Question

determine the inverse Laplace transform of F.
\(F(s)=\frac{e^{-2s}}{(s-3)^3}\)

Answer 1

Step 1
Let f(t) be the inverse Laplace transform of the given function
\(F(s)=\frac{e^{-2s}}{(s-3)^3}\)
Take inverse Laplace transform on both sides:
\(L^{-1}\left[F(s)\right]=L^{-1}\left[\frac{e^{-2s}}{(s-3)^3}\right]\)
\(\Rightarrow f(t)=L^{-1}\left[\frac{e^{-2s}}{(s-3)^3}\right]\)
Use Inverse Laplace Transform Rule :
If \(L^{-1}\left[F(s)\right]=f(t) , \text{ then } L^{-1}\left[e^{-as}F(s)\right]=H(t-a)f(t-a)\) where , H(t) is Heavside step function
Now, we have a = 2 in our case
Hence,
\(f(t)=H(t-2)L^{-1}\left[\frac{1}{(s-3)^3}\right]\)
Step 2
Calculation of inverse Laplace Transform of function on right side:
Apply inverse Laplace Transform rule : If \(L^{-1}[F(s)]=f(t) , \text{ then } L^{-1}[F(s-a)]=e^{at}f(t)\)
\(\Rightarrow L^{-1}\left[\frac{1}{(s-3)^3}\right]=e^{3t}L^{-1}\left[\frac{1}{s^3}\right]=e^{3t}L^{-1}\left[\frac{2}{2s^3}\right]\)
\(\Rightarrow L^{-1}\left[\frac{1}{(s-3)^3}\right]=e^{3t}\frac{1}{2}L^{-1}\left[\frac{2}{s^3}\right]\)
\(\Rightarrow L^{-1}[\frac{1}{(s-3)^3}]=e^{3t}\frac{1}{2t^2} \left\{\text{ Using } L^{-1}\left[\frac{n!}{s^(n+1)}\right]=t^n \right\}\)
\(\Rightarrow L^{-1}\left[\frac{1}{(s-3)^3}\right]=\frac{t^2}{2e^{3t}}\)
Hence, we get:
\(f(t)=H(t-2)\left(\frac{t-2}{2}\right)e^{3(t-2)}\)
Step 3
Answer:
Inverse Laplace Transform is given by:
\(f(t)=H(t-2)\left(\frac{t-2}{2}\right)e^{3(t-2)}\)