homework7a-d_solution - > /T s , where T s is the symbol...

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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 ±rst 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 de±nition 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 pre±x in number of symbols, should be
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Unformatted text preview: > /T s , where T s is the symbol period. All this says is that the duration of the cyclic prex 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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This note was uploaded on 02/25/2011 for the course ECE 5325 taught by Professor Dr.nealpatwari during the Spring '10 term at University of Utah.

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