MHB Calculations with stepfunction(heaviside)

[goohu]

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.

View attachment 9384
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 $$f' * f = ( 2\theta(t) + \delta(t) ) * (2t+1)\theta(t)$$, I don't 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:

View attachment 9385

 

[I like Serena]

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 $$f' * f = ( 2\theta(t) + \delta(t) ) * (2t+1)\theta(t)$$, I don't 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.

 

[soloenergy]

Hi there,

I can understand why you might be confused about the notation used in the solution. The reason why (F' * f)' is written instead of (f * f)' is because the first function, F', is the derivative of the function f. This means that in the convolution, we are finding the derivative of the convolution of F' and f, which is why it becomes F'' * f.

As for your proposed solution, it looks like you have misunderstood the notation used. The * symbol in this context actually represents the convolution operation, not multiplication. So your solution would actually be written as (2\theta(t) + \delta(t)) * (2t+1)\theta(t).

If you're looking for resources to better understand convolution, I would recommend checking out some online tutorials or textbooks on signal processing or Fourier analysis. They often cover convolution in depth and provide helpful explanations and examples.

I hope this helps clarify things for you. Keep at it and don't hesitate to ask for further clarification if needed. Good luck with your calculations!