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Fourier Transform of Hann function

Discussion in 'Mathematics and Physics' started by derick007, Feb 18, 2016.

  1. derick007

    derick007 Member

    Feb 5, 2013
    I am having difficulty completing a fourier transform on the Hann function using the Fourier frequency shifting property and Eulers formula.

    Hann function = h(t) = 0.5 – 0.5 cos2πfct

    When I use Eulers formula and the frequency shifting property I get

    H(F) =

    -A*e-jπ(f +/- fc)T sin (π(f +/- fc)T)
    4*π*(f +/- fc)

    For the – 0.5 cos2πfct term only.

    Unfortunately the minus sign on the cos term in the Hann function (h(t)) carries through to the A term in the Fourier Transform H(F), meaning the two dirac delta functions at +/- fc are both negative – these should be positive.

    When H(F) is plotted the plots turn out to be the inverse of what they should be i.e. they are positive when they should be negative and vice versa. I don’t believe this is simply a case of plotting the magnitude only.

    In order to get the correct answer I believe the minus sign in the Hann time function somehow is cancelled out in the maths when completing the Fourier Transform ?
  2. MrAl

    MrAl Well-Known Member Most Helpful Member

    Sep 7, 2008

    That doesnt seem to be the case if all we are doing is computing the Fourier Transform.

    When i compute the Fourier Transform directly of cos(wt) i get (d here is the curly delta symbol for an impulse):

    but when i compute the Fourier Transform of -cos(wt) directly i get:

    which is the negative of the first solution.

    This looks correct because any constant carries through an integration, and in the next to last step in the limit i get either 1/2 or -1/2 where clearly one is the negative of the other. Multiplied by 2*pi we get either pi or -pi as shown above.

    Also, simpler is two applications of (d as above):
    e^(j*wo*t) <=> 2*pi*d*(w-wo)

    As noted, 'd' above is used in place of the symbol 'δ'.
    Last edited: Feb 24, 2016

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