(Q). Determine the gain, beamwidth, and capture area for a parabolic antenna with 10m diameter dish and dipole feed at 10GHz
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 Author: Rambinay
by 0 (Q). While measuring the gain of a horn antenna the gain oscillator was set for 10GHz frequency and attenuation inserted was found to be 8.9dB. Calculate the gain of horn, the distance between two horns was 45 cm.
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) = …
(Q). A parabolic dish provides a gain of 75dB at a frequency of 150GHz. Calculate the area of the antenna and its 3dB beamwidth.
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 = …
(Q). Paraboloidal reflector has a diameter D=20m and is designed to operate at f=6 GHz with an illumination efficiency η= 0.54. What is the antenna gain in decibels?
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 …
(Q). A parabolic antenna operating at 3GHz has a gain of 26dB. Calculate the diameter of the antenna and half power point beamwidth.
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 = …
(Q).Calculate the attenuation at a frequency of 600MHz when the distance between transmiting and receiving antenna is 40km and gains are 20 dB
by 0 GT = GR =20 dB d = 40 km f = 600 MHz then , (Pr /Pt ) = GT (dB)+GR (dB)-Ls (dB) ———————(a) where, …
(Q). A transmitting antenna mounted at a height of 120m radiates 15kw of power equally in all directions at a frequency of 50MHz.calculate (a) Max LOS range (b) Field strength at a receiving antenna mounted at a height of 16 m at a distance of 12Km
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 …
Q. Find the range of the LOS system when the receiving and transmitting antenna heights are 10m and 100m respectively. Take the effective earth radius into consideration.
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) …
2070 magh (Q). The reflection takes place at a height of 350km and the maximum density in the ionosphere corresponds to a 0.75 refractive index at 10MHz. What will be the range (assume the earth is flat) for which MUF is 12MHz.
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 …
2071 (Q). Calculate free space path loss for 6 GHz Frequency, If transmitter and receiver are 10,500 Km line of sight apart.
by 0 Path loss = 32.45+20logfMHz +20logdkm =32.45+20log*6000+20log10500 Hence : Path loss = 188.43 dB