Inverse Laplace Transforms:     Algebra Review

Find the Inverse Laplace transforms of the given functions.

  1. (a) \(\displaystyle{F(s)= \frac{\sqrt {2}}{s^2-9} }\)    (b) \(\displaystyle{F(s)= \frac{3s}{s^2+4}} \)       solution

  2.  \(\displaystyle{F(s)= \frac{s-3}{s^2-6s+10}} \)       solution

  3.   (a) \(\displaystyle{F(s)= \frac{1}{s^2-4s+5}}\)   (b) \(\displaystyle{F(s)= \frac{2s}{s^2-s-6} }\)       solution

  4.  \(\displaystyle{F(s)= \frac{2s+1}{s^2-2s+2} }\)       solution

  5.   \(\displaystyle{F(s)= \frac{1-2s}{s^2+4s+3} }\)       solution

  6. (a) \( F(s)=\dfrac{s^3}{s^4-1}\)   (b) \(F(s)=\dfrac{s^3+s^2}{s^4-1}\)   (c) \( F(s)=\dfrac{s^3+s}{s^4-1} \) solution

  7.  \(H(s)=\left(\frac{2}{s}+\frac{1}{s^2}\right)+e^{-s}\left(\frac{3}{s}-\frac{1}{s^2}\right)+e^{-3 s}\left(\frac{1}{s}+\frac{1}{s^2}\right)\)  solution

  8.   \(\displaystyle{F(s) = \frac{e^{-2s}}{s^2+s-2}}\)       solution

  9.   \(\displaystyle{F(s) = \frac{e^{-5s}}{s^2-25}}\)      solution

  10.   \(\displaystyle{F(s) = \frac{(s-1)e^{-2s}}{s^2-5s+6}}\)      solution

  11.  \(\displaystyle{F(s)= \frac{1}{s(s+1)} }\)   (using Convolution integral)   solution