Homework_10_solution(1)

# Homework_10_solution(1) - 1 EE113 Homework 10 Problem 17.3...

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Unformatted text preview: 1 EE113 Homework 10 Problem 17.3 Let x ( n ) = δ ( n + 2) − δ ( n ) + δ ( n − 2) . (a) Determine its DTFT. (b) Sample the DTFT to obtain the 6-point DFT of x ( n ) . (c) Obtain the same 6-point DFT using (17.20). (d) Obtain the same 6-point DFT using (17.21). Does aliasing in time occur? Solution: (a) The DTFT of x ( n ) is X ( e jω ) = e j 2 ω − 1 + e − j 2 ω = 2 cos(2 ω ) − 1 (b) The 6-point DFT of x ( n ) is X ( k ) = X ( e jω ) | ω = 2 πk N = 2 cos parenleftbigg 4 πk N parenrightbigg − 1 where N = 6 . (c) The periodic version of the time-domain sequence is x p ( n ) = ∞ summationdisplay m = −∞ x ( n + m · N ) The 6-point DFT of x ( n ) by (17.20) is X ( k ) = N − 1 summationdisplay n =0 x p ( n ) e − j 2 πkn/N = N − 1 summationdisplay n =0 ∞ summationdisplay m = −∞ x ( n + m · N ) e − j 2 πkn/N = ∞ summationdisplay n ′ = −∞ x ( n ′ ) e − j 2 πkn ′ /N n ′ = n + m · N = X ( e jω ) | ω = 2 πk N = 2 cos parenleftbigg 4 πk N parenrightbigg − 1 (d) Note that, within the range ≤ n ≤ N − 1 , x p ( n ) = x ( n ) . Then, the same 6-point DFT by (17.21) is X ( k ) = N − 1 summationdisplay n =0 x ( n ) e − j 2 πkn/N = 2 cos parenleftbigg 4 πk N parenrightbigg − 1 No aliasing occurred. December 2, 2011 DRAFT 2 Problem 17.8 Find the inverse DFTs of the DFT sequences defined below over one period, ≤ k ≤ 5 : (a) X ( k ) = (1 , , − 1 , , 1 , 0) . (b) Y ( k ) = e jπk · X ( k ) . (c) Y ( k ) = e − j π 3 k · X ( k ) . (d) Y ( k ) = sin ( π 3 k ) · X ( k ) . In each case, plot | x ( n ) | and ∠ x ( n ) . Solution: (a) The IDFT of X ( k ) = (1 , , − 1 , , 1 , 0) is x ( n ) = 1 N N − 1 summationdisplay k =0 X ( k ) W − kn N = parenleftBigg 1 6 , 1 − j √ 3 6 , 1 + j √ 3 6 , 1 6 , 1 − j √ 3 6 , 1 + j √ 3 6 parenrightBigg The amplitude of x ( n ) is | x ( n ) | = parenleftbigg 1 6 , 1 3 , 1 3 , 1 6 , 1 3 , 1 3 parenrightbigg The angle of x ( n ) is ∠ x ( n ) = parenleftbigg , 5 π 3 , π 3 , , 5 π 3 , π 3 parenrightbigg (b) The IDFT of Y ( k ) = e jπk · X ( k ) = e − j 2 πk ·− 3 / 6 · X ( k ) , by Table 18.1, is y ( n ) = x [( n + 3) mod 6] = parenleftBigg 1 6 , 1 − j √ 3 6 , 1 + j √ 3 6 , 1 6 , 1 − j √ 3 6 , 1 + j √ 3 6 parenrightBigg The amplitude of y ( n ) is | y ( n ) | = parenleftbigg 1 6 , 1 3 , 1 3 , 1 6 , 1 3 , 1 3 parenrightbigg The angle of y ( n ) is ∠ y ( n ) = parenleftbigg , 5 π 3 , π 3 , , 5 π 3 , π 3 parenrightbigg (c) The IDFT of Y ( k ) = e − j π 3 k · X ( k ) = e − j 2 πk · 1 / 6 · X ( k ) , by Table 18.1, is y ( n ) = x [( n − 1) mod 6] = parenleftBigg 1 + j √ 3 6 , 1 6 , 1 − j √ 3 6 , 1 + j √ 3 6 , 1 6 , 1 − j √ 3 6 parenrightBigg The amplitude of y ( n ) is | y ( n ) | = parenleftbigg 1 3 , 1 6 , 1 3 , 1 3 , 1 6 , 1 3 parenrightbigg The angle of y ( n ) is ∠ y ( n ) = parenleftbigg π 3 , , 5 π 3 , π 3 , , 5 π 3 parenrightbigg (d) The IDFT of Y ( k ) = sin parenleftBig π...
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## This note was uploaded on 01/24/2012 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.

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Homework_10_solution(1) - 1 EE113 Homework 10 Problem 17.3...

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