Finding the inverse transform of a Laplace Transform

64 Views Asked by At

I'm trying to solve an inverse laplace function. The equation is quite simple actually but has a lot of small constants in the initial expression.

The initial expression:

$$\frac{1+5*10^{-3}+100*10^{-9}s^{2} }{400*100*10^{-9}s+5*10^{-3}*100^{-9}s^{2}+1}$$

I tried simplifying a little bit by breaking out common terms. This is the simplified expression:

$$\frac{s^{2}+2*10^{9}}{(s+4*10^{4})^{2}+4*10^{8}}$$

so it is basically on the form

$$\frac{s^{2}+a}{(s+b)^2+c}$$

Any smart idea on how I can break up this to find a good inverse transform?

Thank you in advance

1

There are 1 best solutions below

2
On

Using $s^2 + a = (s+b)^2 + c^2 - 2 b \, (s+b) + a + b^2 - c^2$ then $$ \frac{s^2 + a}{(s+b)^2 + c^2} = 1 - 2 b \, \frac{s+b}{(s+b)^2 + c^2} + \frac{a + b^2 - c^2}{c} \, \frac{c}{(s+b)^2 + c^2}. $$ The inverse of this is $$f(t) = \delta(t) - 2 b \, e^{- b \, t} \, \cos(c t) + \frac{a + b^2 - c^2}{c} \, e^{- b \, t} \, \sin(c t). $$