Thread: Laplace Transform of this signal?

1. Laplace Transform of this signal?

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.

Can someone explain the steps of this solution?

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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 and how to in general solve these kinds of signals?

2. Re: Laplace Transform of this signal?

Originally Posted by nottheface
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.

Can someone explain the steps of this solution?

ImageShack&#174; - Online Photo and Video Hosting

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 and how to in general solve these kinds of signals?
The goal of the algebra is so the can use the t-axis translation theorem. This states that

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

The problem is that you have different time shifts in your unit step function and the other function.

Consider the function

$g(t)=(t-2)^2$

We want to write in with only factors of (t-1) and constants

$(t-2)^2=(t-1-1)^2$

Now for simplicity let $u=t-1$ this gives

$(u-1)^2=u^2-2u+1=(t-1)^2-2(t-1)+1$

Now we need to do something similar with the exponential term $e^{-5t}$ lets just focus on the exponent

$-5t=-5t+\underbrace{5-5}_{\text{add zero}}=-5(t-1)-5$ so

$e^{-5t}=e^{-5(t-1)-5}=e^{-5}e^{-5(t-1)}$

Now lets put everything to gether
$=e^{-5}e^{-5(t-1)}[(t-1)^2-2(t-1)+1]u(t-1)$

Now just distribute and use the t-axis translation theorem to transform each part!