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Fourier series and orthogonality, completeness

  1. Jan 28, 2015 #1
    http://ms.mcmaster.ca/courses/20102011/term4/math2zz3/Lecture1.pdf


    On pg 10, the example says f(x)=/=0 while R.H.S is zero. It is an equations started from the assumption in pg 9; f(x)=c0f(x)0+c1f(x)1…, then how do we get inequality?

    if the system is complete and orthogonal, then (f(x),ϕ_n(x))=0, which makes sense only when f(x)=0.
    but we know for fourier series, we get values for Rhs and Lhs.
     
  2. jcsd
  3. Jan 29, 2015 #2

    DrClaude

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    Staff: Mentor

    That's the key word. The comment on p. 10 is for an arbitrary orthogonal system. That is why completeness is necessary, as explained on p. 11.
     
  4. Jan 29, 2015 #3
    Could you explain the details?

    Also, if (f(x),ϕ_n(x))=0 holds in general for complete series, then fourier series must be also zero since they are complete.



    however we know that for signals we do get values within the interval of pi and -pi. is this because what we usually solve for signal input f(t) is not complete?

    so we are approximating the signal?
     
  5. Jan 29, 2015 #4

    DrClaude

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    Staff: Mentor

    Look at p. 12, where an example is given of an orthogonal system that is not complete. Using only cosines, you could never write an expression for e.g. sin(x).

    But it's the other way around. If ##(f, \phi_n) = 0## for a given ##f(x)## and for all ##\phi_n##, then ##\{\phi_n\}## is not a complete set.

    I'm not sure what you are asking here. If the signal is periodic but not on the interval ##[-\pi,\pi]##, then it is trivial to scale/shift it to that interval. If the signal is finite in time, then it is delt with the same way. If the signal is not finite, or only part of it is known, then indeed thre are approximations being made.
     
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