I have to do this for an upcoming exam, but cannot find anywhere (in the textbook or online) how to do this.
I only really need to know a couple points to plot it… when x = 0, and then the earliest crossing of the x-axis.
[Math] How to sketch a sinc function by hand
calculusfourier analysissignal processing
Related Solutions
Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:
$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega &=& \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\ &=& \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega - \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;. \end{eqnarray}$$
Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get
$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' + \int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$
which is (again using the sufficient decay at infinity)
$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$
on a contour that encloses the pole at the origin, and hence the value is $2\pi$.
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".
Best Answer
You know the value at zero, it goes to zero at $x = n \pi$ and in-between reaches the maximum and minimum at close to $x = (n+ \frac{1}{2}) \pi$ (when $n>0$ or $n<-1$) with the value at these points being $\frac{(-1)^n}{(n+ \frac{1}{2}) \pi}$. So knowing these you should be able to roughly sketch the function.