Find the inverse Laplace transform of $F(s)=\frac{5e^{−6s}}{s^2+4}$

Question

Find the inverse Laplace transform of $F(s)=\dfrac{5e^{−6s}}{s^2+4}$

$f(t)=$ __________?

Here is my work: $L{(5/2) \sin(2t)} = 5/(s^2 + 4)$, we have by the shifting theorem $f(t) = (5/2) \sin(2(t - 6)) u(t - 6)$

Please help me with the correct solution. Thanks

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Ruslan · Accepted Answer · 2013-04-03T23:06:49.760

1

You have already gotten the correct solution if by $u(x)$ you mean Heaviside unit step function $\theta(x)$.
If you want to eliminate Heaviside function you could as well write $$f(t)=\left\{\begin{matrix}0,&\ \text{if}~x<6\\ \frac{5}{2}\sin(2(t-6)),&\ \text{if}~x>6\end{matrix}\right.$$

edited Apr 03 '13 at 23:06

answered Apr 03 '13 at 21:35

Ruslan

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what would the u equal in this case? – Michael Rametta Apr 03 '13 at 21:58
@Michael $u(x)=\theta(x)=\left{\begin{matrix}1: x>0\ 0: x<0\end{matrix}\right.$ – Ruslan Apr 03 '13 at 22:05
so how would i set up f(t) properlly – Michael Rametta Apr 03 '13 at 22:12
Almost the same as you've already written: $$f(t)=\frac{5}{2}sin(2(t-6))\theta(t-6)$$ – Ruslan Apr 03 '13 at 22:14
In fact, your solution was already correct as I now see that $u(x)$ is also a common notation for Heaviside step function. – Ruslan Apr 03 '13 at 22:21
so there is no number that yo can substitue for theta? – Michael Rametta Apr 03 '13 at 22:48
No, theta is a function not a constant. You can instead write $f(t)$ as a piecewise function which in your case would be $0$ if $t<6$ and $\frac{5}{2}\sin(2(t-6))$ if $t>6$. – Ruslan Apr 03 '13 at 22:55

Find the inverse Laplace transform of $F(s)=\frac{5e^{−6s}}{s^2+4}$

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