"Roy Lewallen" bravely wrote to "All" (18 Dec 05 00:52:58)
--- on the heady topic of " Antenna reception theory"

RL From: Roy Lewallen
RL Xref: core-easynews rec.radio.amateur.antenna:221453


RL Richard Harrison wrote:
Reg, G4FGQ wrote:
"Just a number please."

Given 1 volt per m as the field strength, and a 1-m antenna parallel to
the electric vector of the wave, the open-circuit voltage at the end of
the wire is 1 volt.

RL Relative to what? The other terminal has to be extremely close to the
RL end of the wire in order for the voltage to be single valued.

RL The best you can get across the receiver input is
0.5 volt when there is a conjugate match between the receiver and the
antennna.

RL Sorry, that's not just a little wrong, it's wrong by orders of
RL magnitude. For example, a 1 meter long 10 mm diameter dipole,
RL terminated in the complex conjugate of its self impedance (load Z =
RL 0.8855 + j6030 ohms), in a 1 V/m field, has about 1667 volts across
RL the load. Hardly a half volt!


All this talk of high RF voltage has got me a little peckish.
Anyone up for hotdogs and smores?

A*s*i*m*o*v

.... I'm precise. He's discriminating. You're picky.

Reg`s question was:
"What is the voltage between the bottom end and ground of a 1 metre high
vertical antenna, above perfect ground, when the vertically-polarized
field strength is 1 volt per metre, and the antenna height is shorter
than 1/4-wavelength?"

A very clear and succinct question, I thought.

Roy Lewallen wrote:
"Relative to what?"

Reg left little to assume. I inferred that Reg had meant a
ground-mounted 1-meter whip on a small base insulator. That would have
left a short distance between the points of voltage determinarion.

The 1-meter whip directly over flat perfect earth must have a conjugate
match to its receiver to extract all available power and get maximum
voltage at the receiver input. This requires a low-loss coil to tune out
the high capacitive reactannce of a too-short whip. It was specified as
being less than 1/4-wavelength.

Standing wave antennas must be resonant to allow maximum current flow.

An unbalanced whip antenna must consist of two electrical parts just as
a balanced antenna does. The ground or ground plane used with a whip
substitutes for the missing half of a dipole. It provides a virtual
image of the whip above it, to complete a dipole-like antenna. In this
instance, the ground surface intervenes splitting the antenna into real
and virtual parts. Radiation comes from the real part above ground. This
part has only half the impedance of a totally real dipole.

We don`t need to have problems in determining the base voltage of a whip
antenna. We can measure the antenna`s base impedance with a bridge, and
its current with a thermoammeter with good accuracy, then we can
calculate the voltage at the base of the whip.

The answer to Roy`s question is: The r-f voltage at the base of the
antenna is determined with respect to ground at the base of the antenna,

Best regards, Richard Harrison, KB5WZI

Someone mentioned peak values in a posting about computer results. Since
Reg mentioned a discrepncy of 2 to 1, I was alerted to a possibility of
an error source. A-C and R-F instruments are often calibrated in
effective (rms) values. It is also customary to use rms values in
calculations. RMS is 0,707 X the peak value.*Conversely, the peak
value is 1.414 X the rms value.

If you multiply the peak voltge times the peak current, their product is
2X the effective (average) power..

Could Reg`s discrepancy stem from the diference beween peak and
effective values?

Best regards, Richard Harrison, KB5WZI

All I want to know is the voltage. All that is needed is a high
impedance voltmeter.

Theres nn need to do a conjugate match just to measure voltage.
Some folks have conjugate matches on their brains.
----
Reg.

Reg, G4FGQ wrote:
"There`s no need to do a conjugate match just to measure voltage."

True. But, it`s best to measure current and calculate voltage because
voltmeter leads are susceptible to induced voltages.

One benefit of the conjugate match is elimination of a reactive obstacle
to current, but because source and load resistances are equal, exactly
one half of the voltage induced in the antenna appears across the
receiver. An infinite ipedance does not load a source at all.

Reg`s question of how much voltage is induced in 1 m of wire in a field
of 1 V/m is answered clearly by Terman in publications which date back
at least to 1932, the earliest copyright date I saw on "Radio
Engineering". So, many competent and critical reviewers have pored over
Terman`s works that it`s almost certain that any errors have been found
and corrected long ago. Everybody makes mistakes, but now Terman is as
close to infallible on the subject of radio as any author I know. Read
page 2 of Terman`s "Electronnic and Radio Engineering" (1955 edition)
for complete details. The same information appears in some other Terman
authored and edited writings (almost word for word).

Best regards, Richard Harrison, KB5WZI

Terman may or may not be perfectly correct when he states the voltage
INDUCED in a 1 metre high vertical antenna with a field strength of 1
volt per metre.

But Terman is ambiguous. He tells only half of the story.

He FAILS to state between WHICH which pair of points the voltage is
induced.

For the UMPTEEN'th time - what I need to know is the voltage which
can actually be MEASURED between the bottom end of the antenna and
ground ?

I can then draw a circuit and continue with practical calculations.

PLEASE, can no-one put me out of my misery ?
----
Reg, G4FGQ.

Reg Edwards wrote:
Terman may or may not be perfectly correct when he states the voltage
INDUCED in a 1 metre high vertical antenna with a field strength of 1
volt per metre.

But Terman is ambiguous. He tells only half of the story.

He FAILS to state between WHICH which pair of points the voltage is
induced.

For the UMPTEEN'th time - what I need to know is the voltage which
can actually be MEASURED between the bottom end of the antenna and
ground ?

I can then draw a circuit and continue with practical calculations.

PLEASE, can no-one put me out of my misery ?

Hm. Have my postings gone unread? Or just unbelieved?

Roy Lewallen, W7EL

Richard,

There is no problem with Terman's words, but I believe you are missing
his intention.

His point in bringing up the magnetic flux is merely to say that one can
find "exactly the same voltage" in the one meter long conductor by
considered either the electric field directly or by considering the
sweep of the magnetic field. It is just a statement of equivalence of
the two components of the incident plane wave. This same sort of
statement is found in many other textbooks.

Terman's conductor is in free space. He discusses the voltage difference
between one end of the conductor and the other end of the same
conductor. He does not address the question at hand, which is the
voltage between a perfect ground plane and the bottom of a short
conductor near that ground plane.

At least two people have explained why that voltage is not one volt for
an incident field strength of one volt per meter.

73,
Gene
W4SZ


Richard Harrison wrote:

[snip]

Reg`s question of how much voltage is induced in 1 m of wire in a field
of 1 V/m is answered clearly by Terman in publications which date back
at least to 1932, the earliest copyright date I saw on "Radio
Engineering". So, many competent and critical reviewers have pored over
Terman`s works that it`s almost certain that any errors have been found
and corrected long ago. Everybody makes mistakes, but now Terman is as
close to infallible on the subject of radio as any author I know. Read
page 2 of Terman`s "Electronnic and Radio Engineering" (1955 edition)
for complete details. The same information appears in some other Terman
authored and edited writings (almost word for word).

Best regards, Richard Harrison, KB5WZI

"Roy Lewallen" wrote

Hm. Have my postings gone unread? Or just unbelieved?

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

Or, am I trying to find somebody else who believes you ?

Terman, Kraus and Balanis and some computer programs are of no help!
;o)
----
Reg.

Gene Fuller wrote:
"There is no problem with Terman`s words, but I believe you are missing
his intention."

I parsed Terman`s words carefully trying to avoid misinterpretation.

There is a RCA FM Coverage Calculator (special slide rule) pictured and
described on the internet. Text accompanies the rule. This text says the
range of the rule is for a radius of urban coverage of 1000 microvolts
per meter and a radius of rural coverage of 50 microvolts per meter.
They obviously anticipate a much higher urban noise level than found in
rural areas.

The rule has distance scales of 4 to 100 miles, and 16 to 143 miles..

The text says: "If you hold up 1 meter of wire at exactly the right
angle, this is exactly how many millionths of a volt are generated
between its ends.

If you assume that 50 microvolts per meter in the country gives an
acceptable quality signal at the receiver, you`ll be able to calculate
how far away you can reach."

There is much more text dealing with transmitter powers, broadcast
antenna types, and antenna heights. Accuracy is said to be within 10%..

I`ve never seen one of these special slide rules myself, but maybe
Walter Maxwell, Richard Fry, or someone else has and can elaborate. I
quoted the text because it contained in effect the simple statement that
1 microvolt is generated betweens the ends of a well placed 1 meter long
wire when immersed in a 1 microvolt electromagnetic field. I certainly
never expected to see that fact debated.

Best regards, Richard Harrison, KB5WZI