Antenna reception theory
 

Reg, G4FGQ wrote:
"What do photons have to do with winning a contest?"

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

"An Elemental Antenna

Since the length of an antenna is commonly expressed in electrical
degrees and since this allows the convenient use of trigonometric tables
in computing the ratios of currents in various parts of an antenna, let
us choose as our elemental antenna a piece of wire which is 1-degree in
length and in which the current is constant from one end to the other.We
shall first assume that this elemental antenna is located far out in
space, so that its field is not disturbed by reflections from the
surface of the earth or from any other object. This, of course, is a
most improbable set of conditions, but we can certainly imagine that we
have a situation such as this and compute from the field equations its
electromagnetic result.

These computations will show, first of all, that the maximum field
intensity will be produced in the directions which are at right angles
to the direction of current flow. This is a reasonable result, since the
magnetic field which is produced by the current surrounds the wire in
concentric rings and thus gives rise to a radiation field which moves
outward at right angles to the wire. As a matter of fact, the field
intensity, measured according to our standard procedure at a distance of
1 mile in any direction from the radiating element, will be found to be
proportional to the sine of the angle between the direction of the
current flow and the direction in which the measurement is taken. If we
represent the field intensity at 1 mile in any direction by the length
of a vector starting at the center of the element and extending in that
direction, the tips of the vectors will mark out the radiation pattern
of the antenna element. A cross section of the entire radiation pattern
of this element is shown in Fig. 39-2; the entire pattern would be
obtained by rotating the figure about the axis of the antenna element.

But this pattern tells us only the relative signal strength in various
directions; it is a normalized pattern, with the intensity in the
direction of maximum radiation being considered simply as unity. We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. We have said nothing about the frequency, and we
do not need to; as long as the wavelength is shorter than 1 mile, so
that there are no serious induction-ffield effects to upset our
calculations, this figure will be correct at any frequencyfor which the
length of the element is 1/360 wavelength.

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

Richard Harrison wrote:

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

Yes, it should.

. . .
. . . We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. . .

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

A short dipole antenna with 1 amp of current at the center has an
average of 0.5 amp of current along the whole length. So it should be
obvious from the above analysis that the field from a dipole is half the
field from the elemental antenna with uniform current which the author
is discussing. Or, instead of just taking the average, you can integrate
I * delta L to get the same result.

Roy Lewallen, W7EL

Antenna reception theory
 

Richard Harrison wrote:
Reg, G4FGQ wrote:
"What do photons have to do with winning a contest?"

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

"An Elemental Antenna

Since the length of an antenna is commonly expressed in electrical
degrees and since this allows the convenient use of trigonometric tables
in computing the ratios of currents in various parts of an antenna, let
us choose as our elemental antenna a piece of wire which is 1-degree in
length and in which the current is constant from one end to the other.We
shall first assume that this elemental antenna is located far out in
space, so that its field is not disturbed by reflections from the
surface of the earth or from any other object. This, of course, is a
most improbable set of conditions, but we can certainly imagine that we
have a situation such as this and compute from the field equations its
electromagnetic result.

These computations will show, first of all, that the maximum field
intensity will be produced in the directions which are at right angles
to the direction of current flow. This is a reasonable result, since the
magnetic field which is produced by the current surrounds the wire in
concentric rings and thus gives rise to a radiation field which moves
outward at right angles to the wire. As a matter of fact, the field
intensity, measured according to our standard procedure at a distance of
1 mile in any direction from the radiating element, will be found to be
proportional to the sine of the angle between the direction of the
current flow and the direction in which the measurement is taken. If we
represent the field intensity at 1 mile in any direction by the length
of a vector starting at the center of the element and extending in that
direction, the tips of the vectors will mark out the radiation pattern
of the antenna element. A cross section of the entire radiation pattern
of this element is shown in Fig. 39-2; the entire pattern would be
obtained by rotating the figure about the axis of the antenna element.

But this pattern tells us only the relative signal strength in various
directions; it is a normalized pattern, with the intensity in the
direction of maximum radiation being considered simply as unity. We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. We have said nothing about the frequency, and we
do not need to; as long as the wavelength is shorter than 1 mile, so
that there are no serious induction-ffield effects to upset our
calculations, this figure will be correct at any frequencyfor which the
length of the element is 1/360 wavelength.

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

Best regards, Richard Harrison, KB5WZI



How did you get .3253 mv/m at 1 miles?

Dave

Antenna reception theory
 

How did you quote damned near 75 lines of text to ask a 9 word question?

Jim


"EE123" wrote in message
oups.com...

Best regards, Richard Harrison, KB5WZI



How did you get .3253 mv/m at 1 miles?

Dave

Antenna reception theory
 

EEE 123 wrote:
"How do you get .3253 mv/m at 1 mile?"

The formula to calculate the field strength produced by an dlementaarry
doublet is difficult for me to reproduce with my keyboard. It is found
on page 770 of Terman`s 1943 "Radio Sngineers` Handbook. In my prior
posting, Griffith hhas done the work for us.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

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. 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.

Most of the explanations are irrelevant when the field strength is
specified at the antenna.

Terman preached scientific gospel. He had proof to back what he said. In
Terman`s 1943 "Radio Engineers` Handbook" he wrote:

"The strength of a radio wave is expressed in terms of the voltage
stress produced in space by the electric field of the wave, and is
usually expressed in either millivolts or microvolts stress per meter.
The stress expressed this way is exactly the same voltage that the
magnetic flux of the wave induces in a conductor one meter long when rhe
wave sweeps across the conductor with the velocity of light." This is
found on page 770.

There is no qualification or equivocation. Terman also posts the same
statement with no change in substance on page 2 of his 1955 edition of
"Electronic and Radio Engineerinng". It means what it plainly says.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

Richard (Harrison)

I'm sorry Richard, you have not anwered my question. I will repeat it.

"What is the voltage measured between the bottom end and ground of a
1metre high vertical antenna when the field strength is 1 volt per
metre?"

Assume a perfect ground and antenna height is much less than
1/4-wavelength.

A measured voltage is always relative to or is between TWO specified
points.
----
Reg, G4FGQ

Antenna reception theory
 

Reg Edwards wrote:
. . .
A measured voltage is always relative to or is between TWO specified
points.

And when measuring in the presence of a time-varying H field, those two
specified points have to be physically very close together.

Roy Lewallen, W7EL

Antenna reception theory
 

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.

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

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.

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

. . .

Roy Lewallen, W7EL

Antenna reception theory
 

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

For the casual reader, I should emphasize that this analysis, like all
the others on this thread, assumes zero loss. The conditions I described
could never be achieved in real life because of unavoidable loss. But
it's important to first understand how a lossless system behaves before
we add the complicating factor of loss.

Roy Lewallen, W7EL

Antenna reception theory
 

"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.

Antenna reception theory
 

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

Antenna reception theory
 

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

Antenna reception theory
 

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.

Antenna reception theory
 

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

Antenna reception theory
 

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.

Antenna reception theory
 

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

Antenna reception theory
 

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

Antenna reception theory
 

"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.

Antenna reception theory
 

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

Antenna reception theory
 

Gene, W4SZ wrote:
"At least two people have explained why that voltage is not one volt per
meter."

Here are Terman`s exact words again:
"The strength of the wave measured in terms of microvolts per meter of
stress in space is also exactly the same voltage that the magnetic flux
of the wave induces in a conductor 1 m long when sweeping across this
conductor with the velocity of light."

I see Gene`s statemennt as a contradiction within itself. Definition of
field strength is the volts it will generate in a wire 1 meter long.
There is no contradiction in Terman`s statement.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

Richard,

Not only are you misunderstanding Terman, you have twisted my words as well.

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

You removed some of my words and completely changed the meaning.

Once more, Terman is undoubtedly correct with his statement. I agree
completely. However, the configuration described by Terman is NOT the
subject at hand.

73,
Gene
W4SZ


Richard Harrison wrote:
Gene, W4SZ wrote:
"At least two people have explained why that voltage is not one volt per
meter."

Here are Terman`s exact words again:
"The strength of the wave measured in terms of microvolts per meter of
stress in space is also exactly the same voltage that the magnetic flux
of the wave induces in a conductor 1 m long when sweeping across this
conductor with the velocity of light."

I see Gene`s statemennt as a contradiction within itself. Definition of
field strength is the volts it will generate in a wire 1 meter long.
There is no contradiction in Terman`s statement.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

Richard Harrison wrote:

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.


And just how are you going to measure that voltage without getting an
opposite voltage in your voltmeter leads. Or alternatively how can you
present that voltage at the input to a receiver?

You can't. So the above tells us zip about the question Reg asked. Roy
answered the question correctly and you haven't even attempted it. Why
is that?

Andy

Antenna reception theory
 

Reg, G4FGQ wrote:
"Terman, Kraus, and Balanis and some computer programs are of no help!"

My dictionary defines "field strength" as:
"3. The strength of radio waves at a distance from the transmitting
antenna, usually expressed in microvolts-per-meter. This is not the same
as the strength of a radio signal at the antenna terminals of the
receiver."

The definition looks OK to me. The reason the signal is not the same as
the microvolts-per-meter even when the antenna is a 1-meter length of
wire with just the right slant is because the induced voltage gets
divided between the antenna and its load (the receiver).

Maybe Cecil`s IEEE dictionary has something to say about field strength.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

Richard Harrison wrote:
Maybe Cecil`s IEEE dictionary has something to say about field strength.

magnitude of the electric field vector in volts per meter, or
magnitude of the magnetic field vector in amps (or ampere-turns) per meter,
or power flux density P in watts per square meter.
--
73, Cecil http://www.qsl.net/w5dxp

Antenna reception theory
 

On Tue, 20 Dec 2005 14:40:14 GMT, Andy Cowley
wrote:
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.
And just how are you going to measure that voltage without getting an
opposite voltage in your voltmeter leads.

Hi Andy,

You make a loop.

73's
Richard Clark, KB7QHC

Antenna reception theory
 

Andy wrote:
"So the above told us zip about the question Reg asked. Roy answered the
question correctly and you haven`t even attempted it."

I assumed the definition of "field strength" correctly answered Reg`s
question and his descrepancy lay elswhere.

Field strength of an electromagnetic wave is expressed in microvolts per
meter. It is defined as the number of microvolts which would be induced
in a properly placed piece of wire one meter long.

If a field strength of one microvolt per meter does not induce one
microvolt into a piece of wire one meter long, why not? I`ve been
looking for an explanation and found a possible answer in Figure 2-9-1
on page 31 of Kraus` 3rd edition of "Antennas".

The possible explanation is called the "effective height" in meters of
an antenna. It is a "factor", which when multiplied by the microvolts
per meter of the field strength, gives the volts induced in the antenna
at its terminals.

According to Kraus` figure, h is a function of current distribution in
the antenna. The text says that for a dipole 0.1 lambda long, h = 0.5X
the length of the antenna. Reg did not specify a frequency or wavelength
for his antenna, as I recall, and he did specify a ground mounted
vertical whip 1 m long for his receiving antenna.

For a dipole 0.5 lambda long, Kraus gives (h) as 0.64X the length of the
antenna.

For all I know, there is a vertical antenna length in terms of
wavelength for which h=1. If so, the volts between the antenna base and
the ground directly under it would numerically exactly equal the
microvolts per meter of the field strength. All we need to do is pick
the right frequency. The applicable Formula is (1) on page 30.

I suspect that in most cases, h is determined experimentally.

I regret I`ve not had a copy of Kraus nearly as long as I`ve had a copy
of Terman. I`m still using an edition I`ve had for 58 years. It shows
lots of wear and tear.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

On Tue, 20 Dec 2005 15:33:50 -0600, (Richard
Harrison) wrote:

For a dipole 0.5 lambda long, Kraus gives (h) as 0.64X the length of the
antenna.

Hi Richard,

This is exactly (within one percent) of what the NIST uses for their
standard field measurements (described elsewhere in these threads).

73's
Richard Clark, KB7QHC

Antenna reception theory
 

"Richard Clark" wrote
You make a loop.
======================================

I normally reply, if I reply at all to your idiotic statements, with
"Phooey".

But on this occasion, to protect innocent, bystanding, novices from
your deliberate, inexcusible, misleading statement, it should be said
that the voltage induced in a circular loop is altogether different
and very much smaller from that induced in a straight wire of the same
length.

You disgust me! A disgrace to amateur radio!

Have a miserable Christmas!
----
Reg, G4FGQ.

Antenna reception theory
 

Reg, G4FGQ wrote:
"Have a miserable Christmas!"

In the cinema, "The Grinch Who Stole Christmas", the Grinch turned into
a kind, green, Santa Claus. Let`s hope Reg has a change of heart too!

Merry Christmas, Richard Harrison, KB5WZI

Antenna reception theory
 

Richard Clark wrote:
"You make a loop."

There is a difference. The small whip has a high capacitive reactance.
The small loop has a high inductive reacvtance. Both have low radiation
resistance. But, the loop is more often used to determine EM field
strength. You just need the right "fudge factor" to convert antenna
voltage tto field strength or vice versa.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

"Richard Harrison" wrote in message
...
Richard Clark wrote:
"You make a loop."

There is a difference. The small whip has a high capacitive
reactance.
The small loop has a high inductive reacvtance. Both have low
radiation
resistance. But, the loop is more often used to determine EM field
strength. You just need the right "fudge factor" to convert antenna
voltage tto field strength or vice versa.

Best regards, Richard Harrison, KB5WZI
=====================================

For the very last time I will repeat my question :-

"What is the voltage measured between the bottom end of a 1 metre
vertical antenna and ground when the field strength is 1 volt per
metre.

The height (length) of the antenna is much less than 1/4-wavelength.
The bottom end of the antenna is immediately above the ground. The
ground is assumed perfect. The field is vertically polarised.

Frequency, loops, reactance, radiation resistance do not enter into
the argument. No other information is needed.

Terman, Kraus and Balanis' bibles provide answers to a different
question in which I am not interested. Mere mention of these learned
gentlemen only confuses the issue.

The answer is entirely fundamental to e.m. radiation and reception.

All I need is a number of volts. What is it please?
----
Reg, G4FGQ.

Antenna reception theory
 

"Reg Edwards" wrote
For the very last time I will repeat my question :-

Pray it so...

"What is the voltage measured between the bottom end of a 1 metre
vertical antenna and ground when the field strength is 1 volt per
metre."

Of what relevance is this to anyone but (apparently) you? Please elaborate.

RF

Antenna reception theory
 

On Tue, 20 Dec 2005 22:06:56 +0000 (UTC), "Reg Edwards"
wrote:

I normally reply, if I reply at all to your idiotic statements, with
"Phooey".

Ah yes, Punchinello,

As gracious as ever.

it should be said
Well, you generally fill the gap the
that the voltage induced in a circular loop is altogether different
which is redundant to the following:
and very much smaller from that induced in a straight wire of the same
length.

Notably you say nothing of how much.

For the sake of novices for whom you have such paternal feelings (but
absolutely no answers) I would offer that the solution in the loop
(that same wire bent over to touch ground) and loaded with 200 Ohms
(not an open) reveals a voltage that need only be multiplied by 20 to
obtain the correct value. Trivial! And what is more, far simpler to
measure across 200 Ohms than an open at 20MHz.

Thus in response to the question:
On Tue, 20 Dec 2005 14:40:14 GMT, Andy Cowley
wrote:
And just how are you going to measure that voltage without getting an
opposite voltage in your voltmeter leads.
has been responded to fully, with a realizable design (barring this
folderol of perfect ground) and removing the objection for meter
leads. It takes no more software than the free version of EZNEC
(zipped up, but it works).

Have a miserable Christmas!
That will be all too true with a visit to and through several of our
nation's airports here soon. ;-(

XOXOXOXOX,
Richard Clark, KB7QHC

Antenna reception theory
 

Richard Harrison wrote:
Reg, G4FGQ wrote:
"Terman, Kraus, and Balanis and some computer programs are of no help!"

My dictionary defines "field strength" as:
"3. The strength of radio waves at a distance from the transmitting
antenna, usually expressed in microvolts-per-meter. This is not the same
as the strength of a radio signal at the antenna terminals of the
receiver."

The definition looks OK to me. The reason the signal is not the same as
the microvolts-per-meter even when the antenna is a 1-meter length of
wire with just the right slant is because the induced voltage gets
divided between the antenna and its load (the receiver).


No, that's not why. The terminal voltage of an open circuited 1 meter
(electrically short) dipole is 1/2 the field strength in volts/meter.
The terminal voltage when terminated with a conjugately matched load can
be well over a thousand volts (in the theoretical lossless case).

But it's pointless to keep repeating this. Reg keeps asking the same
question, and you keep responding with the same incorrect answers. I
believe I've gotten through to everyone who really wants to know the
answers, so I'll let this be my last repetition.

Roy Lewallen, W7EL

Antenna reception theory
 

"Richard Fry" wrote in message
...
"Reg Edwards" wrote
For the very last time I will repeat my question :-

Pray it so...

"What is the voltage measured between the bottom end of a 1 metre
vertical antenna and ground when the field strength is 1 volt per
metre."

Of what relevance is this to anyone but (apparently) you? Please
elaborate.

RF
=========================================

For starters, Have you never heard of field strength measurements?

Have you ever designed the input stage of a radio receiver?

The topic is fundamental to an understanding of e.m. radiation and
reception.

Can YOU answer the simple question? Or are you entirely dependent on
your gospel faith in 'Bibles'.

On this occasion at least, the Bibles are letting dependent people
down.

The immediate relevance to me is that I have a program which has been
reported to have a calculating error. It was reported by a person who
is not dependent on bibles. He stated that the
conventional/traditional calculating method used in my program was
incorrect. I was not entirely convinced so I posed a related question
on this newsgroup to which only one person has replied with a number.
And he was wrong first time.

Other persons who replied, after consulting their bibles, were unable
even to answer the question, either rightly or wrongly. They just
generated more confusion.

The program concerned is GRNDWAV4 which I think, but not absolutely
certain, has now been corrected. Why not download it, input a very few
standard values, and tell me whether or not it provides the correct
answer to receiver power input? You may, of course, prefer not to
commit yourself.

Is that enough elaboration for you?
----
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Antenna reception theory
 

Reg, G4FGQ wrote:
"All I need is a number of volts."

I`ll guess, because Reg asked, not because I know aanything. I`ve now
discovered Kraus` effective antenna height which may be related to an
Icelandic connection. Reg hasn`t told us everything he knows. One reason
we don`t know is because the effective antenna height is related to the
antenna`s length in terms of wavelength according to Kraus. One of the
examples given by Kraus is a dipole of 1/10 of a wavelength. Kraus tells
us the effective height of this length gives a factor of 0.5. According
to Equation (1) on page 30 of the 3rd edition of "Antennas", Voltage at
the terminals of the antenna = effective height X field strength.

If we guess that a short whip might have the same effective height as a
short dipole, then with a 1 volt per meter field strength X 0.5 as an
effective height factor, their product would be 0.5 volts. I`ll assume
rms because that`s the convention for expression.

I don`t have much confidence in the number because I think you must
determine the effective height experimentally. Terman says on page 991
of his 1943 "Radio Enginneers` Handbook: "If an antenna other than a
loop is used, the effective height must be determined experimentally.
Maybe someone has worked this out since 1943.

Best regards, Richard Harrison, KB5WZI

Antenna reception theory
 

"Richard Harrison" bravely wrote to "All" (20 Dec 05 09:38:02)
--- on the heady topic of " Antenna reception theory"

RH From: (Richard Harrison)
RH Xref: core-easynews rec.radio.amateur.antenna:221565

RH Reg, G4FGQ wrote:
RH "Terman, Kraus, and Balanis and some computer programs are of no
RH help!"
RH My dictionary defines "field strength" as:
RH "3. The strength of radio waves at a distance from the transmitting
RH antenna, usually expressed in microvolts-per-meter. This is not the
RH same as the strength of a radio signal at the antenna terminals of the
RH receiver."

RH The definition looks OK to me. The reason the signal is not the same
RH as the microvolts-per-meter even when the antenna is a 1-meter length
RH of wire with just the right slant is because the induced voltage gets
RH divided between the antenna and its load (the receiver).


Not only that, but also this: the antenna rebroadcasts half of the
intercepted energy.

A*s*i*m*o*v

Antenna reception theory
 

Asimov wrote:
. . .
RH The definition looks OK to me. The reason the signal is not the same
RH as the microvolts-per-meter even when the antenna is a 1-meter length
RH of wire with just the right slant is because the induced voltage gets
RH divided between the antenna and its load (the receiver).


Not only that, but also this: the antenna rebroadcasts half of the
intercepted energy.

But voltage isn't energy. Or power.

Roy Lewallen, W7EL

Antenna reception theory
 

On Tue, 20 Dec 2005 22:08:44 -0600, (Richard
Harrison) wrote:

Reg, G4FGQ wrote:
"All I need is a number of volts."

Hi Richard,

You forgot to quote Reggie's promise this would be the last time he
asked....

I don`t have much confidence in the number because I think you must
determine the effective height experimentally. Terman says on page 991
of his 1943 "Radio Enginneers` Handbook: "If an antenna other than a
loop is used, the effective height must be determined experimentally.
Maybe someone has worked this out since 1943.

Well, in fact it had been known for at least 3 or 4 decades before
that. From the "Standard Handbook for Electrical Engineers,"
1907-1917:

"284. Receiving Antenna. ... When an electric field of intensity
E is produced on the antenna, there is an electromotive force in
the antenna equal to Eh2, where h2 is the height of the receiving
antenna from the earth to the centre of capacity."

As to the term "centre of capacity," this has a variable meaning
according to the structure (top hats were common back then for obvious
reason of the predominance of LF work) and ground conditions:

"In land stations the actual effective height is from 50 to
90% of the measured height. That the height may be made
as high as possible, it is desirable to increase the capacity
of the upper portion...."

73's
Richard Clark, KB7QHC