hw8_sol - seen between two terminals of C E(as discussed...

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

View Full Document Right Arrow Icon
ECE 342 HW#8 Solutions (P9.11) See next page.
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
f L =f P1 +f P2 +f P3 =102.1 Hz Note that, when calculating f P2 , Re should go in parallel with the entire square bracket in equation (9.14), since is is in parallel from the viewpoint of the resistance
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: seen between two terminals of C E (as discussed below the equation 9.14). ' || 2.15 ||1 0.683 1 16.225 2 14.37 0.683 in in S H R R R k f MHz p k π = = = Ω ⇒ = = × × ( ) ( ) → R in = R i || r π =2.15k Ω...
View Full 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