hw4-sol - EEE391 Homework 4 Solutions: 1. 9 e j 10t5 2 x t...

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EEE391 Homework 4 Solutions: 1. x t = 3 2cos 2 1.3 t 2 − sin 15t 2  9 e j 10t 5 x t = 3 cos 2.6 t 4  cos 0 − sin 15t 2  9 e j 10t 5 x t = 3 e 0 1 2 e j4 e j2.6 t 1 2 e j4 e j − 2.6 t e 0 1 2j e j 2 e j15t 1 2j e j 2 e j − 15t 9 e j 10t 5 a) Looking at the frequencies it is seen that the highest frequency components are 1 2j e j 2 e j15t and 1 2j e j 2 e j − 15t therefore in order to have no aliasing after sampling, sampling frequency f s 2 15 2 samples per second. b) If T s = 0.25 sec then f s = 1 0.25 = 4 Hz . For 4 samples per second, any component with frequency higher than 2 Hz, there occurs aliasing. Therefore: 1 2j e j 2 e j15t and 1 2j e j 2 e j − 15t will be aliased. 2. x [ n ]= 25sin 2 0.02n 6  3cos 0.01n 7 x [ n ]= 25cos 2 0.02n 3  3cos 0.01n
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This note was uploaded on 06/18/2010 for the course COMPUTER S EEE 391 taught by Professor Billurbarshan during the Spring '10 term at Bilkent University.

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hw4-sol - EEE391 Homework 4 Solutions: 1. 9 e j 10t5 2 x t...

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