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Fourier Sinusoids

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    The entries of the time-domain vector:
    x(1) = [2 1 -1 -2 -1 1 2 1 -1 -2 -1 1] ; N = 12

    are given by 2cos(ωn) where n = 0:11. what is the value of ω? express x(1) as the sum of two fourier sinusoids. By considering the appropriate columns of the Fourier matrix V, determine the DFT X(1).


    2. Relevant equations
    ω = (2π/N)


    3. The attempt at a solution
    I know that ω = π/6

    But when determining the fourier sinusoid I can only express it as the sum of 7 different parts.

    1/6 + 1/6cos(pi*n/6) - 1/6cos(pi/3*n)...1/6(-1)^n.

    But I says to express it as 2 fourier sinusoids. I don't know how to simplify it or how to decide which columns would lead me to a solution.

    When I fft the above equation (partial shown) It get x(1) back. So the equation is right but does not fully answer the question. Any help would be greatly appreciated!

    DmytriE
     
  2. jcsd
  3. Nov 9, 2012 #2
    UPDATE:
    ω = π/3 because the fundamental period of the time vector is 6.

    What is meant by "express x(1) as the sum of two Fourier sinusoids."
     
  4. Nov 9, 2012 #3

    rude man

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    Correct on w = pi/3.

    I don't understand why express x(1) as a sum of two sinusoids when the original wave is already a pure sinusoid ... 2cos(nπ/3), n = 0, 1,2, ...
     
    Last edited: Nov 9, 2012
  5. Nov 9, 2012 #4
    Using Matlab I did fft of the function you gave above and it does not reproduce the time vector. I am looking for the sinusoidal equivalent I guess using only two sinusoidal functions to show this.
     
  6. Nov 9, 2012 #5
    t = 1/6 + 1/6*cos(pi/6*n) - 1/6*cos(pi/3*n) - 1/3*cos(pi/2*n) - 1/6*cos(2*pi/3*n)+ 1/6*cos(5*pi/6*n) +(-1).^n/6

    ^This the the equation I get that will give me the full time vector. But as far as I can tell it is made up of more than 2 sinusoidal functions....
     
  7. Nov 9, 2012 #6

    rude man

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    I meant n = 0, 1, 2, ..

    The Fourier series of cos(wt) is cos(wt)! So the Fourier series of 2cos(nπ/3) is 2cos(nπ/3), seems like.

    I guess I'm missing something here.
     
  8. Nov 9, 2012 #7

    NascentOxygen

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    I have done no Fourier work since I was a student, so don't take this too seriously.....

    EDIT [strike]Doesn't that 1/6 term indicate DC? There is no DC component here.[/strike] I hadn't read to the end of your series. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon11.gif [Broken]

    Perhaps you are being asked to approximate that stepwise wave using just 2 sinusoids? You'll need the fundamental, together with another to account for the step noise, and by the nature of the symmetry you can see it must be an odd harmonic.
     
    Last edited by a moderator: May 6, 2017
  9. Nov 9, 2012 #8
    If you look at x, you can see that it is symmetric across number 2. Since the real value x is symmetric, then X must show conjugate circular symmetric. Remember that x will be symmetric at N/2. Therefore the elements n and N-n are the same distance from the center of symmetry. this can be expressed as x = e^(m) + e^(m-N). At least this is my understanding of this problem. You can refer to pages 176 through 180 of the following text if that helps.
     
  10. Nov 9, 2012 #9
    Which text are you referring to?
     
  11. Nov 9, 2012 #10
    Yes, thank you Polaris. I will take a look at the text. Haha. I just realized which text you were talking about.
     
  12. Nov 9, 2012 #11
  13. Nov 10, 2012 #12
    Is this a sum of two Fourier Sinusoid?
    n = 0:11;
    e^(j * pi/3*n) + e^(-j*pi/3*n).

    I hope so since I cannot seem to write two sinusoids that will produce the time domain vector.
     
  14. Nov 11, 2012 #13

    rude man

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    Looks good to me if by pi/3*n you mean nπ/3.

    The whole question seems obtuse to me, frankly. A set of discrete numbers is not a sinusoid per se. A sinusoid can generate discrete numbers, to be sure, as in this case. But what is the point of the question, other than to present arcane math?
     
  15. Nov 12, 2012 #14
    Yup, that's exactly what I mean. :approve:
     
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