This question is about the “reflect about the $y$ axis” method of finding $x(-t)$ for a given $x(t)$. How does this method work if the signal is odd to begin with? Let’s say it is a signal whose equation is not obvious ,as the one shown in figure:
Let’s say that we have were only shown the shape in the positive $x$ axis, how do I know that I have to take $-x(t)$ for the reflected version?
Thanks in advance.
reflection of an odd function about y axis
So I added an image which shows the reflected version($x(-t)$), of $x(t)$ about the $y$ axis.
Suppose you have a domain of definition $\Delta = (a,b)$ for the function $x(t)$. To find $x(-t)$ you must consider $-t \in (a,b)$. So it follows that $t \in (-b,-a)$. If you choose a random value for $t,$ let it be $u_0 = -t_0 \in (-b,-a)$ then $$x(-u_o) = x(t_0)$$ and $$t_0 \in \Delta = (a,b)$$ Thus the point $(-t_o, x(t_0))$ of the function $z(t) = x(-t)$ is the symmetric about the $y$-axis of the point $(t_0, x(t_0))$, which belongs to the original signal x(t).
Following this method for every point of $x(t)$, what you manage to accomplish is to reflect the function about the y-axis.
You understand of course all the above I suppose. Until now I have not mentioned anything about the oddness of the signal. It does not make a difference for the method of reflection about the $y$-axis.
However, as you can see, the result of finding $x(-t)$ leads a to the same signal as if we were to take $$x(-t) = - x(t)$$ i.e. the symmetric of the signal about the $x$-axis.
Note: in the photo I have written $t_0$ as $x_0$ and $x(t_0)$ as $f(x_0)$. Sorry for the inconsistency(from habit).