Answers and Discussions for
Problem 2, part a
The $ b $-phase stator winding function is given.
$ \begin{equation} w_{bs}(\phi_{sm}) = 100\sin(2\phi_{sm}) - 10\sin(6\phi_{sm}) \end{equation} $
A symmetric 3-phase machine has a set of winding functions that may be expressed in the following continuous form using a third-harmonic term. The $ a $-phase will precede the $ b $-phase by $ \frac{2\pi}{3} \, \text{rad} $, and likewise the $ b $-phase will precede the $ c $-phase by the same phase shift.
$ \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} $
By matching the given equation to the prescribed form, it is determined that $ \frac{P}{2} = 2 $, $ W_{s1} = 100 $ turns, and $ W_{s3} = 10 $ turns. The $ a $-phase winding function may be written posthaste.
$ \begin{equation} \boxed{w_{as}(\phi_{sm}) = 100 \sin\left(2\phi_{sm} + \frac{2\pi}{3}\right) - 10\sin(6\phi_{sm})} \end{equation} $
