Remember Me?

#1
September 13th 05, 04:31 PM
 Harry Posts: n/a
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

#2
September 13th 05, 05:13 PM
 Tim Wescott Posts: n/a

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
(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
#3
September 13th 05, 05:20 PM
 Bill Turner Posts: n/a

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
#4
September 13th 05, 07:22 PM
 Reg Edwards Posts: n/a

"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

#5
September 13th 05, 08:11 PM
 Harry Posts: n/a

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

#6
September 13th 05, 09:36 PM
 Tom Donaly Posts: n/a

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
#7
September 13th 05, 10:02 PM
 Ham op Posts: n/a

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

#8
September 13th 05, 11:56 PM
 Reg Edwards Posts: n/a

"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.

#9
September 14th 05, 12:12 AM
 Cecil Moore Posts: n/a

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

----== 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 =---
#10
September 14th 05, 12:23 AM
 Cecil Moore Posts: n/a

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

----== 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 =---

 Posting Rules Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On

 Similar Threads Thread Thread Starter Forum Replies Last Post RHF Antenna 25 November 15th 04 09:15 PM RHF Shortwave 22 November 15th 04 09:15 PM J999w Shortwave 10 June 8th 04 07:56 AM Gillis Homebrew 0 February 24th 04 12:07 AM Dr. Slick Antenna 255 July 29th 03 11:24 PM

All times are GMT +1. The time now is 07:18 AM.