Inverse Laplace Transform of $F(s)= e^{-s}\arctan\Big(\frac{s+4}{(s+4)^2+4}\Big)$

Question

I've tried applying theorems to remove $e^{-s}$ and change $(s+4)$ to $s$ but I don't know what to do with the resulting Inverse Laplace transform.

Answer 1

Once you've dealt with the time and frequency shifts, the problem reduces to finding the inverse transform of $$ G(s) = \arctan \left(\frac{s}{s^2+4}\right) $$

The $\arctan$ outer function might suggest you use the derivative property $$ \mathcal L^{-1} \{G'(s)\} = -t g(t) $$

By the chain rule, we have

$$ G'(s) = \frac{1}{1 + \frac{s^2}{(s^2+4)^2}}\frac{4-s^2}{(s^2+4)^2} = \frac{4-s^2}{(s^2+4)^2+s^2} = \frac{4-s^2}{s^4+9s^2+16} $$

The denominator here is a quadratic in $s^2$ with two roots: $\frac{-9\pm\sqrt{17}}{2}$. Since they are both negative we can write

$$ s^4 + 9s^2 +16 = \left(s^2+ \frac{9-\sqrt{17}}{2} \right)\left(s^2+\frac{9+\sqrt{17}}{2}\right) = (s^2+{a_1}^2)(s^2+{a_2}^2) $$

Applying partial fraction decomposition $$ \frac{4-s^2}{(s^2+{a_1}^2)(s^2+{a_2}^2)} = \frac{As+B}{s^2+{a_1}^2} + \frac{Cs+D}{s^2+{a_2}^2} $$

You will quickly find $A=C=0$ since the LHS is an even function. The remaining two constants are up to you to solve.

Taking the inverse transform of $G'(s)$ gives

$$ g(t) = -\frac{B\sin(a_1 t)}{a_1t} -\frac{D\sin(a_2 t)}{a_2t} $$