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Laplace transform converge

  1. Nov 5, 2012 #1
    ##\mathcal{L}\{f(t)\}=F(s)##

    [tex]\mathcal{L}\{e^{at}\}=\frac{1}{s-a},Re(s)>a[/tex]
    [tex]\mathcal{L}\{\sin (at)\}=\frac{a}{s^2+a^2}, \quad Re(s)>0[/tex]
    [tex]\mathcal{L}\{\cos (at)\}=\frac{s}{s^2+a^2},Re(s)>0[/tex]

    If we look at Euler identity ##e^{ix}=\cos x+i\sin x##, how to get difference converge intervals ##Re(s)>a## and ##Re(s)>0##?
     
  2. jcsd
  3. Nov 5, 2012 #2

    jasonRF

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    The problem is that you haven't defined things precisely enough. In this line you are assuming a is real, so the region of convergence only holds for real a. If you allow a to be a general complex number (which is required for your complex exponentials) then you get,
    [tex]\mathcal{L}\{e^{at}\}=\frac{1}{s-a},Re(s)>Re(a).[/tex]

    If [itex]a = i x[/itex] with real [itex]x[/itex], then [itex]Re(a)=0[/itex] so the region of convergence is [itex],Re(s)>0[/itex].


    jason
     
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