# calculations with stepfunction(heaviside)

#### goohu

##### New member
I've some questions in the c) part of the task. The texts in the pictures are loosely translated, so let me know if there is something that is unclear. Is there a specific reason why they wrote (F' * f)' instead of (f * f)' ?

I've posted a solution at the bottom of the post but I don't understand the first step: (F' * f)' = F'' * f.
Is this some standard trick that always applies?

The way I would write is $$\displaystyle f' * f = ( 2\theta(t) + \delta(t) ) * (2t+1)\theta(t)$$, I dont see how it becomes what is written in the solution.

If someone knows of any good sources where I can read up on convolution it would be awesome if you could direct me there. Right now I only vaguely understand what convolution is (I would say: sliding two functions across each other and creating a third one by summing the overlapping area for each time unit).

I'm trying to get started on the mathematical calculations but I find it hard to grasp.

Solution: #### Klaas van Aarsen

##### MHB Seeker
Staff member
I've some questions in the c) part of the task. The texts in the pictures are loosely translated, so let me know if there is something that is unclear.

Is there a specific reason why they wrote (F' * f)' instead of (f * f)' ?
Hi goohu!

Not that I can see. It is the same.

I've posted a solution at the bottom of the post but I don't understand the first step: (F' * f)' = F'' * f.
Is this some standard trick that always applies?
It is indeed a property of a convolution. It is listed on the wiki page about convolutions, including the proof.

We can also just substitute the definition of the convolution and get:
$$(F'\star f)' = \frac{d}{dt}\int_{-\infty}^\infty F'(t-\tau)f(\tau)\,d\tau =\int_{-\infty}^\infty \frac{\partial}{\partial t}F'(t-\tau)f(\tau)\,d\tau =\int_{-\infty}^\infty F''(t-\tau)f(\tau)\,d\tau =F''\star f$$

The way I would write is $$\displaystyle f' * f = ( 2\theta(t) + \delta(t) ) * (2t+1)\theta(t)$$, I dont see how it becomes what is written in the solution.
Another property of a convolution is that $(f+g)\star h = f\star h + g\star h$.
So following your way, we get:
$$f' \star f = (2\theta(t) + \delta(t) ) \star ((2t+1)\theta(t)) = (2\theta(t)) \star ((2t+1)\theta(t)) + \delta(t) \star ((2t+1)\theta(t))$$

Which brings us to yet another property of the convolution:
$$\delta(t)\star g(t) = g(t)$$
so that:
$$f' \star f = (2\theta(t)) \star ((2t+1)\theta(t)) + \delta(t) \star ((2t+1)\theta(t)) = (2\theta(t)) \star ((2t+1)\theta(t)) + (2t+1)\theta(t)$$
That is pretty much the same as what we have in the solution isn't it?

Next is to evaluate $(2\theta(t)) \star ((2t+1)\theta(t))$ by substituting the definition of the convolution.

If someone knows of any good sources where I can read up on convolution it would be awesome if you could direct me there. Right now I only vaguely understand what convolution is (I would say: sliding two functions across each other and creating a third one by summing the overlapping area for each time unit).
See the wiki page that I just mentioned.