presec9_solutions

# presec9_solutions - ECE 2200 Section IX Solutions(Week 10...

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Unformatted text preview: ECE 2200: Section IX Solutions (Week 10 / Spring 2009) 1. Solution 1. (a) y ( t ) = x ( t ) * h ( t ) = (2 u ( t )- 2 u ( t- 2)) * (3 u ( t- 5)- 3 u ( t- 1)). By the linearity of the convolution: y ( t ) = 6 u ( t ) * u ( t- 5)- 6 u ( t ) * u ( t- 1)- 6 u ( t- 2) u ( t- 5) + 6 u ( t- 2) u ( t- 1) . In Example 9- 9 of the textbook, convolution of two unit step function is given as: u ( t ) * u ( t ) = tu ( t ) . (1) Since convolution integral is time invariant ( i.e. , if y ( t ) = x ( t ) * h ( t ), then y ( t- t ) = x ( t- t ) * h ( t )), each term in y ( t ) are equal to: 6 u ( t ) * u ( t- 5) = 6( t- 5) u ( t- 5)- 6 u ( t ) * u ( t- 1) =- 6( t- 1) u ( t- 1)- 6 u ( t- 2) * u ( t- 5) =- 6( t- 7) u ( t- 7) 6 u ( t- 2) * u ( t- 1) = 6( t- 3) u ( t- 3) . Then, y ( t ) is equal to: y ( t ) = 6( t- 5) u ( t- 5)- 6( t- 1) u ( t- 1)- 6( t- 7) u ( t- 7) + 6( t- 3) u ( t- 3) . (b) We will have five different regions: t < 1 , 1 ≤ t < 3 , 3 ≤ t < 5 , 5 ≤ t < 7 , 7 ≤ t ....
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## This note was uploaded on 09/30/2009 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell.

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presec9_solutions - ECE 2200 Section IX Solutions(Week 10...

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