homework3_solutions

homework3_solutions - ECE 220 - Homework 3 Solutions...

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ECE 220 - Homework 3 Solutions February 15, 2007 1 Problem 1 (15pts) Using Euler’s identity, we have cos( θ ) = 1 2 ( e + e - ) (1.1) So that 2cos(ˆ ω 0 )cos( ˆ ω 0 n + φ ) = 1 2 ( e j ˆ ω 0 + e - j ˆ ω 0 )( e j ˆ ω 0 n e + e - j ˆ ω 0 e - ) = 1 2 ( e j ˆ ω 0 ( n +1) e + e - j ˆ ω 0 ( n +1) e - + e j ˆ ω 0 ( n - 1) e + e - j ˆ ω 0 ( n - 1) e - ) = cos(ˆ ω 0 ( n + 1) + φ ) + cos(ˆ ω 0 ( n - 1) + φ ) (1.2) 2 Problem 2 (25pts) n 0 1 2 3 4 x [ n ] 2.4271 2.9002 2.9816 2.6603 1.9798 a cos(ˆ ω 0 n + φ ) a cos( φ ) a cos(ˆ ω 0 + φ ) a cos(2ˆ ω 0 + φ ) a cos( ˆ 3 ω 0 + φ ) a cos(4ˆ ω 0 + φ ) We first wish to find the value of ˆ ω 0 . Using the identity established in problem 1, we have 2 a cos(ˆ ω 0 )cos(ˆ ω 0 + φ ) = a cos(2ˆ ω 0 + φ ) + a cos( φ ) cos(ˆ ω 0 ) = a cos(2ˆ ω 0 + φ ) + a cos( φ ) 2 a cos(ˆ ω 0 + φ ) = 2 . 9816 + 2 . 4271 (2)(2 . 9002) = . 9325 (2.1) Then solving for ˆ ω 0 , from (2.1) ˆ ω 0 = cos - 1 ( . 9325) = . 3695 (2.2) 1
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Next we will solve for φ . a cos(ˆ ω 0 + φ ) a cos( φ ) = 2 . 9002 2 . 4721 = 1 . 1949 a cos(ˆ ω 0 + φ ) a cos( φ ) = cos(ˆ ω 0 )cos( φ ) - sin(ˆ ω 0 )sin( φ ) cos( φ ) = cos(ˆ ω 0 ) - sin(ˆ ω 0 )tan( φ ) = cos(ˆ ω 0 ) - p 1 - cos 2 ω 0 )tan( φ ) (2.3) Then solving the following equation for tan( φ ) cos(ˆ ω 0 ) - p 1 - cos 2 ω 0 )tan( φ ) = 1 . 1949 (2.4) Leading us to the following solution for φ tan( φ ) = . 9325 - 1 . 1949 1 - . 9325 2 = - . 7265 φ = tan - 1 ( - . 7265) = - . 6283 (2.5) Finally, in order to solve for a we have a = 2 . 4271 / cos( φ ) = 2 . 4271 /. 8090 = 3. In summary, we have that
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This note was uploaded on 04/17/2008 for the course ECE 2200 taught by Professor Johnson during the Spring '05 term at Cornell.

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homework3_solutions - ECE 220 - Homework 3 Solutions...

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