 # Use the definition of Laplace Transforms to find Lleft{f(t)right}f(t)=begin{cases}-1 & 0leq t <11 & tgeq 1end{cases}Then rewrite f(t) as a sum of step functions, u_c(t), and show that by taking Laplace transforms, this yields the same answer as your direct computation. Braxton Pugh 2021-02-12 Answered

Use the definition of Laplace Transforms to find $L\left\{f\left(t\right)\right\}$
$f\left(t\right)=\left\{\begin{array}{ll}-1& 0\le t<1\\ 1& t\ge 1\end{array}$
Then rewrite f(t) as a sum of step functions, ${u}_{c}\left(t\right)$, and show that by taking Laplace transforms, this yields the same answer as your direct computation.

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Step 1
Consider the given function:
$f\left(t\right)=\left\{\begin{array}{ll}-1& 0\le t<1\\ 1& t\ge 1\end{array}$
Step 2
Now, by definition of Laplace Transform:
$L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$={\int }_{0}^{1}{e}^{-st}f\left(t\right)dt+{\int }_{1}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$={\int }_{0}^{1}{e}^{-st}\left(-1\right)dt+{\int }_{1}^{\mathrm{\infty }}{e}^{-st}\left(1\right)dt$
$={\left[\frac{{e}^{-st}}{s}\right]}_{0}^{1}+{\left[\frac{{e}^{-st}}{-s}\right]}_{1}^{\mathrm{\infty }}$
$=\left(\frac{{e}^{-s}}{s}-\frac{1}{s}\right)+\left(0+\frac{{e}^{-s}}{s}\right)$
$L\left\{f\left(t\right)\right\}=\frac{2{e}^{-s}}{s}-\frac{1}{s}$
Step 3
Now, Rewrite f(t) as the sum of step functions, ${u}_{c}\left(t\right)$:
$f\left(t\right)=f\left(t\right)=\left\{\begin{array}{ll}-1& 0\le t<1\\ 1& t\ge 1\end{array}$
$=\left(-1\right)\left({u}_{0},1\left(t\right)\right)+\left(1\right)\left({u}_{1}\left(t\right)\right)$
$=\left(-1\right)\left({u}_{0}\left(t\right)-{u}_{1}\left(t\right)\right)+\left(1\right)\left({u}_{1}\left(t\right)\right)$
$=-{u}_{0}\left(t\right)+2{u}_{1}\left(t\right)$
Step 4 Now, use the following formula for Laplace transform of step function:
$L\left\{{u}_{c}\left(t\right)\right\}=\frac{{e}^{-cs}}{s}$
Step 5
Thus, now take Laplace transform:
$f\left(t\right)=-{u}_{0}\left(t\right)+2{u}_{1}\left(t\right)$
$L\left\{f\left(t\right)\right\}=L\left\{-{u}_{0}\left(t\right)\right\}+L\left\{2{u}_{1}\left(t\right)\right\}$
$=-\frac{{e}^{-\left(0\right)s}}{s}+2\frac{{e}^{-\left(1\right)s}}{s}$
$L\left\{f\left(t\right)\right\}=-\frac{1}{s}+2\frac{{e}^{-s}}{s}$
Step 6
Thus, from equation (1) and (2), it can be seen that the Laplace transform of f(t) that is obtained from both the methods is same that is :
$L\left\{f\left(t\right)\right\}=2\frac{{e}^{-s}}{s}-\frac{1}{s}$