How Do You Apply Shifting in Inverse Laplace Transforms?

[Khamul]

Hello everyone, I'm currently enrolled in Control Theory at my University, and part of the coursework requires differential equations; which wouldn't be a problem, if not for the fact it's been 2 years since I've taken D.E. Anyway, over the course of the problem I ran into this little function, and it's giving me a rough time..


F(s) = (1/6) / ((s+2)^2)

I'm attempting to take the inverse Laplace, but I'm not finding any explicit transform pairs that fit this function. I'll be honest, I remember that you're able to shift the function, but I have no recollection of how to do so. Would anyone be so kind as to help me out with this little bugger? I have the rest of the problem complete except for this stickler. Thank you in advance! :)

 

[Khamul]

Just a quick update on my progress; I found an inverse formula fitting the format (a)/((s+a)^2) = a*t*(e^-at)

So, given this, and knowing I have a 1/6 on the top, would I be able to do something similar to this?

f(t)= 1/6 * (L^-1) 1/((s+2)^2) * 2/2

Where I pull the 1/6 out in front of the function, and multiply the top and bottom of the function by 2 to get an a in the numerator, then pull the 2 in the denominator out, creating something like this?

f(t)=1/12 * (L^-1) 2/((s+2)^2)

I think this may be right..if so, I would really like clarification; thank you!

 

[I like Serena]

Yep. That's it.

 

[Khamul]

Great! Thank you for clearing that up, I knew you had to transform the functions, I just wasn't sure on the rules of being able to. I was initially making it a lot more complicated than it actually was I suppose :) cheers!