{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw02fa13-sol - ECE 2026 Fall 2013 Solutions to Problem...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 2026 Fall 2013 Solutions to Problem Set #2 1) Convert the following to polar form: a. 𝑧𝑧 = 𝑒𝑒 ± 𝑒𝑒 ±² / ± 𝑒𝑒 ± ±² / ± = 𝑒𝑒 ± 𝑗𝑗 ( 𝑗𝑗 ) = 𝑗𝑗 2 𝑒𝑒 ± = 2 𝑒𝑒 ± 𝑒𝑒 ±² / ± where in several places we have used the equalities 𝑗𝑗 = 𝑒𝑒 ±² / ± and 𝑗𝑗 = 𝑒𝑒 ± ± ± / ± . b. 𝑧𝑧 = 1 + 𝑒𝑒 ± ± ± . Pull a factor of 𝑒𝑒 ± ± ± out of both terms: 𝑧𝑧 = 𝑒𝑒 ± ± ± ± + 𝑒𝑒 ± ± ± 𝑒𝑒 ± ± ± . Then use Euler’s identity to spot 2 times a cosine inside the parens: 𝑧𝑧 = 2 cos ( 4 𝜔𝜔 ) 𝑒𝑒 ± ± ± . c. Direct approach : Look at the sum and notice that it has three pairs that are negatives of each other. For example, 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ± ( ± ± ± ± / ± ) = 𝑒𝑒 ±² 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ± ± ± / ± . Each of these pairs thus sums to zero and the only term left is 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ±² / ± . Approach using the n’th roots of unity : Note that this is a sum of powers of 𝑒𝑒 ±² / ± , which is one of the 8 ± ± roots of unity. The geometric series has a closed form given by 𝑧𝑧 ± ± ± ± ± ± ± = ( 𝑧𝑧 ± 1 ) ( 𝑧𝑧 1 ) . When you put 𝑧𝑧 = 𝑒𝑒 ±² / ± in this sum, you get zero (verify this!). Note that one term, 𝑒𝑒 ± ± ± / ± , is missing from the series given in this problem. Thus, this series sums to 0 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ±² 𝑒𝑒 ±² / ± = 𝑒𝑒 ±² / ± .
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern