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ranyai12

  • 3 years ago

Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2

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  1. Herp_Derp
    • 3 years ago
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    Ok. So you split up your integrand into two parts, right?

  2. ranyai12
    • 3 years ago
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    I got 2+(2/(1+s)) and it was wrong

  3. ranyai12
    • 3 years ago
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    @Herp_Derp yes I did

  4. amistre64
    • 3 years ago
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    \[L\{f(t)\}=\int_{0}^{inf}e^{-st}f(t)~dt\]

  5. amistre64
    • 3 years ago
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    in this case, you split the integral up into 2 sections

  6. ranyai12
    • 3 years ago
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    I did that and I was wrong :(

  7. amistre64
    • 3 years ago
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    can you type up your work? then we can see where the mistake might have happened

  8. amistre64
    • 3 years ago
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    theres a shifting method that i can never remember, so i tend to do it the splitted way

  9. ranyai12
    • 3 years ago
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    yea 2\[2\int\limits_{0}^{2}e^(-st)+\int\limits_{0}^{infinity}4e^(-st)dt\]

  10. ranyai12
    • 3 years ago
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    it's supposed to be e^(-st)

  11. amistre64
    • 3 years ago
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    0,2; then 2,inf for starters; and dont forget your "t" on the left side

  12. ranyai12
    • 3 years ago
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    f(t)=-2e^(-2s)/s + 2/s

  13. ranyai12
    • 3 years ago
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    oh sorry i actually have that down on my paper but I just typed it wrong

  14. amistre64
    • 3 years ago
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    \[2\int_{0}^{2}e^{-st}t+4\int_{2}^{inf}e^{-st}dt\]

  15. Herp_Derp
    • 3 years ago
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    \[\large \int_0^2 2te^{-st}\,dt+\int_0^{\infty}4e^{-st}\,dt=2\int_0^2 te^{-st}\,dt+4\int_0^{\infty}e^{-st}\,dt\]Just making it look better :)

  16. ranyai12
    • 3 years ago
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    @amistre64 thats what I have. What do I do afterwards but thats where I went wrong

  17. ranyai12
    • 3 years ago
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    Thanks @Herp_Derp but I have that but afterwards is where I'm confused

  18. amistre64
    • 3 years ago
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    2 to inf on the last one, the long right term is simple enough; the left side is by parts

  19. ranyai12
    • 3 years ago
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    I know I got that but afterwards is where I'm confused

  20. ranyai12
    • 3 years ago
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    for the firsr part the integral ends up equaling (3(1-e^(-2s)(2s+1))/(s^2)

  21. ranyai12
    • 3 years ago
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    and the second integral is (e^(-2s))/s

  22. ranyai12
    • 3 years ago
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    so (3(1-e^(-2s)(2s+1))/(s^2) +(e^(-2s))/s

  23. amistre64
    • 3 years ago
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    \[2\int_0^2 te^{-st}\,dt+4\int_2^{\infty}e^{-st}\,dt\] \[2(~\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{1}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\]

  24. amistre64
    • 3 years ago
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    this thing aint latex friendly :) \[2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{2}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\] \[2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\left.\frac{2}{s}\frac{e^{-st}}{-s}\right|_{t=0}^{t=2}+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\] \[2\frac{te^{-2s}}{-s}-2\frac{te^{0}}{-s}+\frac{2}{s}\frac{e^{-2s}}{-s}-\frac{2}{s}\frac{e^{0}}{-s}+4\frac{e^{-s(inf)}}{-s}-4\frac{e^{-2s}}{-s}\]

  25. ranyai12
    • 3 years ago
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    do we stop there?

  26. amistre64
    • 3 years ago
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    id clean it up a little if i was you

  27. ranyai12
    • 3 years ago
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    ok thanks!

  28. amistre64
    • 3 years ago
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    \[-\frac{2te^{-2s}}{s}+\frac{2t}{s}-\frac{2e^{-2s}}{s^2}+\frac{2}{s^2}+\cancel{4\frac{e^{-s(inf)}}{-s}}+\frac{4e^{-2s}}{s}\] stuff like that

  29. ranyai12
    • 3 years ago
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    it's wrong :(

  30. amistre64
    • 3 years ago
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    great time for loading issues eh

  31. amistre64
    • 3 years ago
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    \[2\int_{0}^{2}e^{-st}t+4\int_{2}^{inf}e^{-st}dt\] \[2(\frac{e^{-2s}-1}{s^2})+4\frac{e^{-2s}}{s}\] or some variation of that should work

  32. amistre64
    • 3 years ago
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    i never did get into the heaviside stuff to be competent enough in that method

  33. ranyai12
    • 3 years ago
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    sorry openstuy froze up! an it's still wrong

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