Use the second shifting theorem to find the Laplace transform

Rubik
Messages
95
Reaction score
0

Homework Statement



Use the second shifting theorem to find the Laplace transform of

f(t) = {t2, t<4
{t, t\geq4

Homework Equations



L{f(t - a)u(t - a)} = e-as F(s)

The Attempt at a Solution


Okay so I applied the unit step function to get the equation into the form
f(t) = t2 - t2(u)(t - 4) + t(u)(t - 4)
= t2(1 -u(t - 4) + t(u)(t - 4))

However now I get lost as I know it needs to be in the form f(t - a)u(t - a) and I see that the second part of equation is but the first part is confusing me I think maybe I am suppose to complete the square though I just can't see how?
 
Physics news on Phys.org
You've missplaced a parenthesis. There should be no t^3 power in your expression for f(t).

Anyway, you don't need to complete the square; that would not be appropriate. Think about what g(t1) needs to be so that g(t-a) = t2.
 
Sorry but I am still not sure what to do with the t2 but with my second term, t2(u)(t - 4) is right to do this..

L{t2(u)(t - 4)
= t2 8t +16 - 8t - 16
= (t - 4)2 + 8t - 16)(u)(t - 4)
= ((t - 4)2 _ 8(t - 4) + 16)(u)(t - 4)
= (2/s3 + 8/s2 and here I am not sure what 16 becomes?)e-4s
 
Sorry I posted this before than saw you reply in my other thread so replied with the example I was working with.. should I have linked you to this thread instead is that still okay?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top