I'm having problems understanding how to represent a periodic signal using a Fourier Series, and I'm really struggling to find examples on this that are similar to what I'm working with. For example, the following exercise.
I have this signal:
$\sum_{k=-∞}^∞ x(t-10k)$
And this x(t):
$x(t)= 2\Pi(t/2) + 2 \Pi((t-5)/2) $
Clearly the signal is periodic. At first I thought the fundamental period was 10, tough later I came to realize that given that both rect functions have the same area the fundamental period should be 5. I plotted the signal as a box with a height and width of 2, repeated every 5 t. Not sure if this is correct.
So T=5. Following the formulas:
$x = \sum_{k=-∞}^∞ c_k f(x), e^{(j2πkt)/T}$
$c_k = \dfrac{1}{T}\int_{0}^{T} e^{(j2πkt)/T} \mathrm{dt}$
And here I get lost. I have an idea of what I think I'm supposed to do: take a single period (so one box in this case), get the Fourier Transform of it and that should be my Ck. But I have no idea if this is correct or why is it correct.
Any kind of help would be appreciated.
Best Answer
Given the elementary signal $x(t)= 2\Pi(t/2) + 2 \Pi((t-5)/2)$, where $\Pi(t)$ is the rectangular pulse function.
The pulse train is built as $y(t)=\sum_{k=-\infty}^\infty x(t-10k)$, which can be rewritten as:
$$y(t)=2\sum_{k=-\infty}^\infty\left[\Pi\left(\frac{t-10k}{2}\right) + \Pi\left(\frac{t-10k-5}{2}\right)\right]$$
As you pointed out, the pulse train has fundamental period $T_0=5$. This basically means that the pulse train can be further simplified into:
$$y(t)=\sum_{k=-\infty}^\infty2\Pi\left(\frac{t-5k}{2}\right)$$
We will, therefore, use the elementary pulse centered at $t=0$ and with amplitude $2$, $f(t)=2\Pi\left(\frac{t}{2}\right)$ to perform our Fourier series expansion.
The Fourier series can be written as:
\begin{align} f(t)&=c_{0}+\sum_{n=-\infty}^{-1}\left[c_{n}e^{j\cdot n\omega_0 t}\right]+\sum_{n=1}^{\infty}\left[c_{n} e^{j n \omega_0 t}\right]\\ c_{0}&=\frac{1}{T_0}\int_{T_0}f(t)dt\quad\text{(average value)}\\ c_{n}&=\frac{1}{T_0}\int_{T_0}f(t)e^{-jn \omega_0 t}dt \end{align}
Where $\omega_0$ is the fundamental angular frequency of your signal: $\omega_0=2\pi f_0=\frac{2\pi}{T_0}=\frac{2\pi}{5}$
Let us first calculate the average value:
\begin{align} c_{0}&=\frac{1}{5}\int_{-2.5}^{2.5}2\Pi\left(\frac{t}{2}\right)dt\\ &=\frac{2}{5}\int_{-1}^{1}dt\\ &=\frac{4}{5} \end{align}
Now for the coefficients:
\begin{align} c_{n}&=\frac{1}{5}\int_{-2.5}^{2.5}2\Pi\left(\frac{t}{2}\right)e^{-jn \omega_0 t}dt\\ &=\frac{2}{5}\int_{-1}^{1}e^{-jn \omega_0 t}dt\\ &=\frac{2}{5}\frac{e^{jn \omega_0}-e^{-jn \omega_0}}{jn\omega_0}\\ &=\frac{2}{n\pi} \sin\left(\frac{2n\pi}{5}\right) \end{align}
Then you should just plug the coefficients back into the Fourier series representation and there you have it.
Hope this helps