Page 336 - Integrated Wireless Propagation Models

314 C h a p t e r F i v e

R
,.... - - -�----
-
· - - -
_, .,.. ""'

External wall of building
FIGURE 5.3.4.3.2 Penetration i n to building from external transmitter i n another building.

0
However, the theory and formula shown in reference4 is used for the scenario that
the distance 5 is from the outside transmitter to the external wall of the building. We
may need to modify Eq. (5.3.4.3.1) to cover the internal receiver located away from the
interior wall of the building.

3
5.3.4. . 1 For the Scenario of Either T x Inside and Rx Outside the Building or Vice Versa The
modified equation is
(5.3.4.3.2)

where 51 is the distance from the transmitter to the building wall and 52 is the distance
from the receiver to the building wall. Both 51 and 5 2 are in meters.

5.3.4.3.2 For the Scenario of the lnterbuilding Case {Tx Is Inside in One Building and Rx is
inside in Another Building)
We may have to modify Eq. (5.3.4.3 1 ) before it can fit the situation for which our
.
measured data were collected. First, in our case, the transmitter antenna did not have a
LOS path to the receiver while both the transmitter and receiver were inside the build­
ing, as shown in Fig. 5.3.4.3.2. So we need an extra loss term for the loss of propagation
passing through the external wall from the transmitter's building. Second, in our case,
there is no window on the walls, that the signal would pass through; thus, L, and Le are
e
higher than the given recommended values from reference.9
Based on the above two main observations, we modified Eq. (5.3.4.3.1) to get a new
formula that should fit this case better:

(5.3.4.3.3)

where 51 is the distance from the transmitter to the interior building wall of building A,
52 is the distance between two building walls, and 53 is the distance from the receiver to
the interior building wall of building B. All three segmental distances, 51, 52, and 53 are
in meters.
The reason for doubling L, is due to the consideration of the transmitter's external
wall, and we chose L, = 15 dB and We = 25 dB; thus, the calculating equation is given by
e
2
L = 32.4 + 20 log(2.4) + 20 log(5 1 + 5 2 + 53) + 2 * 15+ 25 * (1-sin 8) (5.3.4.3.4)