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eSpeX

  • 2 years ago

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.

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  1. eSpeX
    • 2 years ago
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    Is the 'del' symbolic of using the delta function?

  2. mahmit2012
    • 2 years ago
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    Absolutely yes.

  3. eSpeX
    • 2 years ago
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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?

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

  5. mahmit2012
    • 2 years ago
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    Do you know extended derivative with delta function?

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

  7. mahmit2012
    • 2 years ago
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    @ali110 is one of them.

  8. eSpeX
    • 2 years ago
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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)}

  9. eSpeX
    • 2 years ago
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    Those should be f(t)...

  10. mahmit2012
    • 2 years ago
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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.

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

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

  13. mahmit2012
    • 2 years ago
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    @ali110 is one of them.

  14. ali110
    • 2 years ago
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    mohan gholami in which of them?

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

  16. eSpeX
    • 2 years ago
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    I believe he was saying that you were an engineer.

  17. mahmit2012
    • 2 years ago
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    No, you have two points not two functions!

  18. ali110
    • 2 years ago
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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

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

  20. eSpeX
    • 2 years ago
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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?

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

  22. eSpeX
    • 2 years ago
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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?

  23. ali110
    • 2 years ago
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    can we take laplace inverse at the end?

  24. mahmit2012
    • 2 years ago
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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.

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

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

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

  28. mahmit2012
    • 2 years ago
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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

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

  30. mahmit2012
    • 2 years ago
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    And Laplace has problem with the single points.

  31. eSpeX
    • 2 years ago
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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.

  32. mahmit2012
    • 2 years ago
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    You're welcome.

  33. mahmit2012
    • 2 years ago
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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)

  34. eSpeX
    • 2 years ago
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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?

  35. ali110
    • 2 years ago
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  36. eSpeX
    • 2 years ago
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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.

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

  38. ali110
    • 2 years ago
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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

  39. ali110
    • 2 years ago
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    @mahmit2012

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

  41. ali110
    • 2 years ago
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    @eSpeX I GUESS

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

  43. eSpeX
    • 2 years ago
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    Okay.

  44. mahmit2012
    • 2 years ago
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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.

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

  46. KenLJW
    • one year ago
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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

  47. KenLJW
    • one year ago
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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

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