Fourier Analysis.

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Presentation transcript:

Fourier Analysis

p. 216, Eq. (5.36) 𝑎 2 𝑑 2 𝑦 𝑑 𝑡 2 + 𝑎 1 𝑑𝑦 𝑑𝑡 + 𝑎 0 𝑦=𝑓 𝑡 p. 246 𝑓 𝑡 = 𝐴 1 cos 𝜔𝑡 + 𝐴 2 sin 𝜔𝑡 𝑦 𝑡 = 𝐵 1 cos 𝜔𝑡 + 𝐵 2 sin 𝜔𝑡 𝑓 𝑡 = 𝑋 𝑚 cos 𝜔𝑡+ 𝜙 𝑥 𝑦 𝑡 = 𝑌 𝑚 cos 𝜔𝑡+ 𝜙 𝑦 𝕏= 𝑋 𝑚 𝑒 𝑗 𝜙 𝑥 ⟹ 𝕐= 𝑌 𝑚 𝑒 𝑗 𝜙 𝑦 𝑋 𝑚 , 𝜙 𝑥 ⟹ 𝑌 𝑚 , 𝜙 𝑦

𝑓 𝑡 = 𝑚=−∞ ∞ 𝑐 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡 = 𝑚=−∞ ∞ 𝐶 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡+ 𝜙 𝑚 = 𝑚=−∞ ∞ 𝕏 𝑚 𝑒 𝑗 𝜔 𝑚 𝑡 𝑦 𝑡 = 𝑚=−∞ ∞ 𝑦 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

𝑎 2 𝑑 2 𝑦 𝑑 𝑡 2 + 𝑎 1 𝑑𝑦 𝑑𝑡 + 𝑎 0 𝑦=𝑓 𝑡 𝑎 2 𝑗 𝜔 𝑚 2 + 𝑎 1 𝑗 𝜔 𝑚 + 𝑎 0 𝕐 𝑚 = 𝕏 𝑚 𝕐 𝑚 = 𝕏 𝑚 𝑎 2 𝑗 𝜔 𝑚 2 + 𝑎 1 𝑗 𝜔 𝑚 + 𝑎 0 = 𝑌 𝑚 𝑒 𝑗 𝜃 𝑚