Question

In Problem $E\left(t\right)=E_0{e}^{-kt},k>0.$$\mathcal{L}\left\{\right(t-1\left)\mathcal{U}\right(t-1\left)\right\}$ems $37-38$find either $F\left(s\right)$or $f\left(t\right)$, as indicated$L^{-1}\left\{\frac{{(1+e^{-2s})}^2}{s+2}\right\}$

Short answer

The Laplace transform is$L^{-1}\left\{\frac{{(1+e^{-2s})}^2}{s+2}\right\}=e^{-2t}+2e^{-2(t-2)}U(t-2)+e^{-2(t-4)}U(t-4)$

Step-by-step solution

Definition of the Unit Step Function 

UNIT STEP FUNCTION:

In engineering, one frequently encounters functions that are either “off” or “on.” It is convenient, then, to define a special function that is the number 0 (off) up to a certain time and then the number 1 (on) after that time. This function is called the unit step function or the Heaviside function.

Distribute the power of numerator. 

$L^{-1}\left\{\frac{{(1+e^{-2s})}^2}{s+2}\right\}=L^{-1}\left\{\frac{1+2e^{-2s}+e^{-4s}}{s+2}\right\}$

Separate numerators.

$L^{-1}\left\{\frac{{(1+e^{-2s})}^2}{s+2}\right\}=L^{-1}\left\{\frac{1}{s+2}+\frac{2e^{-2s}}{s+2}+\frac{e^{-4s}}{s+2}\right\}$

Shifting operator using a unit step function

$L^{-1}\left\{\frac{{(1+e^{-2s})}^2}{s+2}\right\}=e^{-2t}+2e^{-2(t-2)}U(t-2)+e^{-2(t-4)}U(t-4)$

Hence, the answer is$L^{-1}\left\{\frac{{(1+e^{-2s})}^2}{s+2}\right\}=e^{-2t}+2e^{-2(t-2)}U(t-2)+e^{-2(t-4)}U(t-4)$