Question

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

Laplace transform
ANSWERED
asked 2020-11-05
Find the inverse laplace transform of the function
\(Y(s)=\frac{e^{-s}}{s(2s-1)}\)

Answers (1)

2020-11-06
Step 1
The Laplace transform is given \(Y(s)=\frac{e^{-s}}{s(2s-1)}\)
Convert the laplace transform into partial derivative ,
\(\frac{1}{s(2s-1)}=\frac{A}{s}+\frac{B}{2s-1}\)
\(1=A(2s-1)+B(s)\)
\(1=2As-A+Bs\)
\(1=s(2A+B)-A\)
Compare the coefficient of terms,
\(A=-1\)
\(2A+B=0\dots2\)
Step 2 Substitute value of A in equation 2,
\(2(-1)+B=0\)
\(B=2\)
The value of A is -1 and B is 2.
The expression is written as,
\(\frac{1}{s(2s-1)}=\frac{-1}{s}+\frac{2}{2s-1}\)
\(=\frac{-1}{s}+\frac{1}{(s-\frac{1}{2}}\)
For finding the inverse Laplace transform , use theorem below.
\(L^{-1}[e^{-sT}F(s)]=f(t-T)u(t-T)\)
Step 3
Applying the theorem , the given Laplace transform is written as,
\(L^{-1}[e^{-s}(\frac{-1}{s}+\frac{1}{(s-\frac{1}{2}}]=(e^{(\frac{t}{2}-\frac{1}{2})}-1)u(t-1)\)
The inverse Laplace transform of the function \(Y(s)=\frac{e^{-s}}{s(2s-1)}\) is , \((e^{(\frac{t}{2}-\frac{1}{2})}-1)u(t-1)\)
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