两个矩形脉冲的卷积

Question

我试图找到两个矩形脉冲的卷积。在

没有错误被抛出-我得到了一个适当形状的波形输出-然而，我的答案的大小似乎太大了，我也不确定如何将正确的x/时间轴与这个卷积相匹配。在

此外，卷积的大小似乎取决于两个脉冲中的采样数（基本上是采样频率）——这是不正确的。在

当我试图建立一个连续时间信号的模型，而不是离散的，我设置了非常高的采样频率。在

很明显，我做错了什么事——但这是什么，我该如何改正呢？ 非常感谢-如果有些代码不是很“pythonic”（最近的Java转换），请道歉！在

编辑：当我试图用手来评估这一点时，我发现时间轴太小了，只有2倍；同样，我不知道为什么会这样

import numpy as np
import matplotlib.pyplot as plt
from sympy.functions.special import delta_functions as dlta

def stepFunction(t): #create pulses below from step-functions
    retval = 0
    if t == 0:
        retval = 1
    else:
        retval = dlta.Heaviside(t)
    return retval

def hT (t=0, start=0, dur=8, samples=1000):

    time = np.linspace(start, start + dur, samples, True)
    data = np.zeros(len(time))
    hTArray = np.column_stack((time, data))

    for row in range(len(hTArray)):
        hTArray[row][1] = 2 * (stepFunction(hTArray[row][0] - 4) - stepFunction(hTArray[row][0] - 6))
    return hTArray

def xT (t=0, start=0, dur=8, samples=1000):

    time = np.linspace(start, start + dur, samples, True)
    data = np.zeros(len(time))
    hTArray = np.column_stack((time, data))

    for row in range(len(hTArray)):
        hTArray[row][1] = (stepFunction(hTArray[row][0]) - stepFunction(hTArray[row][0] - 4))
    return hTArray    


hTArray = hT() #populate two arrays with functions
xTArray = xT()

resCon = np.convolve(hTArray[:, 1], xTArray[:, 1]) #convolute signals/array data


Xaxis = np.linspace(hTArray[0][0], hTArray[len(hTArray) - 1][0],
     len(resCon), endpoint=True)  # create time axis, with same intervals as original functions
#Plot the functions & convolution    
plt.plot(hTArray[:, 0], hTArray[:, 1], label=r'$x1(t)$')
plt.plot(xTArray[:, 0], xTArray[:, 1], label=r'$x2(t)$')
plt.plot(Xaxis, resCon)

plt.legend(bbox_to_anchor=(0., 1.02, 1., .102), loc=3,
   ncol=2, mode="expand", borderaxespad=0.)

ax = plt.gca()
ax.grid(True)
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
ax.xaxis.set_ticks_position('bottom')
ax.spines['bottom'].set_position(('data', 0))
ax.yaxis.set_ticks_position('left')
ax.spines['left'].set_position(('data', 0))

plt.show()

Answer 1

当您对离散信号进行卷积时，您需要适当缩放以保持信号的能量（积分| x（t）|²）恒定：

import numpy as np
import matplotlib.pyplot as plt


n = 1000
t = np.linspace(0, 8, n)
T = t[1] - t[0]  # sampling width

x1 = np.where(t<=4, 1, 0) # build input functions
x2 = np.where(np.logical_and(t>=4, t<=6), 2, 0)

y = np.convolve(x1, x2, mode='full') * T  # scaled convolution 
ty = np.linspace(0, 2*8, n*2-1)  # double time interval

# plot results:    
fg, ax = plt.subplots(1, 1)
ax.plot(t, x1, label="$x_1$")
ax.plot(t, x2, label="$x_2$")
ax.plot(ty, y, label="$x_1\\star x_2$")
ax.legend(loc='best')
ax.grid(True)

fg.canvas.draw()