# Laplace Transform of a signal?

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.

## The Attempt at a Solution

wukunlin
Gold Member
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 :)