Answers and Discussions for

ECE Ph.D. Qualifying Exam PE-1 August 2014



Problem 2, part b

Recycled Information from part a

A symmetric 3-phase machine has winding functions expressed in the continuous form using a third-harmonic term given below with $ \frac{P}{2} = 2 $, $ W_{s1} = 100 $ turns, and $ W_{s3} = 10 $ turns.

$ \begin{equation} \begin{bmatrix} w_{as}(\phi_{sm}) \\ w_{bs}(\phi_{sm}) \\ w_{cs}(\phi_{sm}) \end{bmatrix} = W_{s1} \begin{bmatrix} \sin\left(\frac{P}{2} \phi_{sm} + \frac{2\pi}{3}\right) \\ \sin\left(\frac{P}{2} \phi_{sm} - 0\right) \\ \sin\left(\frac{P}{2} \phi_{sm} - \frac{2\pi}{3}\right) \end{bmatrix} - W_{s3} \begin{bmatrix} \sin\left(\frac{3P}{2} \phi_{sm}\right) \\ \sin\left(\frac{3P}{2} \phi_{sm}\right) \\ \sin\left(\frac{3P}{2} \phi_{sm}\right) \end{bmatrix} \end{equation} $

Conductor Density Function

A winding function may be calculated by integrating the continuous conductor density function as stated below. Because the winding function is anti-periodic with anti-period $ \frac{2\pi}{P} $, the initial value of the winding function is $ w_{bs}(0) = \frac{1}{2} \int_{\phi_{sm}' = 0}^\frac{2\pi}{P} n_{bs}(\phi_{sm}') \, d\phi_{sm}' $. By the First Fundamental Theorem of Calculus, the conductor density function may be found by differentiating the winding function.

$ \begin{align} w_{bs}(\phi_{sm}) &= -\int_{\phi_{sm}' = 0}^{\phi_{sm}} n_{bs}(\phi_{sm}') \, d\phi_{sm}' + w_{bs}(0) \\ n_{bs}(\phi_{sm}) &= -\frac{d}{d\phi_{sm}} w_{bs}(\phi_{sm}) \end{align} $

The generic, continuous form of the conductor density function is found from the prototypical winding function in vector form.

$ \begin{equation} \begin{bmatrix} w_{as}(\phi_{sm}) \\ w_{bs}(\phi_{sm}) \\ w_{cs}(\phi_{sm}) \end{bmatrix} = -\frac{P}{2} W_{s1} \begin{bmatrix} \cos\left(\frac{P}{2} \phi_{sm} + \frac{2\pi}{3}\right) \\ \cos\left(\frac{P}{2} \phi_{sm} - 0\right) \\ \cos\left(\frac{P}{2} \phi_{sm} - \frac{2\pi}{3}\right) \end{bmatrix} + \frac{3P}{2} W_{s3} \begin{bmatrix} \cos\left(\frac{3P}{2} \phi_{sm}\right) \\ \cos\left(\frac{3P}{2} \phi_{sm}\right) \\ \cos\left(\frac{3P}{2} \phi_{sm}\right) \end{bmatrix} \end{equation} $

The $ b $-phase conductor density function may be written posthaste by substituting values into the prescribed form.

$ \begin{equation} \boxed{n_{bs}(\phi_{sm}) = -200 \cos(2\phi_{sm}) + 60 \cos(6\phi_{sm})} \end{equation} $


Discussion



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