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Describe the Fourier Series

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data
    For he following Fourier series, which of the answers correctly describes the following function


    y(t) = 2 - [itex]\stackrel{1}{π}[/itex]∑1inf1/nsin(n*πt/2)

    a) odd function, period = 2 s
    b) Even function, period = 2s
    c) Odd function, period = 4s
    d) Even functio, period = 4s

    2. Relevant equations



    3. The attempt at a solution

    From the reading my professor assigned I'm pretty sure that it's odd because it has a sin function, though I still don't understand why.

    I'm also confused as to how I find the period. I know that

    T =1/f

    w=2*π*f

    So,

    w = (n*π)/2

    I'm not quite sure what to make of n.

    Any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Feb 20, 2014 #2

    Simon Bridge

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    That would be: $$y(t)=2-\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n\sin(n\pi t/2)}$$

    Check y(-t), if it is the same as -y(t) then y(t) is odd, if it is the same as +y(t) then it is even.

    y(t+T)=y(t), what is T?

    Certainly the period of the sine function is something to do with 2 units.
    But, since it is multi-choice, you don't actually have to find the period - just check to see which of the choices is correct.
     
  4. Feb 21, 2014 #3

    rude man

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    I think you're misreading what the OP meant, or maybe you're trying to point out that he /she miswrote it.

    Your denominator is intended to be the terms in the series, not their inverse. Look at the OP's thumbnail.
     
  5. Feb 21, 2014 #4

    rude man

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    The series is a series of harmonics starting with a dc term ("2") and then a series of sine terms
    So the first sine term must be the fundamental frequency where n = 1.
    So compare sin(nπt/2) with sin(ωt). In other words, compare π/2 with ω since n = 1.

    If you know ω, what is the period T?

    BTW by "period" they mean the periodicity of the lowest frequency component, i.e. for which n=1.
     
  6. Feb 21, 2014 #5

    Simon Bridge

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    ... a bit of both by the looks of things - neither thought completed. My excuse is that dinner was just about ready.

    Yep - should have looked at the thumb.
    $$y(t)=\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n}\sin \left(\frac{n\pi t}{2}\right)$$

    Which would make the RHS a Fourier expansion of something - a saw-tooth wave.
    ... advise still stands.

    ... or the period of y(t) - let OP work out if this is the same thing.

    a pure sine wave would be: ##y(t)=\sin(2\pi t/T)## where T is the period ... comparing with any other sine wave would be how you find T.

    But in this case there is a shortcut. Need only use the definition of the period to check to see if either of the options is the correct one.

    Care needed though, if ##y(t+T)=y(t)## then ##y(t+2T)=y(t)## also.
    But if ##T## is the period of ##y(t)## then ##y(t+T/2)\neq y(t)##
    So it is a matter of putting the numbers into the equation, and thinking about what "period" means.
    Which, I suspect, is the point of the exercise.
     
  7. Feb 21, 2014 #6

    LCKurtz

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    Understandably so. None of the above, even without knowing what "s" is.
     
  8. Feb 21, 2014 #7

    Simon Bridge

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    @Northbysouth: any of this help?
    You should check LCKurtz answer to see why... would y(t) be easier to handle without the "2 -" out the front?
    (i.e. the way I wrote it in post #5?)

    It is likely that there is a context missing from the problem statement you got.
     
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