MUSIC 算法频谱 Python 实现
MUSIC Algorithm Spectrum Python Implementation
我正在做一个小型雷达项目，可以测量心脏和胸部产生的多普勒频移。由于我事先知道源的数量，所以我决定选择 MUSIC 算法进行频谱分析。我正在获取数据并将其发送到 Python 进行分析。但是，我的 Python 代码表示，具有两个频率为 1 Hz 和 2 Hz 的混合正弦波的信号的所有频率的功率是相等的。我的代码与示例输出链接在这里：
from scipy import signal
import numpy as np
from numpy import linalg as LA
import matplotlib.pyplot as plt
import cmath
import scipy
N = 5
z = np.linspace(0,2*np.pi, num=N)
x = np.sin(2*np.pi * z) + np.sin(1 * np.pi * z) + np.random.random(N) * 0.3 # sample signal
conj = np.conj(x);
l = len(conj)
sRate = 25 # sampling rate
p = 2
flipped = [0 for h in range(0, l)]
flipped = conj[::-1]
acf = signal.convolve(x,flipped,'full')
a1 = scipy.linalg.toeplitz(c=np.asarray(acf),r=np.asarray(acf))#autocorrelation matrix that will be decomposed into eigenvectors
eigenValues,eigenVectors = LA.eig(a1)
idx = eigenValues.argsort()[::-1]
eigenValues = eigenValues[idx]
eigenVectors = eigenVectors[:,idx]
idx = eigenValues.argsort()[::-1]
eigenValues = eigenValues[idx]# soriting the eigenvectors and eigenvalues from greatest to least eigenvalue
eigenVectors = eigenVectors[:,idx]
signal_eigen = eigenVectors[0:p]#these vectors make up the signal subspace, by using the number of principal compoenets, 2 to split the eigenvectors
noise_eigen = eigenVectors[p:len(eigenVectors)]# noise subspace
for f in range(0, sRate):
sum1 = 0
frequencyVector = np.zeros(len(noise_eigen[0]), dtype=np.complex_)
for i in range(0,len(noise_eigen[0])):
frequencyVector[i] = np.conjugate(complex(np.cos(2 * np.pi * i * f), np.sin(2 * np.pi * i * f)))#creating a frequency vector with e to the 2pi *k *f and taking the conjugate of the each component
for u in range(0,len(noise_eigen)):
sum1 += (abs(np.dot(np.asarray(frequencyVector).transpose(), np.asarray( noise_eigen[u]) )))**2 # summing the dot product of each noise eigenvector and frequency vector taking the absolute value and squaring
print(1/sum1)
print("\n")
"""
(OUTPUT OF THE ABOVE CODE)
0.120681885992
0
0.120681885992
1
0.120681885992
2
0.120681885992
3
0.120681885992
4
0.120681885992
5
0.120681885992
6
0.120681885992
7
0.120681885992
8
0.120681885992
9
0.120681885992
10
0.120681885992
11
0.120681885992
12
0.120681885992
13
0.120681885992
14
0.120681885992
15
0.120681885992
16
0.120681885992
17
0.120681885992
18
0.120681885992
19
0.120681885992
20
0.120681885992
21
0.120681885992
22
0.120681885992
23
0.120681885992
24
Process finished with exit code 0
"""
这是 MUSIC 算法的公式：
https://drive.google.com/file/d/0B5EG2FEWlIZwYmkteUludHNXS0k/view?usp=sharing
在数学上，问题在于 i 和 f 都是整数。因此，2*π*i*f 是 2π 的整数倍。考虑到一点点舍入误差，余弦值非常接近 1.0，sin 值非常接近 0.0。从一次迭代到下一次迭代，这些值在 frequencyVector 中几乎没有变化。
我还看到一个问题，您设置了 signal_eigen 矩阵 ，但从未使用它。该算法不需要信号本身吗？结果，您所做的就是以 2πi 的间隔对噪声进行采样。
让我们尝试将一个周期分成 sRate 个均匀分布的采样点。这会导致峰值出现在 0.24 和 0.76（超出范围 0.0 - 0.99）。这符合您对它应该如何工作的直觉吗？
signal_eigen = eigenVectors[0:p]
noise_eigen = eigenVectors[p:len(eigenVectors)] # noise subspace
print "Signal\n", signal_eigen
print "Noise\n", noise_eigen
for f_int in range(0, sRate * p + 1):
sum1 = 0
frequencyVector = np.zeros(len(noise_eigen[0]), dtype=np.complex_)
f = float(f_int) / sRate
for i in range(0,len(noise_eigen[0])):
# create a frequency vector with e to the 2pi *k *f and taking the conjugate of the each component
frequencyVector[i] = np.conjugate(complex(np.cos(2 * np.pi * i * f), np.sin(2 * np.pi * i * f)))
# print f, i, np.pi, np.cos(2 * np.pi * i * f)
# print frequencyVector
for u in range(0,len(noise_eigen)):
# sum the squared dot product of each noise eigenvector and frequency vector.
sum1 += (abs(np.dot(np.asarray(frequencyVector).transpose(), np.asarray( noise_eigen[u]) )))**2
print f, 1/sum1
输出
Signal
[[ -3.25974386e-01 3.26744322e-01 -5.24205744e-16 -1.84108176e-01
-7.07106781e-01 -6.86652798e-17 2.71561652e-01 3.78607948e-16
4.23482344e-01]
[ 3.40976541e-01 5.42419088e-02 -5.00000000e-01 -3.62655793e-01
-1.06880232e-16 3.53553391e-01 -3.89304223e-01 -3.53553391e-01
3.12595284e-01]]
Noise
[[ -3.06261935e-01 -5.16768248e-01 7.82012443e-16 -3.72989138e-01
-3.12515753e-16 -5.00000000e-01 5.19589478e-03 -5.00000000e-01
-2.51205535e-03]
[ 3.21775774e-01 8.19916352e-02 5.00000000e-01 -3.70053622e-01
1.44550753e-16 3.53553391e-01 4.33613344e-01 -3.53553391e-01
-2.54514258e-01]
[ -4.00349040e-01 4.82750272e-01 -8.71533036e-16 -3.42123880e-01
-2.68725150e-16 2.42479504e-16 -4.16290671e-01 -4.89739378e-16
-5.62428795e-01]
[ 3.21775774e-01 8.19916352e-02 -5.00000000e-01 -3.70053622e-01
-2.80456498e-16 -3.53553391e-01 4.33613344e-01 3.53553391e-01
-2.54514258e-01]
[ -3.06261935e-01 -5.16768248e-01 1.08027782e-15 -3.72989138e-01
-1.25036869e-16 5.00000000e-01 5.19589478e-03 5.00000000e-01
-2.51205535e-03]
[ 3.40976541e-01 5.42419088e-02 5.00000000e-01 -3.62655793e-01
-2.64414807e-16 -3.53553391e-01 -3.89304223e-01 3.53553391e-01
3.12595284e-01]
[ -3.25974386e-01 3.26744322e-01 -4.97151703e-16 -1.84108176e-01
7.07106781e-01 -1.62796158e-16 2.71561652e-01 2.06561854e-16
4.23482344e-01]]
0.0 0.115397176866
0.04 0.12355071192
0.08 0.135377011677
0.12 0.136669716901
0.16 0.148772917566
0.2 0.195742574649
0.24 0.237792763699
0.28 0.181921271171
0.32 0.12959840172
0.36 0.121070836044
0.4 0.139075881122
0.44 0.139216853056
0.48 0.117815494324
0.52 0.117815494324
0.56 0.139216853056
0.6 0.139075881122
0.64 0.121070836044
0.68 0.12959840172
0.72 0.181921271171
0.76 0.237792763699
0.8 0.195742574649
0.84 0.148772917566
0.88 0.136669716901
0.92 0.135377011677
0.96 0.12355071192
我也不确定实施是否正确；有更多关于公式上下文的论文会有所帮助。我不确定 f 值的范围和采样。当我使用 FFT 软件时，f 以小增量扫过波形，通常为 2π/sRate.
我现在没有得到那些独特的尖峰 -- 不确定我之前做了什么。我做了一个小的参数化更改，添加了一个 num_slice 变量：
num_slice = sRate * N
for f_int in range(0, num_slice + 1):
sum1 = 0
frequencyVector = np.zeros(len(noise_eigen[0]), dtype=np.complex_)
f = float(f_int) / num_slice
当然，你可以随心所欲地计算它，但是随后的循环只运行一个循环。这是我的输出：
0.0 0.136398199883
0.008 0.136583829848
0.016 0.13711117893
0.024 0.137893463111
0.032 0.138792904453
0.04 0.139633157335
0.048 0.140219450839
0.056 0.140365986349
0.064 0.139926689416
0.072 0.138822121693
0.08 0.137054535152
0.088 0.13470609994
0.096 0.131921188389
0.104 0.128879079596
0.112 0.125765649854
0.12 0.122750994163
0.128 0.119976226317
0.136 0.117549199221
0.144 0.115546862203
0.152 0.114021482029
0.16 0.113008398728
0.168 0.112533730494
0.176 0.112621097254
0.184 0.113296863522
0.192 0.114593615279
0.2 0.116551634665
0.208 0.119218062482
0.216 0.12264326497
0.224 0.126873674308
0.232 0.131940131305
0.24 0.137840727381
0.248 0.144517728837
0.256 0.151830000359
0.264 0.159526062508
0.272 0.167228413981
0.28 0.174444818009
0.288 0.180621604818
0.296 0.185241411664
0.304 0.187943197745
0.312 0.188619481273
0.32 0.187445977812
0.328 0.184829467764
0.336 0.181300320748
0.344 0.177396490666
0.352 0.173576190425
0.36 0.170171993077
0.368 0.167379359825
0.376 0.165265454514
0.384 0.163786582966
0.392 0.16280869726
0.4 0.162130870823
0.408 0.161514399035
0.416 0.160719375729
0.424 0.159546457646
0.432 0.157875982968
0.44 0.155693319037
0.448 0.153091632029
0.456 0.150251065569
0.464 0.147402137481
0.472 0.144785618099
0.48 0.14261932062
0.488 0.141076562538
0.496 0.140275496354
0.504 0.140275496354
0.512 0.141076562538
0.52 0.14261932062
0.528 0.144785618099
0.536 0.147402137481
0.544 0.150251065569
0.552 0.153091632029
0.56 0.155693319037
0.568 0.157875982968
0.576 0.159546457646
0.584 0.160719375729
0.592 0.161514399035
0.6 0.162130870823
0.608 0.16280869726
0.616 0.163786582966
0.624 0.165265454514
0.632 0.167379359825
0.64 0.170171993077
0.648 0.173576190425
0.656 0.177396490666
0.664 0.181300320748
0.672 0.184829467764
0.68 0.187445977812
0.688 0.188619481273
0.696 0.187943197745
0.704 0.185241411664
0.712 0.180621604818
0.72 0.174444818009
0.728 0.167228413981
0.736 0.159526062508
0.744 0.151830000359
0.752 0.144517728837
0.76 0.137840727381
0.768 0.131940131305
0.776 0.126873674308
0.784 0.12264326497
0.792 0.119218062482
0.8 0.116551634665
0.808 0.114593615279
0.816 0.113296863522
0.824 0.112621097254
0.832 0.112533730494
0.84 0.113008398728
0.848 0.114021482029
0.856 0.115546862203
0.864 0.117549199221
0.872 0.119976226317
0.88 0.122750994163
0.888 0.125765649854
0.896 0.128879079596
0.904 0.131921188389
0.912 0.13470609994
0.92 0.137054535152
0.928 0.138822121693
0.936 0.139926689416
0.944 0.140365986349
0.952 0.140219450839
0.96 0.139633157335
0.968 0.138792904453
0.976 0.137893463111
0.984 0.13711117893
0.992 0.136583829848
1.0 0.136398199883
我正在做一个小型雷达项目，可以测量心脏和胸部产生的多普勒频移。由于我事先知道源的数量，所以我决定选择 MUSIC 算法进行频谱分析。我正在获取数据并将其发送到 Python 进行分析。但是，我的 Python 代码表示，具有两个频率为 1 Hz 和 2 Hz 的混合正弦波的信号的所有频率的功率是相等的。我的代码与示例输出链接在这里：
from scipy import signal
import numpy as np
from numpy import linalg as LA
import matplotlib.pyplot as plt
import cmath
import scipy
N = 5
z = np.linspace(0,2*np.pi, num=N)
x = np.sin(2*np.pi * z) + np.sin(1 * np.pi * z) + np.random.random(N) * 0.3 # sample signal
conj = np.conj(x);
l = len(conj)
sRate = 25 # sampling rate
p = 2
flipped = [0 for h in range(0, l)]
flipped = conj[::-1]
acf = signal.convolve(x,flipped,'full')
a1 = scipy.linalg.toeplitz(c=np.asarray(acf),r=np.asarray(acf))#autocorrelation matrix that will be decomposed into eigenvectors
eigenValues,eigenVectors = LA.eig(a1)
idx = eigenValues.argsort()[::-1]
eigenValues = eigenValues[idx]
eigenVectors = eigenVectors[:,idx]
idx = eigenValues.argsort()[::-1]
eigenValues = eigenValues[idx]# soriting the eigenvectors and eigenvalues from greatest to least eigenvalue
eigenVectors = eigenVectors[:,idx]
signal_eigen = eigenVectors[0:p]#these vectors make up the signal subspace, by using the number of principal compoenets, 2 to split the eigenvectors
noise_eigen = eigenVectors[p:len(eigenVectors)]# noise subspace
for f in range(0, sRate):
sum1 = 0
frequencyVector = np.zeros(len(noise_eigen[0]), dtype=np.complex_)
for i in range(0,len(noise_eigen[0])):
frequencyVector[i] = np.conjugate(complex(np.cos(2 * np.pi * i * f), np.sin(2 * np.pi * i * f)))#creating a frequency vector with e to the 2pi *k *f and taking the conjugate of the each component
for u in range(0,len(noise_eigen)):
sum1 += (abs(np.dot(np.asarray(frequencyVector).transpose(), np.asarray( noise_eigen[u]) )))**2 # summing the dot product of each noise eigenvector and frequency vector taking the absolute value and squaring
print(1/sum1)
print("\n")
"""
(OUTPUT OF THE ABOVE CODE)
0.120681885992
0
0.120681885992
1
0.120681885992
2
0.120681885992
3
0.120681885992
4
0.120681885992
5
0.120681885992
6
0.120681885992
7
0.120681885992
8
0.120681885992
9
0.120681885992
10
0.120681885992
11
0.120681885992
12
0.120681885992
13
0.120681885992
14
0.120681885992
15
0.120681885992
16
0.120681885992
17
0.120681885992
18
0.120681885992
19
0.120681885992
20
0.120681885992
21
0.120681885992
22
0.120681885992
23
0.120681885992
24
Process finished with exit code 0
"""
这是 MUSIC 算法的公式：
https://drive.google.com/file/d/0B5EG2FEWlIZwYmkteUludHNXS0k/view?usp=sharing
在数学上，问题在于 i 和 f 都是整数。因此，2*π*i*f 是 2π 的整数倍。考虑到一点点舍入误差，余弦值非常接近 1.0，sin 值非常接近 0.0。从一次迭代到下一次迭代，这些值在 frequencyVector 中几乎没有变化。
我还看到一个问题，您设置了 signal_eigen 矩阵 ，但从未使用它。该算法不需要信号本身吗？结果，您所做的就是以 2πi 的间隔对噪声进行采样。
让我们尝试将一个周期分成 sRate 个均匀分布的采样点。这会导致峰值出现在 0.24 和 0.76（超出范围 0.0 - 0.99）。这符合您对它应该如何工作的直觉吗？
signal_eigen = eigenVectors[0:p]
noise_eigen = eigenVectors[p:len(eigenVectors)] # noise subspace
print "Signal\n", signal_eigen
print "Noise\n", noise_eigen
for f_int in range(0, sRate * p + 1):
sum1 = 0
frequencyVector = np.zeros(len(noise_eigen[0]), dtype=np.complex_)
f = float(f_int) / sRate
for i in range(0,len(noise_eigen[0])):
# create a frequency vector with e to the 2pi *k *f and taking the conjugate of the each component
frequencyVector[i] = np.conjugate(complex(np.cos(2 * np.pi * i * f), np.sin(2 * np.pi * i * f)))
# print f, i, np.pi, np.cos(2 * np.pi * i * f)
# print frequencyVector
for u in range(0,len(noise_eigen)):
# sum the squared dot product of each noise eigenvector and frequency vector.
sum1 += (abs(np.dot(np.asarray(frequencyVector).transpose(), np.asarray( noise_eigen[u]) )))**2
print f, 1/sum1
输出
Signal
[[ -3.25974386e-01 3.26744322e-01 -5.24205744e-16 -1.84108176e-01
-7.07106781e-01 -6.86652798e-17 2.71561652e-01 3.78607948e-16
4.23482344e-01]
[ 3.40976541e-01 5.42419088e-02 -5.00000000e-01 -3.62655793e-01
-1.06880232e-16 3.53553391e-01 -3.89304223e-01 -3.53553391e-01
3.12595284e-01]]
Noise
[[ -3.06261935e-01 -5.16768248e-01 7.82012443e-16 -3.72989138e-01
-3.12515753e-16 -5.00000000e-01 5.19589478e-03 -5.00000000e-01
-2.51205535e-03]
[ 3.21775774e-01 8.19916352e-02 5.00000000e-01 -3.70053622e-01
1.44550753e-16 3.53553391e-01 4.33613344e-01 -3.53553391e-01
-2.54514258e-01]
[ -4.00349040e-01 4.82750272e-01 -8.71533036e-16 -3.42123880e-01
-2.68725150e-16 2.42479504e-16 -4.16290671e-01 -4.89739378e-16
-5.62428795e-01]
[ 3.21775774e-01 8.19916352e-02 -5.00000000e-01 -3.70053622e-01
-2.80456498e-16 -3.53553391e-01 4.33613344e-01 3.53553391e-01
-2.54514258e-01]
[ -3.06261935e-01 -5.16768248e-01 1.08027782e-15 -3.72989138e-01
-1.25036869e-16 5.00000000e-01 5.19589478e-03 5.00000000e-01
-2.51205535e-03]
[ 3.40976541e-01 5.42419088e-02 5.00000000e-01 -3.62655793e-01
-2.64414807e-16 -3.53553391e-01 -3.89304223e-01 3.53553391e-01
3.12595284e-01]
[ -3.25974386e-01 3.26744322e-01 -4.97151703e-16 -1.84108176e-01
7.07106781e-01 -1.62796158e-16 2.71561652e-01 2.06561854e-16
4.23482344e-01]]
0.0 0.115397176866
0.04 0.12355071192
0.08 0.135377011677
0.12 0.136669716901
0.16 0.148772917566
0.2 0.195742574649
0.24 0.237792763699
0.28 0.181921271171
0.32 0.12959840172
0.36 0.121070836044
0.4 0.139075881122
0.44 0.139216853056
0.48 0.117815494324
0.52 0.117815494324
0.56 0.139216853056
0.6 0.139075881122
0.64 0.121070836044
0.68 0.12959840172
0.72 0.181921271171
0.76 0.237792763699
0.8 0.195742574649
0.84 0.148772917566
0.88 0.136669716901
0.92 0.135377011677
0.96 0.12355071192
我也不确定实施是否正确；有更多关于公式上下文的论文会有所帮助。我不确定 f 值的范围和采样。当我使用 FFT 软件时，f 以小增量扫过波形，通常为 2π/sRate.
我现在没有得到那些独特的尖峰 -- 不确定我之前做了什么。我做了一个小的参数化更改，添加了一个 num_slice 变量：
num_slice = sRate * N
for f_int in range(0, num_slice + 1):
sum1 = 0
frequencyVector = np.zeros(len(noise_eigen[0]), dtype=np.complex_)
f = float(f_int) / num_slice
当然，你可以随心所欲地计算它，但是随后的循环只运行一个循环。这是我的输出：
0.0 0.136398199883
0.008 0.136583829848
0.016 0.13711117893
0.024 0.137893463111
0.032 0.138792904453
0.04 0.139633157335
0.048 0.140219450839
0.056 0.140365986349
0.064 0.139926689416
0.072 0.138822121693
0.08 0.137054535152
0.088 0.13470609994
0.096 0.131921188389
0.104 0.128879079596
0.112 0.125765649854
0.12 0.122750994163
0.128 0.119976226317
0.136 0.117549199221
0.144 0.115546862203
0.152 0.114021482029
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