Question

Obtenha a transformada de Laplace da função g(t)=sen(2t)t[/tex]

Answer 1

• O resultado da transformada de Laplace da função dada é:$\Large\displaystyle\begin{gathered}\pounds \{\sin(2t)t\} =\frac{4s}{(s^2+4)^2} \end{gathered}$

Desejamos calcular a transformada de Laplace da função $\large\displaystyle\begin{gathered}\mathbf{ g(t)=\sin(2t)t}\end{gathered}$, para isso, temos que:

 $\Large\displaystyle\begin{gathered}\pounds \left\{t^nf(t)\right\} =(-1)^n\frac{d^n}{dx^n}\left(\pounds\{f(t)\} \right) \end{gathered}$

E aplicando isso na função $\large\displaystyle\begin{gathered}\mathbf{ g(t)=\sin(2t)t}\end{gathered}$, ficamos da seguinte forma:

$\Large\displaystyle\begin{gathered}\pounds \{\sin(2t)t\} =-\frac{d}{dx}\left(\pounds\{\sin(2t)\} \right) \end{gathered}$

Agora, vale ressaltar a fórmula da transformada de Laplace:

$\Large\displaystyle\begin{gathered}\pounds\{f(t)\}=\int ^{\infty}_{0^+} e^{-st}f(t)dt=F(s)\end{gathered}$

Agora, vamos calcular a transformada do sin(ωt), para isso, iremos utilizar a Identidade de Euler. Pois se colocarmos na fórmula, ficaria uma integral muito complexa.

 $\Large\displaystyle\begin{gathered}e^{i \omega t}=\cos(\omega t)+i\sin(\omega t) \end{gathered}$

 $\Large\displaystyle\begin{gathered}e^{-i \omega t}=\cos(\omega t)-i\sin(\omega t) \end{gathered}$

Vamos agora eliminar o cos(ωt), para isso, iremos subtrair as duas relações.

 $\Large\displaystyle\begin{gathered}e^{i \omega t}-e^{-i\omega t}=2i\sin(\omega t) \end{gathered}$

Com isso, temos que:

$\Large\displaystyle\text{ \begin{gathered}\boxed{\sin(\omega t)=\frac{e^{i \omega t}-e^{-i\omega t}}{2i} }\end{gathered} }$

Substituindo na transformada:

$\Large\displaystyle\text{ \begin{gathered}\pounds \{\sin(\omega t)\}= \pounds\left\{\frac{e^{i\omega t}-e^{-i\omega t}}{2i} \right\}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\pounds \{\sin(\omega t)\}= \frac{1}{2i}\cdot\left( \pounds\{e^{i\omega t} \}-\pounds\{e^{-i\omega t}\}\right)\end{gathered} }$

Vamos agora calcular a transformada de e^{φt} recorrendo a fórmula:

$\Large\displaystyle\text{ \begin{gathered}\pounds \{e^{\phi t}\}=\int ^{\infty}_{0^+}e^{-st}e^{\phi t}dt\end{gathered} }$

E pela propriedade do expoente de mesma base, temos que:

$\Large\displaystyle\text{ \begin{gathered}\pounds \{e^{\phi t}\}=\int ^{\infty}_{0^+}e^{t(-s+\phi)}dt\end{gathered} }$

• Método da substituição simples:

$\Large\displaystyle\begin{gathered}\star\ u=t(-s+\phi)\end{gathered}$

$\Large\displaystyle\begin{gathered}\star\ \frac{du}{-s+\phi }=dt\end{gathered}$

Desse modo, surge que:

$\Large\displaystyle\text{ \begin{gathered}\pounds \{e^{\phi t}\}= -\int ^{0^+}_{-\infty}e^{u}\frac{du}{-s+\phi}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\pounds \{e^{\phi t}\}= \frac{1}{s-\phi}\cdot \int ^{0^+}_{-\infty}e^{u}du\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\boxed{\pounds \{e^{\phi t}\}= \frac{1}{s-\phi}}\end{gathered} }$

Sabendo disso, a transformada do sin(ωt) fica da seguinte forma:

$\Large\displaystyle\text{ \begin{gathered}\pounds \{\sin(\omega t)\}= \frac{1}{2i}\cdot\left( \pounds\{e^{i\omega t} \}-\pounds\{e^{-i\omega t}\}\right)\end{gathered} }$

$\Large\displaystyle\begin{gathered}\pounds \{\sin(\omega t)\}= \frac{1}{2i}\cdot\left( \frac{1}{s-i\omega} -\frac{1}{s+i\omega} \right)\end{gathered}$

$\Large\displaystyle\begin{gathered}\pounds \{\sin(\omega t)\}= \frac{1}{\diagup\!\!\!2i}\cdot\left( \frac{\diagup\!\!\!2i\omega}{s^2+\omega^2} \right)\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\rightsquigarrow\underline{\boxed{\pounds \{\sin(\omega t)\}= \frac{\omega}{s^2+\omega^2}}} \end{gathered} }$

E como queremos a transformada do sin(2t), logo:

$\Large\displaystyle\begin{gathered}\pounds \{\sin(\omega t)\}= \frac{\omega}{s^2+\omega^2} \end{gathered}$

$\Large\displaystyle\begin{gathered}\pounds \{\sin(2 t)\}=\frac{2}{s^2+4} \end{gathered}$

Agora é só calcular a derivada disso.

$\Large\displaystyle\begin{gathered}\pounds \{\sin(2t)t\} =-\frac{d}{dx}\left(\pounds\{\sin(2t)\} \right) \end{gathered}$

$\Large\displaystyle\begin{gathered}\pounds \{\sin(2t)t\} =-\frac{d}{dx}\left(\frac{2}{s^2+4} \right) \end{gathered}$

Para isso, temos a derivada do quociente, dada da seguinte forma:

$\Large\displaystyle\begin{gathered}\left(\frac{f}{g} \right)'=\frac{f'\cdot g-f\cdot g'}{g^2} \end{gathered}$

Aplicando na questão:

$\Large\displaystyle\begin{gathered}\pounds \{\sin(2t)t\} =-\left(\frac{-4s}{(s^2+4)^2} \right) \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\boxtimes\ \green{\underline{ \boxed{\pounds \{\sin(2t)t\} =\frac{4s}{(s^2+4)^2}}}} \end{gathered} }$

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