Proof of Even Harmonics Absence in Function w/ Odd Symmetry | Fourier Transform

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SUMMARY

A function exhibiting odd symmetry lacks even harmonics in its Fourier spectrum. This conclusion is supported by the mathematical proof involving the calculation of the 2n-th Fourier coefficients, specifically through the integral of the function multiplied by cosine terms. By applying a change of variable and utilizing the properties of odd symmetry, it is demonstrated that the contributions from the even harmonics cancel out, confirming their absence in the spectrum.

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

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)

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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