No two functions have the same Laplace transform. That means that the transform ought to be invertible: we ought to be able to work out the original function if we know its transform.
Indeed we can. The easiest way to do this is, first, to build up a "look-up table" of Laplace transforms of key functions, and then recall the two shift functions: especially the one that says that if $$\mathcal{L}(f(t))=F(s)$$ then $$\mathcal{L}(f(t)\,e^{-\alpha t})=F(s+\alpha).$$ Our task then becomes to represent a function of s as a sum of known transforms. A good first step is usually to reduce the function to partial fractions.
Function |
Laplace transform |
|---|---|
1 |
{\frac{1}{s}} |
t |
{\frac{1}{s^2}} |
| t^n | {\frac{n!}{s^{n+1}}} |
e^{at} |
{\frac{1}{s-a}} |
\cos\,\omega t |
{\frac{s}{s^2+\omega^2}} |
\sin\,\omega t |
{\frac{\omega}{s^2+\omega^2}} |
\cosh\,\omega t |
{\frac{s}{s^2-\omega^2}} |
\sinh\,\omega t |
{\frac{\omega}{s^2-\omega^2}} |
For example, if we're trying to calculate the inverse Laplace transform of $$\frac{2s^3+6s^2-4s-14}{s^4+2s^3-2s^2-6s+5}.$$ The first thing to notice is that if we substitute s=1 into the numerator, we get 0; by the Factor Theorem, it follows that (s-1) is a factor of s^4+2s^3-2s^2-6s+5. Continuing in this way:
\begin{eqnarray} s^4+2s^3-2s^2-6s+5&=&(s-1)(s^3+3s^2+s-5)\\ &=&(s-1)^2(s^2+4s+5), \end{eqnarray}
and (s^2+4s+5) doesn't factorise. Now, $$\frac{2s^3+6s^2-4s-14}{(s-1)^2(s^2+4s+5)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C\,s+D}{s^2+4s+5}.$$ Multiplying both sides by (s-1)^2(s^2+4s+5), $$2s^3+6s^2-4s-14=A\,(s-1)(s^2+4s+5)+B\,(s^2+4s+5)+(C s +D)\,(s-1)^2.$$ Substituting s=1 gives $$-10=10B,$$ hence B=-1. Comparing coefficients of s^3, we obtain $$2=A+C.$$ Comparing coefficients of s, we obtain,
\begin{eqnarray} -4&=&A+4B+C-2D\\ &=&4B+(A+C)-2D\\ &=&-4+2-2D, \end{eqnarray}
whence D=1. Comparing constant terms, we obtain $$-14=-5A+5B+D=-5A-5+1,$$ whence A=2 and hence C=0. So $$\frac{2s^3+6s^2-4s-14}{(s-1)^2(s^2+4s+5)}=\frac{2}{s-1}-\frac{1}{(s-1)^2}+\frac{1}{s^2+4s+5}.$$ Now the inverse Laplace transform of 2/(s-1) is 2e^{1\times t}. Less straightforwardly, the inverse Laplace transform of 1/s^2 is t and hence, by the first shift theorem, that of 1/(s-1)^2 is t\,e^{1\times t}.
Less straightforwardly still, note that $$\frac{1}{s^2+4s+5}=\frac{1}{(s+2)^2+1^2},$$ and recall that the inverse Laplace transform of $$\frac{1}{s^2+1^2}$$ is $$\sin (1\times t).$$ It follows from the shift theorem that the inverse Laplace transform of $$\frac{1}{(s+2)^2+1^2}$$ is $$\sin t\times e^{-2 t}.$$ So, overall, the inverse Laplace transform of $$\frac{2s^3+6s^2-4s-14}{s^4+2s^3-2s^2-6s+5}$$ is $$e^t\,(2-t)+e^{-2t}\sin t.$$