Revision as of 17:45, 20 February 2018 by Mhayashi (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


ECE Ph.D. Qualifying Exam

Power and Energy Devices and Systems (PE)

Question Set 1: Energy Conversion and Reference Frame Theory

August 2017



Problem 1

Problem 1 Text

Solution

Click here to view student answers and discussions

Problem 2

Problem 2 Text

Solution

Click here to view student answers and discussions

Problem 3

Problem 3 Text

Solution

Click here to view student answers and discussions

Trigonometric Identities

  • $ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) $
  • $ \cos\left(x\right) + \cos\left(x - \frac{2\pi}{3}\right) + \cos\left(x + \frac{2\pi}{3}\right) = 0 $
  • $ \sin\left(x\right) + \sin\left(x - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) = 0 $
  • $ \cos\left(x\right) \cos\left(y\right) + \cos\left(x - \frac{2\pi}{3}\right) \cos\left(y - \frac{2\pi}{3}\right) + \cos\left(x + \frac{2\pi}{3}\right) \cos\left(y + \frac{2\pi}{3}\right) = \frac{3}{2} \cos(x - y) $
  • $ \sin\left(x\right) \sin\left(y\right) + \sin\left(x - \frac{2\pi}{3}\right) \sin\left(y - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) \sin\left(y + \frac{2\pi}{3}\right) = \frac{3}{2} \cos(x - y) $
  • $ \sin\left(x\right) \cos\left(y\right) + \sin\left(x - \frac{2\pi}{3}\right) \cos\left(y - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) \cos\left(y + \frac{2\pi}{3}\right) = \frac{3}{2} \sin(x - y) $

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett