Fourier Series Lecture 1
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Fourier Methods • Fourier sine series • Application to the wave equation • Fourier cosine series • Fourier full range series Blue book • Complex form of Fourier series • Introduction to Fourier transforms and the convolution theorem New chapter 12 Dr Mervyn Roy (S6) www2.le.ac.uk/departments/physics/people/academic-staff/mr6 PA214 Waves and Fields Fourier Methods Resources Lecture notes www2.le.ac.uk/departments/physics/people/academic-staff/mr6 214 course texts • Blue book, new chapter 12 available on Blackboard Notes on Blackboard • Notes on symmetry and on trigonometric identities • Computing exercises • Exam tips • mock papers Books • Mathematical Methods in the Physical Sciences (Mary L. Boas) • Library! PA214 Waves and Fields Fourier Methods Introduction The wave equation π2π¦ 1 π2π¦ = 2 2, 2 ππ₯ π ππ‘ for a string fixed at π₯ = 0 and π₯ = πΏ has harmonic solutions ππx ππππ‘ ππππ‘ π¦ π₯, π‘ = sin ππ cos + ππ sin . πΏ πΏ πΏ Superposition tells us that sums of such terms must also be solutions, π¦ π₯, π‘ = π ππx ππππ‘ ππππ‘ sin ππ cos + ππ sin . πΏ πΏ πΏ PA214 Waves and Fields Fourier Methods π¦ π₯, π‘ = π ππx ππππ‘ ππππ‘ sin ππ cos + ππ sin πΏ πΏ πΏ Set coefficients from initial conditions, e.g. string released from 2ππ₯ rest with π¦ π₯, 0 = sin , then ππ = πΏπ2 , ππ = 0. πΏ PA214 Waves and Fields Fourier Methods What happens if the initial shape of the string is something more complex? In general π¦(π₯, 0) can be any function π π₯ , ππx π¦ π₯, 0 = π π₯ = ππ sin . πΏ π The implication is that we can represent any function π(π₯) as a sum of sines … and/or cosines or complex exponentials. This is Fourier’s theorem PA214 Waves and Fields Fourier Methods Fourier sine series (half-range) We will find that a function π(π₯) in the range 0 ≤ π₯ < πΏ can be represented by the Fourier sine series ∞ π π₯ = π=1 πππ₯ ππ sin , πΏ where 2 ππ = πΏ πΏ 0 πππ₯ π π₯ sin ππ₯. πΏ PA214 Waves and Fields Fourier Methods Fourier sine series How does this work ? Need a standard integral (new chapter 12 – A.2)… πΏ 0 πππ₯ πππ₯ πΏ sin sin ππ₯ = πΏππ πΏ πΏ 2 PA214 Waves and Fields Fourier Methods Square wave, π(π) = π ∞ π π₯ = π=1 2 ππ = πΏ πΏ 0 πππ₯ ππ sin , πΏ πππ₯ π π₯ sin ππ₯. πΏ PA214 Waves and Fields Fourier Methods Square wave, π(π) = π ∞ π π₯ = π=1 2 ππ = πΏ πΏ 0 πππ₯ ππ sin , πΏ πππ₯ π π₯ sin ππ₯. πΏ 4 ππ₯ π π₯ ≈ sin π πΏ PA214 Waves and Fields Fourier Methods Square wave, π(π) = π ∞ π π₯ = π=1 2 ππ = πΏ πΏ 0 πππ₯ ππ sin , πΏ πππ₯ π π₯ sin ππ₯. πΏ 4 ππ₯ 4 3πx π π₯ ≈ sin + sin π πΏ 3π πΏ PA214 Waves and Fields Fourier Methods Square wave, π(π) = π ∞ π π₯ = π=1 2 ππ = πΏ πΏ 0 πππ₯ ππ sin , πΏ πππ₯ π π₯ sin ππ₯. πΏ 4 ππ₯ 4 3πx 4 5ππ₯ π π₯ ≈ sin + sin + sin π πΏ 3π πΏ 5π πΏ PA214 Waves and Fields Fourier Methods Square wave, π(π) = π ∞ π π₯ = π=1 2 ππ = πΏ πΏ 0 πππ₯ ππ sin , πΏ πππ₯ π π₯ sin ππ₯. πΏ 4 ππ₯ 4 3πx 4 5ππ₯ 4 7ππ₯ π π₯ ≈ sin + sin + sin + sin π πΏ 3π πΏ 5π πΏ 7π πΏ PA214 Waves and Fields Fourier Methods Square wave, π(π) = π ∞ π π₯ = π=1 2 ππ = πΏ πΏ 0 πππ₯ ππ sin , πΏ πππ₯ π π₯ sin ππ₯. πΏ 4 ππ₯ 4 3πx 4 19ππ₯ π π₯ ≈ sin + sin + β―+ sin π πΏ 3π πΏ 19π πΏ PA214 Waves and Fields Fourier Methods Periodic extension of Fourier sine series ∞ π π₯ = π=1 πππ₯ ππ sin πΏ We know that sine waves have odd symmetry, πππ₯ −πππ₯ sin = − sin , πΏ πΏ and that πππ₯ ππ π₯ + 2πΏ sin = sin . πΏ πΏ PA214 Waves and Fields Fourier Methods Within 0 ≤ π₯ < πΏ can expand any function π(π₯) as a sum of sine waves, ∞ π π₯ = π=1 πππ₯ ππ sin . πΏ How does this expansion behave outside of the range 0 ≤ π₯ < πΏ ? PA214 Waves and Fields Fourier Methods π π₯ = 1, 0 ≤ π₯ < πΏ square wave π π₯ = π₯, 0 ≤ π₯ < πΏ sawtooth wave π π₯ = π −π₯ , 0 ≤ π₯ < πΏ exp wave (odd) PA214 Waves and Fields Fourier Methods The wave equation String fixed at π₯ = 0 and π₯ = πΏ ππx ππππ‘ ππππ‘ sin π΅π cos + π΄π sin . πΏ πΏ πΏ π¦ π₯, π‘ = π Initial conditions π¦(π₯, 0) = π(π₯), and π¦π‘ π₯, 0 = π π₯ . 2 π΅π = πΏ πΏ 0 πππ₯ π π₯ sin ππ₯ πΏ ππππ΄π 2 = πΏ πΏ πΏ 0 πππ₯ π π₯ sin ππ₯ πΏ PA214 Waves and Fields Fourier Methods e.g. π¦(π₯, 0) = π₯, and π¦π‘ π₯, 0 = 0, then π΅π = 2πΏ −1 (see workshop 1, exercises 1 & 2) PA214 Waves and Fields π+1 /(ππ) Fourier Methods e.g. π¦(π₯, 0) = π₯, and π¦π‘ π₯, 0 = 0, then π΅π = 2πΏ −1 (see workshop 1, exercises 1 & 2) PA214 Waves and Fields π+1 /(ππ) Fourier Methods πΏ 2 , and 2 e.g. π¦ π₯, 0 = exp π₯ − π¦π‘ π₯, 0 = 0 (see new chapter 12, exercise 12.5) PA214 Waves and Fields Fourier Methods πΏ 2 , and 2 e.g. π¦ π₯, 0 = exp π₯ − π¦π‘ π₯, 0 = 0 (see new chapter 12, exercise 12.5) PA214 Waves and Fields Fourier Methods Can go through the same procedure with the solutions to other PDEs e.g. Laplace equation (see workshop 1 exercise 3), π2π π2π + 2 = 0. 2 ππ₯ ππ¦ Imagine the boundary conditions are π 0, π¦ = π πΏ, π¦ = 0 and lim π = 0 then π¦→∞ ∞ π π₯, π¦ = π=1 ππx −πππ¦/πΏ π΅π sin π πΏ and can find coefficients π΅π from boundary condition for π(π₯, 0) PA214 Waves and Fields Fourier Methods
