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Which of the signals is not the result of fourier series expansion?

  1. Dec 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Which of the signals is not the result of fourier series expansion?
    options :
    (a) 2cos(t) + 3 cos(3t)
    (b) 2cos([itex]\pi[/itex]t) + 7cos(t)
    (c) cos(t) + 0.5


    2. Relevant equations
    Dirichlet conditions


    3. The attempt at a solution

    From observation, I thought all are periodic and so must be fourier series expansions. But since this was a question in an objective exam, I went for option (b). Although the answer is right, I am not satisfied, by the explanation of Dirichlet conditions. And the plot of the function 2cos([itex]\pi[/itex]t) + 7cos(t) seems periodic too and no discontinuities etc.

    My question is actually is 2cos([itex]\pi[/itex]t) + 7cos(t) a function which cannot be a result of fourier expansion or not?
     

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  3. Dec 25, 2013 #2

    Simon Bridge

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    What are the "Directlet Conditions" conditions for exactly?
    If each of the possible answers are the Fourier series of something - what are they each a Fourier series of?
     
  4. Dec 25, 2013 #3
    f(x) must be absolutely integrable over a period.
    f(x) must have a finite number of extrema in any given interval, i.e. there must be a finite number of maxima and minima in the interval.
    f(x) must have a finite number of discontinuities in any given interval, however the discontinuity cannot be infinite.
    f(x) must be bounded.

    These dirichlet conditions were told not obeying in the case of option (b). But I am confused.
     
  5. Dec 26, 2013 #4

    AlephZero

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    Two questions:

    What is the mathematical definition of a periodic function?

    If you think function (b) is periodic, what is the period?
     
  6. Dec 26, 2013 #5

    Simon Bridge

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    Please answer the questions in post #2.
    Please answer the questions in post #2 - thank you.
    The above merely says what the Direchtlet conditions are. I asked what they were conditions for.
    If an equations satisfied the Directlet conditions - what does it mean?

    Consider - you have a statement that each of the equations are exact fourier transforms of some other function. But for one of them, this statement is false.

    Like AlephZero suggests - you should use the conditions as a test.
     
    Last edited: Dec 26, 2013
  7. Dec 26, 2013 #6
    If a periodic signal meets dirichlet conditions , that mean theoretically I can expand it using FOURIER SERIES EXPANSION THEORY. And a periodic function means self repeating function. Theoretically it expands from -infinity to infinity. For periodic function f(t) = f(t + n T) , n - integer and T time period.
     
  8. Dec 26, 2013 #7

    Simon Bridge

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    What does that mean?
    Surely all functions have a fourier series expansion.
    What is special about the Fourier expansion of those functions that meet the Directlet conditions?

    Great - now, what AlephZero is suggesting is this: for the function in (b), find T.
     
  9. Dec 30, 2013 #8

    LCKurtz

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    Another approach to answering the OP's question. FS have sums of terms ##\cos\frac{n\pi t}p## and ##\sin\frac{n\pi t}p##. What about the ratio of the angular frequencies ##\frac{n\pi} p## for two different values of ##n##?
     
  10. Dec 30, 2013 #9

    Ray Vickson

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    I hope you realize that the Dirichlet conditions are irrelevant to this problem. They give (sufficient) conditions for the Fourier series of a function f(x) to converge to f(x) itself (except at jump discontinituies). In the current problem you need to work backwards from this: you are told a trigonometric series, and you want to know if it could be the Fourier series of some unknown f(x).
     
  11. Dec 30, 2013 #10
    me too thought the same only. One online test was saying that 2cos(πt) + 7cos(t) cannot be a fourier series outcome since it is not obeying the dirichlet conditions. But actually now I think that the question has a mistake. Thanks every one who took effort for this.
     
  12. Dec 31, 2013 #11

    vanhees71

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    The solution is correct. Think about the periodicity of the function!
     
  13. Dec 31, 2013 #12
    All functions here are periodic right, since it has only sinusoids. Am I wrong somewhere? If so please correct me.
     
  14. Dec 31, 2013 #13

    Dick

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    You are wrong. 2cos(πt) is periodic. 7cos(t) is periodic. 2cos(πt)+7cos(t) is NOT periodic.
     
  15. Dec 31, 2013 #14

    Ray Vickson

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    To expand on what Dick said: the term cos(t) is periodic, with period 2π, while cos(πt) is periodic, with period 2. Since π is an irrational number, the two periods are not "commensurate" and that means that when you add the two terms together the result is non-periodic.

    Added note: the above statement seems clear enough, but I will admit it needs proof, and I don't see at the moment how that can be tackled. It might not be easy!
     
    Last edited: Dec 31, 2013
  16. Dec 31, 2013 #15

    LCKurtz

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    At the risk of beating a dead horse: You don't have to analyze the behavior of the original function nor get into a discussion of periodicity or Dirichlet conditions. The ratio of any two angular frequencies in any FS is$$
    \frac{\frac{n\pi}{p}}{\frac{m\pi} p}=\frac n m$$which is rational, and which isn't in the given example.
     
  17. Dec 31, 2013 #16
    Thank you friends. Now that is a point I have to dwell into. Let me try proving it.
    Finally let me ask one more thing. Fourier series representation is only for periodic functions. 2cos(πt) is periodic. 7cos(t) is periodic. 2cos(πt)+7cos(t) is NOT periodic. So 2cos(πt)+7cos(t) must be represented using FOURIER TRANSFORM right?
     
  18. Jan 1, 2014 #17

    Simon Bridge

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    That would be interesting [1]

    Nope.

    Nope.

    Remember - each of the functions, except one, is supposed to be the complete fourier transform for another function ... that is what I was trying to get you to realize about the conditions.
    When you get the fourier transform of a function, it has a characteristic look - especially if the transform is finite. The question is testing that you understand this.
     
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