Matlab到Python代码转换：二进制相移键控（BPSK）

Question

我有这个MATLAB代码：

d=[1 0 1 1 0]; % Data sequence
b=2*d-1; % Convert unipolar to bipolar
T=1; % Bit duration
Eb=T/2; % This will result in unit amplitude waveforms
fc=3/T; % Carrier frequency
t=linspace(0,5,1000); % discrete time sequence between 0 and 5*T (1000 samples)
N=length(t); % Number of samples
Nsb=N/length(d); % Number of samples per bit
dd=repmat(d',1,Nsb); % replicate each bit Nsb times
bb=repmat(b',1,Nsb); dw=dd'; % Transpose the rows and columns
dw=dw(:)'; 
% Convert dw to a column vector (colum by column) and convert to a row vector
bw=bb';
bw=bw(:)'; % Data sequence samples
w=sqrt(2*Eb/T)*cos(2*pi*fc*t); % carrier waveform
bpsk_w=bw.*w; % modulated waveform

% plotting commands follow

subplot(4,1,1);
plot(t,dw); axis([0 5 -1.5 1.5])

subplot(4,1,2);
plot(t,bw); axis([0 5 -1.5 1.5])

subplot(4,1,3);
plot(t,w); axis([0 5 -1.5 1.5])

subplot(4,1,4);
plot(t,bpsk_w,'.'); axis([0 5 -1.5 1.5])
xlabel('time')

这给了我下面的图表：

以下是我使用Numpy / Scipy转换的Python代码

import numpy as np
import scipy
import matplotlib.pylab as plt
plt.clf()
plt.close('all')

d = np.array(np.hstack((1, 0, 1, 1, 0)))
b = 2*d-1.
T = 1
Eb = T/2
fc = 3/T
t = np.linspace(0, 5, 1000)
N = t.shape
Nsb = np.divide(N, d.shape)
dd = np.tile(d.conj().T, Nsb)
bb = np.tile(b.conj().T, Nsb)
dw = dd.conj().T
dw = dw.flatten(0).conj()
bw = bb.conj().T
bw = bw.flatten(0).conj()
w = np.dot(np.sqrt(np.divide(2*Eb, T)), np.cos(np.dot(np.dot(2*np.pi, fc), t)))
bpsk_w = bw*w
plt.subplot(4, 1, 1)
plt.plot(t, dw)
plt.axis(np.array(np.hstack((0, 5, -1.5, 1.5))))
plt.subplot(4, 1, 2)
plt.plot(t, bw)
plt.axis(np.array(np.hstack((0, 5, -1.5, 1.5))))
plt.subplot(4, 1, 3)
plt.plot(t, w)
plt.axis(np.array(np.hstack((0, 5, -1.5, 1.5))))
plt.subplot(4, 1, 4)
plt.plot(t, bpsk_w, '.')
plt.axis(np.array(np.hstack((0, 5, -1.5, 1.5))))
plt.xlabel('time')
plt.show()

但我既没有错误也没有正确的输出：

请告诉我迁移此代码的错误在哪里？

===== UPDATE ==

当我更改Python代码以使用以下行时，我得到了更好的输出：

..............
b = 2.*d-1.
T = 1.
Eb = T/2.
fc = 3./T
...............
w = np.dot(np.sqrt(np.divide(2.*Eb, T)), np.cos(np.dot(np.dot(2.*np.pi, fc), t)))
.............

Answer 1

您的问题源于使用np.tile而不是np.repeat。

举一个两者之间差异的简单例子：

>>> a = np.arange(3)
>>> a
array([0, 1, 2])
>>> np.repeat(a, 4)
array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2])
>>> np.tile(a, 4)
array([0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2])

所以基本上tile采用“平铺数组”并连接它，类似于平铺厨房地板的方式，而repeat在向量之前重复指定次数的向量中的每个元素获取该向量的下一个元素。

现在，使用这些知识，您可以重写matlab示例并结束以下内容：

from __future__ import division

import numpy as np
import scipy
import matplotlib.pylab as plt

unipolar_arr = np.array([1, 0, 1, 1, 0])
bipolar = 2*unipolar_arr - 1
bit_duration = 1
amplitude_scaling_factor = bit_duration/2  # This will result in unit amplitude waveforms
freq = 3/bit_duration  # carrier frequency
n_samples = 1000
time = np.linspace(0, 5, n_samples)

samples_per_bit = n_samples/unipolar_arr.size  # no need for np.divide. Also, use size rather than shape if you want something similar to Matlab's "length"
# 1. Use repeat rather than tile (read the docs)
# 2. No need for conjugate transpose
dd = np.repeat(unipolar_arr, samples_per_bit)  # replicate each bit Nsb times
bb = np.repeat(bipolar, samples_per_bit)  # Transpose the rows and columns
dw = dd
# no idea why this is here
#dw = dw.flatten(0).conj()
bw = bb  # one again, no need for conjugate transpose
# no idea why this is here
#bw = bw.flatten(0).conj()
waveform = np.sqrt(2*amplitude_scaling_factor/bit_duration) * np.cos(2*np.pi * freq * time)  # no need for np.dot to perform scalar-scalar multiplication or scalar-array multiplication
bpsk_w = bw*waveform

f, ax = plt.subplots(4,1, sharex=True, sharey=True, squeeze=True)
ax[0].plot(time, dw)
ax[1].plot(time, bw)
ax[2].plot(time, waveform)
ax[3].plot(time, bpsk_w, '.')
ax[0].axis([0, 5, -1.5, 1.5])
ax[0].set_xlabel('time')
plt.show()

我添加了更多评论以显示根本不需要的东西（这么多的混乱，是你通过转换程序以某种方式向我们展示的代码吗？）并且冒昧地改变了你的大部分1-2字符变量把名字写成更可读的东西，这只是我的一个小小的烦恼。

此外，在Python2.x中，整数除法是默认值，因此5/2将评估为2，而不是2.5。在Python3.x中，这是更好的改进，通过使用行from __future__ import division，您也可以在Python2.x中获得该行为。