## anonymous 5 years ago Inverse laplace transform of (.1s+.9)/(s^2+3.24), I know the answer is .1cos(1.8t)+.5sin(1.8t), but I don't know how to get the answer

1. anonymous

It would help if you would put all the decimals in fraction form... $\frac{.1s+.9}{s^2+3.24} \rightarrow \frac{\frac{1}{10}s+\frac{9}{10}}{s^2+\frac{81}{25}}$ Now split this in to two....... $\frac{\frac{1}{10}s+\frac{9}{10}}{s^2+\frac{81}{25}}\rightarrow \frac{1}{10}(\frac{s}{s^2+\frac{81}{25}})+\frac{\frac{9}{10}}{s^2+\frac{81}{25}}$ Factor out a 1/2 form the second equation.... $\frac{1}{10}(\frac{s}{s^2+\frac{81}{25}})+\frac{1}{2}(\frac{\frac{9}{5}}{s^2+\frac{81}{25}})$ since: $L[\sin(at)]=\frac{a}{s^2+a^2}, L[\cos(at)]=\frac{s}{s^2+a^2}$ notice that $\frac{81}{25}=(\frac{9}{5})^2\rightarrow a^2$ $\frac{1}{10}(\frac{s}{s^2+(\frac{9}{5})^2})+\frac{1}{2}(\frac{\frac{9}{5}}{s^2+(\frac{9}{5})^2})$ Thus the Laplace transform is: $\frac{1}{10}\cos(\frac{9}{5}t)+\frac{1}{2}\sin(\frac{9}{5}t)\rightarrow .1\cos(1.8t)+.5\sin(1.8t)$

2. anonymous