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HW 04_examples - 1 ME261 Solved Examples for Homework#4...

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1 ME261 - Solved Examples for Homework #4 Problem 1: (Discrete-Time Harmonics) Check for the periodicity of the following signals, and compute the common period N if periodic: (a) x [ n ] = cos ° n 2 ± ; (b) x [ n ] = sin ° 4 ± ° 2 cos ° 6 ± ; (c) x [ n ] = 4 ° 3 sin ° 7 4 ± ; (d) x [ n ] = cos ° 5 12 ± + cos ° 4 9 ± ; (e) x [ n ] = e j 0 : 3 ; (f) x [ n ] = 2 e j 0 : 3 + 3 e j 0 : 4 ; (g) x [ n ] = ( j ) n= 2 ; Solution: (a) x [ n ] = cos ° n 2 ± cos (2 °nF ) = cos ° n 2 ± 2 °F = 1 2 ) F = 1 4 ° Since F is not a ratio of integers, the signal is non-periodic. (b) x [ n ] = sin ° 4 ± ° 2 cos ° 6 ± 2 °F 1 = ° 4 ; 2 °F 2 = ° 6 ; F 1 = 1 8 = 1 N 1 ; F 2 = 1 12 = 1 N 2 ; ) both frequencies are a ratio of integers, therefore the signal is periodic. common period: N 1 = 8; N 2 = 12; LCM ( N 1 ; N 2 ) = N = 24 (c) x [ n ] = 4 ° 3 sin ° 7 4 ± 2 °F = 7 ° 4 ) F = 7 8 ) signal is periodic, common period: N = 8 (d) x [ n ] = cos ° 5 12 ± + cos ° 4 9 ± 2 °F 1 = 5 ° 12 ; 2 °F 2 = 4 ° 9 ; F 1 = 5 24 ; F 2 = 2 9 ; ) signal is periodic, N 1 = 24 , N 2 = 9 , LCM ( N 1 ; N 2 ) = N = 72 (e) x [ n ] = e j 0 : 3 x [ n ] = e j 0 : 3 °n = cos (0 : 3 ) + j sin (0 : 3 ) ; 2 °F = 0 : 3 ° ) F = 0 : 3 2 = 3 20 ) signal is periodic, N = 20 (f) x [ n ] = 2 e j 0 : 3 + 3 e j 0 : 4 x [ n ] = 2 cos (0 : 3 ) + 2 j sin (0 : 3 ) + 3 cos (0 : 4 ) + 3 j sin (0 : 4 ) F 1 = 3 20 ; F 2 = 1 5 ; ) signal is periodic, N 1 = 20 , N 2 = 5 , LCM ( N 1 ; N 2 ) = N = 20 (g) x [ n ] = ( j ) n= 2 ( j ) n= 2 = ° e
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