# The Fourier Series, Amplitude and Phase Plot of a Saw Tooth Waveform

I am trying to find the amplitude and phase plots of the saw tooth waveform pictured.I have already computed the Fourier series of the waveform but I don't know how to derive the amplitude and phase plots from the sawtooth's Fourier series.

$$x(t) = \begin{cases} \frac{V}{T}t + V & \text{-T < t < 0} \\ \\ \\ \\ \frac{V}{T}t & \text{0 < t < T  } \\ \end{cases}$$

The Complex Fourier series of this waveform is

$$x(t) = \sum_{n=-\infty}^{n = \infty} C_n e^{jnw_0t}$$

$$x(t) = \sum_{n=\infty}^{n=\infty}\left[ \frac{jV}{nw_0T} e^{jnw_0t} + \left( \frac{jV}{2nw_0T} - \frac{V}{n^2w_0^2T^2} \right) e^{jnw_0(t + \frac{T}{2})} + \left( \frac{V}{n^2w_0^2T^2} + \frac{jV}{nw_0T} \right) e^{jnw_0(t - \frac{T}{2})} \right]$$

where

$$w_0 = 2 \pi f_0 = \frac{2 \pi}{T}$$

I know the amplitude plot is $|C_n|$ vs Frequency and the phase plot is $Arg(C_n)$ vs frequency but how would I do the get the amplitudes and phases from a fourier series such as this one?

Simplifying x(t) by replacing $w_0T = 2\pi$ and $e^{j\pi} = -1$ gives me
$$x(t) = \sum_{n=-\infty}^{n=\infty} \begin{cases} \frac{V}{2\pi n}\left( \frac{2 + 3(-1)^n}{2} \right) e^{j \left( \frac{2 \pi n t}{T} + \frac{\pi}{2} \right)} & \text{if n \ne 0} \\ \\ \\ \\ \\ C_0 = \frac{VT}{2} & \text{ if n = 0 } \end{cases}$$
Still not sure how I can interpret this, but it seems for $f = 0$, amplitude = $\frac{VT}{2}$ and phase = 0, and for f = $\frac{n}{T}$, amplitude = $\frac{5V}{4 \pi n}$ ( for even n ) and $\frac{V}{4 \pi n}$ (for odd n), with the phase = $\frac{2 \pi nt}{T} + \frac{\pi}{2}$ for both even and odd cases, is that right?
• Is this a homework problem? If so, please add a homework or self-study tag. Also, what is $w_0$ and how is it related, if at all, to $T$? Commented Oct 20, 2013 at 12:18
• Further hints: replace $w_0T$ by $2\pi$ and $w_0T/2$ by $\pi$ wherever you find them in your expression, and simplify wherever possible, e.g. $e^{jn\pi}$ equals what? Notice also that $x(t)$ is of period $T$ and cross-check your math by seeing if you get the same answer if you set $$C_n = \frac{1}{T}\int_0^T x(t)e^{j2\pi nt/T}\,dt$$ instead of whatever formula you used. Commented Oct 20, 2013 at 15:00
• I used the standard formula, which is the same formula as yours but with limits running from $- \frac{T}{2}$ to $\frac{T}{2}$, and then verified the $C_n$expression with matlab.It would probably have been much faster if I had used limits going from 0 to T because I wouldn't have have to perform two separate integrations. Commented Oct 20, 2013 at 16:41
• Are you at all concerned that $$\sum_{n=1}^\infty j\frac{V}{2\pi n} = \frac{jV}{2\pi}\sum_{n=1}^\infty j\frac{1}{n}$$ is a divergent sum? Commented Oct 20, 2013 at 20:59