Laplace Transform

  • Thread starter Rob K
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Homework Statement



f(t) = (1/(a^2))(1-e^-at - ate^-at)

Homework Equations



f(t) = (1/a)(1-e^(-at))
F(s) = 1/s(s+a)

f(t) = t e^(-at)
F(s) = 1/(s+a)^2

The Attempt at a Solution



F(s) = (1/s(s+a^2)) - (1/(s+a)^2)


Totally lost now, but I think this is wrong anyway, can anyone help me please. I know the answer is
F(s) = 1/(s + a)^2
but I have no idea why.

Kind Regards

Rob K
 
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Answers and Replies

  • #2
vela
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Homework Statement



f(t) = (1/(a^2))(1-e^-at - ate^-at)

Homework Equations



f(t) = (1/a)(1-e^(-at))
F(s) = 1/s(s+a)

f(t) = t e^(-at)
F(s) = 1/(s+a)^2

The Attempt at a Solution



F(s) = (1/s(s+a^2)) - (1/(s+a)^2)
Can you explain how you came up with this?

Totally lost now, but I think this is wrong anyway, can anyone help me please. I know the answer is
F(s) = 1/(s + a)^2
but I have no idea why.

Kind Regards

Rob K
 
  • #3
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Ah, this is the answer that my lecturer gave in the notes. I just assumed it to be the right answer, but to be honest. Is it wrong?
 
  • #4
vela
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Let's rewrite stuff a bit, so you're not using the same letter to represent every single function.

You're given ##f(t) = \frac{1-e^{-at}-ate^{-at}}{a^2}## and asked to find its Laplace transform, F(s).

It looks like you're also given the following
\begin{align*}
g(t) = \frac{1-e^{-at}}{a} \qquad &\Leftrightarrow \qquad G(s) = \frac{1}{s(s+a)} \\
h(t) = t e^{-at} \qquad &\Leftrightarrow \qquad H(s) = \frac{1}{(s+a)^2}
\end{align*}

What you want to do is write f(t) in terms of g(t) and h(t), and then use the linearity of the Laplace transform to find F(s). By the way, the answer you have is wrong. Note that what you wrote matches what I called H(s), which is the transform of h(t), not f(t).
 
  • #5
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Thank you very much, what you have written is very clear. I was very curious of the answer, I saw that it was the transform of what you are referring to as H(s), my only conclusion was that there was a technique of ignoring a term that we had not been previously taught, but that did not seem right to me. We have just been learning about approximating 3rd order and higher transfer functions by modelling them as 2nd order functions, but this is not one of those, but you can see how my thinking was going and that seemed wrong, hence my post.

Thank you again, I shall now give our head of robotics research a fail!

Kind Regards

Rob K
 
  • #6
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Also just to clarify, could this now be written thus:
\begin{align*}
f(s) = \frac{1}{a^2}(1-e^{-at}\ -\ a\ te^{-at})\qquad &\Leftrightarrow \qquad F(s) = \frac{1}{s(s+a)} - \frac{1}{(s+a)^2}
\end{align*}

Kind regards

Rob K

p.s Just learnt LaTeX through this post, thanks.
 
  • #7
vela
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No, you have to throw some constants in there. That's why you need to find how to write f(t) as a linear combination of g(t) and h(t) first.
 
  • #8
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Thank you, this gives me something I can now work with. I may be back if I get lost.
 

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