Krdytkyu

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: $$x(t+T_2)=x(t+2)=|sin(pi (t+2))|=|sin(pi t+2pi)|=|sin(pi t)|=x(t)$$ 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 $$x(t) =begin{equation}
begin{cases}
sin(pi t), & text{if $,,2nle tle2n+1$}\
-sin(pi t), & text{if $,,2n-1lt tlt 2n$}
end{cases}
end{equation}$$ where $ninmathbb Z$. Then for any $n$, pick any $tin(2n-1,2n)$. (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.