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Solutions Communications technology II WS 2009/2010 - Universität ...

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23 <strong>WS</strong> <strong>2009</strong>/<strong>2010</strong> “<strong>Communications</strong> <strong>technology</strong> <strong>II</strong>” – <strong>Solutions</strong>Solution to exercise 18 (<strong>2009</strong>-10-ofdm):a)b)γ 2 =T ( ) ( ) ( )s 1 1 1 1⇒ T g =T s + T g γ − 1 T 2 s =γ − 1 2 ∆f = 0, 794 − 1 110 kHzT g = 25, 9 µsR b = N · log 2(M)⇒ N = ⌈ R b · (T s + T g )⌉ = ⌈629.5⌉ = 630T s + T g log 2 (M)c) N fft = 1024, f a = N · ∆f = 1024 · 10 kHz = 10, 24 MHzd) Maximal date rate if all subcarriers are modulated:R b = N · log 2(M)T s + T g= 1024 · 2125.9 µs= 16, 27 MBit/se)∆n Pi< 1 ⇒ ∆n Pi < T s= 100 = 3, 86T s τ max T g 25, 9

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