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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 : 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.
- Spring '09