Fourier transform

  • Thread starter magnifik
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  • #1
magnifik
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This isn't really a homework problem, but I am having trouble understanding why this is true:

A function with the following symmetry does not have any even harmonics in its spectrum.
fssym4.gif


I understand the concept based on odd/even symmetry properties, but can anyone provide a mathematical proof?
 

Answers and Replies

  • #2
vela
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I'd just try calculating the 2n-th Fourier coefficients. For example,

[tex]a_{2n} = \frac{1}{T}\int_{-T/2}^{T/2} f(t) \cos 2n\omega t\,dt = \frac{1}{T}\left(\int_{-T/2}^{0} f(t) \cos 2n\omega t\,dt + \int_{0}^{T/2} f(t) \cos 2n\omega t\,dt\right) [/tex]

where ω=2π/T. Try using a change of variable like t' = t + T/2 on the first integral, then use the symmetry of f(t), and see if stuff cancels out.
 

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