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Accepted
Juha P
Asked: 2023-08-21 01:50:48 +0800 CST

为什么使用 freqz 并在 Octave 和 SciPy 之间进行比较时，绘图会翻转？

  • 772

当我绘制这些系数的响应时：

b=[1.01063287e+00, -1.46490341e+01,  9.94030209e+01, -4.19168764e+02, ...
  1.22949513e+03, -2.66000588e+03,  4.39112431e+03, -5.64225597e+03, ...
  5.70320516e+03, -4.55022454e+03,  2.85602975e+03, -1.39550096e+03, ...
  5.20372994e+02, -1.43160328e+02,  2.74037105e+01, -3.26098385e+00, ...
  1.81735269e-01];
a=[1.00000000e+00, -1.45159238e+01,  9.86464912e+01, -4.16614074e+02, ...
  1.22391361e+03, -2.65216678e+03,  4.38533779e+03, -5.64421414e+03, ...
  5.71487734e+03, -4.56742504e+03,  2.87187255e+03, -1.40575405e+03, ...
  5.25150201e+02, -1.44741759e+02,  2.77584882e+01, -3.30950845e+00, ...
  1.84797453e-01];

八度音程（没问题）：

  [h(:,1), w] = freqz(flip(b), flip(a),2048);
  
  plot((w/pi),20*log10(abs(h(:,1))));
  hold on;

在此输入图像描述

在Python中：

    h = np.zeros(2048, float)
    w = np.zeros(2048, float)

    [h[:], w] = freqz(np.flip(b), np.flip(a), 2048)

    plt.plot((w/np.pi),20*np.log10(h[:]));
    plt.show()

在此输入图像描述

，结果在其他方面大致相等，但从图中可以看出，与 Octave 的结果相比，后一个响应的绘制类似于两倍。

发生了什么事以及如何从 Python 中使用 Octave 获得相等的图？我尝试从 Python 代码中删除 Flip 命令，但是响应保持不变。

matplotlib

2 个回答

  1. Best Answer

    scipy.signal.freqz返回 (w,h)，而不是您期望的 (h,w)。

    这应该有效：

    w, h = freqz(np.flip(b), np.flip(a), 2048)

plt.plot(w/np.pi,20*np.log10(np.abs(h)))
plt.show()
    • 1

  2. 来自以下文档scipy.signal.freqz：

    Returns
-------
w : ndarray
    The frequencies at which `h` was computed, in the same units as `fs`.
    By default, `w` is normalized to the range [0, pi) (radians/sample).
h : ndarray
    The frequency response, as complex numbers.


[...]


Notes
-----
Using Matplotlib's :func:`matplotlib.pyplot.plot` function as the callable
for `plot` produces unexpected results, as this plots the real part of the
complex transfer function, not the magnitude.
Try ``lambda w, h: plot(w, np.abs(h))``.
    • 1

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