Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).F(s) = (7e−7s)/(s2 − 49)f(t) =

Respuesta :

Answer:

F(t)[tex]=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}[/tex]

Step-by-step explanation:

We have given [tex]F(S)=\frac{7e^{-7s}}{s^2-49}[/tex]

Now  [tex]F(S)=e^{-7s}G(s)[/tex]

Here [tex]G(S)=\frac{7}{S^2-49}[/tex]

Now first find the Laplace inverse of G(S)

Using partial fraction

[tex]\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}[/tex]

[tex]7=A(S-7)+B(S+7)[/tex]

On comparing the coefficient

[tex]A=\frac{1}{2}[/tex]  and [tex]B=\frac{-1}{2}[/tex]  

On putting the value of A and B  

[tex]G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}[/tex]

Taking inverse Laplace

[tex]G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}[/tex]

Now in G(s) there is onether term [tex]e^{-7s}[/tex]

So F(t)[tex]=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}[/tex]

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