OFDMA_Part14

# OFDMA_Part14 - 2 + E x 1 2 + ... + E x 2 N 2 + E x 2 N 1 N...

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14 The PAPR of a discrete-time signal is given by [15]: PAPR = max n x [ n ] 2 E n x [ n ] 2 …(E9_1) Now consider the OFDM TX from Figure 2; in particular, consider the discrete-time symbols x n generated when the frequency-domain symbols are passed through the FFT -1 block (refer to equation E1). Let us calculate the PAPR for these time-domain symbols that will eventually become a part of the OFDM symbol. First let’s evaluate the denominator of the PAPR expression [15]: E n x [ n ] 2 = E x 0 + x 1 + ... + x N 2 + x N 1 2 N …(E9_2) Since the x n are time-domain symbols corresponding to I.I.D. frequency-domain symbols, x n are also I.I.D. and hence: E n x [ n ] 2 = E x
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Unformatted text preview: 2 + E x 1 2 + ... + E x 2 N 2 + E x 2 N 1 N (E9_3) Each symbol has mean a energy of 1 since the symbols are normalized (see equation E1), and hence: E n x [ n ] 2 = 1 (E9_4) Now, let us evaluate the numerator of the PAPR expression [15]: max n x [ n ] 2 = max x + x 1 + ... + x N 2 + x N 1 N 2 (E9_5) The maximum of the above expression occurs when all x n add up coherently, and hence: max n x [ n ] 2 = N N 2 = N (E9_6) Thus, the: PAPR = N where is the number of subcarriers used by the OFDM system. (E9_7)...
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## This note was uploaded on 09/27/2009 for the course ECE 399 taught by Professor Prof during the Spring '09 term at University of Texas at Austin.

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