Question

make up two functions f₁ and f ₂ that have the same Laplace transform. do not think profound thoughts.

Short answer

Functions f₁ and f₂ that have the same Laplace transform.

$\begin{array}{rcl}f\left(t\right)& =& 1\\ L\left\{f\left(t\right)\right\}& =& \frac{1}{s}\\ L\left\{f\left(t\right)\right\}& =& L\left\{g\left(t\right)\right\}\end{array}$

Step-by-step solution

Step1: definition of Laplace transformation.

A transformation of a function f(x) into the function is particularly useful for reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Consider the following:

$\begin{array}{rcl}f\left(t\right)& =& 1\\ L\left\{f\left(t\right)\right\}& =& \frac{1}{s}\end{array}$

$$\begin{gathered} g\left(t\right)=\left\{\begin{array}{l}1,t\ge 0,t\ne 1\\ 0,t=1\end{array}\right. \\ f\left(t\right)\ne g\left(t\right)butL\left\{g\left(t\right)\right\}=\frac{1}{S} \end{gathered}$$

$L\left\{g\left(t\right)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}g\left(t\right)dt$

 derived the values.

$\begin{array}{rcl}L\left\{g\left(t\right)\right\}& =& \underset{0}{\overset{\infty }{\int }}{e}^{-st}g\left(t\right)dt\\ & =& \underset{0}{\overset{\infty }{\int }}{e}^{-st}.1dt\\ & =& \frac{e^{-st}}{-s}\underset{0}{\overset{\infty }{\int }}=\frac{1}{s}\end{array}$

role="math" localid="1663978309644" $\begin{array}{rcl}f\left(t\right)& \ne & g\left(t\right)\\ L\left\{f\left(t\right)\right\}& =& L\left\{g\left(t\right)\right\}\end{array}$

Hence,

$\begin{array}{rcl}f\left(t\right)& \ne & g\left(t\right),L\left\{f\left(t\right)\right\}=L\left\{g\left(t\right)\right\}\end{array}$