Hw03_Sol_Su09

# Hw03_Sol_Su09 - ECE220 Homework #3 Solutions Solution to...

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ECE220 Homework #3 Solutions Solution to Problem 4.13 z 1 = 2 + j 3 z 2 = 4 - j 2 From the Matlab code below below, clear, close all; z1=2+3j; z2=4-2j; z3=z1+z2 z4=z1-z2 z5=z1*z2 z6=z1/z2 z7=z2/z1 z8=z1^2 z9=z1^(-2) z10=3*z1+2*z2 We can calculate z 3 to z 10 , and they are: z 3 = 6 + j ; z 4 = - 2 + 5 j ; z 5 = 14 + j 8; z 6 = 0 . 1 + j 0 . 8; z 7 = 0 . 15 - j 1 . 23; z 8 = - 5 + j 12; z 9 = - 0 . 03 - j 0 . 07; z 10 = 14 + j 5; Solution to Problem 4.17 3 , - 3 , - j, j, - 2 j, 3 j, - 3 - 4 j . 1 , e - jπ/ 3 , e jπ/ 4 , 5 e jπ/ 6 . 1

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Solution to Problem 4.25 Using Theorem 4.2 on page 147, we express cos( x ) = 1 2 ( e j x + e - j x ) and sin( x ) = 1 j 2 ( e j x - e - j x ) So, 2 cos( x ) sin( x ) = 2 . 1 2 ( e j x + e - j x ) . 1 j 2 ( e j x - e - j x ) = 1 j 2 ( e j x + j x - e j x - j x + e - j x + j x - e - j x - j x ) = 1 j 2 ( e j 2 x - e 0 + e 0 - e - j 2 x ) = 1 j 2 ( e j 2 x - e - j 2 x ) = sin(2 x ). Solution to Problem 4.31 We must solve the equation z 4 = - j 4 = - 1 = e . Using equation 4.35, where we have
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## Hw03_Sol_Su09 - ECE220 Homework #3 Solutions Solution to...

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