Convolution of functions

**jamesdt** · May 14th 2011, 03:28 PM

The convolution of two functions $f_1$ and $f_2$ is defined by
$(f_1 \ast f_2)(t) = \int_{-\infty}^\infty f_1(\tau)f_2(t-\tau) \;dx$
The way you have defined $f_1$ means that it is zero outside the interval $[0,1]$ . So, the product of $f_1$ and $f_2$ is also zero outside the same interval. Therefore it should be sufficient to integrate from 0 to 1.

However, you have not defined the function $g(x)$ , and your notation is very inconsistent. Some places you use the variable $t$ , other places $x$ . So, I'm not sure whether this answers your question.

**Mondreus** · May 14th 2011, 04:31 PM

I suggest you start out with drawing a picture of the two signals. If you mirror and timeshift $f_1$ with x, you end up with these limits ( $f_1$ has the width 1 which means the right end point is x after you mirror and shift it, while the left one is x-1):

$x < -\pi/2 : (f_1 * f_2)(x) = 0$
$-\pi/2 \leq x < 1 - \pi/2: (f_1 * f_2)(x) = \int_{-\pi/2}^{x}f_1(x-\tau)f_2(\tau)d\tau$
$1 - \pi/2 \leq x < \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{x}f_1(x-\tau)f_2(\tau)d\tau$
$\pi/2 \leq x < 1 + \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{\pi/2}f_1(x-\tau)f_2(\tau)d\tau$
$x \geq 1+\pi/2: (f_1 * f_2)(x) = 0$

**token22** · May 15th 2011, 04:27 AM

It is strange, but first and third integral divergent. Could it be right?

**Mondreus** · May 15th 2011, 04:30 AM

Probably not... Either way we can't tell unless you post the complete definition of $f_2$ .

**token22** · May 15th 2011, 04:49 AM

I wrote in first topic, how $f_2$ looks like: for me it's known only:
$f_2(x)$ is tg(x) when x is in [-pi/2 , pi/2] otherwise it is $f_2(x)$ is 0
That's all i know about $f_2$

**Mondreus** · May 15th 2011, 04:57 AM

How exactly did you determine that the first and third integrals are divergent without knowing what $tg(x)$ is? Why is there a t in there when it's a function of x?

**token22** · May 15th 2011, 05:04 AM

Thing with 't' is actually only my attempt to solve it, thus:
for t <-pi/2, -pi/2 + 1> X=<-pi/2, t>
for t <-pi/2 + 1, pi/2> X=<t-1, t>
for t <pi/2, pi/2 + 1> X=<t-1, pi/2>
it shoud be the same as you wrote, but with 't'. Then I tried to solve integral:
$\int f_1(t-y)f_2(y) dy$ with appropriate limits. And it gave me really strange conclusion (for quick calculation i used wolfram)

EDIT: tg(x) is tangens function

**Mondreus** · May 15th 2011, 05:26 AM

I don't know where you got the problem from, but it's poorly written to say the least. I can help you with the last convolution though.

$x\geq 2: (f*f)(x)=\int_1^{x-1}d\tau$

Again, I suggest you draw the two signals. The definition of f here means that it's constantly 1 for $[1, \infty)$ . Mirroring it and translating it with x gives the right endpoint the coordinate x-1.

**token22** · May 15th 2011, 05:34 AM

Alright then i understand this second convolution, but i have problem with first.
By the way, function tg(x) is simply tangens function (i forgot that somebody write it as tan(x))
I got
f1*f2(t) = {-infinity for t in <-pi/2, 1 - pi/2>
ln|cos(t-1)| - ln|cos(t)| for t in <1 - pi/2, pi/2>
infinity for t in <pi/2, pi/2 + 1>
0 otherwise

but i am not sure about infinity

**Mondreus** · May 15th 2011, 07:42 AM

$\tan x$ is not convergent on $[-\pi/2, \pi/2]$ , so $(f_1*f_2)(x)$ is also not convergent.

**token22** · May 15th 2011, 08:13 AM

Hm but second integral is like: ln|cos(x-1)| - ln|cos(x)| does it mean that f1*f2(x) is not convergent in every possible case?

edit: i calculate those integrals over and over again but still do not know what is result
edit2: could you please write result funcion, becasuse i am getting quite desperate now
is result only that the function f1*f2(x) is divergent ? (withou any calculation?)

**Opalg** · May 15th 2011, 10:10 AM

Originally Posted by token22

I got
f1*f2(t) = {-infinity for t in <-pi/2, 1 - pi/2>
ln|cos(t-1)| - ln|cos(t)| for t in <1 - pi/2, pi/2>
infinity for t in <pi/2, pi/2 + 1>
0 otherwise

but i am not sure about infinity

That solution is correct. If $f_1$ is the function defined by $f_1(x) = \begin{cases}1&\text{if }0\leqslant x\leqslant 1\\0&\text{otherwise}\end{cases}$ , and g is any other function, then $(f_1*g)(t)$ is the mean value of g on the interval [t–1,t].

For the function $f_2(x) = \begin{cases}\tan x&\text{if }-\pi/2\leqslant x\leqslant \pi/2\\0&\text{otherwise}\end{cases}$ , its integral over any interval containing $\pm\pi/2$ diverges. Thus $f_1*f_2$ is infinite (or if you prefer, undefined) in those intervals specified in your answer.

**token22** · May 15th 2011, 10:40 AM

Just for sure, so the "undefined state" is caused by divergnet tangent? thus i get some result only in interval [t-1,t], because of $f_1$ function

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