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Laplace transform heaviside function

  • Thread starter vvl92
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I have a question asking for the inverse laplace transform of (e^(-s))/(s^2+pi^2).
I split it up to (e^(-s))/s x s/(s^2+pi^2) and got u(t-1)cos(pi(t-1)),but the correct answer is (sin(pi(t-1)/pi)u(t-1). So here it was split up to (e^(-s))/pi x pi/(s^2+pi^2) and I don't understand where the step function came from. It is the same in all of the questions and on things I have seen online, even though the transform of U(t-a) is (e^(-as))/s.
 

Ray Vickson

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I have a question asking for the inverse laplace transform of (e^(-s))/(s^2+pi^2).
I split it up to (e^(-s))/s x s/(s^2+pi^2) and got u(t-1)cos(pi(t-1)),but the correct answer is (sin(pi(t-1)/pi)u(t-1). So here it was split up to (e^(-s))/pi x pi/(s^2+pi^2) and I don't understand where the step function came from. It is the same in all of the questions and on things I have seen online, even though the transform of U(t-a) is (e^(-as))/s.
The inverse laplace transform of exp(-s)/s * s/(s^2 + pi^2) is the convolution of u(t-1) and cos(pi*t); that is, it equals
[tex] \int_0^t u(\tau - 1) \cos(\pi (t - \tau))\, d \tau.[/tex]
 

LCKurtz

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When trying to inverse transforms that have a factor of ##e^{-as}## I like to use the following idea. Consider the direct transform$$
\mathcal Lf(t-a)u(t-a) = \int_0^\infty e^{-st}f(t-a)u(t-a)\, dt =
\int_a^\infty e^{-st}f(t-a)\, dt $$Now let ##w = t-a## $$
=\int_0^\infty e^{-s(w+a)}f(w)\, dw= e^{-as}F(s)$$
where ##F(s)## is the transform of ##f(t)##. What this says in terms of inversing is that if you have a form ##e^{-as}F(s)## to invert, you can just invert the ##F(s)## to ##f(t)##, then translate and truncate your answer to ##f(t-a)u(t-a)##. This avoids doing the convolution. Try it on your problem.
 

Ray Vickson

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When trying to inverse transforms that have a factor of ##e^{-as}## I like to use the following idea. Consider the direct transform$$
\mathcal Lf(t-a)u(t-a) = \int_0^\infty e^{-st}f(t-a)u(t-a)\, dt =
\int_a^\infty e^{-st}f(t-a)\, dt $$Now let ##w = t-a## $$
=\int_0^\infty e^{-s(w+a)}f(w)\, dw= e^{-as}F(s)$$
where ##F(s)## is the transform of ##f(t)##. What this says in terms of inversing is that if you have a form ##e^{-as}F(s)## to invert, you can just invert the ##F(s)## to ##f(t)##, then translate and truncate your answer to ##f(t-a)u(t-a)##. This avoids doing the convolution. Try it on your problem.
I agree. This is the way I would have chosen to do the problem, but he did ask a direct question, and I felt it important to answer, if only to dispel his misunderstanding about multiplication vs. convolution..
 

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