Laplace transform of $mathcalLtg(t)$

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The original question was to find $mathcalLg(t)$ from



$$g(t) = begincases t, 0 le t < 1 \ 2-t , 1 le t le 2 \ 0 , t>2 endcases $$



I formed a Heaviside unit step function to solve it,



Here it is -



$g(t) = t [ g(t) - g(t-1) ] + (2-t) [g(t-1) - g(t-2)] + 0 [ g(t-2) - g(t- infty) ] $



I ended up getting -



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



Now I have to take Laplace transform of the entire thing, I have to use the formula of $mathcalLU(t-a)f(t-a) = e^-as F(s)$ to solve the 2nd and 3rd one.



But what about $mathcalLtg(t) $ I am not too sure. Can I bring out the t ? For example -
$t mathcal Lg(t) $ I think I $t$ is a variable that I need to transform and it’s not a number (when I can do that).



I searched online and found this formula -



$U(t)f(t) = mathcalLf(t) $



Do I have to use this then ? Is this the correct formula that I need to use ?



so... $mathcal Ltg(t) = mathcalLt = frac 1S^2 $ ?




laplace-transform




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  • 1




    Use math.stackexchange.com/a/2846325/108128
    – Nosrati
    Jul 24 at 12:58










  • Your step-function expression for $g(t)$ should be $g(t) = t U(t) - 2(t-1)U(t-1) - (2-t)U(t-2)$.
    – John Wayland Bales
    Jul 24 at 13:20














up vote
0
down vote

favorite












The original question was to find $mathcalLg(t)$ from



$$g(t) = begincases t, 0 le t < 1 \ 2-t , 1 le t le 2 \ 0 , t>2 endcases $$



I formed a Heaviside unit step function to solve it,



Here it is -



$g(t) = t [ g(t) - g(t-1) ] + (2-t) [g(t-1) - g(t-2)] + 0 [ g(t-2) - g(t- infty) ] $



I ended up getting -



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



Now I have to take Laplace transform of the entire thing, I have to use the formula of $mathcalLU(t-a)f(t-a) = e^-as F(s)$ to solve the 2nd and 3rd one.



But what about $mathcalLtg(t) $ I am not too sure. Can I bring out the t ? For example -
$t mathcal Lg(t) $ I think I $t$ is a variable that I need to transform and it’s not a number (when I can do that).



I searched online and found this formula -



$U(t)f(t) = mathcalLf(t) $



Do I have to use this then ? Is this the correct formula that I need to use ?



so... $mathcal Ltg(t) = mathcalLt = frac 1S^2 $ ?




laplace-transform




share|cite|improve this question















  • 1




    Use math.stackexchange.com/a/2846325/108128
    – Nosrati
    Jul 24 at 12:58










  • Your step-function expression for $g(t)$ should be $g(t) = t U(t) - 2(t-1)U(t-1) - (2-t)U(t-2)$.
    – John Wayland Bales
    Jul 24 at 13:20












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The original question was to find $mathcalLg(t)$ from



$$g(t) = begincases t, 0 le t < 1 \ 2-t , 1 le t le 2 \ 0 , t>2 endcases $$



I formed a Heaviside unit step function to solve it,



Here it is -



$g(t) = t [ g(t) - g(t-1) ] + (2-t) [g(t-1) - g(t-2)] + 0 [ g(t-2) - g(t- infty) ] $



I ended up getting -



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



Now I have to take Laplace transform of the entire thing, I have to use the formula of $mathcalLU(t-a)f(t-a) = e^-as F(s)$ to solve the 2nd and 3rd one.



But what about $mathcalLtg(t) $ I am not too sure. Can I bring out the t ? For example -
$t mathcal Lg(t) $ I think I $t$ is a variable that I need to transform and it’s not a number (when I can do that).



I searched online and found this formula -



$U(t)f(t) = mathcalLf(t) $



Do I have to use this then ? Is this the correct formula that I need to use ?



so... $mathcal Ltg(t) = mathcalLt = frac 1S^2 $ ?




laplace-transform




share|cite|improve this question











The original question was to find $mathcalLg(t)$ from



$$g(t) = begincases t, 0 le t < 1 \ 2-t , 1 le t le 2 \ 0 , t>2 endcases $$



I formed a Heaviside unit step function to solve it,



Here it is -



$g(t) = t [ g(t) - g(t-1) ] + (2-t) [g(t-1) - g(t-2)] + 0 [ g(t-2) - g(t- infty) ] $



I ended up getting -



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



Now I have to take Laplace transform of the entire thing, I have to use the formula of $mathcalLU(t-a)f(t-a) = e^-as F(s)$ to solve the 2nd and 3rd one.



But what about $mathcalLtg(t) $ I am not too sure. Can I bring out the t ? For example -
$t mathcal Lg(t) $ I think I $t$ is a variable that I need to transform and it’s not a number (when I can do that).



I searched online and found this formula -



$U(t)f(t) = mathcalLf(t) $



Do I have to use this then ? Is this the correct formula that I need to use ?



so... $mathcal Ltg(t) = mathcalLt = frac 1S^2 $ ?





laplace-transform





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asked Jul 24 at 12:55









user185692

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  • 1




    Use math.stackexchange.com/a/2846325/108128
    – Nosrati
    Jul 24 at 12:58










  • Your step-function expression for $g(t)$ should be $g(t) = t U(t) - 2(t-1)U(t-1) - (2-t)U(t-2)$.
    – John Wayland Bales
    Jul 24 at 13:20












  • 1




    Use math.stackexchange.com/a/2846325/108128
    – Nosrati
    Jul 24 at 12:58










  • Your step-function expression for $g(t)$ should be $g(t) = t U(t) - 2(t-1)U(t-1) - (2-t)U(t-2)$.
    – John Wayland Bales
    Jul 24 at 13:20







1




1




Use math.stackexchange.com/a/2846325/108128
– Nosrati
Jul 24 at 12:58




Use math.stackexchange.com/a/2846325/108128
– Nosrati
Jul 24 at 12:58












Your step-function expression for $g(t)$ should be $g(t) = t U(t) - 2(t-1)U(t-1) - (2-t)U(t-2)$.
– John Wayland Bales
Jul 24 at 13:20




Your step-function expression for $g(t)$ should be $g(t) = t U(t) - 2(t-1)U(t-1) - (2-t)U(t-2)$.
– John Wayland Bales
Jul 24 at 13:20










1 Answer
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There are various ways to write the Laplace transform you need. It is most commonly written



$$ mathcalLf(t-a)U(t-a)=mathcalLf(t)e^-astag1 $$



An alternate version is



$$ mathcalLf(t)U(t-a)=mathcalLf(t+a)e^-astag2 $$



Using the second rule, for example one has



begineqnarray mathcalL(t-2)U(t-2)&=&mathcalL(t+2)-2e^-2s\
&=&mathcalLte^-2s\
&=&frace^-2ss^2 endeqnarray






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    up vote
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    There are various ways to write the Laplace transform you need. It is most commonly written



    $$ mathcalLf(t-a)U(t-a)=mathcalLf(t)e^-astag1 $$



    An alternate version is



    $$ mathcalLf(t)U(t-a)=mathcalLf(t+a)e^-astag2 $$



    Using the second rule, for example one has



    begineqnarray mathcalL(t-2)U(t-2)&=&mathcalL(t+2)-2e^-2s\
    &=&mathcalLte^-2s\
    &=&frace^-2ss^2 endeqnarray






    share|cite|improve this answer

























      up vote
      1
      down vote













      There are various ways to write the Laplace transform you need. It is most commonly written



      $$ mathcalLf(t-a)U(t-a)=mathcalLf(t)e^-astag1 $$



      An alternate version is



      $$ mathcalLf(t)U(t-a)=mathcalLf(t+a)e^-astag2 $$



      Using the second rule, for example one has



      begineqnarray mathcalL(t-2)U(t-2)&=&mathcalL(t+2)-2e^-2s\
      &=&mathcalLte^-2s\
      &=&frace^-2ss^2 endeqnarray






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        There are various ways to write the Laplace transform you need. It is most commonly written



        $$ mathcalLf(t-a)U(t-a)=mathcalLf(t)e^-astag1 $$



        An alternate version is



        $$ mathcalLf(t)U(t-a)=mathcalLf(t+a)e^-astag2 $$



        Using the second rule, for example one has



        begineqnarray mathcalL(t-2)U(t-2)&=&mathcalL(t+2)-2e^-2s\
        &=&mathcalLte^-2s\
        &=&frace^-2ss^2 endeqnarray






        share|cite|improve this answer













        There are various ways to write the Laplace transform you need. It is most commonly written



        $$ mathcalLf(t-a)U(t-a)=mathcalLf(t)e^-astag1 $$



        An alternate version is



        $$ mathcalLf(t)U(t-a)=mathcalLf(t+a)e^-astag2 $$



        Using the second rule, for example one has



        begineqnarray mathcalL(t-2)U(t-2)&=&mathcalL(t+2)-2e^-2s\
        &=&mathcalLte^-2s\
        &=&frace^-2ss^2 endeqnarray







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 24 at 13:34









        John Wayland Bales

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