Project 11: Fourier Series

Description:

A Fourier series is a way of representing a periodic function as a sum of sine and cosine functions.

In our project, we are going to pick a function and compute it's fourier coefficients analytically

Algorithm and Discussion

In this experiment

Suppose f is a periodic function with a period T = 0.131. Then the Fourier series representation of f is a trigonometric series (that is, it is an infinite series consists of sine and cosine terms of the form.

The coefficient gotten is: $a = n * \frac{\pi}{L}$

Implementation and Code

import matplotlib.pyplot as pl import numpy as np from mpl_toolkits import mplot3d import matplotlib.animation as ani

L = 5 B = 3 N = 1000 x = np.linspace(0, L, N) dx = x[1] - x[0] dt = dx**2/4 y = np.where((x <= L/2), 1., 0.) D = np.zeros([10 * N, N], np.float) D[0:] = y

pl.plot(x, y)

for i in range(1, 10 * N - 1): for j in range(1, N - 1): D[i+1, j] = D[i, j] + dt/dx**2*(D[i, j+1] - 2*D[i, j] + D[i, j-1]) D[i+1, 0] = D[i, 0] + dt/dx**2*(D[i, 1] - 2*D[i, 0] + D[i, 1]) D[i+1, N-1] = D[i, N-1] + dt/dx**2*(D[i, N-2] - 2*D[i, N-1] + D[i, N-2]) yfunc = D[i+1] pl.plot(x, yfunc) pl.grid()

t = 0 def fa_vec(x): return np.where((x <= L/2), 1., 0.) def basis(x, n, t): k = n*np.pi/L if n == 0: return np.sqrt(1/L)*np.ones(len(x)) else: base = np.sqrt(2/L)*np.cos(k*x) * np.exp(-(k**2)*t) return base

When $$y = sin(x) + \frac{sin(3x)}{3}$$

y = np.sin(x) + np.sin(3*x)/3; pl.plot(x,y);

When $$y = sin(x) + \frac{sin(3x)}{3} + \frac{sin(5x)}{5}+ \frac{sin(7x)}{7}$$

y = np.sin(x) + np.sin(3*x)/3 + np.sin(5*x)/5 + np.sin(7*x)/7; pl.plot(x,y);

We can keep adding to increase the function, it adds more waves.

Next: check your answer both graphically and numerically using simpson's rule.

def simpson_array(f, h): if len(f)%2 != 0: raise ValueError("fmust be an array with an even number of elements.") evens = f[2:-2:2] odds = f[1:-1:2] return (f[0] + f[-1] + 2*odds.sum() + 4*evens.sum())*dx/3.0 def braket(n,t): return simpson_array(basis(x,n,t)*fa_vec(x),dx) M=30 coefs = [] for n in range(M): coefs.append(braket(n,0)) # do numerical integral sup = np.zeros(N) def Superpos(t): sup = np.zeros(N) for n in range(M): sup += coefs[n] * basis(x, n, t) return sup T = 0.131 pl.plot(x, Superpos(T)) pl.grid() print("%10s\t%10s" % ('n', 'coef')) print("%10s\t%10s" % ('---','-----')) for n in range(M): print("%10d\t%10.5f" % (n, coefs[n]))

np.sqrt(((D[i+1] - Superpos(T))**2).sum()/len(x))

Conclusion

In this experiment, with period T = 0.131. The Fourier series representation of f is a trigonometric series (that is, it is an infinite series consists of sine and cosine terms of the form. After compting its fourier coefficients analytically. We got the coefficient to be $0.040158442751923665$ which is reasonable.

Project 11: Fourier Series

Description:

Algorithm and Discussion

Implementation and Code

Next: check your answer both graphically and numerically using simpson's rule.

Conclusion

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