hw2slns

hw2slns - Homework #2 Solutions 7.16 Start with the...

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Unformatted text preview: Homework #2 Solutions 7.16 Start with the Discrete-Time Fourier Transform of h [ n ]: h [ n ] = [ n ]- a [ n- k ] , where a = 1 4 H ( ) = 1- ae- jk The DFT H [ k ] is the sampled version of H ( ) evaluated at = 2 N k , where N = 4 k for this problem, and k = 0 ... 4 k- 1: H [ k ] = 1- ae- j 2 4 k k k = 1- ae- j 2 k ,k = 0 ... 4 k- 1 Now consider G ( ) = 1 /H ( ), or more generally, G ( z ) = 1 /H ( z ) (since G ( ) = G ( z ) | z = e j ): G ( z ) = 1 H ( z ) = 1 1- az- k = X l =0 a l z- lk If we then denote the impulse response of G ( z ) by g [ n ], we have G ( z ) = X n =- g [ n ] z- n = X l =0 a l z- lk g [ n ] = X l =0 a l [ n- lk ] = [ n ] + a [ n- k ] + a 2 [ n- 2 k ] + ... We are, however, interested in G [ k ], which can be expressed in terms of G ( ): G [ k ] = 1 H [ k ] = 1 H ( ) | = 2 N k = G ( ) | = 2 N k By the frequency sampling theorem (see class notes), the inverse DFT of G [ k ] is then given by g [ n ] = X l = g [ n + lN ] ! w [ n ] Here, as in class, w [ n ] = 1 ,n = 0 ...N- 1 and 0 everywhere else. Therefore, we can write g [0] = g [0] + g [ N ] + g [2 N ] + ... = 1 + a 4 + a 8 + ... = 1 1- a 4 = 256 255 = g [ k ] = g [ k ] + g [ N + k ] + g [2 N + k ] + ... = a + a 5 + a 9 + ... = a 1- a 4 = 64 255 = a Similarly, g [2 k ] = a 2 1- a 4 = 16 255 = a 2 g [3 k ] =- a 3 1- a 4 = 4 255 = a 3 Note that for 0...
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hw2slns - Homework #2 Solutions 7.16 Start with the...

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