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I have the function below:

$$3-2t+3t^2$$

I tried to get the transfer function, but my solution seems to be wrong, can someone please tell me what I am doing wrong:

$$\frac{1}{s}+\frac{1}{s^3}$$

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Are we just finding $\int_0^\infty (3-2t+3t^2)e^{-st}\,dt$? If so (using integration by parts for the last two terms) I think we get $\frac{3}{s}-\frac{2}{s^2}+\frac{6}{s^3}$. As to what you did wrong, that is hard to know if one does not see what you did. –  André Nicolas May 31 '12 at 19:33
    
You can use the following properties: $ \mathcal{L} \{1\} =\frac{1}{s}$ , $\mathcal{L} \{t^n \} =\frac{n!}{s^{n+1}}$ –  passenger May 31 '12 at 19:37
    
It would be better if you added what was done in a clear way. Indeed, Andre got what you are looking for. –  B. S. May 31 '12 at 19:40
    
Is the function of $t$ meant to be the impulse response? –  copper.hat May 31 '12 at 19:44
    
Thanks guys, this was so simple I was ashamed to continue the question :( got it right now. –  Sean87 May 31 '12 at 20:51
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2 Answers

up vote 2 down vote accepted

Just looking at a table and seeing how $$\mathcal L\{1\}=\frac{1}{s},\quad \mathcal L\{t\}=\frac{1}{s^2},\quad \mathcal L\{t^2\}=\frac{2}{s^3}$$... add them up, combine the constants, hope to get something like $$F(s)=\frac{3}{s}-\frac{2}{s^2}+\frac{6}{s^3}$$

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$$\begin{align*}f(t)&=3-2t+3t^2\\ \\ F(s)&=\mathcal L\{3-2t+3t^2\}\\ \\ &=\mathcal L\{3\}-\mathcal L\{2t\}+\mathcal L\{3t^2\}\\ \\ &=3\mathcal L\{t^0\}-2\mathcal L\{t^1\}+3\mathcal L\{t^2\}\\\\&=\end{align*}$$

use $$\boxed{\begin{align*}\mathcal L\{t^n\}&=\frac{\Gamma(n+1)}{s^{n+1}},\qquad n>-1\\&=\frac{n!}{s^{n+1}},\qquad \qquad n\in\mathbb Z> 0\qquad\end{align*}}$$

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