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as specified by Part 15 of the FCC Rules, Subpart D, which governs the operation of UPCS
devices in the U.S.

The effective interference may be several dB higher than this, if the UPCS device has a frame
structure different than DECT, and/or if it is not synchronized to the FWA frame, as would likely
be the case. Moreover, the analysis described here did not account for adjacent-channel
interference or interference from the FWA uplink. Including these factors would increase the
calculated interference.

Interior RF path loss inside the buildings which house the UPCS systems is not expected to
mitigate the interference problem significantly. As has been shown here, from simple geometry,
the majority of a building’s area is relatively near an exterior wall. Moreover, those in the
management hierarchy of a company who are likely to be provided wireless telephones are also
likely to have offices with windows. Finally, many modern offices use an open “landscape”
interior design that uses soft partitions rather than floor-to-ceiling walls. In such environments,
the radio signal is attenuated relatively little with distance.

Overall, the potential for interference from FWA to UPCS will make spectrum sharing
impractical, except perhaps in rural, low-density areas, where the traffic demand per km2 will be
light for both FWA and UPCS.

5.1.1.6 Annex: CDF of the distance from a point inside a building to nearest outside wall

Consider a square building D meters on a side as shown in Fig. 5-5. If the building area is
divided into four equal triangles as shown, then any point in the building will be inside the
triangle whose base is the outside wall nearest to the point.

D 2s D

D2

s

D
D

Figure 5-5: Assumed building geometry

If the probability that a randomly located point is within some incremental area dA is

proportional to dA (i.e., the spatial distribution is uniform), then the probability that a point is

within some incremental distance ds of the nearest outside wall is k · ds · (D - 2s) ,

where k is a constant to be determined. Therefore, the pdf of s is f s (a ) = k(D - 2a ) .

D2

Since 0 = s = D 2 and total probability must sum to 1, k ? (D - 2a )da = 1, so k = 4 D2 and
0

f s (a) = 4 - 8a . (5-13)
D D2

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