Tuto4_sol - EE3210 Tutorial 4 (Solution) 1. 2. x[ n] ∗ h[...

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Unformatted text preview: EE3210 Tutorial 4 (Solution) 1. 2. x[ n] ∗ h[ n] = 12δ [ n] + 10δ [ n − 1] + 2δ [ n − 2] (a) y1[n] = 2δ [n + 1] + 4δ [n] + 2δ [n − 1] + 2δ [n − 2] − 2δ [n − 4] (b) y 2 [n] = y1 [n + 2] (c) y 3 [n] = y 2 [n] The convolution sum is y[ n] = k = −∞ 3. ∑ x[k ]h[n − k ] = ∑ α k = −∞ ∞ ∞ k u[ k ]β n − k u[ n − k ] = β n k = −∞ ∑ (α / β ) ∞ k u[k ]u[ n − k ] . When n < 0, u[k] u[n-k] = 0 for all k, implying that y[n] = 0. When n ≥ 0, u[k] u[n-k] = 1 for 0 ≤ k ≤ n. Therefore, n ⎡1 − (α / β ) n +1 ⎤ β n +1 − α n +1 y[n] = β n ∑ (α / β ) k = β n ⎢ . ⎥= β −α k =0 ⎣ 1 − (α / β ) ⎦ Combining the above two cases, we have y[n] = β n +1 − α n+1 u[n] . β −α 4. We first observe that these two signals have regions of nonzero overlap, regardless of the value of t. When t – 3 ≤ 0, the product of x(τ) and h(t – τ) is nonzero for –∞ < τ < t – 3, and the convolution integral becomes y (t ) = ∫ t −3 −∞ e 2τ dτ = 1 2 ( t −3) . e 2 For t – 3 ≥ 0, the product of x(τ) and h(t – τ) is nonzero for –∞ < τ < 0, so that the convolution integral is y (t ) = ∫ e 2τ dτ = −∞ 0 1 . 2 5. We need to check whether h(t) is absolutely integrable. Note that e − (1− 2 j )t = e − t . e 2 jt = e − t . Therefore, ∫ ∞ −∞ h(t ) dt = ∫ e −t dt = − e −t 0 ∞ [ ∞ 0 = 1. Hence, the system is stable. ...
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This note was uploaded on 11/22/2009 for the course ELECTRONIC EE3210 taught by Professor Sungchiwan during the Spring '09 term at École Normale Supérieure.

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