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

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

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- Thread starter magnifik
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- #1

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

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

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vela

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[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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