Let f(t) be a function on displaystyle{left[{0},inftyright)}. The Laplace transform of fis the function F defined by the integral displaystyle{F}{left({s}right)}={int_{{0}}^{infty}}{e}^{{-{s}{t}}} f{{left({t}right)}}{left.{d}{t}right.} . Use this definition to determine the Laplace transform of the following function. displaystyle f{{left({t}right)}}={leftlbracebegin{matrix}{1}-{t}&{0}<{t}<{1}{0}&{1}<{t}end{matrix}right.}

Question
Laplace transform
asked 2020-12-27
Let f(t) be a function on \(\displaystyle{\left[{0},\infty\right)}\). The Laplace transform of fis the function F defined by the integral \(\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}\) . Use this definition to determine the Laplace transform of the following function.
\(\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}\)</span>

Answers (1)

2020-12-28
Step 1
We know that , \(\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}\)
Here \(\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}\)</span>
from \(\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{\left( f{{\left({t}\right)}}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{{1}}}{\left({1}-{t}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{\int_{{1}}^{\infty}}{0}\cdot{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\left({1}-{t}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{0}\)
\(\displaystyle{F}{\left({s}\right)}={{\left[\frac{{{\left({1}-{t}\right)}{e}^{{-{s}{t}}}}}{{-{s}}}-\frac{{{\left(-{1}\right)}{e}^{{-{s}{t}}}}}{{{s}^{2}}}\right]}_{{0}}^{{1}}}\)
\(\displaystyle{F}{\left({s}\right)}=\frac{1}{{s}}+\frac{{{e}^{{-{s}}}}}{{s}}-\frac{1}{{s}^{2}}\)
Step 2
This is required Laplace Transform.
0

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