ELEG3301 2015-2016 Second Term Assignment 2 1155047511 Li Man Hin 3 × 108 = 0.1224 m 2.45 × 109 Prad = 1 W = 30 dBm 5 θ = arctan = 26.565◦ 10 2 π sin θ cos 2 = 2.3884 GT = 2 × 1.64 × cos θ 1. λ = PR = Prad GT GR λ 4πr 2 = 1.8142 × 10−6 W = −27.4 dBm 2. r = 3 × 108 × 60 × 60 = 1.08 × 1012 m 3 × 108 = 0.1 m λ= 3 × 109 PT = 20 W GT = 10 dB = 10 2 πd GR = λ Pr,min = −160 dBm = 1 × 10−19 W 2 2 πd λ ⇒ GR = = 9.21 × 106 ⇒ d = 96.5 m Pr,min = PT GT GR 4πr λ 3. d = 0.75λ d ∴ Dn (θ) = sin 2π cos θ = sin2 (1.5π cos θ) λ 2 When Dn (θ) = 0, 1.5π cos θ = −π, 0, π 2 cos θ = 0, ± 3 ◦ θ = ±90 , ±48.19◦ For peaks, Dn0 (θ) = −1.5π sin(1.5π cos θ) cos(1.5π cos θ) sin θ = 0 sin(3π cos θ) = 0 or sin θ = 0 3π cos θ = −π, 0, π or θ = 0 1 θ = 0, ± ⇒ θ = ±90◦ (rej), 70.529◦ 3 1 θ −90◦ Dn (θ) 0 −70.529◦ 1 −48.19◦ 0 0◦ 1 48.19◦ 0 4. (a) f = 2.45 GHz ⇒ λ0 = 0.1224 m λ0 λ0 λg = √ = r 2 λ0 0.1224 L = √ = √ = 0.0306 m = 3.06 cm 2 r 2 4 0.4λg ≤ W ≤ 0.8λg ⇒ 2.45 cm ≤ W ≤ 4.90 cm 2 70.529◦ 1 90◦ 0 (b) Patch Antenna Dimensions 6dB Bandwidth = 2.465 − 2.420 = 0.045 GHz = 450 MHz 3