# Soln4 - EECS 216 Homework 4 Solutions 3.7 1 x(t = 2 2 cos(t 45 2 cos(3t 2 sin(4t 30(a x(t = n= X[n]ejn0 t 0 = cos(t 450 is periodic with T1 = 2 = 2

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3.7 x ( t ) = 2 + 1 2 cos( t + 45 ) + 2 cos(3 t ) - 2 sin(4 t + 30 ) (a) x ( t ) = + n = -∞ X [ n ] e jnω 0 t ω 0 =?? cos( t + 45 0 ) isperiodicwithT 1 = 2 π 1 = 2 π cos(3 t ) isperiodicwithT 2 = 2 π 3 sin(4 t + 30 o ) isperiodicwithT 3 = 2 π 4 = π/ 2 T = LCM ( T 1 ,T 2 ,T 3 ) = 2 π The period of x ( t ) = 2 π ω 0 = 2 π 2 π = 1 C 0 = 2 (since C 0 x ( t ) can be represented in complex exponential form using Euler’s identity, x ( t ) = 2 + 1 4 h e j ( ω 0 t +45 0 ) + e - j ( ω 0 t +45 0 ) i + h e j 3 ω 0 t + e - j 3 ω 0 t i + j h e j (4 ω 0 t +30 0 ) - e - j (4 ω 0 t +30 0 ) i whereω 0 = 1 x ( t ) = 2 + 1 4 e j 45 0 e 0 t + 1 4 e - j 45 0 e - 0 t + e j 3 ω 0 t + e - j 3 ω 0 t + je j 30 o e j 4 ω 0 t - je - j 30 o e - j 4 ω 0 t X [0] = 2 ,X [1] = 1 4 e j 45 ,X [ - 1] = X [1] * = 1 4 e - j 45 0 ,X [3] = 1 ,X [ - 3] = X [3] * = 1 ,X [4] = je j 30 0 ,X [ - 4] = X [4] * = - je - j 30

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## This note was uploaded on 04/03/2008 for the course EECS 216 taught by Professor Yagle during the Fall '08 term at University of Michigan.

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Soln4 - EECS 216 Homework 4 Solutions 3.7 1 x(t = 2 2 cos(t 45 2 cos(3t 2 sin(4t 30(a x(t = n= X[n]ejn0 t 0 = cos(t 450 is periodic with T1 = 2 = 2

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