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ENC3250 Solutions pg 19

ENC3250 Solutions pg 19 - y1 p p y1 1 x1 y12 px1 p 2 2 2...

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-19- tan α = y 1 x 1 p 2 p y 1 1 + y 1 x 1 p 2 p y 1 = y 1 2 px 1 + 1 2 p 2 x 1 y 1 1 2 py 1 + py 1 Because y 1 2 = 2 px 1 , this simplifies to tan α = ( p / y 1 ). 5.7 λ = 30 cm λ = 3 mm Antenna Effective area (m 2 ) Gain Effective area (m 2 ) Gain Isotropic 0.007 1 7.2 × 10 –7 1 Infinitesimal dipole or loop 0.011 1.5 1.1 × 10 –6 1.5 Half-wave dipole 0.012 1.64 1.2 × 10 –6 1.64 Horn 2.54 349 2.54 3.5 × 10 6 Parabolic 1.76 244 1.76 2.4 × 10 6 Turnstile 0.008 1.15 8.2 × 10 –7 1.15 5.8 L dB = 20 log( f MHz ) + 120 +20 log ( d km ) + 60 – 147.56 = 20 log( f MHz ) +20 log ( d km ) + 32.44 5.9 We have P r = [(P t ) (G t ) (G r ) (c) 2 ]/(4 π fd) 2 = [(1) (2) (2) (3 × 10 8 ) 2 ]/[(16) ( π ) 2 (3 × 10 8 ) 2 (10 4 ) 2 ] = 0.76 × 10 –9 W Source: [THUR00] 5.10 a. From Appendix 2A, Power dBW = 10 log (Power W ) = 10 log (50) = 17 dBW Power dBm = 10 log (Power mW ) = 10 log (50,000) = 47 dBm b. Using Equation (5.2), L dB = 20 log(900 × 10 6 ) +20 log (100) – 147.56 = 120 + 59.08 +40 – 147.56 = 71.52 Therefore, received power in dBm = 47 – 71.52 = –24.52 dBm
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