{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hwset5solution

# hwset5solution - ECE 453 Homework Assignment 5 Solution...

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

ECE 453: Homework Assignment 5 Solution 4-13 Solution At 10 MHz, the reactance of the capacitors is j 5 . The branch consisting of C 1 in parallel with R = 50 has Q = 50 / 5 = 10 , and transforms into j 5 in series with 50 / 10 2 = 0 . 5 , as shown in Figure a. jX L j 5 j 5 0 . 5 The two capacitors can be combined in series to yield a branch consisting of j 10 in series with 0 . 5 . This branch transforms into j 10 in parallel with 200 as shown in Figure b. jX L j 10 200 a. The inductor needs to have impedance + j 10 to resonate with the parallel capacitance. b. At resonance, the impedance Z = 200 . c. Q p = 200 / 10 = 20 . 4-20 Solution If a parallel LC circuit is used, then Q = Q p = 100 || 100 / ( ω o L ) . If a series LC circuit is used, then Q = Q s = ω o L/ (100 + 100) . In part a, the largest Q results when a parallel LC is used with the smallest possible inductance. Answer for part a: Parallel LC with L = 10 nH, C = 253 . 3 pF. In part b, the largest Q reults when a series LC is used with the largest possible inductance. Answer for part b: Series LC with L = 100 nH, C = 0 . 253 pF.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern