I have a problem where I have to find the fundamental period and frequency of the following rectified periodic function $$ x(t) = |\sin(\pi t)| $$ From my understanding I know that the fundamental angular frequency is $\omega = (2 \pi) / T$ where $T$ is the fundamental period. My first thought was that $\omega = \pi$ so $T = 2$. When I graph it though I see that $T = 1$.
I have little experience with rectified waves and was just wondering where I'm going wrong and/or what I should take into consideration?
The reason is simple: the question asks you to find the fundamental period. $T_2=2$ can be a period for $x(t)$, but among all possible $T$'s, $T_0=1$ is the smallest period you can use to "cut the graph" so as to obtain a periodic waveform. We can prove that $T_2=2$ is indeed a period: $$x(t+T_2)=x(t+2)=|\sin(\pi (t+2))|=|\sin(\pi t+2\pi)|=|\sin(\pi t)|=x(t)$$ and of course as I said (also from the graph): $$x(t+T_0)=x(t+1)=|\sin(\pi t+\pi)|=|-\sin(\pi t)|=|\sin(\pi t)|=x(t).$$ One approach to find this $T_0$ that you may want to consider:
Note that
$$x(t) = |\sin(\pi t)| =\begin{equation} \begin{cases} \sin(\pi t), & \text{if $\,\,\sin(\pi t)\ge0$}\\ -\sin(\pi t), & \text{if $\,\,\sin(\pi t)\lt0$} \end{cases} \end{equation},$$
or equivalently $$x(t) =\begin{equation} \begin{cases} \sin(\pi t), & \text{if $\,\,2n\le t\le2n+1$}\\ -\sin(\pi t), & \text{if $\,\,2n-1\lt t\lt 2n$} \end{cases} \end{equation}$$ where $n\in\mathbb Z$. Then for any $n$, pick any $t\in(2n-1,2n)$. (Notice the end-points are zero. You can make that an extra line if you want.) We know that $-\sin(\pi t)=\sin(\pi t)$ for that $t$. Therefore $1$ is the smallest integer.