# Tuto1_sol - 3 The power of x t is 4 2 4 2 1 1 2 2 2 = = =...

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EE3210 Tutorial 1 (Solution) 1. = = = = otherwise 0 7 and 2 , 2 , 7 2 / 1 1 and 1 1 0 3 ]} [ { n n n n u Ev = = = = = otherwise 0 2 and 7 2 / 1 7 and 2 2 / 1 1 1 1 1 ]} [ { n n n n n u Od 2. (a) 7, (b) Not periodic, (c) 8 Note that in part (c), the fundamental period is NOT 4. We want to find the smallest integer N so that the following equation holds. ) ) ( 8 cos( ) 8 cos( 2 2 N n n + = π π In other words, we want to find N so that (2 nN + N 2 ) π /8 is an integer multiple of 2 π . The smallest integer that satisfies this condition is 8.
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Unformatted text preview: 3. The power of x ( t ) is 4 2 4 2 1 ) ( 1 2 2 2 = = = â« â« t dt dt t x T T 4. The energy of x [ n ] is equal to 1 + 4 + 4 = 9. 5. Let ) ( ) ( ) ( 2 1 t x t x t x = . If x 1 ( t ) and x 2 ( t ) are both even, then ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 t x t x t x t x t x t x = = â â = â , and x ( t ) is even....
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