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Sec. 8–13 Study-Aid Examples 659
The (CN) dB is evaluated using Eq. (8–55), where
(L FS ) dB = 20 log a 4pd
l b = 20 log a 4p(3.709 * 107 )
b = 205.73 dB
(0.0241)
and
(B) dB = 10 log(B) = 10 log (24 * 10 6 ) = 73.8 dB
Thus,
or
a C N b dB
= (P EIRP ) dBw - (L FS ) dB + a G AR
- k
T dB - B dB
syst
bdB
= 52 - 205.73 + 14.96 - (-228.6) - 73.8
(C>N) dB = 16.03 dB
(E b N 0 ) dB may be evaluated using Eq. (8–44), where B = 24 MHz, and R = 40 Mbs. Then,
a E b
= a C N 0
bdB N b + a B
dB R b = 16.03 - 2.22 = 13.81 dB
dB
An (E b N 0 ) dB of 13.81 dB = 24.05 gives negligible errors for QPSK signaling. That is, if no coding
is used, the QPSK BER is given by Eq. (7–69) as
or one error every 3.4 hours. However, if there is signal fading (because of rain or other atmospheric
conditions), significant errors may occur. This is examined in Example SA8–2.
SA8–2 DSS Link Budget with Fading Repeat Example SA8–1, and assume that there is a
4-dB signal fade of the Ku band signal because of rain. Compute (CN) dB , (E b N 0 ) dB and the
BER with and without coding. For the coding case, assume a 3-dB coding gain.
Solution
P e = Q £ C
2a E b
N 0
b ≥ = Q A12(24.05)B = 2.0 * 10 -12
a C N b fade dB
= a C N b dB
- (L fade ) dB = 16.03 - 4.0 = 12.03 dB
and
Then, the BER with fading and no coding is
a E b
b = 9.81 dB = 9.57
N 0 fade dB
P e = Q A12(9.57)B = 6.04 * 10 -6
or one error in 4.1 ms. This performance is not acceptable; consequently FEC coding is needed.
Using coding with a coding gain of 3dB, compute the effective (E b N 0 ) by referring to Fig. 1–8:
a E b
b = 9.81 + 3.0 = 12.81 dB = 19.10
N 0 dB