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Relationship between Fourier transform and Fourier series?

  1. Nov 7, 2015 #1
    What is the relationship between the Fourier transform of a periodic function and the coefficients of its Fourier series?

    I was thinking Fourier series a special version of Fourier transform, as in it can only be used for periodic function and only produces discrete waves. By this logic, aren't they the same thing then for this case?
     
  2. jcsd
  3. Nov 7, 2015 #2
    Do a Fourier transform of a few short Fourier series (3-5 sin terms), or some simple ones like a square and a triangle wave, and you will see how it works.
     
  4. Nov 7, 2015 #3

    rude man

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    Take a signal g(t) = sin(wt). The Fourier series is of course sin(wt).
    The Fourier integral is quite different: G(f) = (1/j2) [δ(f - f0) - δ(f + f0)]
    with the inversion g(t) = ∫ from -∞ to +∞ of G(f)exp(jωt) df, ω ≡ 2πf.

    You can determine the output of a transfer function H(f) with the Fourier integral: Y(f) = G(f) H(f). Then y(t) = F-1Y(f).
    With the Fourier series of a signal with many harmonics you have to determine the effect of each harmonic separately, then add. Very cumbersome.
    You can also handle a step-sine signal U(t)sin(ω0t) with the transform but not with the series, the latter assuming a signal stretching from -∞ to +∞.
    The series gives an accurate description of an arbitrary periodic function; each coefficient represents the amplitude of each harmonic.
    The integral is called the "spectrum" of the signal. I've always had some problem thinking of a spectrum of a pulse, but there it is.
     
  5. Nov 7, 2015 #4

    HallsofIvy

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    I think of it the other way around! As you take the length of one period going to infinity, the Fourier series goes to the Fourier transform.
     
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