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SolutionsIV_1

# SolutionsIV_1 - ECE 3250 HOMEWORK IV SOLUTIONS Fall 2009...

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Unformatted text preview: ECE 3250 HOMEWORK IV SOLUTIONS Fall 2009 1. (a) By linearity, S ( x ) = S ( x + d ) = y + S ( d ). Now, h is an l 1 signal and d is an l ∞ signal, so, by Criterion 3, k S ( x )- y k ∞ = k S ( d ) k ∞ = k h * d k ∞ ≤ k h k 1 k d k ∞ . Since k d k ∞ = and k h k 1 = ∞ X n =0 | h ( n ) | = 1 / 2 1- 1 / 3 = 3 / 4 , it follows that k S ( x )- y k ∞ ≤ 3 4 . (b) The disturbance d specified by d ( n ) = for all n ∈ Z will work because S ( d )(0) = h * d (0) = ∞ X k =-∞ h ( k ) d (- k ) = ∞ X k =0 h ( k ) = 3 4 , which implies that k S ( d ) k ∞ ≥ (3 / 4) and therefore that k S ( d ) k ∞ = (3 / 4) . 2. This is practice with using convolution-existence criteria. (a) Both signals are right-sided, so by the second criterion (Criterion 5.2 in the mono- graph) their convolution exists. (b) None of the criteria applies so we have to try computing x 1 * x 2 ( t ) for every t ∈ R . x 1 * x 2 ( t ) = Z ∞-∞ x 1 ( t- τ ) x 2 ( τ ) dτ = Z ∞-∞ u ( t- τ ) e- 3 τ u (- τ ) dτ = Z-∞ u ( t- τ ) e- 3 τ dτ = ( R-∞ e- 3 τ dτ when t ≥ R t-∞ e- 3 τ dτ when t < . Neither integral on the last line exists, so x 1 * x 2 does not exist. The equality on the third line happens because u (- τ ) = 0 when τ < 0 and 1 when τ ≥ 0. The equality on the last line happens because u ( t- τ ) = 0 when τ > t and 1 when τ ≤ t . (c) x 1 is bounded (i.e. x 1 ∈ l ∞ ) and x 2 is absolutely integrable (i.e. x 2 ∈ l 1 ), as you can check easily. By the third criterion (Criterion 5.3 in the monograph),can check easily....
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SolutionsIV_1 - ECE 3250 HOMEWORK IV SOLUTIONS Fall 2009...

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