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jenyasd209 [6]
3 years ago
8

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) =
Mathematics
1 answer:
zzz [600]3 years ago
6 0

Answer:

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

Step-by-step explanation:

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

Now  F(S)=e^{-7s}G(s)

Here G(S)=\frac{7}{S^2-49}

Now first find the Laplace inverse of G(S)

Using partial fraction

\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}

7=A(S-7)+B(S+7)

On comparing the coefficient

A=\frac{1}{2}  and B=\frac{-1}{2}  

On putting the value of A and B  

G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}

Taking inverse Laplace

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

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

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

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