Find the laplace transform of the following f(t)=tu_2(t) Ans. F(s)=left(frac{1}{s^2}+frac{2}{s}right)e^{-2s}

Find the laplace transform of the following f(t)=tu_2(t) Ans. F(s)=left(frac{1}{s^2}+frac{2}{s}right)e^{-2s}

Question
Laplace transform
asked 2020-10-31
Find the laplace transform of the following
\(f(t)=tu_2(t)\)
Ans. \(F(s)=\left(\frac{1}{s^2}+\frac{2}{s}\right)e^{-2s}\)

Answers (1)

2020-11-01
Step 1
first calculate Laplace transform of \(u_2(t)\)
\(F_u(s) =\int_-\infty^\infty u_2(t)e^{-st}dt\)
\(=\int_2^\infty e^{-st}dt\)
\(=\frac{e^{-2s}}{s}\)
Step 2
therefore, Laplace transform of \(tu_2(t)\) is :
-\frac{d}{ds}F_u(s)=-\frac{d}{ds}\cdot \frac{e^{-2s}}{s}\)
\(=\frac-{-2se^{-2s}-e^{-2s}}/{s^2}\)
\(=(\frac{1}{s^2}+\frac{2}{s})e^{-2s}\)
Therefore, Laplace transform of \(tu_2(t)\) is
\(=(\frac{1}{s^2}+\frac{2}{s})e^{-2s)}\)
0

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