我尝试根据以下公式实现傅立叶级数函数:

...在哪里...

...和...



这是我解决问题的方法:

import numpy as np
import pylab as py

# Define "x" range.
x = np.linspace(0, 10, 1000)

# Define "T", i.e functions' period.
T = 2
L = T / 2

# "f(x)" function definition.
def f(x):
    return np.sin(np.pi * 1000 * x)

# "a" coefficient calculation.
def a(n, L, accuracy = 1000):
    a, b = -L, L
    dx = (b - a) / accuracy
    integration = 0
    for i in np.linspace(a, b, accuracy):
        x = a + i * dx
        integration += f(x) * np.cos((n * np.pi * x) / L)
    integration *= dx
    return (1 / L) * integration

# "b" coefficient calculation.
def b(n, L, accuracy = 1000):
    a, b = -L, L
    dx = (b - a) / accuracy
    integration = 0
    for i in np.linspace(a, b, accuracy):
        x = a + i * dx
        integration += f(x) * np.sin((n * np.pi * x) / L)
    integration *= dx
    return (1 / L) * integration

# Fourier series.
def Sf(x, L, n = 10):
    a0 = a(0, L)
    sum = 0
    for i in np.arange(1, n + 1):
        sum += ((a(i, L) * np.cos(n * np.pi * x)) + (b(i, L) * np.sin(n * np.pi * x)))
    return (a0 / 2) + sum

# x axis.
py.plot(x, np.zeros(np.size(x)), color = 'black')

# y axis.
py.plot(np.zeros(np.size(x)), x, color = 'black')

# Original signal.
py.plot(x, f(x), linewidth = 1.5, label = 'Signal')

# Approximation signal (Fourier series coefficients).
py.plot(x, Sf(x, L), color = 'red', linewidth = 1.5, label = 'Fourier series')

# Specify x and y axes limits.
py.xlim([0, 10])
py.ylim([-2, 2])

py.legend(loc = 'upper right', fontsize = '10')

py.show()

...这是绘制结果后得到的结果:

我已经阅读了How to calculate a Fourier series in Numpy?,并且已经实现了这种方法。它的效果很好,但是它使用了幂函数方法,在计算a_{n}b_{n}系数的积分时,我想重点介绍三角函数和矩形方法。

先感谢您。

更新(已解决)

最后,这是代码的有效示例。但是,我会花更多的时间在上面,因此,如果有任何可以改进的地方,那就可以了。
from __future__ import division
import numpy as np
import pylab as py

# Define "x" range.
x = np.linspace(0, 10, 1000)

# Define "T", i.e functions' period.
T = 2
L = T / 2

# "f(x)" function definition.
def f(x):
    return np.sin((np.pi) * x) + np.sin((2 * np.pi) * x) + np.sin((5 * np.pi) * x)

# "a" coefficient calculation.
def a(n, L, accuracy = 1000):
    a, b = -L, L
    dx = (b - a) / accuracy
    integration = 0
    for x in np.linspace(a, b, accuracy):
        integration += f(x) * np.cos((n * np.pi * x) / L)
    integration *= dx
    return (1 / L) * integration

# "b" coefficient calculation.
def b(n, L, accuracy = 1000):
    a, b = -L, L
    dx = (b - a) / accuracy
    integration = 0
    for x in np.linspace(a, b, accuracy):
        integration += f(x) * np.sin((n * np.pi * x) / L)
    integration *= dx
    return (1 / L) * integration

# Fourier series.
def Sf(x, L, n = 10):
    a0 = a(0, L)
    sum = np.zeros(np.size(x))
    for i in np.arange(1, n + 1):
        sum += ((a(i, L) * np.cos((i * np.pi * x) / L)) + (b(i, L) * np.sin((i * np.pi * x) / L)))
    return (a0 / 2) + sum

# x axis.
py.plot(x, np.zeros(np.size(x)), color = 'black')

# y axis.
py.plot(np.zeros(np.size(x)), x, color = 'black')

# Original signal.
py.plot(x, f(x), linewidth = 1.5, label = 'Signal')

# Approximation signal (Fourier series coefficients).
py.plot(x, Sf(x, L), '.', color = 'red', linewidth = 1.5, label = 'Fourier series')

# Specify x and y axes limits.
py.xlim([0, 5])
py.ylim([-2.2, 2.2])

py.legend(loc = 'upper right', fontsize = '10')

py.show()

最佳答案

考虑以一种不同的方式(逐块地)开发代码。如果这样的代码在第一次尝试时会起作用,您应该会感到惊讶。正如@ tom10所说,调试是一种选择。另一个选择是在解释器中逐步逐步编写代码原型(prototype),甚至使用ipython更好。

在上面,您期望b_1000不为零,因为输入的f(x)是一个正弦曲线,其中包含1000。您还期望所有其他系数都为零,对吗?

然后,您应该只关注b(n, L, accuracy = 1000)函数。看着它,三件事出了问题。这里有一些提示。

  • dx的乘法在循环内。可以吗
  • 在循环中,i应该是整数吗?真的是整数吗?通过原型(prototype)设计或调试,您将发现此
  • 每次编写(1/L)或类似表达式时都要小心。如果您使用的是python2.7,则可能做错了。如果不是,请至少在源代码的顶部使用from __future__ import division。如果您不知道我在说什么,请阅读this PEP。

  • 如果您解决了这3点,b()将起作用。然后以类似的方式考虑a

    关于python - 用三角法计算傅里叶级数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/27708002/

    10-10 15:05