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homework7a-d_solution

homework7a-d_solution - μ> σ τ/T s where T s is the...

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ECE 5325/6325 Fall 2009: Homework 7(a-d) Solutions 7a: (i) For each subchannel, B/N bandwidth is used. (ii) For each subchannel R/N data rate is carried. Or, if QAM/PSK is used, the bit rate is log 2 M B/N 1+ α . (iii) The center frequency for the k th subchannel, for k = 1 , . . . , N is f 0 +( k - 1) B/N where f 0 is the center frequency of the first subchannel. In other words, each center frequency is spaced by B/N . Given RMS delay spread σ τ and bandwidth B , and B c , the coherence bandwith, is given as 5 τ (to pick one definition for the coherence bandwidth). From the handout, B N < B c implies B/N < 5 τ or equivalently, N > Bσ τ / 5. 7b: The parameter μ , that is, the length of the cyclic prefix in number of symbols, should be
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Unformatted text preview: μ > σ τ /T s , where T s is the symbol period. All this says is that the duration of the cyclic pre±x must be larger than the RMS delay spread, but it also must be an integer number of symbol periods. 7c: The peak to average ratio (PAR) of a signal is the maximum instantaneous power of the signal divided by the average power of the signal. For OFDM signals, it is equal to N , the number of subchannels. 7d: Following the results of section 12.6 in the handout, R = 48 subcarriers 3 / 4 bit 1 coded bit 2 coded bits 1 subcarrier symbol 1 subcarrier symbol 4 × 10-6 sec results in R = 18 Mbps....
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