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 Halfwave
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
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.
 Fall '11
 Dr.TimothyCerner

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