Unsure about Inverse Laplace Heaviside Function question

[1up20x6]

Homework Statement

Find the inverse Laplace transform of
F(s)=5e^(-8s)/(s2+36)

Homework Equations

The Attempt at a Solution

I know that to find the inverse Laplace transform of this function, I start by factoring out (e^(-8s)) to end up with 5/(s^2+36), and that my final answer will be step(t-8)f(t-8), where f(t-8) is the inverse of 5/(s^2+36). I've checked the Laplace table and I can't find the inverse of this function.

 

[jaytech]

It's because your understanding of the Heaviside Step Function is lacking definition. The fact that you have $\int$e^-8s($\frac{5}{s^{2}+36}$)ds is the same as saying you have a square wave with the function F(s) convoluted with it. When you integrate this, it's like integrating a Dirac-Delta function in the sense that if you do $\int$δ(t-a)f(t)dt you get f(t-a). Now imagine you have a bunch of δ(s-c) "glued together". That is one basic way to define the Heaviside Step Function u_c(t).

Now if your integral was just $\int$e^-8sds you'd have δ(t-8). However, if you have $\int$$\frac{e^{-8s}}{s}$ds, you get u_c(t-8). This is because your integral is of the form $\int$e^-8sF(s)ds where your F(s) is $\frac{1}{s}$. Now, the only difference between these two integrals is the F(s). What we are saying is we are performing the convolution between the Dirac Delta Function and a function F(s). The F(s) being $\frac{1}{s}$ though, when brought in the time domain, is just 1. So we are saying (once back in the time domain) that we have a 8 δ(t-a) "glued together", hence a square wave.

Moving on from here, if we have a F(s) that is not $\frac{1}{s}$ , but say $\frac{5}{s^{2}+36}$, then we are saying we have a bunch of F(s) spanning the square wave's domain. So in the time domain this would be represented as u_c(t)f(t-a). This says you have a function spanning the domain of the unit impulse u_c(t).

 

[Ray Vickson]

Homework Statement

Find the inverse Laplace transform of
F(s)=5e^(-8s)/(s2+36)

Homework Equations

The Attempt at a Solution

I know that to find the inverse Laplace transform of this function, I start by factoring out (e^(-8s)) to end up with 5/(s^2+36), and that my final answer will be step(t-8)f(t-8), where f(t-8) is the inverse of 5/(s^2+36). I've checked the Laplace table and I can't find the inverse of this function.

No: the inverse of $5/(s^2 + 36)$ is $f(t)$, not $f(t-8)$.

To find the inverse, either consult a better table, or else do it manually, by writing $1/(s^2 + 36)$ in partial fractions involving $1/(s+ 6i)$ and $1/(s - 6i)$, $i = \sqrt{-1}$. Now invert each term separately and combine the results.

 

[Chestermiller]

Isn't it 5u(t-8) sin(6(t-8))? Check a table of laplace transforms.

Chet

 

[jaytech]

Yes Chet, it is, but not a 5...it's a 5/6 i believe. I was just trying not to give him f(t-a) :p

So: $\frac{5}{6}$u_c(t)sin(6(t-8))

 

[Chestermiller]

Yes Chet, it is, but not a 5...it's a 5/6 i believe. I was just trying not to give him f(t-a) :p

So: $\frac{5}{6}$u_c(t)sin(6(t-8))

Oh boy. You're right. I guess I should learn how to use the tables also.

Chet