Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the function below:


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


share|cite|improve this question
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. – Babak 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
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 $$F(s)=\frac{3}{s}-\frac{2}{s^2}+\frac{6}{s^3}$$

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.