Calculations with stepfunction(heaviside)

In summary: Convolution is a topic that can be studied in more depth, so that might help you understand it better.
  • #1
goohu
54
3
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 \(\displaystyle f' * f = ( 2\theta(t) + \delta(t) ) * (2t+1)\theta(t)\), 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
 

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  • #2
goohu said:
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.

goohu said:
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:
$$(F'\star f)' = \frac{d}{dt}\int_{-\infty}^\infty F'(t-\tau)f(\tau)\,d\tau
=\int_{-\infty}^\infty \frac{\partial}{\partial t}F'(t-\tau)f(\tau)\,d\tau
=\int_{-\infty}^\infty F''(t-\tau)f(\tau)\,d\tau
=F''\star f$$

goohu said:
The way I would write is \(\displaystyle f' * f = ( 2\theta(t) + \delta(t) ) * (2t+1)\theta(t)\), 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.

goohu said:
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.
 
  • #3


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!
 

FAQ: Calculations with stepfunction(heaviside)

What is a step function (Heaviside)?

A step function, also known as a Heaviside function, is a mathematical function that has a value of 0 for all negative inputs and a value of 1 for all positive inputs. It is represented by the symbol H(x) or u(x).

How is a step function (Heaviside) used in calculations?

A step function is commonly used in mathematical and scientific calculations to represent a sudden change or discontinuity in a system. It is also used in differential equations to model real-world phenomena such as population growth or electrical circuits.

What is the difference between a Heaviside function and a unit step function?

The terms Heaviside function and unit step function are often used interchangeably, but technically, a Heaviside function is defined as H(x) = 0 for x < 0 and H(x) = 1 for x ≥ 0, while a unit step function is defined as u(x) = 0 for x < 0 and u(x) = 1 for x ≥ 0. In other words, a Heaviside function has a discontinuity at x = 0, while a unit step function does not.

How do you graph a step function (Heaviside)?

To graph a step function, you can plot a point at (0,0) and then draw a vertical line at x = 0. This represents the discontinuity in the function. For all other values of x, the graph will be a horizontal line at y = 1.

Can a step function (Heaviside) have more than one discontinuity?

Yes, a step function can have multiple discontinuities at different values of x. Each discontinuity will result in a new horizontal line segment on the graph, with a value of 1 for all x values after the discontinuity.

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