boitshupoO

2020-11-05

Find the inverse laplace transform of the function
$Y\left(s\right)=\frac{{e}^{-s}}{s\left(2s-1\right)}$

unett

Expert

Step 1
The Laplace transform is given $Y\left(s\right)=\frac{{e}^{-s}}{s\left(2s-1\right)}$
Convert the laplace transform into partial derivative ,
$\frac{1}{s\left(2s-1\right)}=\frac{A}{s}+\frac{B}{2s-1}$
$1=A\left(2s-1\right)+B\left(s\right)$
$1=2As-A+Bs$
$1=s\left(2A+B\right)-A$
Compare the coefficient of terms,
$A=-1$
$2A+B=0\dots 2$
Step 2 Substitute value of A in equation 2,
$2\left(-1\right)+B=0$
$B=2$
The value of A is -1 and B is 2.
The expression is written as,
$\frac{1}{s\left(2s-1\right)}=\frac{-1}{s}+\frac{2}{2s-1}$
$=\frac{-1}{s}+\frac{1}{\left(s-\frac{1}{2}}$
For finding the inverse Laplace transform , use theorem below.
${L}^{-1}\left[{e}^{-sT}F\left(s\right)\right]=f\left(t-T\right)u\left(t-T\right)$
Step 3
Applying the theorem , the given Laplace transform is written as,
${L}^{-1}\left[{e}^{-s}\left(\frac{-1}{s}+\frac{1}{\left(s-\frac{1}{2}}\right]=\left({e}^{\left(\frac{t}{2}-\frac{1}{2}\right)}-1\right)u\left(t-1\right)$
The inverse Laplace transform of the function $Y\left(s\right)=\frac{{e}^{-s}}{s\left(2s-1\right)}$ is , $\left({e}^{\left(\frac{t}{2}-\frac{1}{2}\right)}-1\right)u\left(t-1\right)$

Do you have a similar question?