problem 4b

# problem 4b - ECE 101 Linear Systems Winter 2009 Problem Set...

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ECE 101 - Linear Systems, Winter 2009 Problem Set # 4 Solutions (send comments/questions to [email protected]) Problem 1 (5.21 (d,f,k) OW2) 5.21 (d) We are given x [ n ] = 2 n sin( πn/ 4) u [ - n ]. Using the Fourier Transform analysis equation obtains X ( e ) = X k = -∞ x [ n ] e - jωn = 0 X k = -∞ 2 n sin( πn/ 4) e - jωn = - X k =0 2 - n sin( πn/ 4) e jωn = - 1 2 j X k =0 2 - n e jπn/ 4 e jωn - e - jπn/ 4 e jωn · = - 1 2 j 1 1 - (1 / 2)exp[( ω + π/ 4) j ] - 1 1 - (1 / 2)exp[( ω - π/ 4) j ] = 4 j - 4 + 5 2cos ω - 3 2 j sin ω 5.21 (f) We are given x [ n ] = n - 3 n 3 0 otherwise The signal can be written as a series of impulse functions x [ n ] = - 3 δ [ n + 3] - 2 δ [ n + 2] - δ [ n + 1] + δ [ n - 1] + 2 δ [ n - 2] + 3 δ [ n - 3] Using the Fourier transform analysis equation (5.9), we obtain X ( e ) = - 3 e 3 - 2 e 2 - e + e - + 2 e - 2 + 3 e - 3 = - 2 j (sin ω + 2 sin2 ω + 3 sin3 ω ) 1

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5.21 (k) We are given x [ n ] = x 1 [ n ] x 2 [ n ], where x 1 [ n ] ± sin( πn/ 5) πn x 2 [ n ] cos ± 7 π 2 n Using Table (5.2) in OW, the Fourier Transform of x 1 [ n ] is X 1 ( e ) = 1 | ω | < π/ 5 0 π/ 5 ≤ | ω | < π Noting that x 2 [ x ] = 0 for odd n values, the function can be rewritten as x 2 [ n ] = cos( πn/ 2) = 1 2 e jπn/ 2 + e - jπn/ 2 · From Table (5.2) the Fourier transform of x 2 [ n ] is X 2 ( e ) = π [ δ ( ω - π/ 2) + δ ( ω + π/ 2)] Using the multiplication property in Table (5.1), we get X ( e ) = 1 2 π Z 2 π X 1 ( e
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## This note was uploaded on 02/24/2010 for the course ECE ECE25 taught by Professor Bill lin during the Spring '09 term at UCSD.

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problem 4b - ECE 101 Linear Systems Winter 2009 Problem Set...

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