Question

Please help using window and step functions, to compute Laplace transform. Easy question, but new concept to learn.

Answer 1

\[g(t)=0;0<t<2 \]

\[g(t)=t+1;2<t\]

Answer 2

I'm not sure of the question -- are you asking how to compute the Laplace transform of that function?

Answer 3

First using Window and Step Functions, yes at the end though

Answer 4

defined sorry

Answer 5

Oops, sorry, misread the question. Here ya go

Answer 6

Okay, well you can use a step function for this, but as you'll see it's really not going to help too much. If we define the step function
\[ \theta(x) = 1 \text{ for x > 0, and } \theta(x) = 0 \text{ for x < 0 } \]

Then that function g can be written

\[ g(x) = (t+1) \theta( t - 2 ) \]
Do you agree?

Answer 7

Sorry, that should be g(t), not g(x)

Answer 8

could you explain how you got (t+t)theta(t-2), this is my first time learning the concept without prior learning sorry

Answer 9

It's okay. Do you see how \[ \theta( t - 2 ) \] is equal to zero of t < 2, but equal to one if t >2?

Answer 10

oh yes yes i get that, in my book theta is a u

Answer 11

Ah, apologies. In physics, we usually use theta. But it doesn't really matter.

Answer 12

The point is that by using theta, we are just satisfying the first part of the requirement on g(t) - that is, that it's equal to zero if t < 2.

Once t > 2, however, the theta function just becomes a 1. Therefore, we multiply the theta function by whatever we'd like g(t) to be when t > 2.

Answer 13

okay I got it, thanks!

Answer 14

So if we do the integral, we could write this:

\[ \mathcal{L}\left( g(t) \right) (s) = \int_0^\infty e^{-s} (t+1)\theta(t - 2) dt\]
do you agree?

Answer 15

yeah I did that and solved it on paper, but using laplace transforms

Answer 16

Wow, I keep making typos - I'm sorry. That should be
\[ e^{-st}\]

You have to be careful - in practice, that theta function just cuts off the lower part of the integral, leaving you with
\[ \int_2^\infty e^{-st} (t+1) dt \]
The 2 means that it's not exactly a laplace transform of t+1 anymore, you see?

Answer 17

yes it's in my textbook

Answer 18

Okay you can make it like a normal laplace transform again by making a substitution:
x= t - 2
or
t = x+ 2

Then the integral becomes
\[ \int_0^\infty e^{-s(x + 2)} (x+2 +1) dx\]
\[ = e^{-2s} \int_0^\infty e^{-sx} (x+3) dx \]

now THAT integral can be done by identifying laplace transforms, because the lower limit is zero again. The result will be your final answer.

Do you understand? I"m not sure if I'm answering your question or not...

Answer 19

yes you're answering my question way further than the book does, thank you

Answer 20

Oh, lovely. Glad to be of assistance. The step or window functions really serve only to limit the domain of integration, like we just did. After you've limited it, you can either solve by directly integrating OR you can make various substitutions to put it in the form of a standard laplace transform, in which case you can look them up in a book or table.

Answer 21

When I said "you'll find that the step function doesn't really help", I mean that if you look at the definition of the function, you could simply say right off the bat that the part of the integral from zero to two doesn't contribute (because the function g(t) is zero on that interval) so you'd just have to start with the piece from two to infinity - and you'd be right where we ended up anyway. But whichever approach you like (or is requested by the book) is fine.

Answer 22

The last thing I'll say (unless you have another question for me) is that if you use a window function or a different step function so that the upper bound is no longer infinity, then you will generally have to integrate directly and won't be able to find an appropriate substitution to use to make it look like a standard Laplace transform.

Answer 23

thank you very much for your time, it was a thorough and concise explanation. I will have more questions in the future!

Answer 24

No problem. I'm not on here every day, but feel free to tag me in a question if you'd like. Otherwise I'll be browsing around.

I rarely get to play with integral transforms anymore, so I thoroughly enjoy this kind of question :)