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 <x< N+0.5 • Peak value: 2N +1 at x =0. The main lobe gets narrower but higher as N increases. • Main Lobe semi-integral: π π N+0.5 N+0.5 sin((N+0.5)x) 1 π sin y x=0 DN (x)dx = x=0 sin 0.5x dx = N+0.5 y=0 y dy[y =(N +0.5)x] R R R sin 2N+1 y y where we substituted y =(N +0.5)x. Now, for large N, we can approximate sin 2N+1 ≈ 2N+1 : π N+0.5 1 π sin y π sin y x=0 DN (x)dx ≈ N+0.5 y=0 y dy =2 y=0 y dy ≈ 3.7038741 ≈ 2π × 0.58949 R R 2N+1 R We see that, for large enough N, the main lobe semi-integral is independent of N. [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).