Thread: Heaviside Function

Got a laplace transformation question involving Heaviside Functions, just need someone to take a quick look over what I've done so far.

Find the Laplace transform of where is the unit Heaviside Function.

Well it's known that the Heaviside function is equal to 1 for , and 0 otherwise.

So for ,

For , the function is equal to

And for , we have .

So as far as I can see, we have for , and otherwise.

So would the Laplace Transform of this equation is just the Laplace Transform of 1, ie ?

Use the UnitStep function on WA.

Thanks for that link again, didn't know you could use the Unit-Step functions on there a well.

Also, I'm not sure where I've gone wrong exactly with my integration?

Your h(t) is fine. To do h(t-1), you have this:



Can you finish?

Originally Posted by Ackbeet

Your h(t) is fine. To do h(t-1), you have this:



Can you finish?

Ahh I see. Not sure why but didn't think that you'd have to integrate the h(t-1) part. Just do the integration as normal, but because of the (t-1), the lower limit is now 1 instead of zero.

Just another query, when you've got your answer , is it good practice to say for what values of t your answer is valid? If so what ones would you use here?

Thanks again for the help.

If it's understood, as it is here, that you're using the one-sided LT, then it's also understood that t is non-negative. So those are the t's for which your result is valid. I don't think it's necessary to state the t-interval, though it certainly isn't wrong to do so. Incidentally, you could also do your whole LT this way:



since the combination of unit steps means the integrand is zero except on the interval (0,1) - it's a box profile.

Merci