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Unformatted text preview: ECE 3250 HOMEWORK V SOLUTIONS Fall 2009 1. In each case, S ( x ) = Conv h ( x ). (a) S ( x ) = Conv h ( x ), so S ( x )( t ) = Z  h ( t ) x ( ) d = Z  e 3( t ) u ( t ) e 5 u ( ) d = Z e 3( t ) u ( t ) e 5 d = e 3 t R t e 2 d if t if t < = (1 / 2)( e 3 t e 5 t ) if t if t < . The equality on the third line holds because of the u ( ) on the second line. The equality on the fourth line holds because of the u ( t ) on the third line. Note that we can write the final answer as S ( x )( t ) = (1 / 2)( e 3 t e 5 t ) u ( t ) for all t R . (b) S ( x ) = Conv h ( x ), so S ( x )( t ) = Z  h ( t ) x ( ) d = Z  e 3( t ) u ( t ) e 5   d = Z t e 3( t ) e 5   d = ( e 3 t R t e 8 d if t < e 3 t R e 8 d + e 3 t R t e 2 d if t = (1 / 8) e 5 t if t < (1 / 8) e 3 t + (1 / 2) e 3 t (1 / 2) e 5 t if t . The equality on the third line holds because of the u ( t ) on the second line. Note that we can write the final answer as S ( x )( t ) = ((5 / 8) e 3 t (1 / 2) e 5 t ) u ( t ) + (1 / 8) e 5 t u ( t ) for all t R . (c) Conv ( x ) = x , so Conv Shift 7 ( ) ( x ) = Shift 7 ( x ). Since the first term in h is Shift 7 of the impulse response of the system in (a), and since x is the same as in (a), the convolution of the first term in h with x is Shift 7 of the answer to (a). Accordingly, S ( x )( t ) = (1 / 2)( e 3( t 7) e 5( t 7) ) u ( t 7) + e 5( t 7) u ( t 7) for all t R = (1 / 2)( e 3( t 7) + e 5( t 7) ) u ( t 7) for all t R . 1 2 (d) S ( x ) = Conv h ( x ), so S ( x )( t ) = Z  h ( t ) x ( ) d = Z  e 3( t ) u ( t ) u ( ) d = Z e 3( t ) u ( t ) d = ( e 3 t R e 3 d if t e 3 t R t e 3 d if t < . = (1 / 3) e 3 t if t 1 / 3 if t < . The equality on the third line holds because of the u ( ) on the second line. The equality on the fourth line holds because of the u ( t ) on the third line. Note that we can write the final answer as (1 / 3) e 3 t u ( t ) + 1 / 3 u ( t ) for all t R . 2. (a) S ( u ) = Conv h ( u ), so S ( u )( t ) = Z  h ( t ) u ( ) d = Z  7( e 3( t ) e 5( t ) u ( t ) u ( ) d = Z 7( e 3( t ) e 5( t ) ) u ( t ) d = 7 e 3 t R t e 3 d 7 e 5 t R t e 5 d if t if t < = (7 / 3)(1 e 3 t ) (7 / 5)(1 e 5 t ) if t if t < The equality on the third line holds because of the u ( ) on the second line. The) on the second line....
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This note was uploaded on 03/26/2010 for the course ECE 3250 at Cornell University (Engineering School).
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