Ing. Cesar Gerónimo Mayor gmayor2002@gmail.com 511 993637230 Lima PERU

Pregunta Se sabe que la relación C/N en recepción es 37.37dB. Calcular la distancia del enlace en Km.

Solución:

dBmPIRE

dBdBdBdBdBmPIRE

dBmC

dBmCdB

NCN

C

dBmN

xdBN

BdBNFTodBmN

3.59

322.135.032

55

)38.92(37.37

38.92

30)1018log(109)293log(106.228

30)log(10)()log(106.228)(6

=+−−−=

−=−−=

−=

−=++++−=

++++−=

Kmd

d

d

MHzddB

fdAo

dBAo

AoPIREIRL

dBmIRL

dBdBdBmIRL

Km

06.25

98.27)log(20

82.794.322.140)log(20

)9800log(20)log(204.322.140

)log(20)log(204.32

2.140

9.80

261.055

==

−−=++=

++=

=−=

−=−+−=

TX 1/2

1/2

32dBm

(0.5 dB)

(0.1 dB)

(1.2dB)

2 2

G = 26dB

G = 32dB

Rx2

d1 = ?

f2 = 9.8GHz

C/N = 37.37 dB B = 18MHz To = 293ºK NF = 9dB