# PS5_sol - EE341 Autumn 2007 Problem Set 5 SOLUTION Reading...

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Unformatted text preview: EE341, Autumn 2007 Problem Set 5 SOLUTION Reading for this problem set: Sections 12.1 - 12.3 of Signals, Systems, and Transforms. 1. Prove that: DT F{nx[n]} = j solution By definition, dX() d X() = n=- x[n]e-jn d d x[n]e-jn X() = d d n=- Because differentiation is linear, we can push the derivative inside the sum. Note that x[n] is constant w.r.t. . d X() = d = j d X() = d n=- x[n] n=- d -jn e d x[n](-jn)e-jn nx[n]e-jn = DT F{nx[n]} n=- 2. Compute the DTFT of the following signals: (a) x[n] = 2n u[-n] (b) x[n] = n sin(0 n)u[n] (c) x[n] = {-2, -1, 0, 1, 2} solution (next page) Copy from Autumn 2006 problem set 5, problem 2. 1 no problem here with ROC issues-unit circle in z plane would be in ROC for corresponding Z transform See text--"time reversal" property of DTFT (assume that values start at time n=0, unless otherwise indicated) You could rewrite, using Euler's rule-to get what might be considered a simpler form 3. Consider a discrete-time periodic function x[n] with DTFT: 2 X() = 4 The values of X0 ( 2k ) 4 are: 0, 2k 6, X0 ( )= 0, 4 6, Determine x[n]. solution Comparing X() with the general form of the DTFT of a discrete-time periodic function in Equation 12.23 of the book: 2 X() = N X0 ( k=- 2k 2k )( - ) 4 4 k k k k =0 =1 =2 =3 X0 ( k=- 2k 2k )( - ) N N We see N = 4, so x[n] has period 4, and there are 4 cases to consider. We calculate each one with the inverse transform: 1 x[n] = N Remember that ej = cos + j sin . n=0 x[0] = 1 2 6 (X0 (0) + X0 ( ) + X0 () + X0 ( )) 4 4 4 1 3 (0 + X0 ( ) + 0 + X0 ( )) = 4 2 2 j20 j20 1 = (6e 4 + 6e 4 ) 4 1 = (6 + 6) = 3 4 6 6j 1 2 2j (0 + X0 ( )e 4 + 0 + X0 ( )e 4 ) 4 4 4 j 3j 1 = (6e 2 + 6e 2 ) 4 6 6 = ((0 + j) + (0 - j)) = 0 = 0 4 4 3 N -1 X0 ( k=0 2k j2kn )e N N n=1 x[1] = n=2 x[2] = 2 4j 1 6 12j (0 + X0 ( )e 4 + 0 + X0 ( )e 4 ) 4 4 4 1 1 j 3j j (6e + 6e ) = 2(6e ) = 4 4 = 3(-1 + j 0) = -3 n=3 x[3] = 1 2 6j 6 18j (0 + X0 ( )e 4 + 0 + X0 ( )e 4 ) 4 4 4 3j j 1 = (6e 2 + 6e 2 ) 4 = x[1] = 0 Therefore x[n] cycles through the values {3, 0, -3, 0}. 4. Copy from Autumn 2006 problem set 5, problem 5. 4 Problem 4 Part (1): (2) ...
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