Wconv1

# Wconv1 - ECE 301 Detailed Solutions Set 3 Continuous...

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Unformatted text preview: ECE 301 Detailed Solutions - Set 3: Continuous Convolution I 1 3-1. A system has an impulse response given by h t = e 2 t u t . Let this system now have input given by x t = 5 δ t 1 3 δ t 4 . Compute the output of the system. Express your answer in three ways: (a) as a sum of scaled and delayed impulse responses, (b) as a piecewise solution, and (c) as a plot of the system output y t over 1 t 6. DETAILED SOLUTION: (a) Scaled and delayed impulse responses : The output y t is the convolution of input x t and impulse response h t . The convolution integral has been derived to be y t = Z ∞ τ = ∞ x τ h t τ d τ a The function x τ is obtain by replacing t with τ in the x t given in the problem statement: x τ = x t ¬ ¬ ¬ t τ = h 5 δ t 1 3 δ t 4 i ¬ ¬ ¬ t τ b The “ t τ ” notation in b signifies “ t replaced by τ ”. Evaluating b then gives x τ = 5 δ τ 1 3 δ τ 4 c Now substitute c into a : y t = Z ∞ τ = ∞ h 5 δ τ 1 3 δ τ 4 i h t τ d τ d Simplify d into two separate integrals: y t = 5 Z ∞ τ = ∞ δ τ 1 h t τ d τ 3 Z ∞ τ = ∞ δ τ 4 h t τ d τ e Next, recall that evaluating an integral containing an impulse is simple: the result is the “other function” evaluated at the location of the impulse. In each of the above integrals the “other function” is h t τ . Thus, equation e simplifies to y t = 5 h t τ ¬ ¬ ¬ τ = 1 3 h t τ ¬ ¬ ¬ τ = 4 f Evaluating f then simplifies to the desired “scaled and delayed impulse response” form y t = 5 h t 1 3 h t 4 g ECE 301 Detailed Solutions - Set 3: Continuous Convolution I 2 3-1 (cont.) (b) Piecewise solution : To obtain the piecewise solution for h t , rewrite the delayed impulse responses in g using the replacement notation: y t = 5 h t ¬ ¬ ¬ t t 1 3 h t ¬ ¬ ¬ t t 4 h Using the h t given in the problem statement, the replacements required in h are given by h t ¬ ¬ ¬ t t 1 = e 2 t u t ¬ ¬ ¬ t t 1 = e 2 t 1 u t 1 i h t ¬ ¬ ¬ t t 4 = e 2 t u t ¬ ¬ ¬ t t 4 = e 2 t 4 u t 4 j Substituting i and j into h then gives y t = 5 e 2 t 1 u t 1 3 e 2 t 4 u t 4 k To obtain the piecewise solution from k , evaluate k over the entire range of time ∞ t ∞ . For t < 1, both unit steps in k are zero, and therefore y t = 0 over this range. As an equation, y t = ; t < 1 l For the range 1 t < 4, the unit step u t 1 in k is one, but the unit step u t 4 in k is still zero. This gives the result y t =...
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## This note was uploaded on 01/25/2010 for the course ECE 301 taught by Professor Hoozturk during the Spring '08 term at N.C. State.

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Wconv1 - ECE 301 Detailed Solutions Set 3 Continuous...

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