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}

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}

Question
Laplace transform
asked 2021-02-04
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) \(F(s)=\frac{e^{-2s}-e^{-3s}}{2}\)

Answers (1)

2021-02-05
Step 1
given that a) \(F(s)=\frac{3e^{-2s}}{s(s+3)}\)
b) \(F(s)=\frac{e^{-2s}}{s(s+1)}\)
find the inverse laplace transform of the above functions
Step 2
a)first lets find the inverse laplace transform of \(\frac{3}{s(s+3)}\) let it write interms of partial fractions
\(\frac{3}{s(s+3)}=\frac{A}{s}+\frac{B}{(s+3)}\)
\(3=A(s+3)+Bs\)
\(\text{put } s=0 \text{ then } 3=3A , A=1\)
\(\text{put } s=-3 \text{ then } 3=-3B , B=-1\)
lets substitute A and B values
\(\frac{3}{s(s+3)}=\frac{1}{s}-\frac{1}{(s+3)}\)
\(L^{-1}\left[\frac{3}{s(s+3)}\right]=L^{-1}\left[\frac{1}{s}-\frac{1}{(s+3)}\right]\)
\(=1-e^{-3t}\)
\(L^{-1}\left[\frac{3e^{-2s}}{s(s+3)}\right]=(1-e^{-3(t-2)})u(t-2)\)
Step 3
b) \(F(s)=\frac{e^{-2s}}{s(s+1)}\)
First lets find the inverse laplace transform of \(\frac{1}{s(s+1)}\) for this lets write interms of partial fractions
\(\frac{1}{s(s+1)}=\frac{A}{s}+\frac{B}{(s+1)}\)
\(1=A(s+1)+Bs\)
\(\text{put } s=0 \text{ then } 1=A , A=1\)
\(\text{put } s=-1 \text{ then } 1=-B , B=-1\)
\(\frac{1}{s(s+1)}=\frac{1}{s}+\frac{1}{(s+1)}\)
\(L^{-1}\left[\frac{1}{s(s+1)}\right]=L^{-1}\left[\frac{1}{s}+\frac{1}{(s+1)}\right]\)
\(=L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{(s+1)}\right]\)
\(=1-e^{-t}\)
next inverse laplace transform of \(\frac{e^{-2s}}{s(s+1)}\) is \((1-e^{-(t-2)})u(t-2)\)
0

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