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:
$$\frac{1}{s}+\frac{1}{s^3}$$
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:
$$\frac{1}{s}+\frac{1}{s^3}$$
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 $$F(s)=\frac{3}{s}-\frac{2}{s^2}+\frac{6}{s^3}$$
$$\begin{align*}f(t)&=3-2t+3t^2\\ \\ F(s)&=\mathcal L\{3-2t+3t^2\}\\ \\ &=\mathcal L\{3\}-\mathcal L\{2t\}+\mathcal L\{3t^2\}\\ \\ &=3\mathcal L\{t^0\}-2\mathcal L\{t^1\}+3\mathcal L\{t^2\}\\\\&=\end{align*}$$
use $$\boxed{\begin{align*}\mathcal L\{t^n\}&=\frac{\Gamma(n+1)}{s^{n+1}},\qquad n>-1\\&=\frac{n!}{s^{n+1}},\qquad \qquad n\in\mathbb Z> 0\qquad\end{align*}}$$