inverse laplace transform of this equation.

Question

I want to know how to find the inverse laplace transform of this function:

$$Y(s)=\frac {1}{\tau s+1}\times \frac{\omega}{s^2+\omega^2}$$

for $\omega$, $\tau$ constant

Answer 1

you can apply convolution theorem

and seperate laplace transform of them are $$\frac {1}{\tau s+1}\times \frac{\omega}{s^2+\omega^2}$$ laplace for 1st is $$g(t)=\frac {1}{\tau}\exp{\frac {-t}{\tau}}$$
and for $$\frac{\omega}{s^2+\omega^2}$$ is $$f(t)=\sin{\omega}{t}$$ hence apply convolution theorem $$\int_0^tg(v)f(t-v)\mathrm dv$$ solve the integral you get the result

Answer 2

$$ y(x)=\frac{e^{-x/\tau} \omega \tau - \omega\tau\cos(\omega x) + \sin(\omega x)}{1+\omega^2\tau^2} $$