How Do You Perform a Laplace Transform with Algebraic Manipulations?

[Chandasouk]

Can someone explain the steps of this solution? In my linear systems class, we are doing Laplace transforms using transform tables and the properties. I can usually do the problems that closely resemble the table, but when they involve heavy algebraic manipulations, I don't know what to do.

http://imageshack.us/photo/my-images/593/40703948.jpg/

Why do the +1 and -1 appear? Where did the extra exponential come from? If someone can go line for line, I would appreciate it.

 

[wukunlin]

well, that looks a little messy, but let's do this step by step:

start from
\mathcal{L} \{ (t-2)^2 e^{-5t} u(t-1) \} = \mathcal{L} \{ (t-1-1)^2 e^{-5t+1-1} u(t-1) \}

basically, t-2= t-1-1 and t = t + 0= t +1 -1, straightforward enough I hope


then
\mathcal{L} \{ (t-1-1)^2 e^{-5t+1-1} u(t-1) \} = \mathcal{L} \{ [ (t-1)^2 - 2 (t-1) +1 ] e ^{-5} e^{-5(t-1)} u(t-1) \}

the first part: (t-1-1)^2 = (t-1)^2 - 2 (t-1) +1 is basically the same as expanding (x-1)^2 seeing x=t-1

the second part: e^{-5t+1-1} = e ^{-5} e^{-5(t-1)} is a property of exponents/logs e^a e^b = e^{a+b}

The rest is basically expanding the terms,
things you need to know is e^{-5} is just a constant, laplace transform is a linear operation so \mathcal{L} \{ e^{-5} f(t) \} = e^{-5} \mathcal{L} \{ f(t) \}
and \mathcal{L} \{ f(t-1) \} = e^{-s} \mathcal{L} \{ f(t) \}

thats basically it, feel free to ask any other questions you may have :)