SOL12

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Signals and Systems Fall 2012 Solution to Homework Assignment #12 1. (a) ( ) ( ) ( ) ( ) 2 1 2 2 3.5 s s s x t u t u t u t = (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 2 2 2 2 2 3 3 4 4 s s s s s s x t t u t t u t u t t u t t u t t u t = + + (c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 2 2 2 2 2 3 3 3 s s s s s s s x t u t u t t u t t u t u t u t t u t = + + 2. (a) Let a b c Λ Ξ Μ Ο = Μ Ο Μ Ο Ν Π v , 1 2 3 ! " # # # \$ % & & & , a b c ! " # # # \$ % & & & = a + 2 b + 3 c = 0 (1) 3 2 1 ! " # # # \$ % & & & , a b c ! " # # # \$ % & & & = 3 a + 2 b + c = 0 (2) (1) (2)=0 a c = (3) Given that a = c , from (3), we obtain 2 b a = . Therefore, any vector 1 2 1 a Ξ Λ Ο Μ = Ο Μ Μ Ο Ν Π v with 0 a is orthogonal to both 1 2 3 Ξ Λ Ο Μ Ο Μ Μ Ο Ν Π and 3 2 1 Ξ Λ Ο Μ Ο Μ Μ Ο Ν Π .

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(b) Let a b c Ξ Λ Ο Μ = Ο Μ Μ Ο Ν Π v 2 1 3 !
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• Fall '09
• MALIKELBULUK
• Vector Space, SIGNALS, #, ⎞, ⎟, 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 k, 1 0 0.2 0.4 0.6 0.8 1 k

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