E-notebook
by 0 λ = 10 cm = 0.1m Gp = 1000 Gp = 6 (D/ λ)^2 D/ λ = √( Gp /6) D = √ (1000/6)*0.1 D …
by 0 D = 20 and r = 10 m, f= 6GHz k = 0.54 λ = 3*10^8/(6*10^9) = 0.05 Now, Antenna gain GP = 4π*k*A/ λ^2 = …
by 0 f = 10GHz λ = 3*10^8/(10*10^9) =0.03 and D = 10 m so, BWFN = 140 λ /D = 140*0.03/10 BWFN = 0.42O GP = …
by 0 8.9 dB = 10log(Pt /Pr ) Pt /Pr = 7.76 Pr/ Pt = 1/7.76 = 0.1288 Given, r = 0.45 m λ = 3*10^8/(10*10^9) = …
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) …