如何计算Numpy的傅立叶级数？

Question

我有周期T的周期函数,想知道如何获得傅立叶系数列表.我尝试从numpy 使用fft模块,但它似乎更专注于傅立叶变换而不是系列.也许它缺乏数学知识,但我看不出如何从fft计算傅里叶系数.

帮助和/或示例表示赞赏.

Answer 1

最后,最简单的事情(用黎曼和计算系数)是解决我问题的最便携/有效/可靠的方法:

def cn(n):
   c = y*np.exp(-1j*2*n*np.pi*time/period)
   return c.sum()/c.size

def f(x, Nh):
   f = np.array([2*cn(i)*np.exp(1j*2*i*np.pi*x/period) for i in range(1,Nh+1)])
   return f.sum()

y2 = np.array([f(t,50).real for t in time])

plot(time, y)
plot(time, y2)

给我:

Answer 2

这是一个老问题,但由于我必须编写代码,我在这里发布使用该numpy.fft模块的解决方案,这可能比其他手工制作的解决方案更快.

该DFT是用于计算达到数值精度的傅立叶级数的函数,定义为参数的解析表达式或作为以上一些离散的点的数值内插函数的系数的工作的工具.

这是实现,它允许通过传递适当的值来计算傅里叶级数或复值系数的实值系数return_complex:

def fourier_series_coeff_numpy(f, T, N, return_complex=False):
    """Calculates the first 2*N+1 Fourier series coeff. of a periodic function.

    Given a periodic, function f(t) with period T, this function returns the
    coefficients a0, {a1,a2,...},{b1,b2,...} such that:

    f(t) ~= a0/2+ sum_{k=1}^{N} ( a_k*cos(2*pi*k*t/T) + b_k*sin(2*pi*k*t/T) )

    If return_complex is set to True, it returns instead the coefficients
    {c0,c1,c2,...}
    such that:

    f(t) ~= sum_{k=-N}^{N} c_k * exp(i*2*pi*k*t/T)

    where we define c_{-n} = complex_conjugate(c_{n})

    Refer to wikipedia for the relation between the real-valued and complex
    valued coeffs at http://en.wikipedia.org/wiki/Fourier_series.

    Parameters
    ----------
    f : the periodic function, a callable like f(t)
    T : the period of the function f, so that f(0)==f(T)
    N_max : the function will return the first N_max + 1 Fourier coeff.

    Returns
    -------
    if return_complex == False, the function returns:

    a0 : float
    a,b : numpy float arrays describing respectively the cosine and sine coeff.

    if return_complex == True, the function returns:

    c : numpy 1-dimensional complex-valued array of size N+1

    """
    # From Shanon theoreom we must use a sampling freq. larger than the maximum
    # frequency you want to catch in the signal.
    f_sample = 2 * N
    # we also need to use an integer sampling frequency, or the
    # points will not be equispaced between 0 and 1. We then add +2 to f_sample
    t, dt = np.linspace(0, T, f_sample + 2, endpoint=False, retstep=True)

    y = np.fft.rfft(f(t)) / t.size

    if return_complex:
        return y
    else:
        y *= 2
        return y[0].real, y[1:-1].real, -y[1:-1].imag

这是一个用法示例:

from numpy import ones_like, cos, pi, sin, allclose
T = 1.5  # any real number

def f(t):
    """example of periodic function in [0,T]"""
    n1, n2, n3 = 1., 4., 7.  # in Hz, or nondimensional for the matter.
    a0, a1, b4, a7 = 4., 2., -1., -3
    return a0 / 2 * ones_like(t) + a1 * cos(2 * pi * n1 * t / T) + b4 * sin(
        2 * pi * n2 * t / T) + a7 * cos(2 * pi * n3 * t / T)


N_chosen = 10
a0, a, b = fourier_series_coeff_numpy(f, T, N_chosen)

# we have as expected that
assert allclose(a0, 4)
assert allclose(a, [2, 0, 0, 0, 0, 0, -3, 0, 0, 0])
assert allclose(b, [0, 0, 0, -1, 0, 0, 0, 0, 0, 0])

以及结果a0,a1,...,a10,b1,b2,...,b10系数的图:

对于两种操作模式,这是该功能的可选测试.您应该在示例之后运行此命令,或者在运行代码之前定义周期函数f和句点T.

# #### test that it works with real coefficients:
from numpy import linspace, allclose, cos, sin, ones_like, exp, pi, \
    complex64, zeros


def series_real_coeff(a0, a, b, t, T):
    """calculates the Fourier series with period T at times t,
       from the real coeff. a0,a,b"""
    tmp = ones_like(t) * a0 / 2.
    for k, (ak, bk) in enumerate(zip(a, b)):
        tmp += ak * cos(2 * pi * (k + 1) * t / T) + bk * sin(
            2 * pi * (k + 1) * t / T)
    return tmp


t = linspace(0, T, 100)
f_values = f(t)
a0, a, b = fourier_series_coeff_numpy(f, T, 52)
# construct the series:
f_series_values = series_real_coeff(a0, a, b, t, T)
# check that the series and the original function match to numerical precision:
assert allclose(f_series_values, f_values, atol=1e-6)

# #### test similarly that it works with complex coefficients:

def series_complex_coeff(c, t, T):
    """calculates the Fourier series with period T at times t,
       from the complex coeff. c"""
    tmp = zeros((t.size), dtype=complex64)
    for k, ck in enumerate(c):
        # sum from 0 to +N
        tmp += ck * exp(2j * pi * k * t / T)
        # sum from -N to -1
        if k != 0:
            tmp += ck.conjugate() * exp(-2j * pi * k * t / T)
    return tmp.real

f_values = f(t)
c = fourier_series_coeff_numpy(f, T, 7, return_complex=True)
f_series_values = series_complex_coeff(c, t, T)
assert allclose(f_series_values, f_values, atol=1e-6)

Answer 3

Numpy不是真正计算傅里叶级数组件的合适工具,因为您的数据必须进行离散采样.你真的想要使用像Mathematica这样的东西,或者应该使用傅里叶变换.

为了粗略地做,让我们看一下简单的周期2pi的三角波,我们可以很容易地计算出傅里叶系数(c_n = -i(( - 1)^(n + 1))/ n,其中n> 0;例如,c_n = { - i,i/2,-i/3,i/4,-i/5,i/6,...}对于n = 1,2,3,4,5,6(使用Sum (c_n exp(i 2 pi nx))作为傅立叶级数).

import numpy
x = numpy.arange(0,2*numpy.pi, numpy.pi/1000)
y = (x+numpy.pi/2) % numpy.pi - numpy.pi/2
fourier_trans = numpy.fft.rfft(y)/1000

如果你看看前几个傅里叶组件:

array([ -3.14159265e-03 +0.00000000e+00j,
         2.54994550e-16 -1.49956612e-16j,
         3.14159265e-03 -9.99996710e-01j,
         1.28143395e-16 +2.05163971e-16j,
        -3.14159265e-03 +4.99993420e-01j,
         5.28320925e-17 -2.74568926e-17j,
         3.14159265e-03 -3.33323464e-01j,
         7.73558750e-17 -3.41761974e-16j,
        -3.14159265e-03 +2.49986840e-01j,
         1.73758496e-16 +1.55882418e-17j,
         3.14159265e-03 -1.99983550e-01j,
        -1.74044469e-16 -1.22437710e-17j,
        -3.14159265e-03 +1.66646927e-01j,
        -1.02291982e-16 -2.05092972e-16j,
         3.14159265e-03 -1.42834113e-01j,
         1.96729377e-17 +5.35550532e-17j,
        -3.14159265e-03 +1.24973680e-01j,
        -7.50516717e-17 +3.33475329e-17j,
         3.14159265e-03 -1.11081501e-01j,
        -1.27900121e-16 -3.32193126e-17j,
        -3.14159265e-03 +9.99670992e-02j,

首先忽略由于浮点精度而接近0的分量(~1e-16,为零).更难的部分是看到3.14159数字(在我们除以1000之前出现的数字)也应该被识别为零,因为函数是周期性的).因此,如果我们忽略了这两个因素:

fourier_trans = [0,0,-i,0,i/2,0,-i/3,0,i/4,0,-i/5,0,-i/6, ...

并且您可以看到傅里叶级数与其他数字一样出现(我没有调查过;但我相信组件对应于[c0,c-1,c1,c-2,c2,...]).我根据wiki使用约定:http://en.wikipedia.org/wiki/Fourier_series.

同样,我建议使用mathematica或能够集成和处理连续函数的计算机代数系统.

Answer 4

正如其他答案所提到的,似乎你所寻找的是一个符号计算包,所以numpy不适合.如果你想使用免费的基于python的解决方案,那么sympy或sage应该满足你的需求.