# (Trigonometric) Fourier series of sawtooth integral

I'm making some mistake in the calculation of the integral for calculating the coefficients of the trigonometric fourier series of a sawtooth function.

The function (one period, simply repeated afterwards) is described as: $$x(t) = 1 + t \qquad \text{for} \ T_0/2 \leq t < T0/2$$

Now the fourier series is given by (this is all theory):

$$\mathcal{F}(x(t)) = a_0 + \sum \limits_{n=1}^{n=\infty} a_n \cos(n \omega_0t) + \sum \limits_{n=1}^{n=\infty} b_n \sin(n \omega_0t)$$

Where $\omega_n = \frac{2\pi}{T_0}$

This results in the follow, integrals over a single period: $$a_0 = \frac{1}{T_0} \int_{T_0}x(t)dt \\ a_n = \frac{2}{T_0} \int_{T_0}x(t)\cos(n \omega_0t)dt \\ b_n = \frac{2}{T_0} \int_{T_0}x(t)\sin(n \omega_0t)dt$$

$a_0 = 1$, this is easy just looking at the graph and noticing the average. Now the difficulty, and probably the error is somewhere here - is in calculating the other coefficients. I'm using a standard definition that $\int t \cos(at)dt = \frac{\cos(t) + a\sin(t)}{a^2}$ and $\int t \sin(at)dt = \frac{\sin(t) - a\cos(t)}{a^2}$ \begin{align} a_n &= \frac{2}{T_0} \int_{-T_0/2}^{T0/2}(1+t)\cos(n \omega t)dt\\ &= \frac{2}{T_0} \left( \sin(n\pi) + \sin(n\pi)\right) + \frac{2}{T_0} \left( \frac{\cos(n\omega t) + n\omega\sin(n\omega t)}{n^2 \omega^2} \middle|_{t=-T_0/2}^{T_0/2} \right) \\ &= 0 + \frac{T_0}{2n^2\pi^2} \left(\cos(n\pi) + n\omega \sin(n\pi) - \cos(-n\pi) - n \omega \sin(-n\pi) \right) \\ &= 0 \end{align}

So there are no cosine components in the resulting fourier series. (As expected, since the function is a constants + an odd function). Now similar the $b_n$ can be calculated.

\begin{align} b_n &= \frac{2}{T_0} \int_{-T_0/2}^{T0/2}(1+t)\sin(n \omega t)dt\\ &= \frac{2}{T_0} \left( -\cos(n\pi) + \cos(-n\pi)\right) + \frac{2}{T_0} \left( \frac{\sin(n\omega t) - n\omega\cos(n\omega t)}{n^2 \omega^2} \middle|_{t=-T_0/2}^{T_0/2} \right) \\ &= 0 + \frac{T_0}{2n^2\pi^2} \left(\sin(n\pi) - n\omega \cos(n\pi) - \sin(-n\pi) + n \omega \cos(-n\pi) \right) \\ &= 0 + \frac{T_0}{2n^2\pi^2} \left(2\sin(n\pi) + 0) \right) \\ &= 0 \end{align}

Which is totally wrong, as that would mean the fourier series results in: $$X(f) = 1$$

The answer should have: $$b_n = \frac{T_0}{n\pi}(-1)^{n+1} \quad \forall M \in \mathbb{N}$$

So where did I lose a term?

The second line for $b_n$ is wrong (and also for $a_n$ btw). In your 'standard' definition there are $t$ factors missing. Correct this and redo the partial integration.
• Oh great, read the definition out of my calculus book wrong. $\int t \sin(at)dt = \frac{\sin(at) - at \cos(at)}{a^2}$ Aug 4, 2016 at 15:45