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Find the Fourier-sequence

  1. Dec 7, 2004 #1
    A function with a period T=2 is defined as [tex]f(x)=\left\{\begin{array}{cc}-1-x,&\mbox{ if }
    -1\leq x<0\\1-x, & \mbox{ if }0\leq x<1\end{array}\right[/tex]
    I'm supposed to find the Fourier-sequence. When trying to find [tex]b_n[/tex] the expression becomes [tex]b_n=\int_{0}^{1} f(x)*sin(n*\omega*x)dx[/tex]
    [tex]\omega=\pi[/tex] My book says [tex]f(x)*sin(n*\omega*x)[/tex] is an even function, which I can't understand. Doesn't that mean that it should be symetrical about the y-axis? I can't see that it is. What am I not doing right here?
     
  2. jcsd
  3. Dec 7, 2004 #2

    Tide

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    A function f(x) is even if f(-x) = f(x) and it is odd if f(-x) = -f(x). Your function satisfies neither of these conditions and so f(x) sin(n*omega*x) is neither even nor odd (the sine function is odd).
     
  4. Dec 7, 2004 #3

    dextercioby

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    Other major defect of f(x) given is that it's not continuous at 0.Restraing things to one of the 2 intervals,i say Tide is right,on any of the 2 intervals the function f(x) sin (n\omega\x) tis neither even,nor odd.
    You should check the text again.
     
  5. Dec 8, 2004 #4

    Maybe i'm the only one seeing this but [tex] f(x) [/tex] is odd, and the product of two odd functions is an even function.

    [tex]f(x)=\left\{\begin{array}{cc}-1-x,&\mbox{ if }
    -1\leq x<0\\1-x, & \mbox{ if }0\leq x<1\end{array}\right[/tex]

    Now take [tex] f(\frac{1}{2}) [/tex]. This is [tex] 1-\frac{1}{2} = \frac{1}{2} [/tex]

    Then take [tex] f(-\frac{1}{2}) [/tex]. This becomes [tex] -1-(-\frac{1}{2}) = -1 + \frac{1}{2} = -\frac{1}{2} [/tex].

    More generally, [tex] f(x) [/tex], [tex] x > 0 [/tex] is [tex] 1-x [/tex]. Then [tex] f(-x) [/tex], [tex] x > 0 [/tex] is [tex] -1-(-x) = -1 + x = x - 1 = -(1-x) = -f(x) [/tex]. The same will hold for [tex] x < 0 [/tex]
     
  6. Dec 8, 2004 #5
    Also, your formula for [tex] b_n [/tex] is wrong. You have to integrate across the entire period, so you should get :

    [tex] b_n = \int_{-1}^{0} f(x)*sin(n*\omega*x) dx + \int_{0}^{1} f(x)*sin(n*\omega*x) dx [/tex]

    And input the correct form of [tex] f(x) [/tex] based on the relevant interval.
     
  7. Dec 8, 2004 #6
    Thanks to all of you! I might come back for some more help in a while... :smile:
     
  8. Dec 8, 2004 #7
    If I draw the line 1-x then this alone would be neither an even or an odd function, right? But if I add the line -1-x, then the two lines are symmetrical about origo. Is this enough to claim that f(x) is an odd function?
     
  9. Dec 8, 2004 #8
    The two lines, I believe, are antisymmetrical. Which is enough to claim f(x) is odd.
    The best thing to do, however, is what Tide did.
     
  10. Dec 8, 2004 #9
    I'm also having some trouble finding out whether this function
    [tex]f(x)=\left\{\begin{array}{cc}0 &\mbox{ if }
    -2\leq x<0\\(1/2)x &\mbox{ if }0\leq x<2\end{array}\right[/tex]
    is even or odd (or neither). Could someone guide me through a bit step by step please? I'm trying to use the definitions for even and odd functions, but I haven't gotten the hang of it yet...
     
  11. Dec 8, 2004 #10
    REad my post, it is an odd function. No Palindrome, Tide did it wrong.
     
  12. Dec 8, 2004 #11
    [tex]f(x)=\left\{\begin{array}{cc}0 &\mbox{ if }
    -2\leq x<0\\(1/2)x &\mbox{ if }0\leq x<2\end{array}\right[/tex]

    That function is neither odd nor even.

    f(1) = 1/2.
    f(-1) = 0

    IF it was even then f(-1) = 1/2, and if it was odd f(-1) = -1/2.
     
  13. Dec 8, 2004 #12
    Actually, I meant you, on your post "maybe I'm the only one...".
    I just the wrong name... :uhh:
     
  14. Dec 8, 2004 #13
    While dealing with Fourier question, always keep in mind that whatever function that can be expressed in Fourier series has a transformed form of [tex]a_0 + \Sigma(a_n\cos\frac{n\Pi x}{L} + b_n\sin\frac{n\Pi x}{L})[/tex]

    Then if you can't remember the formula for each coefficient, we can do it by the conventional way, which is tedious as well.
    First,
    [tex]\int f(x)dx=\int a_0 + \Sigma(a_n\cos\frac{n\Pi x}{L} + b_n\sin\frac{n\Pi x}{L}) dx[/tex]
    We know that any other terms besides [tex]a_0[/tex] would be cancelled. Thus this enable us to find [tex]a_0[/tex]
    Then, to find the coefficient [tex]a_n[/tex] or [tex]b_n[/tex], we can time the LHS and RHS with their respective trigonometric function. One at a time.
    i.e. [tex]\int f(x)\cos\frac{n\Pi x}{L}dx=\int a_0\cos\frac{n\Pi x}{L} + \Sigma(a_n(\cos\frac{n\Pi x}{L})^2 + b_n\sin\frac{n\Pi x}{L}\cos\frac{n\Pi x}{L}) dx[/tex]
    Then only [tex]a_n(\cos\frac{n\Pi x}{L})^2[/tex] will survive in this integration. (Consider the graph of [tex](\cos\frac{n\Pi x}{L})^2[/tex] which is always [tex]\geq{0}[/tex]). To find [tex]b_n[/tex], we can do the same thing. For piece function, we just integrate them at their respective interval.
    This method is super tedious and is highly not recommended during exam. In the exam, just memorize the formulas and apply them.
     
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