HW06f09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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Unformatted text preview: GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025 SOLUTIONS Problem Set # 6 Fall 2009 Problem 6.1 (a) If x [ n ] = 10cos(0 . 13 n + / 13), and f s = 100, then among infinitely many solutions we single out 2 f 1 f s = 0 . 13 and 2 f 2 f s = 0 . 13 + 2 These give respectively f 1 = . 13 2 100Hz and f 2 = 2 . 13 2 100Hz corresponding to the continuous time signals x 1 ( t ) = 10cos(13 t + / 13) and x 2 ( t ) = 10(213 t + / 13) . (b) The input corresponding to the given spectrum is x ( t ) = 26cos(2 200 t + / 4) + 14cos(2 500 t + 3 / 4) . Let f 1 = 200 Hz and f 2 = 500 Hz. The corresponding digital frequencies are b 1 = 2 f 1 f s = 2 200 800 = 2 and b 2 = 2 f 2 f s = 2 500 800 = 5 4 . This is not in the range (- , ). Noting that b 2- 2 = 2 (- 300) 800 , The second term in x ( t ) gets folded to 300 Hz, thus yielding the output signal y ( t ) = 26cos(2 (200) t + / 4) + 14cos(2 (300) t- 3 / 4) ....
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This note was uploaded on 11/24/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Institute of Technology.

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HW06f09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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