BMACC Computational Fourier Analysis Mathematics, Computing and - PowerPoint PPT Presentation

฀ ฀ ฀Computational฀Physics฀Curricula ฀฀฀Blended฀Multimodal฀Access฀to ฀ ฀ ฀ ฀ BMACC ฀ Computational Fourier Analysis Mathematics, Computing and Nonlinear Oscillations Rubin H Landau Sally Haerer, Producer-Director Oregon State University with National Science Foundation Support Course: Computational Physics I (Prerequisite: Intro Computational Science ) 1 / 1

Outline Text: Three Fourier Units Unit I: Fourier series: Review & DFT (Only part of I today!) Unit II: Signal filtering to reduce noise Unit III: Fast Fourier transform (FFT) ⇒ real time DFT Today’s Lecture: 2 / 1

Problem: Amount of sin ω t in Nonlinear Oscillation? Example Example (in green): Sawtooth function 1 1 Y( ) y(t) 0 0 -1 -1 0 20 t ω What “frequencies” present? (What does this mean?) Meaning: “Fourier decomposition (right)” Periodic, many sinusoids; yet, one period Only steady state, let transients die out 3 / 1

Physics Problem (see ODE): Nonharmonic Oscillator Nonharmonic (nontranditional) oscillator Harmonic Anharmonic V ( x ) = k | x | p , p � = 2 p = 2 p = 6 V V large p ⇒ square well x x square well ⇒ sawtooth Linear Nonlinear Perturbed oscillator V ( x ) = 1 2 kx 2 � 1 − 2 � V(x) 3 α x Harmonic Anharmonic Linear: 1st approximation x All V s → constant period T Linear Unbound Periodic not sinusoidal Nonlinear 4 / 1

Math (Theory): Fourier Series Fourier’s Theorem: Function Representation Single-valued periodic function ("signal") Finite number discontinuities ∞ y ( t ) = a 0 � 2 + ( a n cos n ω t + b n sin n ω t ) (1) n = 1 a n ↔ cos n ω t “in” y ( t ) Intensity ( ω ) , Power ( ω ) ∝ a 2 n + b 2 n True frequency ω ≡ ω 1 = 2 π T � = HO frequency Need know period T , y ( t + T ) = y ( t ) Period may depend on amplitude 5 / 1

Math: Fourier Series Fit Fourier Series ∞ y ( t ) = a 0 � 2 + ( a n cos n ω t + b n sin n ω t ) (2) n = 1 Theorem: Series = “Best Fit” i [ y ( t i ) − y i ] 2 � Least-squares sense: minimizes ⇒ � ∞ a n cos n ω t ≃ � y ( t ) � i ⇒ miss discontinuities Need infinite number of terms Not exact math fit (power series) Not good numerical solution ( ∞ terms) Not closed form analytic solution 6 / 1

Fourier Integrals (Transforms) Nonperiodic Functions 1.0 s = 1, τ = 0 Nonperiodic ≡ T → ∞ 0.5 0.0 ⇒ continuous ω –0.5 Math: � → � –1.0 –6 –4 –2 0 2 4 6 t � Numerically: → � y ( t ) = Numerically: the same! b 0 + b 1 cos ω 1 t + b 2 cos 2 ω 1 t + · · · Fourier Semantics ω 1 = 2 π T = fundamental b 0 : DC component � normal modes n ω 1 = n th harmonic ω > ω 1 : overtones Modes: see next slide 7 / 1

Fourier Series for Nonlinear Oscillations? ∞ y ( t ) = a 0 � 2 + ( a n cos n ω t + b n sin n ω t ) (3) n = 1 OK by Theorem if periodic, but Nonlinear oscillators: ω 1 = ω ( A ) = 2 π/ T Individual terms � = solution Linear eqtn ↔ principle of linear superposition y 1 ( t ) , y 2 ( t ) = solutions (4) ⇒ α y 1 ( t ) + β y 2 ( t ) = solution (5) Useful if not too nonlinear Broadband spectrum ⇒ chaos Many strong overtones 8 / 1

Determine Fourier Coefficients a n , b n Project Out Components � 2 π � � ∞ a 0 � � ω | y � = dt cos n ω t 2 + ( a n cos n ω t + b n sin n ω t ) 0 n = 1 (6) � T � a n � = 2 � cos n ω t � ⇒ dt y ( t ) (7) b n T sin n ω t 0 a 0 = 2 � y ( t ) � Be smart: use Symmetry Odd y ( − t ) = − y ( t ) ⇒ a n ≡ 0 Even y ( − t ) = y ( t ) ⇒ b n ≡ 0 � Realistic calculations: → � , y ( t < 0 ) =? ⇒ small b n 9 / 1

Determine Fourier Coefficients a n , b n Project Out Components � 2 π � � ∞ a 0 � � ω | y � = dt cos n ω t 2 + ( a n cos n ω t + b n sin n ω t ) 0 n = 1 (6) � T � a n � = 2 � cos n ω t � ⇒ dt y ( t ) (7) b n T sin n ω t 0 a 0 = 2 � y ( t ) � Be smart: use Symmetry Odd y ( − t ) = − y ( t ) ⇒ a n ≡ 0 Even y ( − t ) = y ( t ) ⇒ b n ≡ 0 � Realistic calculations: → � , y ( t < 0 ) =? ⇒ small b n 10 / 1

Determine Fourier Coefficients a n , b n Project Out Components � 2 π � � ∞ a 0 � � ω | y � = dt cos n ω t 2 + ( a n cos n ω t + b n sin n ω t ) 0 n = 1 (6) � T � a n � = 2 � cos n ω t � ⇒ dt y ( t ) (7) b n T sin n ω t 0 a 0 = 2 � y ( t ) � Be smart: use Symmetry Odd y ( − t ) = − y ( t ) ⇒ a n ≡ 0 Even y ( − t ) = y ( t ) ⇒ b n ≡ 0 � Realistic calculations: → � , y ( t < 0 ) =? ⇒ small b n 11 / 1

Determine Fourier Coefficients a n , b n Project Out Components � 2 π � � ∞ a 0 � � ω | y � = dt cos n ω t 2 + ( a n cos n ω t + b n sin n ω t ) 0 n = 1 (6) � T � a n � = 2 � cos n ω t � ⇒ dt y ( t ) (7) b n T sin n ω t 0 a 0 = 2 � y ( t ) � Be smart: use Symmetry Odd y ( − t ) = − y ( t ) ⇒ a n ≡ 0 Even y ( − t ) = y ( t ) ⇒ b n ≡ 0 � Realistic calculations: → � , y ( t < 0 ) =? ⇒ small b n 12 / 1

Determine Fourier Coefficients a n , b n Project Out Components � 2 π � � ∞ a 0 � � ω | y � = dt cos n ω t 2 + ( a n cos n ω t + b n sin n ω t ) 0 n = 1 (6) � T � a n � = 2 � cos n ω t � ⇒ dt y ( t ) (7) b n T sin n ω t 0 a 0 = 2 � y ( t ) � Be smart: use Symmetry Odd y ( − t ) = − y ( t ) ⇒ a n ≡ 0 Even y ( − t ) = y ( t ) ⇒ b n ≡ 0 � Realistic calculations: → � , y ( t < 0 ) =? ⇒ small b n 13 / 1

Example: Sawtooth Function, Analytic a n , b n Example 1 1 y(t) Y( ) t 0 ≤ t ≤ T � T / 2 , ( 7 ) 2 y ( t ) = 0 0 t − T T T / 2 , 2 ≤ t ≤ T -1 -1 0 20 t ω Periodic, nonharmonic, discontinuous, sharp corners Odd ( ⇒ sine series), obvious shift to left: t − T 2 ≤ t ≤ T y ( t ) = T / 2 , (8) 2 � + T / 2 2 t b n = dt sin n ω t (above) (9) T T / 2 − T / 2 2 � sin ω t − 1 2 sin 2 ω t + 1 � ⇒ y ( t ) = 3 sin 3 ω t · · · (10) π 14 / 1

Exercise: Sawtooth Function Synthesis Sum Fourier series for n = 2 , 4 , 10 , 20 terms. 1 Plot two periods & compare to signal. 2 Check that series gives the mean value (0) at discontinuity 3 Gibbs overshoot: Check that series overshoots 4 discontinuity by about 9 % , and that this persists even when summing over a large number of terms. You deserve a break right now! 5 15 / 1

Math (Theory): Fourier Transforms Fourier series = right tool for periodic functions Fourier Integral for Nonperiodic Functions 1.0 s = 1, τ = 0 0.5 Wave packets, pulses 0.0 –0.5 Continuous frequencies –1.0 –6 –4 –2 0 2 4 6 t � + ∞ d ω Y ( ω ) e i ω t Signal: y ( t ) = √ (11) 2 π −∞ Think linear superposition: exp ( i ω t ) = cos ω t + i sin ω t Complex transform Y ( ω ) ∼ ( a n , b n ) ∼ coefficients √ 1 / 2 π, + i ω t = physics conventions 16 / 1

Fourier Transform: y ( t ) → Y ( ω ) 1 1 y(t) Y( ) 0 0 -1 -1 0 20 t ω Fourier Amplitudes ( a n ) → Transform � + ∞ d ω Y ( ω ) e + i ω t √ Signal, function: y ( t ) = (12) 2 π −∞ � + ∞ dt y ( t ) e − i ω t Transform, spectral: Y ( ω ) = √ (13) 2 π −∞ Power Spectrum ∝ | Y ( ω ) | 2 , maybe log 10 | Y | 2 17 / 1

Math Identity: Inversion of Transform Substitute for y ( t ) and rearrange � + ∞ � + ∞ dt e − i ω t d ω ′ e i ω ′ t √ √ Y ( ω ′ ) Y ( ω ) = (14) 2 π 2 π −∞ −∞ � + ∞ �� + ∞ � dt e i ( ω ′ − ω ) t d ω ′ Y ( ω ′ ) Y ( ω ) = (15) 2 π −∞ −∞ Depends on noncomputable Dirac delta function: � + ∞ dt e i ( ω ′ − ω ) t = 2 πδ ( ω ′ − ω ) (16) −∞ Don’t try on computer! 18 / 1

Summary Most any periodic function “represented” by Fourier series. Most any nonperiodic function “represented” by Fourier integral. Infinite series or integral not practical algorithm or in experiment. Must know actual period. Period depends on amplitude in general. DFT (next): when done is simple, elegant and powerful Fast Fourier transform (FFT) is faster. That’s all folks! 19 / 1