As i remember the factor L (cable length) in the equation is an simplification of the real value sin(L). As one may remember , sin(L)=L for L very small. As the formula was meant for small dipoles, that is close to correct. I remember also that already L in sinus(L) was simplified from a factor called "wave number" ???? Anyone knows the details ???
Gert Gremmen -----Original Message----- From: owner-emc-p...@majordomo.ieee.org [mailto:owner-emc-p...@majordomo.ieee.org]On Behalf Of Wan Juang Foo Sent: maandag 11 maart 2002 4:36 To: emc-p...@majordomo.ieee.org Subject: Re: Electric field radiation from rf currents Ken, Um, my apologies, I typed too fast and read too little. > E (V/m) = 1.26 10^-4 (f*L*Icm) > > In this equation, f is frequency in MHz, L is cable length in meters, and > Icm is current in milliamperes. If f is in MHz, and Icm is in mA, then the constant should be OK as it is. Regarding You said it, that's the assumption behind the electrically short radiating element (Hertzian Dipole). The assumption is that the current distribution along the radiating element is constant. If the radiating element is electrically short then the RE should be proportional to the length of the radiating element. This assumption starts to fall apart as l approaches lambda/4. In the example cited: > MHz, 1.5 meter cable length and 2 mA current to arrive at a predicted field > intensity of 92 dBuV/m. Note that 1.5 meters is one half wavelength at 100 > MHz, so this is a half-wave dipole. I could instead use: > > Pd = Pt G/4 pi r^2 > > and substitute E^2/120 pi = Pd. If I further assume I am driving 2 mA into > the antenna which looks like 72 Ohms and has a numeric gain of 1.64, then I > get 81 dBuV/m, which is 11 dB less than predicted by Mr. Armstrong's As l approaches lambda/4, the model will over estimate the radiated emission as the current distribution is more like a quarter of a cycle of a cosine wave. This model will have to be replaced by another model (equation). > Can anyone comment on this? Or offer other expressions and their > derivations (or references)? Thank you in advance. Referring to page 187 of CR Paul, (Eq 5.22) you can use: E= 60 I/r for fields at broadside theta=90 degrees for an electrically lambda/4 monopole that is thin. if I= 2 mA, r=10 meters, then therefore E= 0.012 V/m or 81 dBuV/m which is the same as your isotropic model. sincerely, Tim Foo |---------+-----------------------------> | | Ken Javor | | | <ken.javor@emccomp| | | liance.com> | | | | | | 03/11/02 10:59 AM | | | | |---------+-----------------------------> >--------------------------------------------------------------------------- ---------------------------------------------| | | | To: Wan Juang Foo <f...@np.edu.sg>, emc-p...@majordomo.ieee.org | | cc: | | Subject: Re: Electric field radiation from rf currents | >--------------------------------------------------------------------------- ---------------------------------------------| I believe the original equation posted by Mr. Armstrong had numerically taken into account the 1/r factor. At least it works out right if you assume that the 1.26 e-4 part derives from 4 pi * 1e-7 Henry/m, the permeability of free space. My problem was the assumption implicit in the equation that radiation efficiency would monotonically increase with increasing cable length, which is counter-intuitive unless the cable is electrically short. ---------- >From: "Wan Juang Foo" <f...@np.edu.sg> >To: Ken Javor <ken.ja...@emccompliance.com>, <emc-p...@majordomo.ieee.org> >Subject: Re: Electric field radiation from rf currents >Date: Sun, Mar 10, 2002, 8:55 PM > > > Ken, > The 1/r term is missing from that equation, I belive that it was factor > into the constant errornously. The constant should be 1.26 e-7. > :-) > The equation for the worst case (with ground plane reflection) commom mode > current radiating from a 'electriaclly short' radiating element ( Herzian > dipole) and note in this case cable is too long to be considered > electrically short anyway. :-) > The Equation should/could also be expressed as > [ 4 pi x 10 -7 f I l ] > E = --------------------- V/m > r > > where (4pi x 10^7)/10 = 1.257e-7 > > AFAIK, this is a simplified equation and deriable from 1st principle. <snip> ------------------------------------------- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. 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