Wconv2

# Wconv2 - ECE 301 Detailed Solutions - Set 4: Continuous...

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Unformatted text preview: ECE 301 Detailed Solutions - Set 4: Continuous Convolution II 1 4-2. Let a system have impulse response h t = e 10 t u t and input x t = e 20 t u t . The output can be written in the form y t = K h e at e bt i u t . Use the convolution integral to compute K , a , and b . DETAILED SOLUTION: For the causal h t and x t , the “reflect-and-shift” form of the convolution intergral is given by y t = Z τ = t τ = h τ x t τ d τ a For the causal h t and x t given in the problem statement, the functions h τ and x t τ are given by h τ = e 10 τ u τ ; x t τ = e 20 t τ u t τ b ; c For t > 0, the functions in b and c are shown in the plots below. Remember the “ t τ ” first reflects the “ x function”, and then shifts the leading edge to the right by t units. τ .367 1 .367 1 τ τ = e x( - ) t τ τ =t τ =t x( - ) t τ h( ) τ h( ) τ-10 τ = e τ u( )-20 (t-τ29 u(t-τ29 ECE 301 Detailed Solutions - Set 4: Continuous Convolution II 2 4-2. (cont.) You can see in the previous plots that the leading edge of the shifting function x t τ has beed shifted to the right to the value τ = t . Hence the product function, shown in the bottom plot, is non-zero only over τ t . Therefore, using b and c , the convolution integral in a becomes y t = Z τ = t τ = e 10 τ e 20 t τ d τ d Since the integral in d is over τ , the e 20 t may be moved outside the integral, giving y t = e 20 t Z τ = t τ = e 10 τ e 20 τ d τ = e 20 t Z τ = t τ = e 10 τ d τ e Evaluating the integral in e then gives y t = e 20 t ¡ 1 10 e 10 τ ¬ ¬ ¬ τ = t τ = = : 1 e 20 t h e 10 t 1 i f Equation f may be rewritten in the form desired in the problem statement : y t = : 1 h e 10 t e 20 t i f ECE 301 Detailed Solutions - Set 4: Continuous Convolution II 3 4-3. A continuous system having impulse response h t = 5 e 4 t u t has an input x t = e 2 t u t . Compute y : 5 , the output y t evaluated at t = 0.5 seconds. DETAILED SOLUTION: Note that you do not have to compute the function solution y t . You only have to compute y t at one specific value of time. For the causal h t and x t in this problem the causal convolution integral of equation (3) in the Notes can be used: y t = Z τ = t τ = h τ x t τ d τ = Z τ = t τ = 5 e 4 τ e 2 t τ d τ a Now evaluate a at the specific value t = : 5: y t = : 5 = Z τ = : 5 τ = 5 e 4 τ e 2 : 5 τ d τ b Note that the integral in b is over τ , giving y t = : 5 = 5 e 2 : 5 Z τ = : 5 τ = e 4 τ e 2 τ d τ c Now evaluate the integral in c : y t = : 5 = 1 : 839 ¢ 1 2 e 2 τ ¬ ¬ ¬ τ = τ = : 5 = : 920 h 1 e 1 i d Completing the computation in d then gives the desired answer: y : 1 = : 581 e ECE 301 Detailed Solutions - Set 4: Continuous Convolution II 4 4-5. Let a system have impulse response h t and input x t given by h t = 5 ; t 1 ; else ; x t = 3 ; t 1 ; else Compute the analytical form (math form) for the output...
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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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Wconv2 - ECE 301 Detailed Solutions - Set 4: Continuous...

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