Solutions Communications Technology II WS 2011/2012

23 WS 2011/2012 “Communications Technology II” – SolutionsSolution to exercise 18 (ofdm09):a)b)γ 2 = T ( ) ( ) ( )s 1 1 1 1 1⇒ T g =T s +T g γ −1 T 2 s =γ −1 2 ∆f = 0.794 −1 10kHzT g = 25.9µsR b = N ·ld(M) ⇒ N = ⌈ R b ·(T s +T g )⌉ = ⌈629.5⌉ = 630T s +T g ld(M)c) N FFT = 1024, f a = N FFT ·∆f = 1024·10kHz = 10.24MHzd) Maximal data rate, if all subcarriers are modulated:R b = N ·ld(M)T s +T g= 1024·2125.9µs = 16.27MBit/se)∆n PiT s< 1τ max⇒ ∆n Pi < T sT g= 10025.9 = 3.86