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Fourier Analysis.

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Presentation on theme: "Fourier Analysis."— Presentation transcript:

1 Fourier Analysis

2 p. 216, Eq. (5.36) 𝑎 2 𝑑 2 đ‘Ļ 𝑑 𝑡 2 + 𝑎 1 𝑑đ‘Ļ 𝑑𝑡 + 𝑎 0 đ‘Ļ=𝑓 𝑡 p. 246 𝑓 𝑡 = 𝐴 1 cos 𝜔𝑡 + 𝐴 2 sin 𝜔𝑡 đ‘Ļ 𝑡 = đĩ 1 cos 𝜔𝑡 + đĩ 2 sin 𝜔𝑡 𝑓 𝑡 = 𝑋 𝑚 cos 𝜔𝑡+ 𝜙 đ‘Ĩ đ‘Ļ 𝑡 = 𝑌 𝑚 cos 𝜔𝑡+ 𝜙 đ‘Ļ 𝕏= 𝑋 𝑚 𝑒 𝑗 𝜙 đ‘Ĩ ⟹ 𝕐= 𝑌 𝑚 𝑒 𝑗 𝜙 đ‘Ļ 𝑋 𝑚 , 𝜙 đ‘Ĩ ⟹ 𝑌 𝑚 , 𝜙 đ‘Ļ

3 𝑓 𝑡 = 𝑚=−∞ ∞ 𝑐 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡 = 𝑚=−∞ ∞ đļ 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡+ 𝜙 𝑚 = 𝑚=−∞ ∞ 𝕏 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡
đ‘Ļ 𝑡 = 𝑚=−∞ ∞ đ‘Ļ 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡 = 𝑚=−∞ ∞ 𝑌 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡+ 𝜃 𝑚 = 𝑚=−∞ ∞ 𝕐 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡 𝕏 𝑚 ⟹ 𝕐 𝑚 𝑋 𝑚 , 𝜙 𝑚 ⟹ 𝑌 𝑚 , 𝜃 𝑚

4 𝑎 2 𝑑 2 đ‘Ļ 𝑑 𝑡 2 + 𝑎 1 𝑑đ‘Ļ 𝑑𝑡 + 𝑎 0 đ‘Ļ=𝑓 𝑡 𝑎 2 𝑗 𝜔 𝑚 𝑎 1 𝑗 𝜔 𝑚 + 𝑎 0 𝕐 𝑚 = 𝕏 𝑚 𝕐 𝑚 = 𝕏 𝑚 𝑎 2 𝑗 𝜔 𝑚 𝑎 1 𝑗 𝜔 𝑚 + 𝑎 0 = 𝑌 𝑚 𝑒 𝑗 𝜃 𝑚

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