Help with inverse Laplace transform

Question

Please help me to find the inverse Laplace transform of :

$$ \frac{1}{S(aS^2 + bS + c)} \left( 1 - \exp (-TS) \right) $$

where a,b,c,T are constants

thanks

Answer 1

This looks like a good candidate for the convolution theorem, which states that the inverse transform of a product of two transforms is the convolution of the inverse transforms. In this case, the inverse transform of

$$F(s) = \frac1{a s^2+b s+c} \implies f(t) = \frac{e^{-b t/(2 a)}}{\sqrt{b^2-4 a c}} \left (e^{\frac{\sqrt{b^2-4 a c}}{2 a} t}-e^{-\frac{\sqrt{b^2-4 a c}}{2 a} t}\right )$$

$$G(s) = \frac{1-e^{-s T}}{s} \implies g(t) = \theta(T-t)$$

where $\theta$ is the Heaviside step function.

The inverse transform is then

$$\int_0^t dt' \, f(t') g(t-t') $$

Now, generally,

$$\int_0^t dt' \, e^{-p t'} \theta(T-(t-t')) = \int_0^t dt' \, e^{-p t'} \theta(t'-(t-T)) $$

The integral is then, because $t' \gt 0$,

$$ \int_{\max(0,t-T)}^{t} dt' \, e^{-p t'}\theta(t-T) = \begin{cases}\displaystyle \left (\frac{e^{p T}-1}{p} \right ) e^{-p t} & t\gt T \\ \displaystyle\left (\frac{e^{p t}-1}{p} \right )e^{-p t} & t\lt T \end{cases}$$

Therefore

$$\int_0^t dt' \, e^{-p t'} \theta(T-(t-t')) =\left [\left (\frac{e^{p T}-1}{p} \right ) \theta(t-T) + \left (\frac{e^{p t}-1}{p} \right ) \theta(T-t)\right ] e^{-p t} $$

Now plug the expression for $f(t)$ above into this expression, i.e., $p=(b\pm\sqrt{b^2-4 a c})/(2 a)$ and you are done.