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Wire and Wireless Communication Applications Chap. 8
is located on the top of the subscriber’s TV set [Thomson, 1994]. The LNB has a gain of 40 dB and
a noise figure (NF) of 0.6 dB. The RG6U coaxial transmission line is 110 feet in length and has a
loss of 8 dB100 ft in the 950–1450-MHz band. The receiver detects the data packets, decodes
them, and converts the data to analog video and audio signals using the built-in digital-to-analog
converters. The receiver has a NF of 10 dB and an IF bandwidth of 24 MHz. Assume that the antenna
source temperature is 20 K.
Compute the (CN) dB , (E b N 0 ) dB , and the BER for this receiving system.
Solution. The receiving antenna pointing parameters (azimuth and elevation) from Gainesville,
FL, to the satellite are evaluated using Eqs. (8–47) to (8–54).
The azimuth is
The elevation is
Using Eq. (8–49), we find that the slant range is
Thus,
b = cos -1 [cos(29.71) cos(101 - 82.43)] = 34.58°
E = tan -1 c
A = 360 - cos -1 a-
The overall NF for the receiving system is evaluated using Eq. (8–34), where F 1 = 0.6
dB = 1.15, G 1 = 40 dB = 10 4 , F 2 = 110 ft × 8 dB100 ft = 8.8 dB = 7.59, G 2 =-8.8 dB = 0.13,
and F 3 = 10 dB = 10. Then,
F = F 1 + F 2 - 1
G a1
+
1
tan (34.58)
-
F 3 - 1
G a1 G a2
= 1.15 +
tan (29.71)
tan (34.58) b = 214.13°
3963
(26,205) sin (34.58) d = 49.82°
d = 3(26,205) 2 + (3963) 2 - 2(3693)(26,205) cos (34.58)
d = 23,052 miles, or 3.709 * 10 7 m
7.59 - 1
10 4 +
10 - 1
(10 4 )(0.13)
or F = 1.15 + 6.59 × 10 -4 + 6.83 × 10 -3 = 1.15 = 0.6 dB. Thus, T e = (F - 1)T 0 = (1.15 - 1)
(290) = 43.18K.
Note: The gain of the LNB is designed to be sufficiently large so that the NF contributions of the
transmission line and the receiver are negligible.
The receiving antenna is computed using Table 8–4 for an 18–in = 0.46-m-diameter
parabola, where l = cf = 3 × 10 8 12.45 × 10 9 = 0.0241 m.
Thus,
(G AR ) dB = 10 log c 7p(0.46>2)2
(0.0241) 2 d = 32.96 dB
(T syst ) dB = 10 log (T AR + T e ) = 10 log (20 + 43.18) = 18.01 dBK
a G AR
= 32.96 - 18.01 = 14.96 dB>K
T syst
bdB