1. ## 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 $x(t) = h(t) - h(t-1)$ where $h(t)$ is the unit Heaviside Function.

Well it's known that the Heaviside function is equal to 1 for $t>0$, and 0 otherwise.

So for $t<0$, $x(t) = 0$

For $0 < t < 1$, the function is equal to $1-0 = 1$

And for $t>1$, we have $x(t) = 1-1 = 0$.

So as far as I can see, we have $x(t) = 1$ for $0 < t < 1$, and $x(t) = 0$ otherwise.

So would the Laplace Transform of this equation is just the Laplace Transform of 1, ie $\frac{1}{s}$?

3. ## Re: Heaviside Function

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?

4. ## Re: Heaviside Function

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

$\mathcal{L}[h(t-1)]=\int_{0}^{\infty}h(t-1)e^{-st}\,dt=\int_{1}^{\infty}e^{-st}\,dt\dots$

Can you finish?

5. ## Re: Heaviside Function

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

$\mathcal{L}[h(t-1)]=\int_{0}^{\infty}h(t-1)e^{-st}\,dt=\int_{1}^{\infty}e^{-st}\,dt\dots$

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 $\frac{1-e^{-s}}{s}$, 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.

6. ## Re: Heaviside Function

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:

$\mathcal{L}[h(t)-h(t-1)]=\int_{0}^{1}e^{-st}\,dt,$

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

Merci