Bird is negligible, i.e. the same net voltage and net current exists
whether the Bird is in or out of the circuit.

STOP!

you can not assume this, especially if you consider the actual results of
the experiment. the bird is a 50 ohm transmission line segment and it is
seeing a 50 ohm load. the 50 ohm load plus 1/2 wave of anything is 50 ohms.
so as far as the bird knows there is no reflected power and the real world
reading is correct.

Dave wrote:

"Cecil Moore" wrote in message
Also assume that the impedance bump caused by the insertion of the
Bird is negligible, i.e. the same net voltage and net current exists
whether the Bird is in or out of the circuit.

you can not assume this, especially if you consider the actual results of
the experiment. the bird is a 50 ohm transmission line segment and it is
seeing a 50 ohm load. the 50 ohm load plus 1/2 wave of anything is 50 ohms.
so as far as the bird knows there is no reflected power and the real world
reading is correct.

Remove the Bird, reconnect the two pieces of 75 ohm coax, and I'll
bet you will measure 70.7 volts and 1.414 amps at that point, 1/2
WL back from the 50 ohm load. If so, the Bird is NOT changing the
conditions when it is inserted. And we know there has to be reflected
power all up and down the 75 ohm coax. There is no place in the system
on the load side of the tuner where reflected power is zero. The
reflected power exists and the Bird reports a bogus reading for it.

With or without the Bird in the circuit the ratio of net voltage
to net current at the measurement point, 1/2WL away from the 50
ohm load, will be 50 ohms according to transmission line theory.
That's all the Bird is seeing - a V/I ratio of 50 which is irrelevant
to SWR unless Z0 is known and Z0 is known to be 75 ohms, NOT 50 ohms.

The ratio of net voltage to net current is All THE BIRD EVER SEES.
"As far as the bird knows ..." it is embedded in a 50 ohm environment.
But the Bird is not all-knowing. In this case, it is embedded in a
75 ohm environment and is giving bogus readings because all it sees
(samples) is a net voltage to net current whose ratio is 50 ohms.

It is what the Bird doesn't know that is important. The Bird doesn't
know that it is not embedded in a 50 ohm environment, and in its
ignorance of that fact, reports bogus results.

The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
.. .
Dave wrote:

"Cecil Moore" wrote in message
Also assume that the impedance bump caused by the insertion of the
Bird is negligible, i.e. the same net voltage and net current exists
whether the Bird is in or out of the circuit.

you can not assume this, especially if you consider the actual results of
the experiment. the bird is a 50 ohm transmission line segment and it is
seeing a 50 ohm load. the 50 ohm load plus 1/2 wave of anything is 50
ohms. so as far as the bird knows there is no reflected power and the
real world reading is correct.

Remove the Bird, reconnect the two pieces of 75 ohm coax, and I'll
bet you will measure 70.7 volts and 1.414 amps at that point, 1/2
WL back from the 50 ohm load. If so, the Bird is NOT changing the
conditions when it is inserted. And we know there has to be reflected
power all up and down the 75 ohm coax. There is no place in the system
on the load side of the tuner where reflected power is zero. The
reflected power exists and the Bird reports a bogus reading for it.

With or without the Bird in the circuit the ratio of net voltage
to net current at the measurement point, 1/2WL away from the 50
ohm load, will be 50 ohms according to transmission line theory.
That's all the Bird is seeing - a V/I ratio of 50 which is irrelevant
to SWR unless Z0 is known and Z0 is known to be 75 ohms, NOT 50 ohms.

The ratio of net voltage to net current is All THE BIRD EVER SEES.
"As far as the bird knows ..." it is embedded in a 50 ohm environment.
But the Bird is not all-knowing. In this case, it is embedded in a
75 ohm environment and is giving bogus readings because all it sees
(samples) is a net voltage to net current whose ratio is 50 ohms.

It is what the Bird doesn't know that is important. The Bird doesn't
know that it is not embedded in a 50 ohm environment, and in its
ignorance of that fact, reports bogus results.

The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?
--
73, Cecil http://www.qsl.net/w5dxp

obviously at 1/4 wave from the load the impedance seen by the bird is not 50
ohms. you just can't ignore that internally the bird is a 50 ohm line, even
though it is only a couple inches long it is 50 ohms, put a 50 ohm load on
it and it will read no reflected power. this is just how it works, you
can't wish it to be anything else, especially when the actual measurements
prove me correct. you may try to assume away the bird and its
characteristic impedance, but it just won't go away as evidenced by the real
world measurements. you have to realize that the bird is not measuring what
is going on outside it's case, the only thing it can measure is the voltage
and current in its little 50 ohm internal world. so look at it from the
meter's point of view... it looks out the load side and sees 50 ohms, it
can't know that it is really a 75 ohm line, in fact it can't know there is
any line there at all, it just sees a 50 ohm load, therefore no reflection
into the meter, and no reflected power reading. qed.

Dave wrote:
obviously at 1/4 wave from the load the impedance seen by the bird is not 50
ohms.

That's the whole point, Dave. The poor ignorant Bird doesn't know what
is going on external to itself. You and I have to be smarter than the
Bird and not report its bogus results as "correct". The Bird reports
a 9:1 SWR on a matched-line 450 ohm ladder-line. Is the SWR really 9:1?

this is just how it works, you
can't wish it to be anything else, especially when the actual measurements
prove me correct.

The actual measurements prove you wrong! The forward power reported by
the Bird is NOT the actual forward power. The reflected power reported
by the Bird is NOT the actual reflected power. The SWR reported by the
Bird does not exist anywhere in the system on the load side of the tuner.
The Bird is totally confused because it is being abused by the operator.
The operator is totally ignorant when he reports the readings are "correct".

you have to realize that the bird is not measuring what
is going on outside it's case, the only thing it can measure is the voltage
and current in its little 50 ohm internal world.

That's exactly the point and that's why some readings reported by the
Bird are obviously bogus when compared to the broader knowledge of
actual external conditions. The operator needs to be smarter than the
Bird.

so look at it from the
meter's point of view... it looks out the load side and sees 50 ohms, it
can't know that it is really a 75 ohm line, in fact it can't know there is
any line there at all, it just sees a 50 ohm load, therefore no reflection
into the meter, and no reflected power reading. qed.

But you and I know that 75 ohm reflected energy is flowing through the
Bird and the Bird is ignoring it. Of course, we cannot blame the Bird
for the bogus readings. We can only blame the ignorant operator who reports
that the bogus readings are "correct" when they are obviously false.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:


The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?

You need to keep thinking about that. What if they are both right? Is
it really one continuous, uniform transmission line? Is SWR the same
everywhere in a tee stub circuit?

ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
The actual SWR on a lossless line doesn't change. Yet, in another
posting, I showed that moving the Bird 1/4WL closer to the load
caused a reported SWR change by the Bird from 1:1 to 2.25:1. How
could both results possibly be right?

You need to keep thinking about that. What if they are both right?

Here's the example sans the Bird. Between the tuner output
and the load, where exactly is the actual SWR = 1:1 and where
exactly is the actual SWR = 2.25:1? Answer: nowhere!

XMTR--tuner---1 WL 75 ohm coax---50 ohm load
--
73, Cecil http://www.qsl.net/w5dxp