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eSpeX

Laplace Transform: Evaluate L{f(t)} \[\begin {align*} f(t) &= 1 ,\ t \ge 0, \quad t \neq 1, \ t \neq 2 \\ &= 3,\ t = 1\\ &= 4,\ t = 2\end {align*} \] Would appreciate someone explaining how to set this up and evaluate.

  • one year ago
  • one year ago

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  1. eSpeX
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    Is the 'del' symbolic of using the delta function?

    • one year ago
  2. mahmit2012
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    Absolutely yes.

    • one year ago
  3. eSpeX
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    Could you please explain the logic you approach this with and how you handled the intervals where the function equaled a constant?

    • one year ago
  4. eSpeX
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    In my textbook, the delta function is not introduced for 4 more sections.

    • one year ago
  5. mahmit2012
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    Do you know extended derivative with delta function?

    • one year ago
  6. mahmit2012
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    Mathematician said there is no derivative in uncontinious point but engineers said it has.

    • one year ago
  7. mahmit2012
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    @ali110 is one of them.

    • one year ago
  8. eSpeX
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    So far what I know of piecewise functions is that L{f(x)} = L{f_1(x)} + L{f_2(x)} + L{f_3(x)}

    • one year ago
  9. eSpeX
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    Those should be f(t)...

    • one year ago
  10. mahmit2012
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    no, there is no information in an alone point for Laplace, but I made derivative to make an information then used the Laplace.

    • one year ago
  11. eSpeX
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    How did you make a derivative of a function that has only a constant value?

    • one year ago
  12. ali110
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    L(f(t))=L(1)=1/s if for f(t)=t then L(f(t))=1/s^2

    • one year ago
  13. mahmit2012
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    @ali110 is one of them.

    • one year ago
  14. ali110
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    mohan gholami in which of them?

    • one year ago
  15. eSpeX
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    If it never equals 't', then do I only have \[\frac{1}{s} +\frac{1}{s} +\frac{1}{s}\] ?

    • one year ago
  16. eSpeX
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    I believe he was saying that you were an engineer.

    • one year ago
  17. mahmit2012
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    No, you have two points not two functions!

    • one year ago
  18. ali110
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    oh i am an electrical engineering student of 5th semester who got 71 marks out of 100 in laplace transform in his 4th semster:))) @eSpeX

    • one year ago
  19. mahmit2012
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    Unfortunately Laplace transform doesn't sense points unless with delta function.

    • one year ago
  20. eSpeX
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    Laplace does not make sense to me on how to handle them, and with respect to this piecewise I do not see how it will be done if we have not been shown the delta function. Is it something (or similar) to the heavyside step function?

    • one year ago
  21. ali110
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    L(f(t))=L(3)=3L(1)=3*1/s=3/s

    • one year ago
  22. eSpeX
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    According to the book, the answer is 1/s. Does this mean that the laplace of t=1 and t=2 equate to 0?

    • one year ago
  23. ali110
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    can we take laplace inverse at the end?

    • one year ago
  24. mahmit2012
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    No the answer is just 1/s because Laplace transform can not sense limit points, and it just follow the infinity points which defined with delta function.

    • one year ago
  25. eSpeX
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    So you would have: L{f(t)} = L{1} + L{3} + L{4} -> 1/s + 0 + 0 ?

    • one year ago
  26. mahmit2012
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    ok I solve it with integral. int(0 inf) f(t)=1/s

    • one year ago
  27. mahmit2012
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    You know the integral change the limit points to continues function and never sense them.

    • one year ago
  28. mahmit2012
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    1 f'(t)=3del(t-1)-3del(t-1)+4del(t-2)-4del(t-2) and f(0)=1 L(f'(t))=3e^-s-3e^-s+4e^-2s-4e^-2s=sL(f(t))-1 L(f(t))=1/s(0+0+1)=1/s

    • one year ago
  29. mahmit2012
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    I thought in two points 1 and 2 it has jumped so they were two alone points.

    • one year ago
  30. mahmit2012
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    And Laplace has problem with the single points.

    • one year ago
  31. eSpeX
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    Ah. Okay, I will see if I can't apply this to the rest of my problems. Thank you very much.

    • one year ago
  32. mahmit2012
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    You're welcome.

    • one year ago
  33. mahmit2012
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    if the points in 1 and 2 was jumped so the solution was : 1 f'(t)=3del(t-1)+4del(t-2) and f(0)=1 L(f'(t))=3e^-s+4e^-2s=sL(f(t))-1 L(f(t))=1/s(3e^-s+4e^-2s+1)

    • one year ago
  34. eSpeX
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    But at this point I would have needed to use the integral approach since we have not reached the delta function?

    • one year ago
  35. ali110
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    • one year ago
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  36. eSpeX
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    It appears that all of those examples have a range that the integral is evaluated over. So none of the laplace methods evaluate a point.

    • one year ago
  37. mahmit2012
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    But @ali110 there is no unit step and delta function! I guess Openhiem is the best refrence. Isn't it?

    • one year ago
  38. ali110
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    check page 11 and every problem will solve as writer show that for t not equal to 1 And 2 as in above question F=0 and Agha! i love alan V openheim as i take all his video lectures about signals and systems but in our engineerig college we study Indian professor Ghosh sumarjit check his book on Signal and system and about fourier series more intersesting then oppenheim

    • one year ago
  39. ali110
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    @mahmit2012

    • one year ago
  40. ali110
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    L{f(t)} = L{1} + L{3} + L{4} -> 1/s + 0 + 0=1/s

    • one year ago
  41. ali110
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    @eSpeX I GUESS

    • one year ago
  42. ali110
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    @eSpeX CHECK laplace transform linearity property (in which one to one property)

    • one year ago
  43. eSpeX
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    Okay.

    • one year ago
  44. mahmit2012
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    Ok. And I should mention that don't use L(3)=0 because it is not true. You can write L*(3)=0 and define L* means Laplace for limit points.

    • one year ago
  45. eSpeX
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    Alright, I'll have to look up limit points then because I have not seen them yet as I recall.

    • one year ago
  46. KenLJW
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    You have to use the unit step function for 2 and 3 3u(t-1) 4u(t-2) for the first one I'd break it up u(t-1) - u(t-1minus) + u(t-1 plus) -u(t-2 minus) ect With these there's direct transformations

    • 11 months ago
  47. KenLJW
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    In EE the slope of the step function is an indication of bandwidth, if there was infinite bandwidth it would be a unit step

    • 11 months ago
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