The Inverse Laplace transform is used to go back from the Laplace domain to the real-values function. In the article Laplace Transform there is shown that by bringing problems in the s-domain computations can be performed more easily. I.e. a convolution becomes a multiplication and an ode turns into an algebraic expression.
The official way to compute an Inverse Laplace transform is by the use of the following integral;
L−1{F(s)}=2πi1∫γ−i∞γ+i∞estF(s)ds Again, the integral is the formal way of computing L−1{F(s)}, however, we can use the same lookup as with the conversion from f(t) --> F(s), but now it is simply the other way around, F(s) --> f(t)
| Frequency Domain F(s) | Time Domain L−1{F(s)}=f(t) |
|---|
| s1 | 1 |
| 1 | δ(t) |
| e−τs | δ(t−τ) |
| sn+1n! | tn |
| (s+α)n+1n! | tne−αt |
| s−α1 | eαt |
| s2+ω2ω | sin(ωt) |
| (s−α)2+ω2ω | eαtsin(ωt) |
| s2+ω2s | cos(ωt) |
| (s−α)2+ω2s−α | eαtcos(ωt) |
| ... | ... |
Example Lookup Table
Let's say we need to transfer the following function in the s-domain back toward the time domain;
F(s)=s2+15s−1=s2+15s−s2+11 By using the properties of the lookup table we can find f(t), which equals (check yourself);
f(t)=L−1{F(s)}=5cos(t)−sin(t) Of course, many different examples can be thought of, in general, it is good practice to recognize particular forms present in the lookup table. The same is done in the example. For example, rewriting the fraction forms can appear in a more clear structure.
In this article, there is shown how to compute L−1{F(s)}. In further articles, we dive deeper into the computation of the formal integrals.
