Fourier series of the elementary waveforms

In general, given a repeating waveform $X[n]$, we can evaluate its Fourier series coefficients $A[k]$ by directly evaluating the Fourier transform:

$$A[k] = {1 \over N} {\cal FT}\{X[n]\}(k)$$

$$= {1 \over N} \left [ X[0] + {U^{-k}} X[1] + \cdots + {U^{-(n-1)k}} X[n-1] \right ]$$

but doing this directly for sawtooth and parabolic waves will require massive algebra (somewhat less if we were willing resort to differential calculus). Instead, we rely on properties of the Fourier transform to relate the transform of a signal $x[n]$ with its first difference, defined as $x[n] - x[n-1]$. The first difference of the parabolic wave will turn out to be a sawtooth, and that of a sawtooth will be simple enough to evaluate directly, and thus we'll get the desired Fourier series.

In general, to evaluate the strength of the $k$th harmonic, we'll make the assumption that $N$ is much larger than $k$, or equivalently, that $k/N$ is negligible.

We start from the Time Shift Formula for Fourier Transforms (page ) setting the time shift to one sample:

$${\cal FT}\{ x[n-1] \} = \left [ \cos(k \omega) - i \sin (k \omega) \right ] {\cal FT}\{ x[n] \}$$

$$\approx (1 + i \omega k) {\cal FT}\{ x[n] \}$$

Here we're using the assumption that, because $N$ is much larger than $k$, $\omega k = 2\pi k / N$ is much smaller than unity and we can make approximations:

$$\cos(\omega k) \approx 1 \; , \; \sin(\omega k) \approx \omega k$$

which are good to within a small error, on the order of $(k/N)^2$. Now we plug this result in to evaluate:

$${\cal FT}\{ x[n] - x[n-1] \} \approx i \omega k {\cal FT}\{ x[n] \}$$