## Figure 222 Fresnel Zone Calculation

To calculate Fresnol Zone diameter in either English or Metric units:

ro calculate ^resr.eE Zone diameter in English units diam {ft.) = 72.1 (dl)fd2) d 1 d2 = mites . f(d1td2) f = GHz

(-nesinel Zone

(-nesinel Zone

 d1 d2

The amount of Fresnel zone clearance is determined by the wavelength of the signal, the total path length, and the distance to the obstacle. The Fresnel zone is always widest in the middle of the path, between the two antennas. At least 60 percent of the calculated Fresnel zone must be clear to avoid sig nificant si gnal atten uation. I n Figure; 2-22, the top of the hill extends so far into the Fresnel zone that 60 percent of the Fresnel zone is not clear; therefore, part of the signal will be attenuated. Figuro 2-23 s hows the calculatipn of F resnei zone cleara nce for a green light wave oveu a two -mile path with a hill located at the middle of the path.

Figure 2-23. Fresnel Zone Clearance for a Green Light

Fresöef Zone Dsain^ter in Feet dl, dP - 1 rnile f = 577,000 GHz (Gnaen Lighl}

=r0\$671 = in 60% Oí rz = 0.6 (0.8) In. - 0.46 In. [12.0 mm)

Fnesnol Zone

 d1 d2

Figure 4. s hows the calcul;ation of Fresn el zone eleara nee for a 2.4- GHz signal over the same two-mile path.

Figure 2-24. Fresnel Zone Clearance for 2.4 GHz

Fresne) Zone Diameter in Feat d 1. d? - 1 rnile f GHZ

Fresnel Zona

Fresnel Zona

 di da

The green lighit wave in Figure; 2-23 must: have a clear F resnel zone d i ameter that is at least 0.48 in. (1.22 cm), or 60 percent of the calculated Fresnel zone diameter, to avoid being partially attenuaee8- The required cle aran ce 2bove the l-ill (the radius of the calcul ated 60 percene Fresnel zone diameter) i e one-lnalf'of the diameter, so the green light wave must clear the hill by one-half of 0.48 in., or by 0.24 in. (0.61 cm).

The 2.4-GHz wireless wave in Figure 2-24 must have a clear Fresnel zone diameter that is at least by 19.7 ft (6 m), or 60 percent of the calculated Fresnel zone diameter, to avoid being partially- attenuated. The required clearance atwve the hill (the -adius of the calculated 60 percent Fresnel zone diamet er) is on e-half of the diameter, eo the wireless wave must clear the hill by one-hai f of 19.7 ft, or by 9.85 ft ( 3 m).

You can see from Figures 2-23 and 2-24 how a visual LOS path can exist that allows you to see from Point A to Point B with no attenuation, whereas a wireless wave traveling the same path will experience significant additional attenuation. Many times in your life, you have heard the expression, "Seeing is believing." Figures 2-23 and 2-24 provide a graphic example that, when you are working with wireless, "Seeing is not believing." In other words, just because you can see to the other end of a wireless path, do not believe that you have a clear LOS wireless path.

The clear, visual LOS path does not mean that you have an attenuation-free wireless path. You must calculate the size of the Fresnel zone and confirm that the clearance above any obstacle(s) is at least equal to one-half of 60 percent of the Fresnel zone diameter.

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