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Math Help - Math Exam- Laplace Transform

  1. #1
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    Math Exam- Laplace Transform

    Hello, Im currently revising for my university maths exam and have found the following question:
    F(s) denotes the laplace transform of the causal signal f(t), with the region of convergence Re(s) >σc. Prove the first shift theorem, using the basic definition of the laplace transform, by showing that the transform of e^at f(t) is given by L{f(t)}=F(s-a), with the region of conversion Re(s)>σc +Re(a).

    Im not really sure what its asking. I know that the first shift theorem is:
    L{e^at f(t)}=F(s-a), where F(s)=L{f(t)}

    and i know that the definition of a laplace transform is:
    F(s)=L{f(t)}=(meaning integral from 0- infinity in this case)of e^-st f(t) dt.

    How do i use these to come up with the answer that is expected? In the question, this is worth 5 marks out of a total 20.
    All help appreciated, Thank you.
    Andy
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  2. #2
    Senior Member yeKciM's Avatar
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     F(S) = \displaystyle \int _{-0} ^{\infty} e^{at} e^{-st} dt = \frac {1}{S-a} with  ROC : Re(S)>-Re(a)
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  3. #3
    MHF Contributor chisigma's Avatar
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    From definition of the LT is...

    \displaystyle F(s)= \int_{0}^{\infty} f(t)\ e^{- s t} dt (1)

    ... and the integral (1) converges for \Re (s) > \sigma. From (1) it follows that...

    \displaystyle \int_{0}^{\infty} f(t)\ e^{a t} e^{- s t} dt = \int_{0}^{\infty} f(t)\ e^{- (s-a) t} dt = F(s-a) (2)

    ... and the integral (2) converges for \Re (s-a) > \sigma...

    Kind regards

    \chi \sigma
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