Hi Richard/all, Table 2 shows horizontal Edmax for receive heights of 1 to 4 meters, not just at 4 meters. Up to about 100-200 MHz the receive height for horizontal Edmax should be at 4m or higher, so the value at 4m would be valid. For higher frequencies the height for horizontal Edmax will be lower. You will need to compute for multiple receive heights and find the max values.
I hope this helps, Dennis Camell National Institute of Standards and Technology 325 Broadway, MS 818.02 Boulder, CO 80305 USA From: [email protected] [mailto:[email protected]] On Behalf Of Richard Georgerian Sent: Sunday, January 13, 2008 11:29 AM To: IEEE emc-pstc Subject: ANSI C63.5: Determination of EdMax; Radio Frequency Principles and Applications Greetings All, I've been reading Albert A. Smith's book, Radio Frequency Principles and Applications, and ANSI C63.5 trying to work out the formula of the determination of maximum received field, EdMax. In Smith's book, the one of the formula's I am trying to work out is on page 51, equation (3.21) and in ANSI C63.5, similar formulas are in Annex A, equation (A.1). I am getting the same EdMax values up to 100MHz in Table 2 of ANSI 63.5, but for whatever reason, for frequencies greater than 100MHz, EdMax's are completely different. I am not sure what is happening to get such erroneous results. For my own edification, I've been attempting to work through the equations to see if I can get the same values as in ANSI C63.5 Table 2, with little success. Hopefully, someone with a little more math experience can help me. For those who don't have the book or ANSI C63.5 the equations are as follows. I have used the Greek variable names instead of their symbols, to reduce the chance of them not being formatted correctly in different browsers. For the horizontal plane - EdhMax = ((49.2)^0.5 * {(d2^2 + d1^2*|rho,h|^2 + 2*d1*d2|rho,h|*cos[(phi) - (beta)*(d2-d1)]}^0.5)/(d1*d2) d1 = [R^2 + (h1 - h2)^2]^0.5 d2 = [R^2 + (h1 + h2)^2]^0.5 h1 is the height of the transmit antenna, h1 = 2 meters h2 is the height of the receiving antenna, h4 = 4 meters R is the antenna distance between the transmit and receive antenna, R = 10 meters rho,h = {sin(gamma) - [( K - j60*(lambda)*(sigma) - (cos(gamma))^2)]^0.5}/{sin(gamma) + [( K - j60*(lambda)*(sigma) - (cos(gamma))^2)]^0.5} = |rho,h|*(exp)^j(phi) |rho,h| is the magnitude K is the relative dielectric constant of the ground plane. K =1 for perfect conductor sigma is the conductivity of the ground plane, siemens per meter (S/m). sigma = infinity for perfect conductor gamma is the grazing angle. gamma = arccos[(h1 + h2)/R] phi is the phase angle of reflection coefficient beta = 2*pi/(lambda) lambda is the wavelength, meters. Many thanks in-advance, Richard ===== Richard Georgerian Compliance Engineer email: [email protected] ===== - This message is from the IEEE Product Safety Engineering Society emc-pstc discussion list. Website: http://www.ieee-pses.org/ To post a message to the list, send your e-mail to [email protected] Instructions: http://listserv.ieee.org/request/user-guide.html List rules: http://www.ieee-pses.org/listrules.html For help, send mail to the list administrators: Scott Douglas [email protected] Mike Cantwell [email protected] For policy questions, send mail to: Jim Bacher: [email protected] David Heald: [email protected] All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc - This message is from the IEEE Product Safety Engineering Society emc-pstc discussion list. Website: http://www.ieee-pses.org/ To post a message to the list, send your e-mail to [email protected] Instructions: http://listserv.ieee.org/request/user-guide.html List rules: http://www.ieee-pses.org/listrules.html For help, send mail to the list administrators: Scott Douglas [email protected] Mike Cantwell [email protected] For policy questions, send mail to: Jim Bacher: [email protected] David Heald: [email protected] All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc
