Find the Inverse Laplace transforms of the given functions.
- (a) \(\displaystyle{F(s)= \frac{\sqrt {2}}{s^2-9} }\)
(b) \(\displaystyle{F(s)= \frac{3s}{s^2+4}} \) solution
- \(\displaystyle{F(s)= \frac{s-3}{s^2-6s+10}} \) solution
- (a) \(\displaystyle{F(s)= \frac{1}{s^2-4s+5}}\) (b) \(\displaystyle{F(s)=
\frac{2s}{s^2-s-6} }\) solution
- \(\displaystyle{F(s)= \frac{2s+1}{s^2-2s+2} }\) solution
- \(\displaystyle{F(s)= \frac{1-2s}{s^2+4s+3} }\) solution
- (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
- \(H(s)=\left(\frac{2}{s}+\frac{1}{s^2}\right)+e^{-s}\left(\frac{3}{s}-\frac{1}{s^2}\right)+e^{-3 s}\left(\frac{1}{s}+\frac{1}{s^2}\right)\) solution
- \(\displaystyle{F(s) = \frac{e^{-2s}}{s^2+s-2}}\) solution
- \(\displaystyle{F(s) = \frac{e^{-5s}}{s^2-25}}\) solution
- \(\displaystyle{F(s) = \frac{(s-1)e^{-2s}}{s^2-5s+6}}\) solution
- \(\displaystyle{F(s)= \frac{1}{s(s+1)} }\) (using Convolution integral) solution