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How was this equation simplified? (Fourier coefficients)

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    http://puu.sh/gGhdb.jpg [Broken]

    Solution:

    http://puu.sh/gGh3E.jpg [Broken]

    2. Relevant equations


    3. The attempt at a solution
    How did they get that solution for the Fourier coefficient? When I evaluate the integral I can only seem to get it to:

    (1/-jk2π)[2*exp(-jkπt)-exp(-jk2πt)-1]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 19, 2015 #2

    DrClaude

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    Staff: Mentor

    It's a definite integral. You can't have a result that is a function of t.
     
  4. Mar 19, 2015 #3
    True, my mistake. Take out the t's and that's what I have. I did the integration right I think (plugged in for t) but forgot to remove the t also:

    (1/-jk2π)[2*exp(-jkπ)-exp(-jk2π)-1]

    Still can't seem to get it in the form they have, which is necessary for the next part of the problem.
     
  5. Mar 19, 2015 #4

    DrClaude

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    What values can k take?
     
  6. Mar 19, 2015 #5
    Sorry I'm not sure what you mean by that, but I'm looking at a table of properties for coefficients and it says if x(t) is real and odd, then ak should be jIm(ak). Using that with Euler's equation, I get:

    (1/kπ)[cos(kπ)-(1/2)cos(k2π)-(1/2)], which doesn't seem anywhere near right.
     
  7. Mar 19, 2015 #6

    DrClaude

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    My question goes back to the definition of Fourier series. What does the ##k## in ##a_k## stand for? What can be its value?

    The answer in the solution is in terms of an exponential, so don't use Euler's equation.
     
  8. Mar 19, 2015 #7
    The k's are the harmonics, and they can be from 0 to 8 because of the function's range?
     
  9. Mar 19, 2015 #8

    DrClaude

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    We need to go back to basics. Whats is the general for ##x(t)## written as a Fourier series?
     
  10. Mar 19, 2015 #9
    x(t) = a summation from negative infinity to positive infinity of ak*exp(jkwot)

    sorry about my ignorance on this subject, it's really difficult for me to grasp this stuff just from reading my textbook
     
  11. Mar 19, 2015 #10

    DrClaude

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    We're almost there: summation over what?

    Don't worry, and don't think I am trying to give you a hard time. I simply think you'll learn more if you find it by yourself with a bit of guidance.
     
  12. Mar 19, 2015 #11
    Summation over k, the harmonic components
     
  13. Mar 19, 2015 #12

    DrClaude

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    Ok, so you have
    $$
    \frac{1}{-jk2\pi} \left[ 2 \exp(-jk\pi) - \exp(-jk2\pi)-1 \right]
    $$
    where ##k## can take any integer value. Use this fact to simplify the equation.
     
  14. Mar 19, 2015 #13
    This is where I thought it'd be helpful to expand it with Euler's equation. The term in the brackets becomes:

    2coskπ - j2sinkπ - cosk2π + jsink2π - 1

    If k can be any integer value, all the sines should become 0 right? so i'm left with:

    2coskπ - cosk2π - 1

    in the brackets.

    But like you said before the simplified equation is in exponential terms so what I have doesn't help. I don't know how to simplify the equation as it is.
     
  15. Mar 19, 2015 #14

    DrClaude

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    You're missing a factor 1/2 here.

    Right

    You can use the same kind of identities directly in the exponential. In any case, you should be able to simplify ##\cos(k 2\pi)## further. You can at the end convert the cosines back into exponentials.
     
  16. Mar 19, 2015 #15
    Okay so distributing that (1/2) from the term outside the brackets, I have:

    cos(kπ) - (1/2)cos(k2π) - (1/2)

    Then using the identity cos(2x) = 1 - 2sin²(2x), I have:

    cos(kπ) - 1, since the sine term becomes 0

    Using Euler's equation, I then have:

    (1/2)exp(jkπ) + (1/2)exp(-jkπ) - 1
     
  17. Mar 20, 2015 #16
    Bump.

    That's still not where it should be though.
    Right now I have:

    (1/jπk)[(1/2)exp(jkπ) + (1/2)exp(-jkπ) - 1]

    but I'm not sure how to simplify it further.
     
  18. Mar 20, 2015 #17

    DrClaude

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    After looking in more details at your solution, it appears that you can't simply convert back to an exponential using Euler formula, even though your solution is formally equivalent to the one in the book. To obtain the one in the book, when you have
    $$
    \frac{1}{-jk2\pi} \left[ 2 \exp(-jk\pi) - \exp(-jk2\pi)-1 \right]
    $$
    you take that ##\exp(-jk2\pi) = 1 \forall k##, and then you get
    $$
    \frac{1}{jk\pi} \left[ 1 - \exp(-jk\pi) \right]
    $$
    This is formally equivalent to your answer,
    $$
    \frac{1}{jk\pi} \left[ 1 - \cos(k\pi) \right]
    $$
    because both ##\exp(-jk\pi)## and ##\cos(k\pi)## equal 1 when ##k## is even and -1 when ##k## is odd.
     
  19. Mar 20, 2015 #18
    Okay this right here makes sense. Thanks!
     
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