 # Find the inverse Laplace transforms of the functions given. Accurately sketch the time functions. a) F(s)=frac{3e^{-2s}}{s(s+3)} b) F(s)=frac{e^{-2s}}{s(s+1)} c) F(s)=frac{e^{-2s}-e^{-3s}}{2} FizeauV 2021-02-04 Answered
Find the inverse Laplace transforms of the functions given. Accurately sketch the time functions.
a) $F\left(s\right)=\frac{3{e}^{-2s}}{s\left(s+3\right)}$
b) $F\left(s\right)=\frac{{e}^{-2s}}{s\left(s+1\right)}$
c) $F\left(s\right)=\frac{{e}^{-2s}-{e}^{-3s}}{2}$
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Step 1
given that a) $F\left(s\right)=\frac{3{e}^{-2s}}{s\left(s+3\right)}$
b) $F\left(s\right)=\frac{{e}^{-2s}}{s\left(s+1\right)}$
find the inverse laplace transform of the above functions
Step 2
a)first lets find the inverse laplace transform of $\frac{3}{s\left(s+3\right)}$ let it write interms of partial fractions
$\frac{3}{s\left(s+3\right)}=\frac{A}{s}+\frac{B}{\left(s+3\right)}$
$3=A\left(s+3\right)+Bs$

lets substitute A and B values
$\frac{3}{s\left(s+3\right)}=\frac{1}{s}-\frac{1}{\left(s+3\right)}$
${L}^{-1}\left[\frac{3}{s\left(s+3\right)}\right]={L}^{-1}\left[\frac{1}{s}-\frac{1}{\left(s+3\right)}\right]$
$=1-{e}^{-3t}$
${L}^{-1}\left[\frac{3{e}^{-2s}}{s\left(s+3\right)}\right]=\left(1-{e}^{-3\left(t-2\right)}\right)u\left(t-2\right)$
Step 3
b) $F\left(s\right)=\frac{{e}^{-2s}}{s\left(s+1\right)}$
First lets find the inverse laplace transform of $\frac{1}{s\left(s+1\right)}$ for this lets write interms of partial fractions
$\frac{1}{s\left(s+1\right)}=\frac{A}{s}+\frac{B}{\left(s+1\right)}$
$1=A\left(s+1\right)+Bs$

$\frac{1}{s\left(s+1\right)}=\frac{1}{s}+\frac{1}{\left(s+1\right)}$
${L}^{-1}\left[\frac{1}{s\left(s+1\right)}\right]={L}^{-1}\left[\frac{1}{s}+\frac{1}{\left(s+1\right)}\right]$
$={L}^{-1}\left[\frac{1}{s}\right]-{L}^{-1}\left[\frac{1}{\left(s+1\right)}\right]$
$=1-{e}^{-t}$
next inverse laplace transform of $\frac{{e}^{-2s}}{s\left(s+1\right)}$ is $\left(1-{e}^{-\left(t-2\right)}\right)u\left(t-2\right)$