# Exercise about convolution

1. Feb 16, 2015

### bznm

1. The problem statement, all variables and given/known data
I have found the following exercise just solved, but I haven't understood some steps...

$$f(t)=\begin {cases}1, t\in [-1,1] \\ 0, t\not \in [-1,1]\end{cases}$$

We have to do the convolution $$f*f=\int^{+\infty}_{-\infty}f(\tau)f(\tau)f(t- \tau) d\tau=\int^{+1}_{-1}f(t- \tau) d\tau$$

2. The suggested solution is:

We know that $$-1\ge t-\tau \le 1$$ and $$-1\ge \tau \le 1$$
so we can write: $$f*f=\begin {cases}0, t<-2\\?, t \in[-2,0]\\?, > t\in[0,2]\\0, t>2 \end {cases}$$

1) If $$t<0 \rightarrow t+1>-1 \rightarrow t>-2$$

Thus $$f*f=\int_{-1}^{t+1}dt=t+2$$

2) while if $$t>0 \rightarrow t+1>-1 \rightarrow t>-2$$
Thus $$f*f=\int_{t-1}^{1}dt=2-t$$

3. My doubts:
1) I don't understand why the last two integrals are in "dt", while in the first integral the integration variable is $$\tau$$.. I have thought to a typing error, but I'm not sure
2) I haven't understood in which way the condition on t are chosen (i.e. $$t<0 \rightarrow t+1>-1, if t>0 \rightarrow t-1<1$$). They are "true", but I haven't understood why they are put in this way and not in another one.
3) I haven't understood in which way the "extremes" of integration are chosen. I suspect that there is a trick, but I'm not sure..
4) There is another way to carry on the correct result?

A lot of thanks!

2. Feb 16, 2015

### Ray Vickson

You need to sit down and think it through.

In the integral $F(t) = \int_{-1}^1 f(t-\tau) \, d \tau$, what would you get for $t < -2$? Don't "guess"; work it out logically. Next: what happens when $t$ is between -2 and 0? Again, don't "guess" and don't try to jump right away to the solution. Take your time and reason it out logically and carefully, taking as much time as you need. Draw pictures of the integration region in 2-dimensional $(\tau,t)$-space if you need to; of course, the integration is one-dimensional for any fixed $t$, but making a two-dimensional drawing can help to clarify what is happening.

Keep going like that for other values of $t$.

3. Feb 16, 2015

### Zondrina

Your convolution looks wrong, it should read:

$$(f * f)(t) = f(t) * f(t) = \int_{-\infty}^{\infty} f(t-x) f(x) dx$$

Looking at the definition of your function, it is equal to $1$ for $t \in [-1, 1]$, and $0$ everywhere else. So replacing $t$ by $x$ in the function, the integral reduces to:

$$(f * f)(t) = f(t) * f(t) = \int_{-\infty}^{\infty} f(t-x) f(x) dx = \int_{- 1}^{1} f(t-x) f(x) dx$$

So for $x \in [-1, 1]$ we know $f(x) = 1$ and so:

$$(f * f)(t) = f(t) * f(t) = \int_{-\infty}^{\infty} f(t-x) f(x) dx = \int_{- 1}^{1} f(t-x) f(x) dx = \int_{- 1}^{1} f(t-x) dx$$

Let $u = t - x$, then $x = t - u \Rightarrow dx = -du$.

The limits of integration also need to be changed due to the substitution. If $u = t - x$, then $x = 1 \Rightarrow u = t - 1$. Similarly, $x = -1 \Rightarrow u = t + 1$.

So the integral takes the form:

$$\int_{- 1}^{1} f(t-x) dx = - \int_{t + 1}^{t-1} f(u) du$$

I believe this will provide an alternate method to obtain the answer.

4. Feb 16, 2015

### bznm

If $t < -2$ or $t>2$, $F(t)=0$, while for -2<t<0 and 0<t<2, $f(t-\tau)=1$

The thing that I struggle to understand is the choice of the expression of the extremes of integration:
ok, $t-1<\tau<t+1$ and if t<0, the inferior extreme must be -1 (ok, I understand it!) but why the next step is writing $\int_{-1}^{t+1} d\tau$? The superior extreme must be 1!... The only reason that I have thought to write the integral in that form is that doing that I obtain a relation in t (and I woldn't obtain it if I do $\int_{-1}^{1} d\tau$...).

The reasoning is similar if t>0...

When I spoke of "trick" in the question, I was thinking to a "trick" like this... "If t<0, I have to express the inferior value in a numeric form, while the superior extreme had to be in parametric form. If t>0, I have to express the superior extreme in numeric form, while the inferior one has to be in parametric form. So I obtain a F(t) in which t appears..." Can it be a good trick? :)

A fast question: $dx = -du$ because t is fixed, isn't it?

But when I do the integration... t disappers! :(
Many thanks for your help, I thought too a change variable... :)

5. Feb 16, 2015

### Zondrina

We have $u$ as function of $x$.

$u = t - x \Rightarrow \frac{du}{dx} = -1 \Rightarrow du = - dx$.

Perhaps if you show the working we can see what went wrong.

6. Feb 16, 2015

I think these

are typos on your part: $-1 \ge \tau \le 1$ is not a proper inequality (same for the one involving $t - \tau$)

7. Feb 16, 2015

### bznm

oh, you are right!! :D I'm sorry, but I can't edit the original post! :(

Last edited: Feb 16, 2015
8. Feb 16, 2015

### bznm

Probably I have expressed not very well my doubt.. I asked clarification about $u = t - x \Rightarrow \frac{du}{dx} = -1 \Rightarrow du = - dx$ My doubt was: I obtain the relation that you wrote and not $du=dt-dx$ because the variable of integration is $\tau$ and t is like a constant... isn't it?

Then, about the sencence in my previuos post: "But when I do the integration... t disappers!": it was related to: $- \int_{t+1}^{t-1} du= -[t-1-t-1]=2$
So, if I have to obtain a relation in which t appears explicitally, I can't use this method... Do you agree?

9. Feb 16, 2015

### Zondrina

The variable of integration was $x$, so when the $u$ substitution was made, $u$ is a function of $x$ regardless of the other variables. So $t$ acts like a constant when differentiating.

You are assuming $f(u) = 1$ on the whole interval, rather you want to establish what is $f(u)$ for $u \in [t + 1, t - 1]$.

10. Feb 16, 2015

### bznm

I'm sorry, I have confused t with x, because in my sheets I have used "t" instead of your x :)

I have thought that the condition $f(u)=1$ was guaranteed by the extremes of integration t-1 and t+1... where am I wrong?

11. Feb 17, 2015

### Zondrina

Ask yourself, what is $f(t + 1)$ and $f(t - 1)$?

Looking at the original function, $f(t + 1)$ corresponds to the exact same function you have in your first post, just on a different interval.

Namely, $f(t + 1) = 1$ when $-1 \leq t + 1 \leq 1$. Re-arranging the interval a bit: $-2 \leq t \leq 0$.

Similarly, $f(t - 1) = 1$ when $-1 \leq t - 1 \leq 1$. Re-arranging the interval a bit: $0 \leq t \leq 2$.

Notice these are the intervals from your first post in the alternate solution, and $f(u) = 0, \forall t \notin [-2, 2]$.

What can you conclude? What should you do to this integral:

$$- \int_{t+1}^{t-1} f(u) \space du$$

12. Feb 17, 2015

### bznm

I'm really sorry, Zondrina, but I can't "see" the solution for this exercise.... I have tried to solve it since yesterday morning... My room is full of sheets full of this exercise, but every time that I arrive at that integral, I don't know how to proceed.. because I always obtain the result "2", and it is a wrong solution...

I can't understand how to collect the conditions about t and about f(u)....

If you show me the correct proceeding, I'll deeply meditate on it... I'm discouraged, and probably it blocks me even more...

So many thanks for your kind help!