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-   -   73 Ohms, How do you get it? (https://www.radiobanter.com/antenna/78184-73-ohms-how-do-you-get.html)

Cecil Moore September 15th 05 12:12 AM

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


Equation 4-37 is from section "4.3 SMALL DIPOLE", i.e. shorter
than 1/2WL. The dipole is so short that its current distribution
is triangular, not sinusoidal. Quoting section 4.3: "The radiation
resistance of the antenna is strongly dependent upon the current
distribution." The "1/2" on the diagram does NOT mean 1/2WL.

Equation 4-93 is from section "4.6 HALF-WAVELENGTH DIPOLE". The
current distribution on the thin-wire 1/2WL dipole is considered
to be sinusoidal.

See Figure 4.8. Note the triangular current distribution for
L = 1/4WL and the sinusoidal current distribution for L = 1/2WL.
--
73, Cecil http://www.qsl.net/w5dxp

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Harry September 15th 05 02:50 AM

See Figure 4.8. Note the triangular current distribution for
L = 1/4WL and the sinusoidal current distribution for L = 1/2WL.


Very good,..... thanks a lot!
Your figure 4.8 (second edition) is figure 4.7 in the first edition.
These current distribution curves (1/4, 1/2, 1, 3/2, and 2 Lamda
length)
are interesting to study.

-- Harry


news-z-non-exist-z September 15th 05 07:26 PM

| "Roy Lewallen"
| wrote in message ...
|
| Just about any antenna textbook will show you the calculation
| [...]
| ln(gamma * x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . .
|
| Roy Lewallen, W7EL

[... ] and after all that,
we finally draw a figure such the one at:

http://antennas.ee.duth.gr/Antennas_...II_2005/23.htm

pezSV7BAXdag



Allodoxaphobia September 15th 05 07:51 PM

On Thu, 15 Sep 2005 21:26:04 +0300, news-z-non-exist-z wrote:
| "Roy Lewallen"
| wrote in message ...
|
| Just about any antenna textbook will show you the calculation
| [...]
| ln(gamma * x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . .
|
| Roy Lewallen, W7EL

[... ] and after all that,
we finally draw a figure such the one at:

http://antennas.ee.duth.gr/Antennas_...II_2005/23.htm

pezSV7BAXdag


Yes, but:
http://antennas.ee.duth.gr/Antennas_...II_2005/22.htm

explains it best of all! HI!HI!

73
Jonesy
--
Marvin L Jones | jonz | W3DHJ | linux
Pueblo, Colorado | @ | Jonesy | OS/2 __
38.24N 104.55W | config.com | DM78rf | SK

news-z-non-exist-z September 15th 05 09:04 PM

"Allodoxaphobia" wrote in message ...
| On Thu, 15 Sep 2005 21:26:04 +0300, news-z-non-exist-z wrote:
| [...]
| Yes, but:
| http://antennas.ee.duth.gr/Antennas_...II_2005/22.htm
|
| explains it best of all! HI!HI!
|
| 73
| Jonesy

d;^)



Ian Jackson September 17th 05 08:24 AM

In message , Jim Kelley
writes


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


What is a 'terminated dipole'?
And why 600-800 ohms?

No, Mine will be just a normal dipole (but long). I'll be in a
spaceship, miles from anywhere, and I intend to put out a really long
antenna so I can work the universe on all amateur bands. (I don't care
about the polar diagram - there's bound to be someone out there in one
of the major lobes).

I intend to throw out an infinitely long wire either side of the ship,
and use a balanced tuner connected directly to the antenna (no feeder
required). Because of weight restrictions, I can only take one tuner,
and I want to make sure that the one I do take will cope with the feed
impedance of the antenna.

I think the impedance will be the same at all frequencies (maybe even at
'zero Hz'). But what will it be?

Ian.

--


Cecil Moore September 17th 05 02:52 PM

Ian Jackson wrote:
What is a 'terminated dipole'?


In an anechoic chamber, the dipole has a resistor to chassis at
both ends. When the value of the resistor equals the characteristic
impedance of the dipole, the dipole becomes a traveling wave
antenna instead of a standing wave antenna. Your infinite dipole
in free space will be a traveling wave antenna, i.e. no reflections.

And why 600-800 ohms?


That's a ballpark Z0 for traveling wave antennas not located
near ground. It's the Z0 of a 1/2WL dipole during the transient
state before the arrival of the first reflections from the ends.

The resonant feedpoint impedance of a center-fed 1/2WL dipole is
(the difference between the values of the forward voltage and the
reflected voltage) divided by (the sum of the values of the forward
current and reflected current). Note: the two voltages are out of
phase and the two currents are in phase.

This is the (B) definition of "impedance" in the "IEEE Dictionary",
i.e. impedance as a *result*, not a cause. Destructive wave interference
at the feedpoint of a center-fed 1/2WL dipole is what causes a Z0-match
to a low resistive value, e.g. 50 ohms or 70 ohms.
--
73, Cecil http://www.qsl.net/w5dxp


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Reg Edwards September 17th 05 03:24 PM

What is the impedance at the centre of an infinitely long dipole
(in
free space)?


===============================
Its not very different from -

Zin = 120 * Ln( Wavelength / d ) ohms.

where d = conductor diameter, both measured in metres.

Thus, at wavelength = 80 metres with 14 gauge copper wire, input
impedance = 1300 ohms approx.

If you don't believe me, just measure it.
----
Reg.



Ham op September 17th 05 08:20 PM

Ian Jackson wrote:

In message , Jim Kelley
writes



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.


Now#1, an infinitely long wire has an infinite self inductance.
Now#2, an infinitely long wire has an infinite self capacitance.
Now#3, frequency does not enter into the equation

Zo = SQRT[L(infinite)/C(infinite)]

Now, from my old college days, infinity divided by infinity is
undefined. The SQRT of an undefined value is by definition undefined.

Conclusion: the impedance of an infinitely long wire in free space is
not the SAME as a terminated dipole in an anechoic chamber.

NOTE: an infinitely long wire in free space cannot be terminated. A
terminated wire in an anechoic chamber is not the same as a free space
model.

A terminated wire in in an anechoic chamber is a standard math model
with Zo varying as the ratio of the wire diameter and the distance to
the 'space cloth/cones' [which are not perfect rf absorbers ... hence
some reflection]. Also, the termination destroys the symmetry.



ac6xg


What is a 'terminated dipole'?
And why 600-800 ohms?

No, Mine will be just a normal dipole (but long). I'll be in a
spaceship, miles from anywhere, and I intend to put out a really long
antenna so I can work the universe on all amateur bands. (I don't care
about the polar diagram - there's bound to be someone out there in one
of the major lobes).

I intend to throw out an infinitely long wire either side of the ship,
and use a balanced tuner connected directly to the antenna (no feeder
required). Because of weight restrictions, I can only take one tuner,
and I want to make sure that the one I do take will cope with the feed
impedance of the antenna.

I think the impedance will be the same at all frequencies (maybe even at
'zero Hz'). But what will it be?

Ian.



Reg Edwards September 17th 05 09:00 PM

You had better go back to your old college days and start again with
new teachers.

The input impedance of an infinite dipole is as I have already simply
mathematically described. It changes a little with frequency and wire
diameter.

Anything less than infinite length will be altogether different.

Now you can stop trying to pull our legs.
----
Reg.




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