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Laplace Transform of a signal?

  • Thread starter Chandasouk
  • Start date
  • #1
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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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
wukunlin
Gold Member
406
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[itex][/itex]well, that looks a little messy, but let's do this step by step:

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

basically, [itex]t-2= t-1-1[/itex] and [itex] t = t + 0= t +1 -1 [/itex], straightforward enough I hope


then
[tex] \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) \} [/tex]

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

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

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

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

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