Othogonality of Discrete Sine and Cosine Series
The specific case of interest is the inner product:
\[ \label{forceEq} f = p' \Lambda q \] where \(p\) and \(q\) are vectors defining the \(m^{th}\) and \(n^{th}\) harmonic of flux around a magnetic bearing rotor and
\[ \label{lambdax} \Lambda = \mbox{diag}(\cos(\Theta) ) \] or \[ \label{lambday} \Lambda = \mbox{diag}(\sin(\Theta)) \] are diagonal matrices that define the contributions of each pole to the force in a particular force direction where pole angle \(\Theta\) is defined as:
\[ \label{poleAngle} \Theta = 360^o \left( \frac{k}{N} \right)_{k=0 \ldots N-1} \] Inner product \(\ref{forceEq}\) can be re-written as an explicit summation with a single index, since \(\Lambda\) is diagonal:
\[ \label{appEq0} f = \sum_{k=0}^{N-1} Z_k \] In all, there are six cases to consider, spanning all combinations of cosine or sine distributions for \(\Lambda\), \(p\) and \(q\):
\[ \label{appEq1} Z_k= \cos \left(\frac{2 \pi k}{N} \right) \cos \left(\frac{2 \pi m k }{N} \right) \cos \left(\frac{2 \pi n k }{N} \right) \] \[ \label{appEq2} Z_k= \cos \left(\frac{2 \pi k}{N} \right) \cos \left(\frac{2 \pi m k }{N} \right) \sin \left(\frac{2 \pi n k }{N} \right) \] \[ \label{appEq3} Z_k= \cos \left(\frac{2 \pi k}{N} \right) \sin \left(\frac{2 \pi m k }{N} \right) \sin \left(\frac{2 \pi n k }{N} \right) \] \[ \label{appEq4} Z_k= \sin \left(\frac{2 \pi k}{N} \right) \cos \left(\frac{2 \pi m k }{N} \right) \cos \left(\frac{2 \pi n k }{N} \right) \] \[ \label{appEq5} Z_k= \sin \left(\frac{2 \pi k}{N} \right) \cos \left(\frac{2 \pi m k }{N} \right) \sin \left(\frac{2 \pi n k }{N} \right) \] \[ \label{appEq6} Z_k= \sin \left(\frac{2 \pi k}{N} \right) \sin \left(\frac{2 \pi m k }{N} \right) \sin \left(\frac{2 \pi n k }{N} \right) \]
The goal is to obtain a closed-form solution for \(\ref{appEq0}\) so that the series does not have to be explicitly evaluated for particular values of \(N\), \(m\) and \(n\). A closed-form solution can be obtained by following an approach similar to that used for proving orthogonality of Digital Fourier Transform sinusoids from [1]:
- Convert \(\ref{appEq0}\) to exponential form so that the series can be considered a geometric series with respect to index \(k\);
- Use the closed-form expression for the sum of a geometric series to write a closed-form result for \(\ref{appEq0}\)
The procedure will be demonstrated in detail for the first case. Only the result will be presented for the remaining cases because the algebraic manipulation involved in obtaining a closed-form solution is laborious. For the purposes of this analysis, \(N\) is assumed to be an even number.
For the first case, trigonometric angle sum identities [2] can be applied to convert \(\ref{appEq1}\) to:
\[ \label{appEq7} \begin{array}{ccl} Z_k & = & \cos \left(\frac{2 \pi k}{N} \right) \cos \left(\frac{2 \pi m k }{N} \right) \cos \left(\frac{2 \pi n k }{N} \right) \\ & = & \frac{1}{4} \cos \left(\frac{2 \pi k (m-n+1)}{N}\right)+\frac{1}{4} \cos \left(\frac{2 \pi k (-m+n+1)}{N}\right)+\frac{1}{4} \cos \left(\frac{2 \pi k (m+n-1)}{N}\right)+\frac{1}{4} \cos \left(\frac{2 \pi k (m+n+1)}{N}\right) \end{array} \] Then, Euler's Formula can be used to convert the cosines into exponential form:
\[ \label{appEq8} Z_k= \frac{1}{8} e^{-\frac{2 i \pi k (m-n+1)}{N}}+\frac{1}{8} e^{\frac{2 i \pi k (m-n+1)}{N}}+\frac{1}{8} e^{-\frac{2 i \pi k (-m+n+1)}{N}}+\frac{1}{8} e^{\frac{2 i \pi k (-m+n+1)}{N}}+\frac{1}{8} e^{-\frac{2 i \pi k (m+n-1)}{N}}+\frac{1}{8} e^{\frac{2 i \pi k (m+n-1)}{N}}+\frac{1}{8} e^{-\frac{2 i \pi k (m+n+1)}{N}}+\frac{1}{8} e^{\frac{2 i \pi k (m+n+1)}{N}}\] Now, \(\ref{appEq0}\) can be written as the sum of eight geometric series:
\[ \label{appEq9} \sum_{k=0}^{N-1} Z_k = \frac{1}{8} \left( \sum_{k=1}^{N-1} e^{-\frac{2 i \pi k (m-n+1)}{N}}+ \sum_{k=1}^{N-1}e^{\frac{2 i \pi k (m-n+1)}{N}}+ \sum_{k=1}^{N-1} e^{-\frac{2 i \pi k (-m+n+1)}{N}}+ \sum_{k=1}^{N-1}e^{\frac{2 i \pi k (-m+n+1)}{N}}+ \sum_{k=1}^{N-1} e^{-\frac{2 i \pi k (m+n-1)}{N}}+ \sum_{k=1}^{N-1} e^{\frac{2 i \pi k (m+n-1)}{N}}+ \sum_{k=1}^{N-1}e^{-\frac{2 i \pi k (m+n+1)}{N}}+ \sum_{k=1}^{N-1}e^{\frac{2 i \pi k (m+n+1)}{N}} \right)\]
Each geometric series can be summed using the closed-form solution for the sum of a geometric series from [1]:
\[\label{appEq10} \sum_{k=0}^{N-1} z^k = \frac{1-z^N}{1-z} \] to yield
\[ \label{appEq11} \sum_{k=0}^{N-1} Z_k = \frac{1}{8} \left( \frac{1-e^{-2 i \pi (m-n+1)}}{1-e^{-\frac{2 i \pi (m-n+1)}{N}}}+\frac{1-e^{2 i \pi (m-n+1)}}{1-e^{\frac{2 i \pi (m-n+1)}{N}}}+\frac{1-e^{-2 i \pi (-m+n+1)}}{1-e^{-\frac{2 i \pi (-m+n+1)}{N}}}+\frac{1-e^{2 i \pi (-m+n+1)}}{1-e^{\frac{2 i \pi (-m+n+1)}{N}}}+\frac{1-e^{-2 i \pi (m+n-1)}}{1-e^{-\frac{2 i \pi (m+n-1)}{N}}}+\frac{1-e^{2 i \pi (m+n-1)}}{1-e^{\frac{2 i \pi (m+n-1)}{N}}}+\frac{1-e^{-2 i \pi (m+n+1)}}{1-e^{-\frac{2 i \pi (m+n+1)}{N}}}+\frac{1-e^{2 i \pi (m+n+1)}}{1-e^{\frac{2 i \pi (m+n+1)}{N}}} \right)\] In the case where \(|m-n| \neq 1\), the numerator of each term in \(\ref{appEq11}\) is zero, leading to a result of zero for the complete sum. In the case where \(n=m+1\), substitution for \(n\) readily shows that the numerators of the last six terms are zero. However, if \(m+1\) is simply substituted for \(n\), both the numerator and denominator of the first two terms are zero, leading to an indeterminate solution. By applying L'Hospital's Rule [4] where differentiation of the numerator and denominator is with respect to \(n\), it can be shown that the result for each of the first two terms is \(N\), leading to a series sum of \(N/4\) for the case where \(|m-n|=1\) with \(m \neq \frac{N}{2}\) and \(N/2\) for the case where \(|m-n|=1\) with \(m = \frac{N}{2}\).
Evaluation of the series sum results for the other cases involves a similarly laborious procedure. The results for the other cases are presented below in Table 1. Note that these results can be readily confirmed by evaluating \(\ref{appEq0}\) for any particular choice of case, \(N\), \(m\), and \(n\).
| Case | \(|m-n| \neq 1\) | \(m-n = 1, \, m \neq \frac{N}{2} \) | \(m-n = 1, \, m = \frac{N}{2} \) | \(n-m = 1, \, n \neq \frac{N}{2} \) | \(n-m = 1, \, n = \frac{N}{2} \) |
| 1 | 0 | \(\frac{N}{4}\) | \(\frac{N}{2}\) | \(\frac{N}{4}\) | \(\frac{N}{2}\) |
| 2 | 0 | 0 | 0 | 0 | 0 |
| 3 | 0 | \(\frac{N}{4}\) | 0 | \(\frac{N}{4}\) | 0 |
| 4 | 0 | 0 | 0 | 0 | 0 |
| 5 | 0 | -\(\frac{N}{4}\) | -\(\frac{N}{2}\) | \(\frac{N}{4}\) | 0 |
| 6 | 0 | 0 | 0 | 0 | 0 |
Table 1: Series sum results for each case of sine/cosine combination assuming an even number of magnetic bearing poles.
Table 1 shows that for every case, the sum is zero if \(|m-n| \neq 1\). This is the orthogonality property need to assign even and odd harmonics as bias and control current spaces for the purposes of generalized bias current linearization.
References
[1] J. Smith, Mathematics of the Discrete Fourier Transform (DFT): with Audio Applications, 2nd ed., W3K Publishing, 2007.
[2] "List of trigonometric identities", Wikipedia, accessed 05Oct2019.
[3] "Euler's formula", Wikipedia, accessed 05Oct2019.
[4] "L'Hospital's rule", Wikipedia, accessed 05Oct2019.