Taking the convolution using the inverse Laplace transform

[arctic mirage]

I was trying to take the convolution of the function in the picture using the convolution theorem sort of in the opposite way it's meant to be used and my answer was t but the given answer is u of t times t I suspect the difference in answers has something to do with the functions needing to be causal but I'm not sure exactly where my error is as I applied the formulas in a way I think is correct.

[prisma moth]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[arctic mirage]

Guess this one is too hard for anyone 😭

[empty flume]

which picture is the problem statement?

[empty flume]

i suspect it's more about it being unclear what you are asking, and you expect the reader to know what your notation means

[arctic mirage]

https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/09%3A_Transform_Techniques_in_Physics/9.09%3A_The_Convolution_Theorem

The notation is standard for convolutions and Laplace transforms, although my handwriting is shit

The problem statement is in the last pic

It simply asks for the convolution of the unit step function with itself

[empty flume]

oh, heaviside

let H(t) be the heaviside function

1. H(t-1/2) is nonzero only for t > 1/2 and H(t-(x-1/2)) is nonzero only when x-t > -1/2

these two conditions appear at the same time only for t >= 0, hence the convolution is 0 for t<0

and then all that's left is to calculate

$$ H(t-1/2) * H(t+1/2) = \int _{-\infty} ^\infty H(x-1/2)H(t-(x-1/2))dx $$

[maiden niche]

aL

[empty flume]

$$ = \int _{1/2}^{t+1/2} H(x-1/2)H(t-(x-1/2))dx = \int _{1/2}^{t+1/2}dx = t $$

[maiden niche]

aL

[empty flume]

so conclude

$$ H(t-1/2)*H(t+1/2) = \begin{cases} t, &t\geqslant 0 \ 0, &t<0 \end{cases} = tH(t) $$

[maiden niche]

aL

[empty flume]

now repeat the process for H(t-eps) * H(t+eps)

for however small epsilon

@arctic mirage

[arctic mirage]

Nice! Yes that is the correct proof that I have seen. I'm wondering why using the convolution theorem and instead going through the laplace domain does not work and instead returns simply t and not the correct answer

[empty flume]

nothing's wrong with your approach, you are neglecting the domains

and hence won't conclude the thing you want

[arctic mirage]

I'm a little confused what you mean sorry. Could you explain more in detail about how the domains affect this and how I would apply them.

[empty flume]

what's the laplace transform of heaviside?

[arctic mirage]

1/s if I recall

Well there's an etern if your c is not zero

[empty flume]

yes, if c=0 then 1/s

[arctic mirage]

$$e^{-cs}/s$$

[maiden niche]

Stronk

[arctic mirage]

Is there a valid domain that I am forgetting?

[empty flume]

what do you think s is?

arbitrary real number? complex number? any restrictions?

[arctic mirage]

I thought s could be any complex number. Of course it's not valid when s is equal to zero...

[empty flume]

$$ F(s) = \int_0^\infty \exp (-st)f(t)dt $$

[maiden niche]

aL

[empty flume]

you need this integral to converge

$$|exp (-st)| = |\exp (\mathrm{Re}(s)t) \exp (i\mathrm{Im}(s)t)| = |\exp (\mathrm{Re}(s)t)| $$

[maiden niche]

aL

[arctic mirage]

Oh so it won't converge if s or t is somehow negative?

[empty flume]

if Re(s) is positive, then the integral converges

[arctic mirage]

In the case of a non-time shifted unit step right?

[empty flume]

f is arbitrary here

well, integrable of course

[arctic mirage]

But then what if t is negative, wouldn't the exponential not converge?

Unless s was negative as well

[empty flume]

if s was negative the whole thing would blow up fast

[arctic mirage]

But if they're both negative then it goes back to being positive right? So converge for negative time?

[empty flume]

t is positive by definition

[arctic mirage]

Is that in general or just for control systems?

[empty flume]

you want the exponent to be "negative" if you're only considering real s

more generally, it turns out the integral converges whenever Re(s) is positive

[arctic mirage]

Okay I'll explore this a little bit more tomorrow. But how does the range of s affect the method that I used in the screenshots for finding the convolution?

[empty flume]

this is correct

but you are assuming t nonnegative here

[arctic mirage]

Oh interesting

[empty flume]

if t < 0 then the thing is 0

[arctic mirage]

So if I use that method for other functions, do I just have to slap a heavy side function on the result if I want it to be valid for all t's

[empty flume]

which is the same as t u(t)

[empty flume]

i have no idea, follow the definition and things will work out :/

[arctic mirage]

Does the definition say that the Laplace transform of arbitrary functions is always 0 for time less than zero

[empty flume]

if you're referring to this, like I said, follow the definition of the function