final exam solution

final exam solution - EE321 - Spring 04 - Final Exam -...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
W c 12.5 = W c2 10 = W c2 0 2 i 2 14 i 2 + d = W c1 5 2 = W c1 0 1 i 1 5i 1 d = Trajectory 1. λ 2 i 1 4i 2 + = λ 1 1 3i 2 + = Problem 2 R l µ A = F ab R Φ = F ab l Φ µ A = B Φ A = Φ BA = Φ S B d = F ab Bl µ = H B µ = F ab Hl = F ab a b l H d = Problem 1 EE321 - Spring 04 - Final Exam - Solution
Background image of page 1

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

View Full DocumentRight Arrow Icon
i f 1000 15.4 = i f v dc v fsw r f i fbar L ff d 1 f sw := i fbar 7.325 = i fbar v fbar r f := v fbar 14.65 = v fbar v dc v fsw () d v fd 1d + := v fd 0.75 := f sw 10 10 3 := r f 2 := v fsw 1.5 := d 0.8 := L ff 20 10 3 := v dc 20 := Problem 3 The co-energy is a function of trajectory - therefore it is not conservative W c 14.5 = W c 11 2 9 + = W c 0 1 t 11t d 0 1 2t 9t d + = t will go from 0 to 1 i 2 = i 1 t = Tractory 2
Background image of page 2
i bs α sin β t () = i as cos β t = Now we have that W bs 5 4 sin 4 φ sm = W as 5 4 cos 4 φ sm = First we compute the winding functions.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/19/2012 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University.

Page1 / 6

final exam solution - EE321 - Spring 04 - Final Exam -...

This preview shows document pages 1 - 4. Sign up to view the full document.

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