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HW06f09_soln

# HW06f09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025 SOLUTIONS Problem Set # 6 Fall 2009 Problem 6.1 (a) If x [ n ] = 10 cos(0 . 13 πn + π/ 13), and f s = 100, then among infinitely many solutions we single out 2 πf 1 f s = 0 . 13 π and 2 πf 2 f s = 0 . 13 π + 2 π These give respectively f 1 = 0 . 13 2 × 100 Hz and f 2 = 2 . 13 2 × 100 Hz corresponding to the continuous time signals x 1 ( t ) = 10 cos(13 πt + π/ 13) and x 2 ( t ) = 10(213 πt + π/ 13) . (b) The input corresponding to the given spectrum is x ( t ) = 26 cos(2 π 200 t + π/ 4) + 14 cos(2 π 500 t + 3 π/ 4) . Let f 1 = 200 Hz and f 2 = 500 Hz. The corresponding digital frequencies are b ω 1 = 2 πf 1 f s = 2 π 200 800 = π 2 and b ω 2 = 2 π f 2 f s = 2 π 500 800 = 5 π 4 . This is not in the range ( - π, π ). Noting that b ω 2 - 2 π = 2 π ( - 300) 800 , The second term in x ( t ) gets folded to 300 Hz, thus yielding the output signal y ( t ) = 26 cos(2 π (200) t + π/ 4) + 14 cos(2 π (300) t - 3 π/ 4) . 1

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Problem 6.2 (a) The highest frequency in the input is 120 Hz. Since the original signal is recovered faithfully at the output, the sampling frequency is at least equal to the correspondning Nyquist frequency, 240 Hz.
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HW06f09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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