Laplace transform of a ramp function with a $2$ second delay












1












$begingroup$


Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.



$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$



This is what I get. Where am I going wrong?






laplace-transform







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  • 2




    $begingroup$
    You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:13












  • $begingroup$
    Now you have changed from $ds$ to $dt$ which is it?
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:17










  • $begingroup$
    $, mathrm{d}t$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:18










  • $begingroup$
    I saw your comment! @RodrigodeAzevedo
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:18










  • $begingroup$
    Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
    $endgroup$
    – Keilara
    Oct 16 '16 at 21:29
















1












$begingroup$


Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.



$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$



This is what I get. Where am I going wrong?






laplace-transform







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:13












  • $begingroup$
    Now you have changed from $ds$ to $dt$ which is it?
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:17










  • $begingroup$
    $, mathrm{d}t$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:18










  • $begingroup$
    I saw your comment! @RodrigodeAzevedo
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:18










  • $begingroup$
    Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
    $endgroup$
    – Keilara
    Oct 16 '16 at 21:29














1












1








1





$begingroup$


Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.



$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$



This is what I get. Where am I going wrong?






laplace-transform







share|cite|improve this question











$endgroup$




Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.



$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$



This is what I get. Where am I going wrong?






laplace-transform




laplace-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 16 '16 at 18:14









Rodrigo de Azevedo

13.1k41960




13.1k41960










asked Oct 16 '16 at 18:07









KeilaraKeilara

558




558








  • 2




    $begingroup$
    You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:13












  • $begingroup$
    Now you have changed from $ds$ to $dt$ which is it?
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:17










  • $begingroup$
    $, mathrm{d}t$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:18










  • $begingroup$
    I saw your comment! @RodrigodeAzevedo
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:18










  • $begingroup$
    Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
    $endgroup$
    – Keilara
    Oct 16 '16 at 21:29














  • 2




    $begingroup$
    You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:13












  • $begingroup$
    Now you have changed from $ds$ to $dt$ which is it?
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:17










  • $begingroup$
    $, mathrm{d}t$
    $endgroup$
    – Rodrigo de Azevedo
    Oct 16 '16 at 18:18










  • $begingroup$
    I saw your comment! @RodrigodeAzevedo
    $endgroup$
    – Olba12
    Oct 16 '16 at 18:18










  • $begingroup$
    Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
    $endgroup$
    – Keilara
    Oct 16 '16 at 21:29








2




2




$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13






$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13














$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17




$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17












$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18




$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18












$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18




$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18












$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29




$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29










1 Answer
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$begingroup$

The 2 second delayed ramp function with slope 1 is given by



$$ (t - 2) u(t -2) $$



The Laplace transform of this is:



$$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$



This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:



$$ = 0 + int_2^infty f(t-2), e^{-st} dt $$



This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.



To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that



$$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$



$$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$



But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:



$$ = e^{-2s} F(s) $$



If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with



$$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$






share|cite|improve this answer









$endgroup$













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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

    votes









    0












    $begingroup$

    The 2 second delayed ramp function with slope 1 is given by



    $$ (t - 2) u(t -2) $$



    The Laplace transform of this is:



    $$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$



    This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:



    $$ = 0 + int_2^infty f(t-2), e^{-st} dt $$



    This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.



    To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that



    $$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$



    $$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$



    But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:



    $$ = e^{-2s} F(s) $$



    If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with



    $$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The 2 second delayed ramp function with slope 1 is given by



      $$ (t - 2) u(t -2) $$



      The Laplace transform of this is:



      $$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$



      This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:



      $$ = 0 + int_2^infty f(t-2), e^{-st} dt $$



      This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.



      To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that



      $$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$



      $$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$



      But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:



      $$ = e^{-2s} F(s) $$



      If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with



      $$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The 2 second delayed ramp function with slope 1 is given by



        $$ (t - 2) u(t -2) $$



        The Laplace transform of this is:



        $$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$



        This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:



        $$ = 0 + int_2^infty f(t-2), e^{-st} dt $$



        This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.



        To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that



        $$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$



        $$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$



        But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:



        $$ = e^{-2s} F(s) $$



        If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with



        $$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$






        share|cite|improve this answer









        $endgroup$



        The 2 second delayed ramp function with slope 1 is given by



        $$ (t - 2) u(t -2) $$



        The Laplace transform of this is:



        $$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$



        This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:



        $$ = 0 + int_2^infty f(t-2), e^{-st} dt $$



        This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.



        To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that



        $$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$



        $$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$



        But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:



        $$ = e^{-2s} F(s) $$



        If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with



        $$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 '17 at 2:40









        rhodyrhody

        1134




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