I ended up with the following system of equations in the frequency domain: \begin{equation} \mathcal{L}\{f\}(s)=\frac{\alpha (s \beta+ \gamma)}{(s \delta + \alpha+\gamma)(s\beta+\gamma)-\gamma \gamma} \end{equation}

\begin{equation} \mathcal{L}\{g\}(s)=\frac{\alpha \gamma}{(s \beta +\gamma)(s \delta + \alpha+\gamma)-\gamma \gamma} \end{equation}

where $\alpha, \beta, \gamma, \delta$ are constants.

I want to factorise them first so I can transform them back to the time domain.

Can anyone give a hint how to factorise the above equations?


Those are transfer functions of a second order system and integrate as a product of $sin$, $cos$ and $exp$ or $sinh$, $cosh$ and $exp$. You can start from this : $$G(s)=\frac{\alpha\gamma}{s^2\delta\beta+s(\alpha\beta+\gamma\beta+\gamma\delta)+\gamma\alpha}$$ $$=\frac{\alpha\gamma}{(s\sqrt{\delta\beta}+\frac{1}{2}(\alpha\beta+\gamma\beta+\gamma\delta))^2-(\alpha\beta+\gamma\beta+\gamma\delta)^2+\gamma\alpha}$$ Still you need to have some information about the sign of $-(\alpha\beta+\gamma\beta+\gamma\delta)^2+\gamma\alpha$ to use a table of transforms.

If you have no clue about this term the idea to go further would be : $$G(s)=\frac{\alpha\gamma/\delta\beta}{(s+\frac{1}{2}(\alpha\beta+\gamma\beta+\gamma\delta)/\sqrt{\delta\beta})^2-(\alpha\beta+\gamma\beta+\gamma\delta)^2/\delta\beta+\gamma\alpha/\delta\beta}$$ Then let $K=\alpha\gamma/\delta\beta$, $A=\frac{1}{2}(\alpha\beta+\gamma\beta+\gamma\delta)/\sqrt{\delta\beta}$, $\Omega=\sqrt{-(\alpha\beta+\gamma\beta+\gamma\delta)^2/\delta\beta+\gamma\alpha/\delta\beta}$

Note that you have the relation : $$\Omega=\sqrt{K-4A^2}$$ then : $$G(s)=\frac{K}{(s+A)^2+\Omega^2}=\frac{K}{\Omega}\frac{\Omega}{(s+A)^2+\Omega^2}=\mathcal{L}\left[\frac{K}{\Omega}e^{-At}\sin(\Omega t)\cdot u(t)\right]$$ You just have to hope that $\Omega\in\mathbb{R}$ which corresponds to the nice condition $K-4A^2\geq 0$

  • $\begingroup$ I don't think I have information about that. We can't simplify this any further? $\endgroup$
    – user105627
    Oct 26 '16 at 17:45
  • $\begingroup$ You can certainly simplify but you will have possibly complex expressions in the time domain (which can be transformed into $sin$, $cos$) $\endgroup$ Oct 27 '16 at 8:22

Factor out the $\beta$ and $\delta$ from the denominator, then you will be left with something of the form $(s+a)(s+b) -c$. Try writing this as $(s-d)^2 - e$. (so I'm working towards 25 in this table.)

Can you work it out from here? You need to write out $(s+a)(s+b)$ and $(s-d)^2$ out and try to find $d$ and $e$ to get the trick done.

  • $\begingroup$ I can't see the table. The link doesn't seem to be working. Can you please explain it a little bit more. I still can't figure out how to do it. $\endgroup$
    – user105627
    Oct 26 '16 at 10:28
  • $\begingroup$ You are trying to get to the form of transform of $\exp(at) \cosh(bt)$ and $\exp(at)\sinh(bt)$. Generally, when dealing with Laplace transforms it's a bad idea to go guns blazing and factorize your fraction to smaller parts. Try remembering the shape of sin & cos transforms as, those will come about often. $\endgroup$ Oct 26 '16 at 13:13
  • $\begingroup$ @user105627 I added other link to a Laplace table. $\endgroup$ Oct 26 '16 at 13:16
  • $\begingroup$ I thought that by factorising the above it will be possible to end up with something of the form $e^{at}-e^{bt}/a-b$ for the $g$. $\endgroup$
    – user105627
    Oct 26 '16 at 17:42
  • $\begingroup$ You will... in the end. Factorizing is a long road usually. It's way more handy to rewrite in the form of cosh times exp. (note cosh is a sum of two exps itself!). $\endgroup$ Oct 26 '16 at 19:04

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