Using the flip-and-drag method to draw a convolution

Question

So, we are asked to graphically convolve (using the flip-and-drag method) the following pair of signals: $$x_1(t) = \delta(t+1)-2\delta(t-1) \ \ \ \ \ \ \ \ \ \ \ \ \ x_2(t) = \begin{cases} -1-2t, & -\frac{1}{2}\leq t\leq 0 \\ -1+2t, & 0\leq t\leq \frac{1}{2} \\ 0 & |t|>\frac{1}{2}\end{cases}$$

where $\delta(t)$ is the dirac delta function.

I don't know how to draw using Latex, so I am going to try and describe you what I've drawn so far using words (as best as I can). So, I've successfully been able to draw $x_2(t)$ (which gives me an inverted triangle whose summits coordinates are : $\left(-\frac{1}{2}, 0\right), \left(\frac{1}{2}, 0 \right)$ and $\left(0, -1 \right)$).

My problem starts when I have to draw $x_1(t)$. After doing some research, I have found that we represent the dirac delta function using "arrows", but I still don't know how to draw $x_1(t)$ .

Also, if we assume I was able to draw it (i.e I would have a graph with arrows), how am I supposed to draw the convolution (using the flip and drag method)?

Answer 1

Convolution is a linear operator, so let me explain just how to convolve $x_2(t)$ with $\delta(t-t_0)$. The only thing you would have to do afterwards is to convolve the same signal $x_2(t)$ with the other Dirac delta, multiply the function you get by some amplitude, and then add both convolution results.

The Dirac delta has what is known as sifting property:

$$\int_{-\infty}^\infty f(t)\delta(t-t_0)\,\mathrm{d}t=f(t_0)$$

This means that if we multiply a function by an impulse located at $t=t_0$, the area under that differential along the $t$-axis corresponds to the value of the function at $t_0$.

To solve your convolution graphically, imagine the following. Leave the impulses where they are. Then flip your convoluting signal, $x_2(t)$, and take it to start at $-\infty$. Start sliding this function to the right. At first, the convolution (the integral) returns $0$ for all $t$ because the product of $x_2(t)$ and $\delta(t+1)$ is $0$, as they are not intersecting anywhere. But, at some point, they both come across each other. This will continue to happen as long as the sliding version of $x_2(t)$ takes some non-zero value in $t=t_0$. When the triangle goes through entirely, then the convolution will be zero for every other $t$ again, as they will not encounter again.

I'll put some (extremely rudimentary) drawings to make this easier to see.

The first picture corresponds to the cases where the triangle and the impulse don't take any non-zero values at common points. The second one, where the triangle and the impulse are both non-zero at some value (in this case, always at $t_0$). The third picture corresponds to the moment where the triangle has passed through entirely, and thus the convolution is $0$ again.

Try to think about it, and do it on your own. As a rule of thumb, convoluting with deltas is by far the easiest convolution that there can be. In general, if we have something like this (with $A$ being some constant that represents the amplitude of the impulse): $$g(t)=f(t)*A\delta(t-t_0)$$ Then $g(t)$ will just be $f(t)$ but centered at $t_0$ and multiplied by $A$. That means, you just shift $f(t)$ and invert it or scale it depending on the value of $A$.