简短摘要：如何快速计算两个数组的有限卷积？
问题描述
我试图获得由定义的两个函数 f(x)、g(x) 的有限卷积
为了实现这一点，我对函数进行了离散样本并将它们转换为长度数组steps：
xarray = [x * i / steps for i in range(steps)]
farray = [f(x) for x in xarray]
garray = [g(x) for x in xarray]
然后我尝试使用该scipy.signal.convolve函数计算卷积。此函数给出的结果与此处conv建议的算法相同。但是，结果与分析解决方案有很大不同。修改算法以使用梯形规则可以得到所需的结果。conv
为了说明这一点，我让
f(x) = exp(-x)
g(x) = 2 * exp(-2 * x)
结果是：
这里Riemann表示一个简单的黎曼和，trapezoidal是使用梯形法则的黎曼算法的修改版本，scipy.signal.convolve是 scipy 函数，analytical是解析卷积。
现在让g(x) = x^2 * exp(-x)结果变为：
这里的“比率”是从 scipy 获得的值与分析值的比率。上面说明了这个问题不能通过重新归一化积分来解决。
问题
是否可以使用 scipy 的速度但保留梯形规则的更好结果，还是我必须编写 C 扩展才能达到预期的结果？
一个例子
只需复制并粘贴下面的代码即可查看我遇到的问题。steps通过增加变量可以使这两个结果更接近一致。我相信这个问题是由于右手黎曼和的伪影造成的，因为积分在增加时被高估，而在它减少时再次接近解析解。
编辑：我现在已经将原始算法2作为比较，它给出了与函数相同的结果scipy.signal.convolve。
import numpy as np
import scipy.signal as signal
import matplotlib.pyplot as plt
import math
def convolveoriginal(x, y):
'''
The original algorithm from http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P, Q, N = len(x), len(y), len(x) + len(y) - 1
z = []
for k in range(N):
t, lower, upper = 0, max(0, k - (Q - 1)), min(P - 1, k)
for i in range(lower, upper + 1):
t = t + x[i] * y[k - i]
z.append(t)
return np.array(z) #Modified to include conversion to numpy array
def convolve(y1, y2, dx = None):
'''
Compute the finite convolution of two signals of equal length.
@param y1: First signal.
@param y2: Second signal.
@param dx: [optional] Integration step width.
@note: Based on the algorithm at http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P = len(y1) #Determine the length of the signal
z = [] #Create a list of convolution values
for k in range(P):
t = 0
lower = max(0, k - (P - 1))
upper = min(P - 1, k)
for i in range(lower, upper):
t += (y1[i] * y2[k - i] + y1[i + 1] * y2[k - (i + 1)]) / 2
z.append(t)
z = np.array(z) #Convert to a numpy array
if dx != None: #Is a step width specified?
z *= dx
return z
steps = 50 #Number of integration steps
maxtime = 5 #Maximum time
dt = float(maxtime) / steps #Obtain the width of a time step
time = [dt * i for i in range (steps)] #Create an array of times
exp1 = [math.exp(-t) for t in time] #Create an array of function values
exp2 = [2 * math.exp(-2 * t) for t in time]
#Calculate the analytical expression
analytical = [2 * math.exp(-2 * t) * (-1 + math.exp(t)) for t in time]
#Calculate the trapezoidal convolution
trapezoidal = convolve(exp1, exp2, dt)
#Calculate the scipy convolution
sci = signal.convolve(exp1, exp2, mode = 'full')
#Slice the first half to obtain the causal convolution and multiply by dt
#to account for the step width
sci = sci[0:steps] * dt
#Calculate the convolution using the original Riemann sum algorithm
riemann = convolveoriginal(exp1, exp2)
riemann = riemann[0:steps] * dt
#Plot
plt.plot(time, analytical, label = 'analytical')
plt.plot(time, trapezoidal, 'o', label = 'trapezoidal')
plt.plot(time, riemann, 'o', label = 'Riemann')
plt.plot(time, sci, '.', label = 'scipy.signal.convolve')
plt.legend()
plt.show()
感谢您的时间！
