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#1




73 Ohms, How do you get it?
I am a newbi in antennas.
Here's my question: I know that a halfwave dipole in free space has a feedpoint 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




Harry wrote:
I am a newbi in antennas. Here's my question: I know that a halfwave dipole in free space has a feedpoint 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 twosemester juniorlevel 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 zerosize 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 antennaspecific 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




Harry wrote:
I know that a halfwave dipole in free space has a feedpoint 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




"Harry" wrote I know that a halfwave dipole in free space has a feedpoint 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




Hi Tim and Reg,
Thank you for your valuable information. Is there any website or textbook that actually shows the stepbystep 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




Harry wrote:
Hi Tim and Reg, Thank you for your valuable information. Is there any website or textbook that actually shows the stepbystep 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 0471592684. 73, Tom Donaly, KA6RUH 
#7




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 stepbystep 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




"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




Harry wrote:
I am a newbi in antennas. Here's my question: I know that a halfwave dipole in free space has a feedpoint 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 (493) 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  UnlimitedUncensoredSecure Usenet News== http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups = East/WestCoast Server Farms  Total Privacy via Encryption = 
#10




Harry wrote:
Thank you for your valuable information. Is there any website or textbook that actually shows the stepbystep 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  UnlimitedUncensoredSecure Usenet News== http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups = East/WestCoast Server Farms  Total Privacy via Encryption = 
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