73 Ohms, How do you get it?
 

I am a newbi in antennas.

Here's my question:

I know that a half-wave dipole in free space has
a feed-point impedance of approximately 73 ohms.

Can anyone tell me **exactly** how this number is calculated.

(or tutorial webpages)

I mean I like to see a general formula (a function of
antenna height above earth....) and all the detailed steps
that will get the impedance number.

Thanks!

-- Harry

Harry wrote:
I am a newbi in antennas.

Here's my question:

I know that a half-wave dipole in free space has
a feed-point impedance of approximately 73 ohms.

Can anyone tell me **exactly** how this number is calculated.

(or tutorial webpages)

I mean I like to see a general formula (a function of
antenna height above earth....) and all the detailed steps
that will get the impedance number.

Thanks!

-- Harry

That calculation comes about as the culmination of a two-semester
junior-level college course in Electrodynamics. Because of their
mathematical intensity most people suffer through such courses with grim
determination rather than greeting them with joy.

The 73 ohm number assumes an antenna in free space with a magic
zero-size current source at it's center and no wires going to the
antenna. Then some handwaving simplifications are made, a current
gradient is assumed, a voltage gradient is calculated, and the impedance
(and antenna pattern) is calculated. If you want to avoid the
handwaving simplifications you take _another_ year of antenna-specific
E&M and/or you write a program like NEC to do the calculation numerically.

This is why the antenna chapters in the Handbook start with very basic
theory then take a very long jump to a compendium of results, without
trying to fill in all the space in between.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Harry wrote:

I know that a half-wave dipole in free space has
a feed-point impedance of approximately 73 ohms.

Can anyone tell me exactly how this number is calculated.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Simple Ohm's law: If you apply 73 volts of RF, 1 amp of current will
flow.

To calculate the effects of nearby ground a program like EZNEC is
useful, although you almost never know the exact parameters of the
ground, so the results should not be taken as precise.

73, Bill W6WRT

"Harry" wrote
I know that a half-wave dipole in free space has
a feed-point impedance of approximately 73 ohms.

Can anyone tell me **exactly** how this number is calculated.

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

All you have to do is integrate the power flowing outwards from a
dipole at the centre of an arbitrary sphere with a surface area of x
square metres and equate it to the current flowing in the dipole,
taking into account the distribution of current along the dipole, and
you will obtain the radiation resistance referred to its feedpont.
OK?

But in your case, all you can do is just accept the hearsay value of
73 ohms as being good enough.

Actually it depends on the diameter of the dipole relative to its
length and at HF it is a few percent less. Not that anybody ever
notices such minor discrepancies.
----
Reg, G4FGQ

Hi Tim and Reg,

Thank you for your valuable information. Is there any website or
textbook that actually shows the step-by-step calculation of this magic
number which has been quoted so often in the cable industry?

You know most video cables and connectors have characteristic
impedance, 75 Ohms.

I am not afraid of math. I just like to understand the details of its
derivation.

-- Harry

Harry wrote:
Hi Tim and Reg,

Thank you for your valuable information. Is there any website or
textbook that actually shows the step-by-step calculation of this magic
number which has been quoted so often in the cable industry?

You know most video cables and connectors have characteristic
impedance, 75 Ohms.

I am not afraid of math. I just like to understand the details of its
derivation.

-- Harry


What you're seeking is in the book _Antenna Theory, Analysis and Design_
by Constantine A. Balanis, ISBN 0-471-59268-4.
73,
Tom Donaly, KA6RUH

Transmission line calculations are much easier than antenna calculations.

To a first approximation: Zo = SQRT(L/C); where L = inductance per unit
length, and C = capacitance per unit length.

Harry wrote:
Hi Tim and Reg,

Thank you for your valuable information. Is there any website or
textbook that actually shows the step-by-step calculation of this magic
number which has been quoted so often in the cable industry?

You know most video cables and connectors have characteristic
impedance, 75 Ohms.

I am not afraid of math. I just like to understand the details of its
derivation.

-- Harry

"Ham op" wrote
Transmission line calculations are much easier than antenna
calculations.

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

Antenna conductors ARE transmission lines and the same sort of
calculations apply.
----
Reg.

Harry wrote:
I am a newbi in antennas.

Here's my question:

I know that a half-wave dipole in free space has
a feed-point impedance of approximately 73 ohms.

Can anyone tell me **exactly** how this number is calculated.

From "Antenna Theory" by Balanis:

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

Prad is found by integrating the Poynting Vector over a
certain radius. Io is the current maximum magnitude.

The ASCII limitation prevents much more than this.
--
73, Cecil http://www.qsl.net/w5dxp


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Harry wrote:
Thank you for your valuable information. Is there any website or
textbook that actually shows the step-by-step calculation of this magic
number which has been quoted so often in the cable industry?

"Antenna Theory" by Balanis, second edition, Chapter 4.
--
73, Cecil http://www.qsl.net/w5dxp


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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.
--
73, Cecil http://www.qsl.net/w5dxp


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It is equally correct to say that the radiation resistance of a dipole
is 2*73 = 146 ohms, since this is the value when the radiation
resistance is assumed to be uniformly distributed along its length as
is the conductor resistance.

For calculating purposes the radiation and conductor resistances can
simply be added together.

And when referred to the end of a dipole, at HF the radiation
resistance is of the order of Q-squared * 73/2 = 3000 ohms. Although
this is somewhat indeterminate because of the difficulty of feeding an
isolated dipole at one end. The conductor diameter also plays a
significant part.
----
Reg.

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

Harry wrote:
Hi Tim and Reg,

Thank you for your valuable information. Is there any website or
textbook that actually shows the step-by-step calculation of this magic
number which has been quoted so often in the cable industry?

You know most video cables and connectors have characteristic
impedance, 75 Ohms.

I am not afraid of math. I just like to understand the details of its
derivation.

-- Harry

"Cecil Moore" wrote
Prad is found by integrating the Poynting Vector over a
certain radius.

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

Cec, what's the Poynting Vector?
---
Reg.

Reg Edwards wrote:

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

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

73, Bill W6WRT

Reg Edwards wrote:

"Cecil Moore" wrote

Prad is found by integrating the Poynting Vector over a
certain radius.

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.

http://www.cartage.org.lb/en/themes/...oynting/1.html
--
73, Cecil http://www.qsl.net/w5dxp


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"Reg Edwards" wrote
It is equally correct to say that the radiation resistance of a
dipole
is 2*73 = 146 ohms, since this is the value when the radiation
resistance is assumed to be uniformly distributed along its length
as
is the conductor resistance.

For calculating purposes the radiation and conductor resistances can
simply be added together.
=====================================

Hints and tips -

By the same arithmetical reasoning -

The centre feedpoint input resistance of a 1/2-wave dipole is 73 ohms
plus HALF of the conductor end-to-end HF resistance, which to be
precise sometimes matters. This is exact insofar as 73 is exact - not
an approximation.

The HF resistance of a copper wire at 20 degrees C is -

Rhf = Sqrt( F ) / 12 / d ohms per metre,

Where F is the frequency in MHz and d is the wire diameter in
millimetres.

For example, the end-to-end loss resistance of a 1/2-wave dipole at
1.9 MHz using 16-gauge copper wire is 7.07 ohms, which increases the
feedpoint resistance to 73 + 3.53 = 76.53 ohms, to give a radiating
efficiency of 95.4 percent. Some people would consider that's enough
to lose a contest!

Why not start a little notebook to record useful, simple, little
formulae such as above which don't appear in Terman et al? Or if they
do appear then you can never find the right page in the right volume.
----
Reg.

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.

-- Harry

Hi Roy,

Thanks for all the valuable information. Are you a professor?

You may want to write a book titled "Story of the 73 Ohms".

-- Harry

Harry wrote:
Hi Roy,

Thanks for all the valuable information. Are you a professor?

Not by a long shot. I'm way too poor a student to ever be considered for
a job as a teacher.

You may want to write a book titled "Story of the 73 Ohms".

It's already been written, by many people who know a great deal more than I.

Roy Lewallen, W7EL

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


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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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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

| "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

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

"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;^)

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.

--

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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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.

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.

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.