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anonymous

  • 5 years ago

Show the procedure for finding the transform of f(t)=cos5t-sin5t

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  1. anonymous
    • 5 years ago
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    Fourier?

  2. anonymous
    • 5 years ago
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    It doesn't say but thats what I'm guessing since thats what we've covered.

  3. anonymous
    • 5 years ago
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    You haven't covered Laplace?

  4. anonymous
    • 5 years ago
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    Series and Sequences are my absolute weakness.

  5. anonymous
    • 5 years ago
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    I have covered LaPlace as well.

  6. anonymous
    • 5 years ago
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    hmm

  7. anonymous
    • 5 years ago
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    That was our last week of my class

  8. anonymous
    • 5 years ago
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    Its a short answer type question though so I'm not sure what it means by transform.. the only 2 I can recall are fourier and laplace

  9. anonymous
    • 5 years ago
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    When's it due?

  10. anonymous
    • 5 years ago
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    3 hours

  11. anonymous
    • 5 years ago
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    ouch

  12. anonymous
    • 5 years ago
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    Yeah I have like 20 more questions, but they are much more simple.

  13. anonymous
    • 5 years ago
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    The answer's going to depend on whether it's Fourier or Laplace, that's the problem.

  14. anonymous
    • 5 years ago
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    Well wouldn't fourier be a series? Which would give me an interval?

  15. anonymous
    • 5 years ago
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    I have to leave for a bit - bad timing. I'll try and be back before your three hours is up.

  16. anonymous
    • 5 years ago
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    Cool thanks dude. No problem

  17. anonymous
    • 5 years ago
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    kk

  18. anonymous
    • 5 years ago
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    I can do it quickly if it's Laplace transform.

  19. anonymous
    • 5 years ago
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    I haven't done any Fourier transform in a while.

  20. anonymous
    • 5 years ago
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    I think its LaPlace

  21. anonymous
    • 5 years ago
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    I don't think we did Fourier transforms, we did Fourier Series.

  22. anonymous
    • 5 years ago
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    They are different right?

  23. anonymous
    • 5 years ago
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    yes, they are.

  24. anonymous
    • 5 years ago
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    Ok well I guess its a laplace because I haven't done any transform other than LaPlace

  25. anonymous
    • 5 years ago
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    It would be easy then. do you use a Laplace transform table? or should I solve using the definition of Laplace transform?

  26. anonymous
    • 5 years ago
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    We use transform tables usually....

  27. anonymous
    • 5 years ago
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    that's good.. we will just use two formulas, which are: \[1) L[\sin kt]={k \over s^2+k^2}\] \[2) L[\cos kt]= {s \over s^2+k^2}\] where L[f(t)] denotes the Laplace transform of f(t).

  28. anonymous
    • 5 years ago
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    do you think you can apply these two formulas?

  29. anonymous
    • 5 years ago
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    Hmm Let me see... I'm really not fluent in using the table as it was our last topic discussed in the class.

  30. anonymous
    • 5 years ago
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    Wouldn't I just substitute K for the cosine and sine values?

  31. anonymous
    • 5 years ago
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    yeah k in this case is what?

  32. anonymous
    • 5 years ago
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    5

  33. anonymous
    • 5 years ago
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    OK.. show me what you get.

  34. anonymous
    • 5 years ago
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    \[5-s/s^2+25\]

  35. anonymous
    • 5 years ago
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    but I can factor the bottom if I wanted

  36. anonymous
    • 5 years ago
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    are you sure about factorizing the bottom?

  37. anonymous
    • 5 years ago
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    Nevermind I can;t

  38. anonymous
    • 5 years ago
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    f(t)=5-s/s^2+25 final solution.

  39. anonymous
    • 5 years ago
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    1) well what is the Laplace transform of f(t)? 2) is it 5-s or s-5?

  40. anonymous
    • 5 years ago
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    s-5 I read it inccorectly

  41. anonymous
    • 5 years ago
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    ok.. the notation that's usually used to denote a function after being transformed is uppercase letter, in our case it will be: \[F(s)= {s-5 \over s^2+25}\]

  42. anonymous
    • 5 years ago
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    Oh right... because you're transforming it in terms of f(t) and s... I recall that now.

  43. anonymous
    • 5 years ago
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    if you're asking about how you can get the formulas (which is the most important part of the answer), you can just use the definition and integrate for the first one: \[\int\limits_{0}^{\infty}{\cos (kt)} e ^{-st} dt\] the result of this integral (can be integrated using integration by parts) is, by definition, the Laplace transform of cos (kt).

  44. anonymous
    • 5 years ago
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    Right which will help me build a transform table if I didn't have one available?

  45. anonymous
    • 5 years ago
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    Exactly.

  46. anonymous
    • 5 years ago
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    Sweet tthanks a lot :)

  47. anonymous
    • 5 years ago
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    You're welcome!

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