Power of signal $x(t) = 2\operatorname{rect}(\sin(2\pi t))$ Assuming that  $\operatorname{rect}(x)$ is the rectangular function:
$$ \operatorname{rect}(t) = \Pi(t) = \begin{cases} 0 &|t| >\frac{1}{2} \\ \frac{1}{2} & |t| = \frac{1}{2} \\ 1 & |t|< \frac{1}{2} \end{cases} $$
Find the power of the signal $x(t) = 2\operatorname{rect}(\sin(2\pi t))$
The power of a periodic signal $x(t)$ is given by the formula $$P = \frac{1}{T_0}\int_{T_0}{|(x(t))|^2dt}$$ where $T_0$ is the period, which in our case is $T_0 = 1$
Upon trying to use the formula I came across the problem that I don't know how to handle the $\operatorname{rect}^2(t)$ component given that in the case of $\operatorname{rect}(t)$ the Area can be easily calculated as $\frac{1}{a^2}, a \in \Re$. In this case due to the periodicity of $\sin(2\pi t)$ I tried to find the values of $t$ for which $|\sin(2\pi t)| \le 1/2\space$ so that I manually plug in the values and calculate the area, but I failed to do so. Mathematica outputs $\frac{4}{3}$ as a result. Any ideas on how to approach this one?
 A: First thing to note is that ${\rm rect}(t)^2$, at least for the purposes of integration, is equal to ${\rm rect}(t)$. This is trivially true for all $|t|\neq1/2$, and the value at the single points $t=\pm1/2$ has no effect on the value of the integral, as long as no dirac functions are involved. So our integral becomes:
$$P=\frac{1}{T_0}\int_{T_0}(2{\rm rect}(\sin(2\pi t)))^2\ {\rm d}t=
\frac{1}{T_0}\int_{T_0}4{\rm rect}(\sin(2\pi t))\ {\rm d}t$$
Now, the way I'm familiar with handling ${\rm rect}$ and unit step functions in integrals, is to change the bounds. But first, we need to find which values of $t$ over one period satisfy $|\sin(2\pi t)|<1/2$. First note that for one period of the function, it we get two intervals of $t$ that satisfy the inequality (one with $\sin(2\pi t)$ increasing, one decreasing). The endpoints of these intervals can be found by solving the equality:
$$\sin(2\pi t)=\frac{1}{2}\quad\Rightarrow\quad t=\frac{\sin^{-1}(1/2)}{2\pi}=\frac{\pi/6}{2\pi}=\frac{1}{12}$$
$$\sin(2\pi t)=-\frac{1}{2}\quad\Rightarrow\quad t=\frac{\sin^{-1}(-1/2)}{2\pi}=\frac{-\pi/6}{2\pi}=-\frac{1}{12}$$
Remembering that $y=\pi-\sin^{-1}(x)$ is also a solution to $\sin(y)=x$, we get the endpoints of the other interval:
$$\sin(2\pi t)=\frac{1}{2}\quad\Rightarrow\quad t=\frac{\pi-\sin^{-1}(1/2)}{2\pi}=\frac{5\pi/6}{2\pi}=\frac{5}{12}$$
$$\sin(2\pi t)=-\frac{1}{2}\quad\Rightarrow\quad t=\frac{\pi-\sin^{-1}(-1/2)}{2\pi}=\frac{7\pi/6}{2\pi}=\frac{7}{12}$$
Plugging in $T_0=1$ and choosing our period to be $[-1/4,3/4)$, we get:
$$P=
\frac{1}{T_0}\int_{T_0}4{\rm rect}(\sin(2\pi t))\ {\rm d}t=\frac{1}{1}\left(\int_{-1/12}^{1/12}4\ {\rm d}t+\int_{5/12}^{7/12}4\ {\rm d}t\right)\\
=\left(\frac{1}{12}-\frac{-1}{12}\right)4+\left(\frac{7}{12}-\frac{5}{12}\right)4\\
=\left(\frac{2}{12}+\frac{2}{12}\right)4\\
=\frac{4}{3}$$
