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hw2slns

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

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Homework #2 Solutions

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7.16 Start with the Discrete-Time Fourier Transform of h [ n ]: h [ n ] = δ [ n ] - [ n - k 0 ] , where a = 1 4 H ( ω ) = 1 - ae - jωk 0 The DFT H [ k ] is the sampled version of H ( ω ) evaluated at ω = 2 π N k , where N = 4 k 0 for this problem, and k = 0 . . . 4 k 0 - 1: H [ k ] = 1 - ae - j 2 π 4 k 0 k 0 k = 1 - ae - j π 2 k , k = 0 . . . 4 k 0 - 1 Now consider G ( ω ) = 1 /H ( ω ), or more generally, G ( z ) = 1 /H ( z ) (since G ( ω ) = G ( z ) | z = e ): G ( z ) = 1 H ( z ) = 1 1 - az - k 0 = X l =0 a l z - lk 0 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 0 ˜ g [ n ] = X l =0 a l δ [ n - lk 0 ] = δ [ n ] + [ n - k 0 ] + a 2 δ [ n - 2 k 0 ] + . . . We are, however, interested in G [ k ], which can be expressed in terms of G ( ω ): G [ k ] = 1 H [ k ] = 1 H ( ω ) | ω = 2 π N k = G ( ω ) | ω = 2 π 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 = β g [ k 0 ] = ˜ g [ k 0 ] + ˜ g [ N + k 0 ] + ˜ g [2 N + k 0 ] + . . . = a + a 5 + a 9 + . . . = a 1 - a 4 = 64 255 = Similarly, g [2 k 0 ] = a 2 1 - a 4 = 16 255 = a 2 β g [3 k 0 ] = - a 3 1 - a 4 = 4 255 = a 3 β Note that for 0 n N - 1, when n is not an integer multiple of k 0 , g [ n ] = 0. This follows from ˜ g [ n ] only having values at multiples of k 0 , and g [ n ] being sums of periodic

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

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