Solutions Communications Technology II WS 2011/2012
Solutions Communications Technology II WS 2011/2012 Solutions Communications Technology II WS 2011/2012
“Communications Technology II” – Solutions WS 2011/2012 16Solution to exercise 12 (mobrad02):a) BER for BPSK and AWGNP b = 1 2 erfc (√EbN 0)BER for BPSK and AWGN with instantaneous SNR(√ )P b (h) = 1 2 erfc |h| 2E bN 0BER for the given channel coefficentsP b (h 1 ) = 1 2 erfc (√P b (h 2 ) = 1 2 erfc (√)|0.5·exp(jπ/4)| 2E b= 1 erfc(1.12) = 0.0569N 0 2)|0.8·exp(jπ/6)| 2E b= 1 erfc(1.789) = 0.0057N 0 2P b (h 3 ) = 1 ( )2 erfc |0.1+0.2j| 2E b= 1 erfc(0.5) = 0.2398N 0 2Uniform Prob. of occurence: P 1 = P 2 = P 3 = 1/3, E b /N 0 = 7dB ≈ 5¯P b = 1 3 (P b(h 1 )+P b (h 2 )+P b (h 3 )) = 0.1008b)¯P b = 0.6P b (h 1 )+0.3P b (h 2 )+0.1P b (h 3 ) = 0.0598c) The strongest channel coefficient is h 2P b,min = P b (h 2 ) = 0.0057d) data rate R b = 1/T Baud = 1/(50 ns) = 20 Mbit/sTransmitted in 30% of all cases: ¯Rb = 0.3·R b = 6 Mbit/s
17 WS 2011/2012 “Communications Technology II” – SolutionsSolution to exercise 13 (mobrad03):a)τ 0 = l 0 /c 0⎛ ⎞h(0) = ρ 0 ·exp⎝−j2π f 0 ·τ } {{ } 0⎠f 0 ·τ 0 = f 0l 0c 0= 2000 → h(0) = 1l 1f 0 ·τ 1 = f 0 = 4666.¯6 → h(1) =ρ 1 ·exp(−j2π (4666+0.¯6)) = 0.5·[−0.5+j ·0.866]c 0=−0.25+j ·0.433h(t) = δ(t)+(−0.25+j ·0.433)δ(t−(τ 1 −τ 0 ))b)|H(jω)| = |1+0.5·exp(−j ·2π ·0.¯6)·exp(−j ·ω ·∆τ)|∆τ = τ 1 −τ 0 = l 1− l 0= 1400mc 0 c 0 3·10 8 m/s − 600m3·10 8 m/s = 2.¯6µsMaximum, if exponent is a multiple of 2πψ 1 −ω max ∆τ = n2π ⇒ f max =−2π ·0.¯6−n2π2π∆τ=−2π ·0.¯62π ·2.¯6µs − n2.¯6µsf max = −250 kHz+n·375 kHz,n = 0,±1,±2,±3·Minimum, if exponent is an odd multiple of πψ 1 −ω min ∆τ = nπ ⇒ f min = ψ 1 −nπ2π∆τ=−2π ·0.¯62π·2.¯6µs − n2·2.¯6µsf min = −250 kHz+n·187.5 kHz,n = ±1,±3,...c)f D = f 0 · vc 0·cos(α)f D0 = f Dmax = f 0 · 150Km/h3·10 8 m/s = 138.8889Hz( πf D1 = f Dmax ·cos = √4)1 ·f Dmax = 98.2093Hz 2r(t) = s(t−τ 0 )·exp(j2πf Dmax ·t)+s(t−τ 1 )·exp(j2π0.707·f Dmax ·t)
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17 <strong>WS</strong> <strong>2011</strong>/<strong>2012</strong> “<strong>Communications</strong> <strong>Technology</strong> <strong>II</strong>” – <strong>Solutions</strong>Solution to exercise 13 (mobrad03):a)τ 0 = l 0 /c 0⎛ ⎞h(0) = ρ 0 ·exp⎝−j2π f 0 ·τ } {{ } 0⎠f 0 ·τ 0 = f 0l 0c 0= 2000 → h(0) = 1l 1f 0 ·τ 1 = f 0 = 4666.¯6 → h(1) =ρ 1 ·exp(−j2π (4666+0.¯6)) = 0.5·[−0.5+j ·0.866]c 0=−0.25+j ·0.433h(t) = δ(t)+(−0.25+j ·0.433)δ(t−(τ 1 −τ 0 ))b)|H(jω)| = |1+0.5·exp(−j ·2π ·0.¯6)·exp(−j ·ω ·∆τ)|∆τ = τ 1 −τ 0 = l 1− l 0= 1400mc 0 c 0 3·10 8 m/s − 600m3·10 8 m/s = 2.¯6µsMaximum, if exponent is a multiple of 2πψ 1 −ω max ∆τ = n2π ⇒ f max =−2π ·0.¯6−n2π2π∆τ=−2π ·0.¯62π ·2.¯6µs − n2.¯6µsf max = −250 kHz+n·375 kHz,n = 0,±1,±2,±3·Minimum, if exponent is an odd multiple of πψ 1 −ω min ∆τ = nπ ⇒ f min = ψ 1 −nπ2π∆τ=−2π ·0.¯62π·2.¯6µs − n2·2.¯6µsf min = −250 kHz+n·187.5 kHz,n = ±1,±3,...c)f D = f 0 · vc 0·cos(α)f D0 = f Dmax = f 0 · 150Km/h3·10 8 m/s = 138.8889Hz( πf D1 = f Dmax ·cos = √4)1 ·f Dmax = 98.2093Hz 2r(t) = s(t−τ 0 )·exp(j2πf Dmax ·t)+s(t−τ 1 )·exp(j2π0.707·f Dmax ·t)
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