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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 convolutionexistence criteria. (a) Both signals are rightsided, 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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