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]