$mathcalL ( (t+5)U(t-1) ) $

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$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $



a= 1



I am having trouble dealing with $f(t-a)$ and finding $F(s)$



$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $



$f(t) = e^t-2 $



How do I Laplace transform this ?







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  • Are you looking for Laplace transform of $(t+5)U(t-1)$?
    – Alla Tarighati
    Aug 29 at 6:31











  • @AllaTarighati yes
    – user185692
    Aug 29 at 6:37














up vote
0
down vote

favorite












$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $



a= 1



I am having trouble dealing with $f(t-a)$ and finding $F(s)$



$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $



$f(t) = e^t-2 $



How do I Laplace transform this ?







share|cite|improve this question




















  • Are you looking for Laplace transform of $(t+5)U(t-1)$?
    – Alla Tarighati
    Aug 29 at 6:31











  • @AllaTarighati yes
    – user185692
    Aug 29 at 6:37












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $



a= 1



I am having trouble dealing with $f(t-a)$ and finding $F(s)$



$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $



$f(t) = e^t-2 $



How do I Laplace transform this ?







share|cite|improve this question












$mathcalL ( (t+5)U(t-1) ) = U(t-a)f(t-a) = e^-as F(s) $



a= 1



I am having trouble dealing with $f(t-a)$ and finding $F(s)$



$f(t-1) = e^t-3 = e^(t-1)-3+1 = e^(t-1)-2 $



$f(t) = e^t-2 $



How do I Laplace transform this ?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 29 at 6:21









user185692

1536




1536











  • Are you looking for Laplace transform of $(t+5)U(t-1)$?
    – Alla Tarighati
    Aug 29 at 6:31











  • @AllaTarighati yes
    – user185692
    Aug 29 at 6:37
















  • Are you looking for Laplace transform of $(t+5)U(t-1)$?
    – Alla Tarighati
    Aug 29 at 6:31











  • @AllaTarighati yes
    – user185692
    Aug 29 at 6:37















Are you looking for Laplace transform of $(t+5)U(t-1)$?
– Alla Tarighati
Aug 29 at 6:31





Are you looking for Laplace transform of $(t+5)U(t-1)$?
– Alla Tarighati
Aug 29 at 6:31













@AllaTarighati yes
– user185692
Aug 29 at 6:37




@AllaTarighati yes
– user185692
Aug 29 at 6:37










1 Answer
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Hint:



$$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$



Where we know:
$$ mathcalLU(t) = frac1s,$$
and
$$mathcalLtU(t)=frac1s^2.$$






share|cite|improve this answer




















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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    Hint:



    $$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$



    Where we know:
    $$ mathcalLU(t) = frac1s,$$
    and
    $$mathcalLtU(t)=frac1s^2.$$






    share|cite|improve this answer
























      up vote
      0
      down vote



      accepted










      Hint:



      $$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$



      Where we know:
      $$ mathcalLU(t) = frac1s,$$
      and
      $$mathcalLtU(t)=frac1s^2.$$






      share|cite|improve this answer






















        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        Hint:



        $$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$



        Where we know:
        $$ mathcalLU(t) = frac1s,$$
        and
        $$mathcalLtU(t)=frac1s^2.$$






        share|cite|improve this answer












        Hint:



        $$ mathcalLleft[ (t+5)U(t-1)right]= mathcalL left[(t-1+6)U(t-1)right] = mathcalLleft[ (t-1)U(t-1) + 6U(t-1)right] = mathcalL left[(t-1)U(t-1)right] + 6mathcalLleft[ U(t-1)right]$$



        Where we know:
        $$ mathcalLU(t) = frac1s,$$
        and
        $$mathcalLtU(t)=frac1s^2.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 29 at 6:40









        Alla Tarighati

        2623




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