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Math Help - Heaviside Function

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    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}?
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    Re: Heaviside Function

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    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?
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    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?
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    Super Member craig's Avatar
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    Re: Heaviside Function

    Quote Originally Posted by Ackbeet View Post
    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.
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    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.
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    Re: Heaviside Function

    Merci
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