How to plot this signals or functions? i´m taking a course in signals and systems, and i really don´t know how to plot this signals
Consider the pulse signal written as $p(t)=u(t+0.5)-u(t-0.5).$ where $u(t)$ is the Heaviside step function, given that i need to plot or graph $p(t)$, now set $x(t)=p(t-1)$ and $y(t)=p(t+1)$, i need to plot both signals and find and plot the product $[x*y](t)$, i really one´t have any idea about plotting signals, and i don´t know if is more convenient to ask in the electrical engineering blog.
I really appreciate your lhelp..
 A: Think carefully about the definition of the step function
$$
u(t) = \begin{cases} 1 : t\geq 0 \\ 0 : t < 0\end{cases}
$$
so, for $u(t + 0.5)$, if $t + 0.5 > 0$ (i.e. $t > -0.5$), then $u(t + 0.5) = 1$. Writing this down as a piecewise definition:
$$
u(t + 0.5) = \begin{cases} 1 : t\geq -0.5 \\ 0 : t < -0.5\end{cases}
$$
Equivalently, you can just remember that the effect of "$+ \ 0.5$" will shift the original graph to the left $0.5$ units.
Similarly for the next function, it is shifted to the right, and reflected across the  horizontal axis:
$$
-u(t - 0.5) = \begin{cases} -1 : t\geq 0.5 \\ 0 : t < 0.5\end{cases}
$$
Next, we want to add these together. You could do this visually, by drawing the graphs of $u(t+0.5)$ and $-u(t-0.5)$, or try and go straight to the piecewise definition, then graph that. Specifically, we have three cases to look at:
1) $t < -0.5$: in this case, both functions are 0, so $p(t) = 0$.
2) $-0.5 \leq t < 0.5$: in this case, $u(t + 0.5) = 1$, but $-u(t - 0.5) = 0$, so $p(t) = 1$.
3) $0.5 \leq t$: here $u(t + 0.5) = 1$, and $-u(t - 0.5) = -1$, so $p(t) = 0$.
If you draw this, you'll see it is a step function, that steps up to 1 at t = -0.5, and back down to 0 at t = 0.5.
For x(t) and y(t), I suggest thinking about them as horizontal shifts of p(t).
For the convolution, I suggest starting by writing down the definition:
$$
[x*y](t) = \int_{-\infty}^\infty x(t - \tau)y(\tau) d \tau
$$
and thinking about this in terms of "reflecting x (across vertical axis), shifting x, multiplying by y, and then integrating". Since $x$ and $y$ are both step functions, a lot of the time, they won't "overlap" after the shifting, and so the integral will be zero. Try and determine for exactly what values of $t$ this occurs. For what value of $t$ do they overlap completely? What about the in-between cases?
Wikipedia also has an animation that I think might be helpful in developing a sense of what convolution "does".
