Hw03_Sol_Su09

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online