Page 396 - Integrated Wireless Propagation Models

374 C h a p t e r S i x

6.4.4. 1 Finding RSSI For Points 12 and 14

6.4.4. 1 . 1 RSSI at Point 12 For point 12, the mobile distance is 1 mile plus 0.373 mile,
and it is in LOS. The effective antenna height is 90 m, while the height of actual antenna
h/ is 45 m. Thus:

G (h = 20 log (h/h/) = 20 log (90/45) = 6 dB
)
eflli
45
= ( 6 - 6 ) + ( 0 - 0 ) + 2 0 log + 0 = 3 . 1 dB
31.5
2 0°
A = 20 log( � ) = 9 dB
1
8 0
L = 0 (no-shadow condition)

1
Then the received signal strength is found from Eq. (3.2. . 1 ) :
P, = P," - y · log(�) - A 1 + G fl11(h,) - L + a
,
· 3
= - 61.7 dBm- 3 4.8 logCr ) - 9 + 6 - 0 + 3.1 = -66.39 dBm

6.4.4. 1.2 RSSI at Point 14 For point 14, the terrain height is 25 m, and mobile distance
1
is . 437 miles. In this case, the signal is blocked by the peak of hill and is under a shadow
condition. The base station antenna height as well as the ERP of point 14 is the same as
that of Point 12.
From Fig. 6.4.4.2, we have found that r = 2200 m, r2 = 100 m, and h = 1 m . We use
P
1
)
�
the diffraction parameter v = ( -h ) ( i ( * + ) , and the diffraction factor v i s 0.4. From
P
1
Fig. . 9.2.2.1.2, the diffraction loss curve shows the loss is about 10 dB at v = 0.4.
The maximum effective antenna height gain defined in Eq. (3 1 .2.4.2) is
.
Max. G (h, ') = 20 log [ �: J
eflli

where the maximum effective antenna height h_' is the same as h of point 12 since
,
point 12 is on the top of the hill. Therefore, h , ' = 90 m, and Max. G (h , ') = 6 dB. From
,ffh
.
Eq. (3 1 . 2.4.2), the diffraction loss is
L = L 0 - 2 0 log[�: J = 1 0 - 6 = 4 d B

G (h , ) = 0
eflli
45
a = ( 6 - 6 ) + ( 0 - 0 ) + 2 0 log + 0 = 3 . 1 dB
31.5