I am asked to draw the phase of $x(t)=2i\sin(\pi t)e^{-i\pi t}$. The requested plot should look like:
When $t\in\mathbb{Z}$, the graph jumps from $-\pi/2$ to $\pi/2$. Both values mean that the function is 'fully' imaginary if $t\in\mathbb{Z}$, but $x(t)=0$. And why does the phase value decrease linearly between integers? Could you give me some hints? Thanks in advance!
From properties of phase we have that:
$$\angle x(t) = \angle 2i + \angle \sin(\pi t) + \angle e^{-i\pi t} = \frac{\pi}{2} + \pi k - \pi t$$
$\sin(\pi t)$ is always a real number, so its contribution to the phase will always be some integer multiple of $\pi$. Then using the convention that the phase is in the principal branch, i.e. $\angle x(t) \in (-\pi,\pi]$, look at the behavior of the $\pi k$ term. $\sin(\pi t)$ changes phase for every $t\in\mathbb{Z}$.
Now take the interval $(0,1)$ for example. On $(0,1)$, $-\frac{\pi}{2}\leq\frac{\pi}{2}-\pi t\leq \frac{\pi}{2}$ and $\sin(\pi t) > 0$, therefore $k$ must be an even integer. The only choice of even integer that keeps our phase in the principal branch is $k=0$. Playing with more cases, we get the following phase function as our final solution:
$$\angle x(t) = \frac{\pi}{2} - \pi t + \pi\lfloor t \rfloor = \frac{\pi}{2} - \pi \{t\}$$
where $\{t\}\equiv t - \lfloor t \rfloor$ is the fractional part of $t$.