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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..

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From the definition of the Heaviside function, figure out the values of $u(t)$ for $t = -2, -1.5, -1, -0.5, 0, 0.5, 1.0, 1.5, 2.0$. Make a neat table with two columns labeled $t$ and $u(t)$. Next, add a column labeled $u(t+0.5)$. If you think about it, you should be able to figure most the entries in this column from the entries in the first two columns. For example, if $t = -1$, $u(t+0.5) = u(-1+0.5) = u(-0.5)$ should be there somewhere, right? Repeat for $u(t-0.5)$ and then add yet another column $u(t+0.5)-u(t-0.5)$. Now plot points $(t,u(t+0.5)-u(t-0.5))$ on rectangular axes $\ldots$. – Dilip Sarwate Sep 7 '12 at 1:08
up vote 2 down vote accepted

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".

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