Question

laplace for (e^(-2t))/((s^2)+4)

Answer 1

Hm. I felt like the t in e^(-2t) might have been a mistake.
I'll assume it is not, and go on from there.

If we took the Laplace transform of e^(-2t) / (s^2 + 4), note that the denominator is a function of s already. But the Laplace transform is defined as this:
\(\displaystyle \mathcal{L} \left[ f(t) \right] (s) = \int_{0}^{\infty} e^{-st} f(t) \ dt \)
What I believe you would have to do here is translate the 1/(s^2 + 4) part out of the integral because it is the integral in t and not s, and find the Laplace transform of e^(-2t) alone so that:
\( \displaystyle \int_{0}^{\infty} e^{-st} \frac{e^{-2t}}{s^2 + 4} \ dt = \dfrac{1}{s^2 + 4} \int_{0}^{\infty} e ^{-st} e^{-2t} \ dt \\ \qquad \qquad \qquad \ \ \ \ \quad \displaystyle = \frac{1}{s^2 + 4 }\mathcal{L} \left(e^{-2t} \right) \)

Do you know how to find the Laplace transform of that function now?

Answer 2

thank you
you helped me but i think it would be (\[.5*\sin(2*(t-2))*u2\]

Answer 3

Okay, so I think you meant to have this:
Laplace transform is e^(-2s) / (s^2 + 4) , the original post has e^(-2t) with a "t" which confused me.

In which case, we do get the following:
0.5 * sin (2 (t - 2)) * u(t - 2)

Answer 4

yes i think we all are right

Answer 5

Great! Did you need any more help or are you good here? :)

Answer 6

yes

Answer 7

ln \[((s^2 +1)/(s*(s+1)))\]

Answer 8

Have you considered anything so far? I was thinking about long division / partial fraction decomposition to get simpler fractions...

Answer 9

Or more simply to reduce the numerator to a degree less than the denominator's
\( \dfrac{s^2 + 1}{s^2 + s} = \dfrac{s^2 \color{green}{+ s - s} + 1}{s^2 + s} = \dfrac{\cancel{s^2 \color{green}{+ s}}}{\cancel{s^2 + s}} \ \color{blue}1 + \dfrac{\color{green}{-s} + 1}{s^2 + s} \)
Then use partial fractions to split up (-s + 1) / (s^2 + s) and a table might lead the way from there.

Answer 10

its ln

Answer 11

it is ln ((s^2)+1) /(( s^2)+s)

Answer 12

Ohh! Sorry, I didn't notice it!

Well currently we have:
\( F(s) = \ln \left( \dfrac{s^2 + 1}{s(s + 1)} \right) \)
First we could break it up using the logarithm properties for product and quotients, summarized here::
\(\ln \left( \dfrac{a*b}{c*d} \right) = \ln a + \ln b - \left( \ln c + \ln d \right) \)
Note the minus sign distributes to both ln c and ln d...
\( F(s) = \ln \left(s^2 + 1 \right) - \left( \ln s + \ln (s + 1) \right) \)
\( F(s) = \ln \left( s^2 + 1 \right) - \ln s - \ln (s + 1) \)
The definition for derivative of logarithm converts these into rational functions which we better understand, right? So we might take the derivative of both sides and see waht to do with that.
\( F '(s) = \dfrac{1}{s^2 + 1} * 2s - \dfrac{1}{s} - \dfrac{1}{s+1} \)
At last, we can find the inverse Laplace transform.
On the left side, we have the derivative of a Laplace transform. We would benefit greatly from this identity:
\( \mathcal{L} \left[ t f(t) \right] = - F'(s) \)
\( \mathcal{L}^{-1} \left[ F '(s) \right] = -t f(t) \)
On the right side, notice that the first term is similar to the transform of cosine. The second and third terms appear to be exponentials! This all we can check on a Laplace transform chart such as http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf .

So if we put all these together, we would have something in this form:
\( - t f(t) = cos(at) + e^{bt} + e^{ct} \)
Where we simply solve for f(t) now by dividing off -t.

Answer 13

thank you

Answer 14

glad to help! :)