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 182.5125 kHz21.5|H(jω)| 210.5−62.5 kHz0−200 −150 −100 −50 0 50 100 150 200f in kHz
19 WS 2011/2012 “Communications Technology II” – Solutions4 OFDMSolution to exercise 14 (ofdm03):a) Kernel symbol length: T s = 1∆f = 4msBandwidth: B = N ·∆f = 512kHzData rate: R = ld(M)·NT s+T g= N T = 20486ms = 341kBit/sb) FFT length: N FFT = 4096c)N FFT samples account for the interval with the length of T s , since exactly one FFT isused for the generation of the kernel symbol.Sample rate: f A = N FFTT s= 1024kHzSamples within guard interval: N g = N FFT·T sT g= 2048Transmitter power:N 0 = 2· N02 = 1.2·10−4 WsE b = E OFDMld(M)·N = 1.4Ws2048 = 6.836·10−4 WsE b= 6.836·10−4 Ws= 5.7 ≈ 7.55dBN 0 1.2·10 −4 Ws(γg 2 = 1− T )g= 2 T g +T s 3(√ )P b = 1 2 erfc Eb·γgN 2 0= 1 2 erfc(1.9488)= 3·10 −3P = E OFDM= 1.4WsT s +T g 6ms= 233.3W
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19 <strong>WS</strong> <strong>2011</strong>/<strong>2012</strong> “<strong>Communications</strong> <strong>Technology</strong> <strong>II</strong>” – <strong>Solutions</strong>4 OFDMSolution to exercise 14 (ofdm03):a) Kernel symbol length: T s = 1∆f = 4msBandwidth: B = N ·∆f = 512kHzData rate: R = ld(M)·NT s+T g= N T = 20486ms = 341kBit/sb) FFT length: N FFT = 4096c)N FFT samples account for the interval with the length of T s , since exactly one FFT isused for the generation of the kernel symbol.Sample rate: f A = N FFTT s= 1024kHzSamples within guard interval: N g = N FFT·T sT g= 2048Transmitter power:N 0 = 2· N02 = 1.2·10−4 WsE b = E OFDMld(M)·N = 1.4Ws2048 = 6.836·10−4 WsE b= 6.836·10−4 Ws= 5.7 ≈ 7.55dBN 0 1.2·10 −4 Ws(γg 2 = 1− T )g= 2 T g +T s 3(√ )P b = 1 2 erfc Eb·γgN 2 0= 1 2 erfc(1.9488)= 3·10 −3P = E OFDM= 1.4WsT s +T g 6ms= 233.3W
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