Harry wrote:
Hi Cecil,

From "Antenna Theory" by Balanis:

Rr = 2*Prad/|Io^2| = 73 ohms (4-93)

I got Balanis' book and found the above formula.
Thanks a lot! (Tom also mentioned the same book.)

4-94 (I have the 1982 edition) is from formula 4-70.
The math is OK to me. I happen to have a PH.D. degree in math.

For your "passing-out" test, calculate the radiation pattern of a
two-radial groundplane antenna directly from Maxwell's equations
(including the two small cross-polarized lobes that nobody else
mentions), using pencil and paper only, in a pub.

I've seen it done by a math professor from W6. (Nah... it must have been
a party trick... he must have rehearsed that a million times... mustn't
he?)

When he'd finished, Charlie sat back, smiled asked for comments. The
rest of us passed out.


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

"Harry" wrote about Roy Lewallen:
You may want to write a book titled "Story of the 73 Ohms".
___________

It should include the point that the 73 ohm radiation resistance value
applies to a physical 1/2-wave, thin-wire, linear dipole in free space, and
that a reactance term of + j42.5 ohms also applies to such a configuration
(Kraus 3rd Edition, p. 182).

The dipole length needs to be shorted by a few percent in order to zero out
the reactance term, at which time the resistance term will be about 65 ohms.

RF

"Roy Lewallen" wrote in message
...
Just about any antenna textbook will show you the calculation of a half
wavelength, infinitely thin dipole in free space. For that special case,
the answer ends up being simply 30 * Cin(2 * pi), where Cin is a
modification of the cosine integral Ci[*]. That's where the "magic number"
comes from. Kraus' _Antennas_ is just one of the many textbooks which give
the derivation for this. It takes about 3 pages and 23 equations for Kraus
to derive.

One assumption made in calculation of this value is that the current
distribution is sinusoidal, an assumption that's true only for an
infinitely thin antenna. For finite thickness wire, the calculation
becomes much more difficult. The radiation resistance changes only slowly
with wire diameter, however, so sinusoidal distribution is a reasonably
good approximation provided that the antenna is thin. Feedpoint reactance,
though, varies much more dramatically with both antenna length and
diameter. Calculating its value exactly requires solution of a triple
integral equation which can't be solved in closed form. That's why
computer programs are used to solve it numerically.

To include the effect of ground, you need to calculate the mutual
impedance between the antenna and its "image". If the antenna is about 0.2
wavelength above ground or higher (for a half wave antenna -- the height
must be greater if the antenna is longer), you can assume that the ground
is perfect and get a pretty good result. Below that height, the
calculation again becomes much more complicated because the quality of the
ground becomes a factor. If you're interested in the numerical methods
used, locate the NEC-2 manual (available on the web), which describes it.

If you're satisfied with approximate results, the work by S.I. Shelkunoff
provides formulas for free-space input impedance of antennas with finite
diameter wire which can be solved with a programmable calculator or
computer. They're detailed in "Theory of Antennas of Arbitrary Size and
Shape", in Sept. 1941 Proceedings of the I.R.E. The formulas for R and X
contain many terms involving sine and cosine integrals, which can be
approximated with numerical series. You'll find additional information in
his book _Advanced Antenna Theory_. For approximate calculations of mutual
impedance of thin linear antennas, see "Coupled Antennas" by C.T. Tai, in
April 1948 Proceedings of the I.R.E. Those also involve multiple terms of
sine and cosine integrals.

Before numerical calculations became possible, many very good
mathematicians and engineers devised a number of approximation methods of
varying complexity and accuracy. You'll find their works in various
journals primarily in the 1940s - 1960s.

The complexity and difficulty of the problem is why virtually all antenna
calculations are done today with computers, using numerical methods such
as the moment method.

In summary, here are your choices:

1. You can calculate the approximate radiation resistance but not
reactance of a thin, free-space antenna by assuming a sinusoidal current
distribution and using the method Reg described. To include the effect of
ground, you have to calculate or look up from a table the mutual impedance
between the antenna and its "image", and modify the feedpoint impedance
accordingly by applying the mesh equations for two coupled antennas. This
method of including the effect of ground becomes inaccurate below around
0.2 wavelength, if the antenna is over typical earth.

2. You can use various approximation methods to calculate reactance, and
resistance with better accuracy. But for the effect of ground, you're
still limited to being greater than about 0.2 wavelength high.

3. To accurately include the effect of real ground with low antennas,
and/or to get resistance and reactance values with arbitrarily good
accuracy requires numerical methods. A computer program is the only
practical way to do this. A very good basic description of the moment
method can be found in the second and later editions of Kraus' _Antennas_.

[*] Cin(x) = ln(gamma * x) - Ci(x), where gamma = Euler's constant, 0.577.
. . Ci(x) = the integral from -infinity to x of [cos(v)/v dv] = ln(gamma *
x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . .

Roy Lewallen, W7EL

Very interesting references. In particular Kraus' 2nd ed. pp 359 - 408.
Also Stutzman and Thiele, 1st ed. pp 306 - 374. Pretty much grad level,
needs effort, (for me) even though I took Stutzman and Thiele's antenna
course in 1998 (NCEE).

Regards,

Frank

Cec, what's the Poynting Vector?

Same thing as a power flow vector. The dimensions are
energy per second per unit area. If the power flow vector
is integrated over the entire surface of a sphere of radius
r, the result is total radiated power. I thought maybe you
and Mr. Poynting might have worked together at some time in
the 20th century. :-)

Poynting, John Henry (1852-1914)
English physicist, mathematician, and inventor. He devised an equation by
which the rate of flow of electromagnetic energy
(now called the Poynting vector) can be determined.

S = E X H

In message , Cecil Moore
writes
Reg Edwards wrote:
Antenna conductors ARE transmission lines and the same sort of
calculations apply.

You're right, Reg. Without reflections from the ends of a
dipole, the feedpoint impedance would be hundreds of ohms.
A standing wave antenna is like a lossy transmission line
where the loss is to radiation. And the SWR on a 1/2WL dipole
standing wave antenna is quite high - in the neighborhood of
20:1.

Just a quick question.
What is the impedance at the centre of an infinitely long dipole (in
free space)?
Ian.
--

Hi Cecil,

I find another formula in Balnais book (section 4.3, 1982 edition)
for Rr:

Rr = 20 * (pi)^2 * ( l / lamda)^2 (4-37)

If l = lamda/2, then this formula gives Rr .=. 50 Ohms.

(4-93) the one you quoted is

Rr = (120*pi) / (4*pi) * Cin( 2*pi ) = 73 Ohms, (for l = lamda/2)

where 120*pi is the intrinsic impedance for a free-space medium.

Would you please explain the difference between these
two formulas (4-37 and 4-93)?


Thanks!

-- Harry

Anyway, reading the op carefully, he asked for an exact _explanation_, not
an exact calculation result..
Steve, K;9.D,C'I

"Bill Turner" wrote in message
...
Reg Edwards wrote:

There's no such value as 'exact'.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Nonsense. I have exactly two pencils in my pencil holder. :-)

73, Bill W6WRT

Cecil Moore wrote:
Ian Jackson wrote:

Just a quick question.
What is the impedance at the centre of an infinitely long dipole (in
free space)?


Same as a terminated dipole in an anechoic chamber? 600-800 ohms?

Not at zero Hz.

ac6xg

Harry wrote:

Hi Cecil,

I find another formula in Balnais book (section 4.3, 1982 edition)
for Rr:

Rr = 20 * (pi)^2 * ( l / lamda)^2 (4-37)

If l = lamda/2, then this formula gives Rr .=. 50 Ohms.

(4-93) the one you quoted is

Rr = (120*pi) / (4*pi) * Cin( 2*pi ) = 73 Ohms, (for l = lamda/2)

where 120*pi is the intrinsic impedance for a free-space medium.

Would you please explain the difference between these
two formulas (4-37 and 4-93)?


Thanks!

-- Harry

4-37 should apparently read:

Rr = 30 * (pi)^2 * (1 / lambda)^2.

ac6xg

Ian Jackson wrote:
Just a quick question.
What is the impedance at the centre of an infinitely long dipole (in
free space)?

Same as a terminated dipole in an anechoic chamber? 600-800 ohms?
--
73, Cecil http://www.qsl.net/w5dxp


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---