HW 06_examples - 1 ME261 - Solved Examples for Homework #6...

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Unformatted text preview: 1 ME261 - Solved Examples for Homework #6 Problem 1: Show that DT harmonics form an orthogonal basis. Solution: We need to show that e j! o kn & e j! o mn = & ;k 6 = m some real number, k = m Let x [ n ] = e j! o kn and y [ n ] = e j! o mn ; ! o = 2 & N ) P = N & 1 P n =0 x [ n ] y [ n ] = N & 1 P n =0 e j! o kn e & j! o mn = N & 1 P n =0 e j! o ( k & m ) n = N P n =0 e j! o ( k & m ) n e j! o ( k & m ) N = N P n =0 e j! o ( k & m ) n e j! o N (notice when k = m , e j! o ( k & m ) = 1 ) Apply identity: N P n =0 & n = ( 1 & N +1 1 & for & 6 = 1 N + 1 for & = 1 Now, P = ( 1 & ( e j!o ( k & m ) ) N +1 1 & e j!o ( k & m ) e j! o ( k & m ) N = 0 ; k 6 = m N + 1 e j! o ( k & m ) N = N + 1 1 = N; k = m Aside: 1 & ( e j!o ( k & m ) ) N +1 1 & e j!o ( k & m ) e j! o ( k & m ) N = 1 & e j!o ( k & m )( N +1) 1 & e j!o ( k & m ) e j!o ( k & m ) N ( 1 & e j!o ( k & m ) ) 1 & e j!o ( k & m ) = = 1 & e j!o ( k & m )( N +1) & e j!o ( k & m ) N + e j!o ( k & m )( N +1) 1 & e j!o ( k & m ) = 1 & e j!o ( k & m ) N 1 & e j!o ( k & m ) = 1 & e j 2 & N ( k & m ) N 1 & e j!o ( k & m ) = 1 & e j 2 & ( k & m ) 1 & e j!o ( k & m ) = = 1 & 1 1 & e j!o ( k & m ) = 0 Problem 2: x ( t ) = 5 cos 2 (10) t + & 4 + 3 cos 2 (20) t + & 5 + 2 cos 2 (30) t + & 6 a) Compute 2 times undersampled, b) 4 times oversampled DT versions of the CT time. Determine if aliasing takes place and fold the aliased signal into the principal range. Indicate DFS of each sampled function. Solution: a) Compute 2 times undersampled (aliasing occurs) By observation, f o = 10; f b (bandlimit) = 3 f o = 30; (highest frequency = 30 Hz) S N = 2 f b = 60 Hz (critical sampling rate) For 2 times undersampling, S f = (60) = 2 = 30 Hz F 1 = 10 30 = 1 3 ; F 2 = 20 30 = 2 3 ! 1 3 ; F 3 = 30 30 = 1 ! ( into the principal range, i.e. folding ) x [ n ] = 5 cos 2 1 3 n + & 4 + 3 cos 2 1 3 n + & 5 + 2 cos 2 n + & 6 = 5 cos 2 1 3 n + & 4 + 3 cos 2 1 3 n & 5 + 2 cos 2 3 n + & 6 For DFS: x [ n ] = (2 : 5 e j & 4 e j 2 & 1 3 n + 2 : 5 e & j & 4 e & j 2 & 1 3 n ) + (1 : 5 e & j & 5 e j 2 & 1 3 n + 1 : 5 e j & 5 e & j 2 & 1 3 n ) + ( e j & 6 e j 2 & 3 n + e & j & 6 e & j 2 & 3 n ) = (2 : 5 e j & 4 + 1 : 5 e & j &...
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HW 06_examples - 1 ME261 - Solved Examples for Homework #6...

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