Laplace Transform - Step Function

twiztidmxcn
Messages
43
Reaction score
0

Homework Statement



What is the Laplace transform of f(t) = t^2 - 18 for 0 < t < 3 and f(t) = (t-3)^2 for t>3?

Homework Equations



Laplace Transforms

3. Work

Using Heaviside/step function, made equation into:

f(t) = t^2 - 18 + u(t-3)( (t-3)^2 - (t^2-18) )

Then, using Laplace transforms, found:

L{f(t)} = ( 2 / s^3 ) - ( 18 / s ) + e^(-3s)*L{27-6t}

I know that I have to put t in the L{27-6t} into either t-3 or t+3, but not sure which.

I went under the assumption that it t+3 would substitute for the t, so applying Laplace transforms I found:

L{27-6t} = L{27-6(t+3)} = L{27-6t-18} = L{9-6t} = 9/s - 6/s^2

Leaving me with the final answer:

L{f(t)} = 2/s^3 - 18/s + e^-3s*(9/s - 6/s^2)

...any chance this is close to right?
 
Last edited:
Physics news on Phys.org
em,you're quite right
Let
u(t-3)( (t-3)^2 - (t^2-18) )=u(t-3)g(t-3)
You're using the property of time shifting. Put t into t+3 and you get u(t)g(t).
Then
L{u(t-3)g(t-3)}=e^(-3s) L{u(t)g(t))}

ps:
g(t-3)=27-6t
g(t)=9-6t
 
Last edited:
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