Question

In problems 37-48 find either F(s) or f(t), as indicated.

$\mathit{L}\mathbf{\left\{}\mathbf{\right(}\mathit{t}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathit{U}\mathbf{\right(}\mathit{t}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathbf{\right\}}$

Short answer

The f(t) of $\mathcal{L}\left\{\right(t-1\left)\mathcal{U}\right(t-1\left)\right\}$is obtained as$\mathcal{L}\left\{\right(t-1\left)U\right(t-1\left)\right\}=\frac{e^{-s}}{s^2}$

Step-by-step solution

Define the property

If $F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}$ and$a>0,then$$\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Replace the function with the values

We replace the function f(t) with t and a with 1 in the formula we have defined in the first step.

$\begin{array}{l}\mathcal{L}\left\{\right(t-1\left)U\right(t-1\left)\right\}=e^{-s}\mathcal{L}\left\{t\right\}\\ =\frac{e^{-s}}{s^2}\end{array}$ Since we know $\left(\mathcal{L}\left\{t\right\}=\frac{1}{s^2}\right)$

Thus, the solution is$L\left\{\right(t-1\left)U\right(t-1\left)\right\}=\frac{e^{-s}}{s^2}$