Reg Edwards wrote:

"Ian Jackson" wrote -

Are you sure it's as high as that, Reg? I once did a Smith Chart

plot of

the impedance at the centre of a dipole, the valued being taken from

a

table 'compiled by Wu' (LK Wu?). These only catered for a lengths up

to

a few wavelengths. As the plot progressed round and round the Smith
Chart, it seemed to be heading for something around 350 to 400 ohms.

I've just done a search on 'Wu+dipole+impedance', and one of the

results

is
http://www.fars.k6ya.org/docs/antenn...nce-models.pdf
I'll have a read of it today.

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

The characteristic impedance of an infinitely long wire is Zo.

If we cut the line and measure between the two ends we obtain an input
impedance of twice Zo. Which is the answer to our problem.

Zo is a function of wavelength, conductor diameter and conductor
resistance R where R includes the uniformly distributed radiation
resistance. On a high Zo line the radiation resistance is small
compared with Zo and the only effect of the radiation resistance is to
give Zo a small negative angle. Which when estimating Zo can be
ignored. (It is conductor resistance which at HF gives Zo of ALL
lines a very small negative angle).

In the problem posed, the current is also uniformly distributed along
the low-loss line and radiation resistance is not the value we are
familiar with and what we might do with it.

And so we get approximately -

Rin = 120 * ( Ln( Wavelength / 2 / d ) - 1 )

At a wavelength of 2 metres and a conductor diameter of 10mm the input
resistance = 433 ohms.

I cannot guarantee the above formula to be correct. But is it low
enough for you? ;o)

Mr Wu calculates radiation resistance which is not the same as input
impedance unless correctly referenced. It is usual in technical papers
to calculate Radres at one end of the antenna. Or it may be the
distributed value. I havn't the time to find and study the full text.
From past experience, with me, it usually ends up as a wild goose
chase.
----
Reg.



Sounds reasonable, Reg. To put it for simple people like me, it would
mean it's a transmission line of diameter x with an infinite diameter
shield. Then we feed 2 of them, balanced, colinear, and that's our R sub r.

Did I misunderstand?

tom
K0TAR

Tom Ring wrote:

Sounds reasonable, Reg. To put it for simple people like me, it would
mean it's a transmission line of diameter x with an infinite diameter
shield. Then we feed 2 of them, balanced, colinear, and that's our R
sub r.

Did I misunderstand?


Minus the radiation, of course.

Now, the question is, how much does that change things.

"Tom Ring" wrote in message
. ..
Tom Ring wrote:

Sounds reasonable, Reg. To put it for simple people like me, it
would
mean it's a transmission line of diameter x with an infinite
diameter
shield. Then we feed 2 of them, balanced, colinear, and that's
our R
sub r.

Did I misunderstand?


Minus the radiation, of course.

Now, the question is, how much does that change things.

==================================
Tom,

Twas not I who posed the original "input impedance of an infinitely
long dipole" question.

The radiation resistance does not enter very much into the solution.
But in any case the radiation resistance is not the one we are
accustomed to because the current distribution along the wire is not
of sinewaveform but decays rather slowly exponentially.

I've just remembered the name of the real antenna which best fits the
problem. It is the terminated Beverage which is just a very long
horizontal wire some distance above ground.

The Zo of the Beverage is 60 * Ln( 4 * Height / d ) ohms, where d is
wire diameter. When terminated its input resistance at LF is Zo (see
the Bible, the ARRL Antenna Book). A typical value of Zo is 550 ohms.

Note that radiation resistance does not enter the formula although it
cannot be denied radiation does occur. The formula is a close
approximation which serves present purposes.

Putting two Beverages back-to-back to make a dipole we get an input
impedance of 1100 ohms. The infinite dipole is in the same high
impedance ballpark.

To calculate Zo of an isolated infinite dipole we shall have to change
dimensions Height above ground disappears and is replaced by
wavelength (or frequency).

Zo = 60 * ( Ln( Wavelength / 2 / d ) - 1 ) ohms, approximately.

A more exact formula involves inverse hyperbolic functions and
wavelength, height, wire length, and wire diameter, but nobody ever
uses it. You won't find it in Terman.

On the favourite American 40-meter band with a 14 AWG infinitely long
wire Zo = 505 ohms.

Which makes the dipole input impedance = 1010 ohms but a nice, round
1000 ohms is near enough for me.

Your 'infinite shield' is a fair description for the return path but
the end-effect is fairly large. I prefer 'the rest of the Universe'.
But the nearest point is still the Earth's surface.

I leave the people, who attempted to use Smith Charts and Eznec to
solve the interesting problem, to fathom out where they (or the charts
or Eznec) went wrong.
----
Reg.

Reg Edwards wrote:
The Zo of the Beverage is 60 * Ln( 4 * Height / d ) ohms, where d is
wire diameter. When terminated its input resistance at LF is Zo (see
the Bible, the ARRL Antenna Book). A typical value of Zo is 550 ohms.

Note that radiation resistance does not enter the formula although it
cannot be denied radiation does occur. The formula is a close
approximation which serves present purposes.

Putting two Beverages back-to-back to make a dipole we get an input
impedance of 1100 ohms. The infinite dipole is in the same high
impedance ballpark.

To calculate Zo of an isolated infinite dipole we shall have to change
dimensions Height above ground disappears and is replaced by
wavelength (or frequency).

Zo = 60 * ( Ln( Wavelength / 2 / d ) - 1 ) ohms, approximately.

A more exact formula involves inverse hyperbolic functions and
wavelength, height, wire length, and wire diameter, but nobody ever
uses it. You won't find it in Terman.

On the favourite American 40-meter band with a 14 AWG infinitely long
wire Zo = 505 ohms.

Which makes the dipole input impedance = 1010 ohms but a nice, round
1000 ohms is near enough for me.

Your 'infinite shield' is a fair description for the return path but
the end-effect is fairly large. I prefer 'the rest of the Universe'.
But the nearest point is still the Earth's surface.

I leave the people, who attempted to use Smith Charts and Eznec to
solve the interesting problem, to fathom out where they (or the charts
or Eznec) went wrong.
----
Reg.

In message , Reg
Edwards writes

"Tom Ring" wrote in message
...
Tom Ring wrote:

Sounds reasonable, Reg. To put it for simple people like me, it
would
mean it's a transmission line of diameter x with an infinite
diameter
shield. Then we feed 2 of them, balanced, colinear, and that's
our R
sub r.

Did I misunderstand?


Minus the radiation, of course.

Now, the question is, how much does that change things.

==================================
Tom,

Twas not I who posed the original "input impedance of an infinitely
long dipole" question.

The radiation resistance does not enter very much into the solution.
But in any case the radiation resistance is not the one we are
accustomed to because the current distribution along the wire is not
of sinewaveform but decays rather slowly exponentially.

I've just remembered the name of the real antenna which best fits the
problem. It is the terminated Beverage which is just a very long
horizontal wire some distance above ground.

The Zo of the Beverage is 60 * Ln( 4 * Height / d ) ohms, where d is
wire diameter. When terminated its input resistance at LF is Zo (see
the Bible, the ARRL Antenna Book). A typical value of Zo is 550 ohms.

Note that radiation resistance does not enter the formula although it
cannot be denied radiation does occur. The formula is a close
approximation which serves present purposes.

Putting two Beverages back-to-back to make a dipole we get an input
impedance of 1100 ohms. The infinite dipole is in the same high
impedance ballpark.

To calculate Zo of an isolated infinite dipole we shall have to change
dimensions Height above ground disappears and is replaced by
wavelength (or frequency).

Zo = 60 * ( Ln( Wavelength / 2 / d ) - 1 ) ohms, approximately.

A more exact formula involves inverse hyperbolic functions and
wavelength, height, wire length, and wire diameter, but nobody ever
uses it. You won't find it in Terman.

On the favourite American 40-meter band with a 14 AWG infinitely long
wire Zo = 505 ohms.

Which makes the dipole input impedance = 1010 ohms but a nice, round
1000 ohms is near enough for me.

Your 'infinite shield' is a fair description for the return path but
the end-effect is fairly large. I prefer 'the rest of the Universe'.
But the nearest point is still the Earth's surface.

I leave the people, who attempted to use Smith Charts and Eznec to
solve the interesting problem, to fathom out where they (or the charts
or Eznec) went wrong.
----
Reg.



I'd like to say thanks to all who responded to my original question.
There have been a variety of answers.

The only answers I feel might be wrong are those where there is a
reactive component. Surely, this can be removed by lengthening the
antenna by the appropriate amount - but as it is already infinite, this
might be difficult.

Anyway, all very interesting.
Cheers,
Ian.

--