Sawtooth wave

First we apply this to the sawtooth wave $s[n]$. For $0 \le n < N$ we have:

$$s[n] - s[n-1] = -{1 \over N} + \left \{ \begin{array}{ll} {1} & {n = 0} \\ 0 & \mbox{otherwise} \end{array} \right .$$

Ignoring the DC offset of $-{1 \over N}$, this gives an impulse, zero everywhere except one sample per cycle. The summation in the Fourier transform only has one term, and we get:

$${\cal FT}\{ s[n] - s[n-1] \} (k) = 1 , \; k \neq 0, \; -N < k < N\$$

We then apply the difference formula backward to get:

$${\cal FT}\{ s[n] \} (k) \approx {1 \over {i \omega k}} = {{-iN} \over {2 \pi k}}$$

valid for integer values of $k$, small compared to $N$, but with $k \neq 0$ . (To get the second form of the expression we plugged in $\omega=2\pi/N$ and $1/i = -i$.)

This analysis doesn't give us the DC component ${\cal FT}\{ s[n] \}(0)$, because we would have had to divide by $k=0$. Instead, we can evaluate the DC term directly as the sum of all the points of the waveform: it's approximately zero by symmetry.

To get a Fourier series in terms of familiar real-valued sine and cosine functions, we combine corresponding terms for negative and positive values of $k$. The first harmonic ($k = \pm 1$) is:

$${1 \over N}\left [ {\cal FT}\{ s[n] \} (1) \cdot {U^n} + {\cal FT}\{ s[n] \} (-1) \cdot {U^{-n}} \right ]$$

$$\approx {{-i} \over {2 \pi}} \left [ U^n - U^{-n} \right ]$$

$$= {{\sin ( \omega n)} \over {\pi}}$$

and similarly the $k$th harmonic is

$${{\sin ( k \omega n)} \over {k \pi}}$$

so the entire Fourier series is:

$$s[n] \approx {1 \over \pi} \left [ {\sin ( \omega n )} + ... ...\over 2} + {{\sin ( 3 \omega n)} \over 3} + \cdots \right ]$$