5: Gibbs ⊲ Phenomenon Discontinuities Discontinuous Waveform Gibbs Phenomenon Integration Rate at which coefficients decrease with m Differentiation Periodic Extension t2 Periodic 5: Gibbs Phenomenon Extension: Method (a) t2 Periodic Extension: Method (b) Summary
E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 1 / 11 Discontinuities
5: Gibbs Phenomenon A function, v(t), has a discontinuity of amplitude b at t = a if ⊲ Discontinuities Discontinuous Waveform lime→0 (v(a + e) − v(a − e)) = b 6= 0 Gibbs Phenomenon Integration Rate at which Conversely, v(t), is continuous at t = a if the limit, b, equals zero. coefficients decrease with m Differentiation Periodic Extension Examples: t2 Periodic Extension: Method b b (a) u(t) v(t) t2 Periodic 0 0 Extension: Method a – e a a + e a – e a a + e (b) Time (t) Time (t) Summary Continuous Discontinuous
We will see that if a periodic function, v(t), is discontinuous, then its Fourier series behaves in a strange way.
E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 2 / 11 Discontinuous Waveform
5: Gibbs Phenomenon Pulse: T = 1 = 20, width= 1 T , height A = 1 1 Discontinuities F 2 0.5 Discontinuous . T 0 ⊲ Waveform 1 0 5 −i2πmFt 0 5 10 15 20 Gibbs Phenomenon Um = Ae dt T 1 0 max(u )=0.500 Integration 0 0.5T 0.5 i −i2πmFt N=0 Rate at which = R e 0 coefficients decrease 2πmFT 0 0 5 10 15 20 with m m i i 1 −iπm ((−1) −1) max(u )=1.137 Differentiation e 1 = 2πm − 1 = 2πm 0.5 N=1 Periodic Extension 0 t2 Periodic 0 5 10 15 20 Extension: Method 0 m 6= 0, even 1 max(u )=1.100 (a) 3 0.5 2 = 0.5 m = 0 N=3 t Periodic 0 Extension: Method −i 0 5 10 15 20 (b) mπ m odd 1 max(u )=1.094 Summary 5 0.5 1 2 1 N=5 So, u(t) = + sin 2πFt + sin 6πFt 0 2 π 3 0 5 10 15 20 1 1 + sin 10πFt + ... max(u )=1.089 5 41 0.5 N=41 0 N i2πmFt 0 5 10 15 20 Define: uN (t) = m N Ume 1 =− max(u )=1.089 41 0.5 uN (0) = 0.5 ∀N 0 P -1 -0.5 0 0.5 1 1 1 π sin t maxt uN (t) −→ + π t dt ≈ 1.0895 N→∞ 2 0 [Enlarged View: u41(t)] R
E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 3 / 11 [Powers of −1 and i]
Expressions involving (−1)m or, less commonly, im arise quite frequently and it is worth becoming familiar with them. They can arise in several guises:
m e−iπm = eiπm = eiπ = cos(πm)=(−1)m m i 1 πm i 1 π m e 2 = e 2 = i m −i 1 πm −i 1 π m e 2 = e 2 =(−i) As m increases these expressions repeat with periods of 2 or 4. Simple expressions involving these quantities make useful sequences.
m −4 −3 −2 −1 0 1 2 3 4 (−1)m = cos πm = eiπm 1 −1 1 −1 1 −1 1 −1 1 im = ei0.5πm 1 i −1 −i 1 i −1 −i 1 (−i)m = e−i0.5πm 1 −i −1 i 1 −i −1 i 1 1 m 2 (1+(−1) ) 1 0 1 0 1 0 1 0 1 1 m 2 (1 − (−1) ) 0 1 0 1 0 1 0 1 0 1 m m 2 (i +(−i) ) = cos0.5πm 1 0 −1 0 1 0 −1 0 1 1 m m m 4 (1+(−1) + i +(−i) ) 1 0 0 0 1 0 0 0 1
E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – note 1 of slide 3 Gibbs Phenomenon
5: Gibbs Phenomenon N i2πmFt Discontinuities Truncated Fourier Series: uN (t) = m=−N Ume Discontinuous Waveform If u(t) has a discontinuity of height b at t = a then: ⊲ Gibbs Phenomenon P Integration u(a−e)+u(a+e) (1) uN (a) −→ lime→0 Rate at which N→∞ 2 coefficients decrease with m (2) uN (t) has an overshoot of about 9% of b at the discontinuity. For Differentiation Periodic Extension large N the overshoot moves closer to the discontinuity but does 2 t Periodic not get smaller (Gibbs phenomenon). In the limit the overshoot Extension: Method π (a) equals − 1 + 1 sin t dt b ≈ 0.0895b. t2 Periodic 2 π 0 t Extension: Method −1 (b) (3) For large m, theR coefficients, Um decrease no faster than |m| . Summary Example: 1 uN (0) −→ 0.5 0.5 N→∞ 0 0 5 10 15 20
maxt uN (t) −→ 1.0895 ... 1 max(u )=1.089 N 41 →∞ 0.5 N=41 0 0 m 6= 0, even 0 5 10 15 20 1 max(u )=1.089 41 Um = 0.5 m = 0 0.5 i 0 − -1 -0.5 0 0.5 1 mπ m odd
E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 4 / 11 [Origin of Gibbs Phenomenon]
This topic is included for interest but is not examinable.
Our first goal is to express the partial Fourier series, uN (t), in terms of the original signal, u(t). We begin by substituting the integral expression for Un in the partial Fourier series +N i2πnFt +N 1 T −i2πnFτ i2πnFt uN (t)= − Une = − u(τ)e dτ e Pn= N Pn= N T R0 Now we swap the order of the integration and the finite summation (OK if the integral converges ∀n) 1 T +N i2πnF (t−τ) uN (t)= u(τ) − e dτ T R0 Pn= N Now apply the formula for the sum of a geometric progression with z = ei2πF (t−τ): − − +N n z N −zN+1 z (N+0.5)−zN+0.5 − z = = − Pn= N 1−z z 0.5−z0.5 − − − 1 T ei2π(N+0.5)F (τ t)−e i2π(N+0.5)F (τ t) uN (t)= u(τ) − − − dτ T R0 ei2π0.5F (τ t)−e i2π0.5F (τ t) − = 1 T u(τ) sin π(2N+1)F (τ t) dτ T R0 sin πF (τ−t) sin((N+0.5)x) So if we define the Dirichlet Kernel to be DN (x)= sin 0.5x , and set x =2πF (τ − t), we obtain 1 T uN (t)= u(τ)DN (2πF (τ − t)) dτ T R0 So what we have shown is that uN (t) can be obtained by multiplying u(τ) by a time-shifted Dirichlet Kernel and then integrating over one period. Next we will look at the properties of the Dirichlet Kernel.
E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – note 1 of slide 4 [Dirichlet Kernel]
This topic is included for interest but is not examinable.
+N inx N sin((N+0.5)x) Dirichlet Kernel definition: DN (x)= − e =1+2 cos nx = Pn= N Pn=1 sin 0.5x DN (x) is plotted below for N = {2, 5, 10, 21}. The vertical red lines at ±π mark one period.
D (x) 10 D (x) 20 D (x) 40 D (x) 4 2 5 10 21 2 5 10 20 D2(x) D5(x) 0 0 D10(x) 0 D21(x) 0 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 x x x x • Periodic: with period 2π • Average value: hD (x)i = 1 +π D (x)dx =1 N 2π R−π N π π π • First Zeros: DN (x)=0 at x = ± N+0.5 define the main lobe as − N+0.5 [In MATLAB DN (x)=(2N + 1) × diric(x, 2N + 1)] E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – note 2 of slide 4 [Gibbs Phenomenon Overshoot] 1 This topic is included for interest but is not examinable. u (t) 41 0.5 T=20 The partial Fourier Series, uN (t), can be obtained by multiplying u(t) 0 -1 -0.5 0 0.5 1 1.5 2 by a shifted Dirichlet Kernel and integrating over one period: u (t)= 1 T u(τ)D (2πF (τ − t)) dτ N T 0 N 80 t=0 R 60 40 20 For the special case when u(t) is a pulse of height 1 and width 0.5T : 0 -20 1 0.5T -1 -0.5 0 0.5 1 1.5 2 uN (t)= T 0 DN (2πF (τ − t)) dτ 80 t=0.24 R 60 40 Substitute x =2πF (τ − t) 20 0 − − -20 1 πFT 2πFt 1 π 2πFt -1 -0.5 0 0.5 1 1.5 2 uN (t)= 2πFT −2πFt DN (x) dx = 2π −2πFt DN (x) dx R R 80 t=0.96 60 • For t =0 (the blue integral and the blue circle on the upper graph): 40 1 π 20 u (0) = D (x) dx =0.5 0 N 2π 0 N -20 R -1 -0.5 0 0.5 1 1.5 2 T • For t = 2N+1 (the red integral and the red circle on the upper graph): π− π π− π T 1 N+0.5 1 0 1 N+0.5 uN 2N+1 = 2π − π DN (x) dx= 2π − π DN (x) dx + 2π 0 DN (x) dx R N+0.5 R N+0.5 R For large N, we replace the first term by a constant (since it is the semi-integral of the main lobe) and replace the upper limit of the second term by π: ≈ 0.58949 + 1 π D (x) dx =1.08949 2π R0 N • For 0 ≪ t ≪ 0.5T , uN (t) ≈ 1 (the green integral and the green circle on the upper graph). E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – note 3 of slide 4 Integration 5: Gibbs Phenomenon ∞ i2πmFt u t Ume Discontinuities Suppose ( ) = m=−∞ Discontinuous Waveform Define v(t) to be the integral of u(t) [boundedness requires U0 = 0] Gibbs Phenomenon P t t ∞ i πmFτ ⊲ Integration 2 v(t) = u(τ)dτ = m=−∞ Ume dτ Rate at which coefficients decrease ∞ t i2πmFτ with m =R m=−∞ UmR Pe dτ [assume OK to swap and ] Differentiation ∞ i πmFt Periodic Extension 1 2 = c + m=−∞ Um i πmF e t2 Periodic P R 2 R P Extension: Method ∞ i2πmFt c Vme c (a) = + Pm=−∞ where is an integration constant t2 Periodic −i Extension: Method Hence Vm = 2πmF Um except for V0 = c (arbitrary constant) (b) P Summary Example: −2i Square wave: Um = mπ for odd m (0 for even m) −i −2i −1 Triangle wave: Vm = 2πmF × mπ = π2m2F for odd m (0 for even m) 1 5 u (t) v (t) 7 7 0 0 -1 -5 0 5 10 15 20 0 5 10 15 20 1 Convergence: v(t) always converges if u(t) does since Vm ∝ m Um vN (t) is a good approximation even for small N E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 5 / 11 Rate at which coefficients decrease with m 5: Gibbs Phenomenon −2i −1 Discontinuities Square wave: Um = π m for odd m (0 for even m) Discontinuous −1 −2 Waveform Triangle wave: Vm = π2F m for odd m (0 for even m) Gibbs Phenomenon 1 5 u (t) v (t) Integration 7 7 0 0 Rate at which coefficients -1 -5 ⊲ decrease with m 0 5 10 15 20 0 5 10 15 20 Integrating Differentiation −i −1 Periodic Extension u(t) multiplies the Um by 2πF × m ⇒ they decrease faster. t2 Periodic Extension: Method (a) The rate at which the coefficients, Um, decrease with m depends on the t2 Periodic Extension: Method lowest derivative that has a discontinuity: (b) Summary • Discontinuity in u(t) itself (e.g. square wave) −1 For large |m|, Um decreases as |m| • Discontinuity in u′(t) (e.g. triangle wave) −2 For large |m|, Um decreases as |m| • Discontinuity in u(n)(t) −(n+1) For large |m|, Um decreases as |m| • No discontinuous derivatives −|m| For large |m|, Um decreases faster than any power (e.g. e ) E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 6 / 11 Differentiation 5: Gibbs Phenomenon −i Um Discontinuities Integration multiplies by 2πmF . Discontinuous Waveform 2πmF Um i πmF Gibbs Phenomenon Hence differentiation multiplies by −i = 2 Integration Rate at which du(t) coefficients decrease If u(t) is a continuous differentiable function and w(t) = dt then, with m provided that w(t) satisfies the Dirichlet conditions, its Fourier coefficients ⊲ Differentiation Periodic Extension are: t2 Periodic Extension: Method 0 m = 0 (a) Wm = . t2 Periodic (i2πmFUm m 6= 0 Extension: Method (b) Summary Since we are multiplying Um by m the coefficients Wm decrease more slowly with m and so the Fourier series for w(t) may not converge (i.e. w(t) may not satisfy the Dirichlet conditions). d d −→dt −→dt −2 −1 −0 Um ∝|m| Um ∝|m| Um ∝|m| Differentiation makes waveforms spikier and less smooth. E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 7 / 11 Periodic Extension 5: Gibbs Phenomenon Suppose y(t) is only defined over a finite interval (a, b). Discontinuities Discontinuous Waveform You have two reasonable choices to make a periodic version: Gibbs Phenomenon Integration Rate at which (a) T = b − a, u(t) = y(t) for a ≤ t < b coefficients decrease with m Differentiation y(t) a ≤ t ≤ b ⊲ Periodic Extension (b) T = 2(b − a), u(t) = t2 Periodic (y(2b − t) b ≤ t ≤ 2b − a Extension: Method (a) t2 Periodic Example: Extension: Method 2 (b) y(t) = t for 0 ≤ t < 2 Summary 4 4 4 2 2 2 0 0 0 -2 0 2 4 -2 0 2 4 -2 0 2 4 y(t) (a) T = 2 (b) T = 4 Option (b) has twice the period, no discontinuities, no Gibbs phenomenon overshoots and if y(t) is continuous the coefficients decrease at least as fast as |m|−2. E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 8 / 11 t2 Periodic Extension: Method (a) 5: Gibbs Phenomenon y t t2 t < Discontinuities ( ) = for 0 ≤ 2 Discontinuous Waveform Method (a): T = 1 = 2 Gibbs Phenomenon F Integration 1 T 2 −i2πmFt 1 T 2 4 Rate at which Um = t e dt U0 = t dt = coefficients decrease T 0 T 0 3 with m − − − T Differentiation 1 R t2e i2πmFt 2te i2πmFt 2e i2πmFt R Periodic Extension = T −i2πmF − (−i2πmF )2 + (−i2πmF )3 t2 Periodic 0 Extension: Method h i πmF i πmFT i ⊲ (a) Substitute e− 2 0 = e− 2 = 1 [for integer m] t2 Periodic Extension: Method T 2 T (b) 1 2 = T i πmF − 2 Summary − 2 (−i2πmF ) 2ih 2 i = πm + π2m2 4 4 4 K=1 K=3 K=6 2 2 2 0 0 0 -2 0 2 4 -2 0 2 4 -2 0 2 4 U0:3 = [1.333, 0.203 + 0.637i, 0.051 + 0.318i, 0.023 + 0.212i] E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 9 / 11 t2 Periodic Extension: Method (b) 5: Gibbs Phenomenon y t t2 t < Discontinuities ( ) = for 0 ≤ 2 Discontinuous Waveform Method (b): T = 1 = 4 Gibbs Phenomenon F Integration 1 0.5T 2 −i2πmFt 1 0.5T 2 4 Rate at which Um = t e dt U0 = t dt = coefficients decrease T −0.5T T −0.5T 3 with m − − − 0.5T Differentiation 1 R t2e i2πmFt 2te i2πmFt 2e i2πmFt R Periodic Extension = T −i2πmF − (−i2πmF )2 + (−i2πmF )3 t2 Periodic −0.5T Extension: Method h m i (a) Substitute e±iπmFT = e±iπm =(−1) [for integer m] t2 Periodic Extension: Method m ⊲ (b) (−1) −2T = 2 [all even powers of t cancel out] Summary T (−i2πmF ) (−1)mTh2 (−1)m8i = 2π2m2 = π2m2 4 4 4 K=1 K=3 K=6 2 2 2 0 0 0 -2 0 2 4 -2 0 2 4 -2 0 2 4 U0:3 = [1.333, −0.811, 0.203, −0.090] [u(t) real+even ⇒ Um real] Convergence is noticeably faster than for method (a) E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 10 / 11 Summary 5: Gibbs Phenomenon • Discontinuity at t = a Discontinuities Discontinuous ◦ Gibbs phenomenon: uN (t) overshoots by 9% of iump Waveform Gibbs Phenomenon ◦ uN (a) → mid point of iump Integration t −i Rate at which • Integration: If v(t) = u(τ)dτ, then Vm = Um coefficients decrease 2πmF with m and V0 = c, an arbitrary constant. U0 must be zero. Differentiation R du(t) Periodic Extension • Differentiation: If w(t) = , then Wm = i2πmFUm provided t2 Periodic dt Extension: Method w(t) satisfies Dirichlet conditions (it might not) (a) t2 Periodic • Rate of decay: Extension: Method k (b) −(k+1) d u(t) ◦ For large n, Un decreases at a rate |n| where is ⊲ Summary dtk the first discontinuous derivative 2 2 ◦ Error power: (u(t) − uN (t)) = |n|>N |Un| • Periodic ExtensionDof finite domainE signalP of length L ◦ (a) Repeat indefinitely with period T = L ◦ (b) Reflect alternate repetitions for period T = 2L no discontinuities or Gibbs phenomenon For further details see RHB Chapter 12.4, 12.5, 12.6 E1.10 Fourier Series and Transforms (2014-5559) Gibbs Phenomenon: 5 – 11 / 11