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by 0 GdB =75dB , f = 15GHz λ = 3*10^8/(15*10^9) = 0.02 m 75= 10logG G = 31622776.6 Now, Capture Area = Directivity(G)* λ^2/4π Ae = …
by 0 The effective area = πd^2*η = π*10^2*0.54 = 54 and Wavelength, λ = 3*10^8/(6*10^4) = 0.05 m Now, the gain of the reflector antenna is …
by 0 λ = c/f = 3*10^8/(3*10^9) =10 cm The gain Ap = 6D^2/ λ ^2 & Ap = 10log Ap 26 = 10log Ap Ap = …
by 0 GT = GR =20 dB d = 40 km f = 600 MHz then , (Pr /Pt ) = GT (dB)+GR (dB)-Ls (dB) ———————(a) where, …
by 0 (a) LOS range = √2r‘ (√ht +√hr )m = √(2*4/3*r)(√120 +√16) m = √(2*4/3*6370*103 )(√120 +√16) d = 61.61 Km (b) Power(p) =15 kw λ = 3*108 /(50*106 …
by 0 Light of sight range (LOS) is given by d= 4.12(√ht +√hr ) Km =4.12(√10 + √100) Km LOS (d) = 54.24 Km OR, LOS (d) …
by 0 μ = √(1-81N/f2 ) 0.75 = √(1-81N/(10*106 ) Nmax =5.4*1011 fc = 9√(Nmax ) =6.614*106 Hz and Dskip = 2h √((fmuf /fc)2 -1) =2*350√((12*106 /(6.614*106 …
by 0 Path loss = 32.45+20logfMHz +20logdkm =32.45+20log*6000+20log10500 Hence : Path loss = 188.43 dB