Communication Systems Engineering Episode 1 Part 8 pdf

Lecture 10: Link Budget Analysis and Design Eytan Modiano AA Dept. Eytan Modiano Slide 1 Signal attenuation Channel P T Rx + n(t) P R L • The signal suffers an attenuation loss L – Received power P R = P T /L – Received SNR = E b /N 0 , E b = P R /R b • Antennas are used to compensate for attenuation loss – Capture as much of the signal as possible G T G R Tx P T L = (4πd/λ) 2 Rx P R = P T G T G R /L L = free space loss, d = distance between Tx and Rx Eytan Modiano Slide 2 λ = signal wavelength Tx Antenna Gains G R = A R 4π/λ 2 A R is the effective area of the antenna For Parabolic antenna A R = πηD 2 /4 η = illumination efficiency factor, 0.5 < η < 0.6 D = dish diameter => G R = η(πD/λ) 2 => PR = P T G T D 2 η/(4d) 2 Eytan Modiano Slide 3 θθ θ λλ λ θθ θ Antenna Beamwidth θ B • Beamwidth is a measure of the directivity of the antenna – Smaller beamwidth concentrated power along a smaller area • Free space loss assumes that power is radiated in all directions • An antenna with a smaller beamwidth concentrates the power hence yields a gain – For parabolic antenna, θ B ~ 70 λ /D – Gain (G T ) s proportional to ( θ B ) -2 – Hence a doubling of the diameter D increases gain by a factor of 4 Eytan Modiano Slide 4 Example (GEO Satellite) d = 36,000 km = 36,000,000 meters f c = 4 Ghz => λ = 0.075m P T = 100w, G T = 18 dB Receiver antenna is parabolic with D = 3 meters A) What is PR? B) Suppose (E b /N 0 ) req = 10 dB, what is the achievable data rate R b ? Eytan Modiano Slide 5 Repeaters Tx P T2 + A P T3 + A + Rx P N P N P N • A repeater simply amplifies the signal to make up for attenuation P R1 = P T /L, P T2 = P R1 A, P R2 = P T2 /L, … P N1 = P N , P N2 = P N1 A/L + P N , … Let A = L => P RK = P T /L, P NK = KP N P RK /P NK = P T /LKP N = 1/K (P R1 /P N1 ) Received SNR is reduced by a factor of K Eytan Modiano (E b /N 0 )k = 1/K (E b /N 0 ) Slide 6 Regenerators Tx P T P N + P N + Rx Tx P T Rx Tx P T • A regenerator demodulates, detects and retransmits the signal – Each segment has the same P R /P N and the same received E b /N 0 – P b = probability of error on a segment (independent between segments) – P b (overall) = 1 - P(no error) = 1 - (1-P b ) K ~ KP b • Now compare repeater to regenerator (e.g. PAM) P b = Q ( E N 2 0 b / ) For repeater : P b ( overall ) = Q ( E KN / 2 0 b ) For regenerator : P b ( overall ) = KQ ( E N / 2 0 b ) Eytan Modiano KQ ( E N b 2 0 / ) < Q ( E KN / b 2 0 ) Slide 7 Satellite example • Uplink received (E b /N 0 ) u = downlink received (E b /N 0 ) d = 10dB • PAM modulation P b = Q( E N2 0 b / ) • Repeater: Received (E b /N 0 ) u/d = 1/2 (E b /N 0 ) u = 10 dB - 3dB = 7dB – => Pb = 5x10 -4 from table 7.55 or 7.58 • Regenerator: P b (up) = P b (down) = 3x10 -6 – (from table with (E b /N 0 ) d = 10dB) – Hence P b (up/down) ~ 2 P b (up) ~ 6x3x10 -6 • Two orders of magnitude difference between repeaters and regeneration – Greater difference with more segments Eytan Modiano Slide 8 . attenuation P R1 = P T /L, P T2 = P R1 A, P R2 = P T2 /L, … P N1 = P N , P N2 = P N1 A/L + P N , … Let A = L => P RK = P T /L, P NK = KP N P RK /P NK = P T /LKP N = 1/ K (P R1 /P N1 ) Received. (E b /N 0 ) d = 10 dB • PAM modulation P b = Q( E N2 0 b / ) • Repeater: Received (E b /N 0 ) u/d = 1/ 2 (E b /N 0 ) u = 10 dB - 3dB = 7dB – => Pb = 5x10 -4 from table 7.55 or 7. 58 • Regenerator:. f c = 4 Ghz => λ = 0.075m P T = 10 0w, G T = 18 dB Receiver antenna is parabolic with D = 3 meters A) What is PR? B) Suppose (E b /N 0 ) req = 10 dB, what is the achievable data rate