ENC3250 Solutions pg 19

ENC3250 Solutions pg 19 - y1 p − p y1 1 x1 − y12 −...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: y1 p − p y1 1 x1 − y12 − px1 + p 2 2 2 tan α = y1 p = 1 1+ p y1 x1 y1 − 2 py1 + py1 x1 − 2 2 Because y1 = 2 px1 , this simplifies to tan α = (p/y1 ). 5.7 Antenna Isotropic λ = 30 cm Effective area (m2 ) 0.007 Gain 1 λ = 3 mm Effective area (m2 ) 7.2 × 10 –7 Gain 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 LdB = 20 log(fMHz ) + 120 +20 log (dkm) + 60 – 147.56 = 20 log(fMHz ) +20 log (dkm) + 32.44 5.9 We have P r = [(P t) (Gt) (Gr ) (c)2 ]/(4πfd)2 = [(1) (2) (2) (3 × 108 )2 ]/[(16) (π)2 (3 × 10 8 )2 (104 )2 ] = 0.76 × 10–9 W Source: [THUR00] 5.10 a. From Appendix 2A, PowerdBW = 10 log (PowerW) = 10 log (50) = 17 dBW PowerdBm = 10 log (PowermW) = 10 log (50,000) = 47 dBm b. Using Equation (5.2), LdB = 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 c LdB = 120 + 59.08 +80 – 147.56 =111.52; Pr,dBm = 47 – 111.52 = –64.52 dBm d The antenna gain results in an increase of 3 dB, so that Pr,dBm = –61.52 dBm Source: [RAPP02] 5.11 a. From Table 5.2, G = 7A/λ2 = 7Af2 /c 2 = (7×π ×(0.6) 2 ×(2× 109 )2 ]/(3×108 )2 = 351.85 GdB = 25.46 dB b. 0.1 W x 351.85 = 35.185 W c. Use LdB = 20 log (4π) + 20 log (d) + 20 log (f) – 20 log (c) – 10 log(G r ) – 10 log (G t) LdB = 21.98 + 87.6 + 186.02 – 169.54 – 25.46 – 25.46 = 75.14 dB The transmitter power, in dBm is 10 log (100) = 20. The available received signal power is 20 – 75.14 = –55.14 dBm -19- ...
View Full Document

Ask a homework question - tutors are online