ECE Ph.D. Qualifying Exam

Power and Energy Devices and Systems (PE)

Question Set 1: Energy Conversion and Reference Frame Theory

August 2016



Problem 1

Problem 1 Text

Solution

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Problem 2

Problem 2 Text (Side A)
Problem 2 Text (Side B)

Solution

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Click here to view student answers and discussions
Click here to view student answers and discussions

Trigonometric Identities

  • Single-phase
    • $ \sin\left(\frac{\pi}{2} - x\right) = \cos x $
    • $ \sin\left(x + \pi\right) = -\sin x $
    • $ \cos\left(x + \pi\right) = -\cos x $
    • $ \cos\left(a \pm b\right) = \cos a \cos b \mp \sin a \sin b $
    • $ \sin\left(a \pm b\right) = \sin a \cos b \pm \cos a \sin b $
    • $ \cos\theta + \cos\phi = 2 \cos\left(\frac{\theta + \phi}{2}\right) \cos\left(\frac{\theta - \phi}{2}\right) $
    • $ \cos\theta - \cos\phi = -2 \sin\left(\frac{\theta + \phi}{2}\right) \sin\left(\frac{\theta - \phi}{2}\right) $
    • $ \sin\theta - \sin\phi = 2 \cos\left(\frac{\theta + \phi}{2}\right) \sin\left(\frac{\theta - \phi}{2}\right) $
    • $ 2 \sin a \cos b = \sin(a + b) + \sin(a - b) $
    • $ 2 \cos a \cos b = \cos(a + b) + \cos(a - b) $
    • $ -2 \sin a \sin b = \cos(a + b) - \cos(a - b) $
    • $ 2 \cos^2 x = \cos 2x + 1 $
    • $ 2 \sin^2 x = 1 - \cos 2x $
    • $ \int_{0}^{2\pi} \cos^2 x \, dx = \int_{0}^{2\pi} \sin^2 x \, dx = \pi $
  • Three-phase
    • $ \cos x + \cos\left(x - \frac{2\pi}{3}\right) + \cos\left(x + \frac{2\pi}{3}\right) = 0 $
    • $ \sin x + \sin\left(x - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) = 0 $
    • $ \cos^2 x + \cos^2\left(x - \frac{2\pi}{3}\right) + \cos^2\left(x + \frac{2\pi}{3}\right) = \frac{3}{2} $
    • $ \sin^2 x + \sin^2\left(x - \frac{2\pi}{3}\right) + \sin^2\left(x + \frac{2\pi}{3}\right) = \frac{3}{2} $
    • $ \sin x \cos x + \sin\left(x - \frac{2\pi}{3}\right) \cos\left(x - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) \cos\left(x + \frac{2\pi}{3}\right) = 0 $
    • $ \sin x \cos y + \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) $
    • $ \sin x \sin y + \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) $
    • $ \cos x \cos y + \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 x \cos y + \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) $
    • $ \sin x \sin y + \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) $
    • $ \cos x \cos y + \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) $

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Sean Hu, ECE PhD 2009