Question

In Problems 3 - 24 fill in the blanks or answer true or false.

12.$\mathit{L}\left\{sin\left(2t\right)U\left(t-\pi \right)\right\}$=…….

Short answer

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

Step-by-step solution

Consider the function

The function is,

$\begin{array}{l}L\left\{\sin \left(2t\right)\mathrm{U}\left(t-\pi \right)\right\}\\ Thes\tan dardformulaisgivenas,\\ L\left\{\sin \left(at\right)\right\}=\frac{a}{s^2+a^2}\end{array}$

Compute the Laplace transform

Consider the function,

$\begin{array}{l}L\left\{\sin \left(2t\right)\mathrm{U}\left(t-\pi \right)\right\}\\ Thes\tan dardformulais,\\ L\left\{\mathrm{U}\left(t-c\right)f\left(t\right)\right\}=e^{-cs}L\left\{f\left(t+c\right)\right\}\\ Here,\\ \mathrm{For} \sin \left(2t\right)\mathrm{U}\left(t-\pi \right):f\left(t\right)=\sin \left(2t\right), c=\pi \\ AndL\left\{\sin \left(2\left(t+\pi \right)\right)\right\}=L\left\{\sin \left(2t\right)\right\}=\frac{2}{s^2+4}\\ L\left\{U\left(t-\pi \right)\right\}=e^{-\pi s}\\ SubstitutethevalueandfindtheLaplacetransform.\\ L\left\{\sin \left(2t\right)\mathrm{U}\left(t-\pi \right)\right\}=\frac{2}{s^2+4}{e}^{-\pi s}\\ \Rightarrow L\left\{\sin \left(2t\right)\mathrm{U}\left(t-\pi \right)\right\}=\frac{2e^{-\pi s}}{s^2+4}\\ Therefore,theLaplacetransformofL\left\{\sin \left(2t\right)\mathrm{U}\left(t-\pi \right)\right\}is,\\ \therefore L\left\{\sin \left(2t\right)\mathrm{U}\left(t-\pi \right)\right\}=\frac{2e^{-\pi s}}{s^2+4}\end{array}$