在算法仿真中，由于是全精度的计算，曲线往往比较理想。但在把算法写入硬件时，由于资源限制，必须要进行量化。由此带来的误差，将在本节用零极点，以及下一节用频率响应进行演示。 系统为二阶的数字滤波器，方程为： H(z) = b / a 其中，b = 0.05，a = 1 + 1.7*z^-1 + 0.745*z^-2 N为量化位数，可以任意设置。N越多，误差越小，当然消耗的硬件资源也会随之增加。 下面是量化bit为4位和8位时的情况。可以看到，当量化bit为4时，系统严重失真。极点跑到了单位圆外面，导致频率响应不存在了；而8bit量化时，量化后的性能接近量化前的理想情况。


代码如下： import matplotlib.pyplot as plt import numpy as np from scipy import signal from matplotlib.patches import Circle N = 8 # quantization bits Fs = 1000 # sampling frequency a = np.array([1, 1.7, 0.745]) # denominator b = np.array([0.05, 0, 0]) # numerator # z zero, p pole, k gain z1, p1, k1 = signal.tf2zpk(b,a) # zero, pole and gain c = np.vstack((a,b)) Max = (abs(c)).max() #find the largest value a = a / Max #normalization b = b / Max Ra = (a * (2**((N-1)-1))).astype(int) # quantizan and truncate Rb = (b * (2**((N-1)-1))).astype(int) z2, p2, k2 = signal.tf2zpk(Rb,Ra) fig, ax = plt.subplots() circle = Circle(xy = (0.0, 0.0), radius = 1, alpha = 0.9, facecolor = 'white') ax.add_patch(circle) for i in p1: ax.plot(np.real(i),np.imag(i), 'bx') #pole before quantization for i in z1: ax.plot(np.real(i),np.imag(i),'bo') #zero before quantization for i in p2: ax.plot(np.real(i),np.imag(i), 'rx') #pole after quantization for i in z2: ax.plot(np.real(i),np.imag(i),'ro') #zero after quantization ax.set_xlim(-1.8,1.8) ax.set_ylim(-1.2,1.2) ax.grid() ax.set_title("%d bit quantization" %N) plt.show()