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ENC3250 Solutions pg 20

ENC3250 Solutions pg 20 - 5.12 From Equation 2.2 the ratio...

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-20- 5.12 From Equation 2.2, the ratio of transmitted power to received power is P t /P r = (4 π d/ λ ) 2 If we double the frequency, we halve λ , or if we double the distance, we double d, so the new ratio for either of these events is: P t /P r2 = (8 π d/ λ ) 2 Therefore: 10 log (P r /P r2 ) = 10 log (2 2 ) = 6 dB 5.13 By the Pythagorean theorem: d 2 + r 2 = ( r + h ) 2 Or, d 2 = 2 rh + h 2 . The h 2 term is negligible with respect to 2 rh , so we use d 2 = 2 rh . Then, d km = 2 r km h km = 2 r km h m /1000 = 2 × 6.37 × h m = 3.57 h m 5.14 For radio line of sight, we use d = 3.57 K h , with K = 4/3, we have 80 2 = (3.57) 2 × 1.33 × h . Solving for h, we get h = 378 m. 5.15 N = –228.6 dBW + 10 log T + 10 log B We have T = 273.15 + 50 = 323.15
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