Question

假设我们有下面的图片，它代表功率谱图片

enter image description here

我的目标是：

1.检测此功率谱图像的峰值

2.检测它的频率

首先，我从以下命令获得了这张照片

[Pxx,f]=periodogram(B,[],[],100);
plot(f,Pxx);

其中B是输入信号，100是采样频率，我试图在matlab中使用findpeaks命令，就像这样

[pxx_peaks,location]=findpeaks(Pxx);

然后找

f(location)

但它似乎不适合实际频率，所以请告诉我如何从给定峰值中找到频率？非常感谢

示例如下：

峰值如下：

0.417543614272817
0.389922187581014
0.381603315802419
0.601652859233616
0.396925294300794
0.369200511917405
0.477076452316346
0.792431584110476
0.612598437936600
0.564751537228850
0.940538666131177
0.600215481734847
0.985881201195063
0.950077461673118
1.24336273213410
1.84522775800633
1.73186592736729
3.46075557122590
4.93259798197976
8.47095716918618
25.2287636895831
1422.19492782494
60.8238733887811
11.3141744831953
8.65598040591953
3.92785491164888
2.51086405960291
2.27469230188760
1.90435488292485
1.25933693517960
1.52851575480462
0.933543409438383
1.21157308704582
0.821400666720535
1.28706199713640
1.19575886464231
0.736744959694641
0.986899895695809
0.758792061180657
0.542782326712391
0.704787750202814
0.998785634315287
0.522384453408780
0.602294251721841
0.525224294805813
0.624034405807298
0.498659616687732
0.656212420735658
0.866037361133916
0.624405636807668
0.435350646037440
1.22960953802959
0.891793067878849
1.06358076764171
1.34921178081181
1.02878577330537
1.93594290806582
1.14486512201656
2.01004088022982
2.24124811810385
2.15636037453584
4.81721425534142
4.87939454466131
10.5783535493504
27.0572221453758
1490.03057130613
62.3527480644562
13.6074209231800
9.85304975304259
16.3163128995995
74.1532966917877
1510.37374385290
27.7825124315786
8.66382951478539
7.72195587189507
6.06702456628919
3.35353608882459
4.90341095941571
5.07665716731356
4.47635486149688
9.79494608790444
22.9153086380666
1119.97978883924
57.0699524267842
15.2791339483160
5.36617545130941
3.90480316969632
2.58828964019220
1.16385064506181
1.55998411282069
1.14803074836796
0.468260146832541
0.467641715366303
0.698088976126660
0.504713663418641
0.375910057283262
0.331115262928959
0.204555648718379
0.182936666944843
0.293075999812128
0.272993318570981
0.280495615619829
0.148399626645134

位置：

和f（位置）

f(location)

ans =

    0.3906
    0.9766
    1.3672
    1.9531
    2.3438
    2.9297
    3.3203
    3.7109
    4.1016
    4.6875
    5.0781
    5.6641
    6.0547
    6.4453
    7.0313
    7.4219
    8.0078
    8.3984
    8.7891
    9.3750
    9.7656
   10.5469
   11.1328
   11.7188
   12.1094
   12.5000
   13.0859
   13.4766
   14.0625
   14.4531
   14.8438
   15.2344
   15.8203
   16.4063
   16.7969
   17.1875
   17.5781
   18.1641
   18.5547
   19.1406
   19.5313
   19.9219
   20.5078
   20.8984
   21.4844
   21.8750
   22.2656
   22.8516
   23.2422
   23.8281
   24.2188
   24.6094
   25.0000
   25.5859
   25.9766
   26.5625
   26.9531
   27.3438
   27.9297
   28.3203
   28.7109
   29.2969
   29.6875
   30.2734
   30.6641
   31.4453
   32.0313
   32.4219
   33.2031
   33.7891
   34.1797
   34.7656
   35.5469
   35.9375
   36.5234
   36.9141
   37.5000
   37.8906
   38.2813
   38.6719
   39.2578
   39.6484
   40.4297
   41.0156
   41.4063
   41.9922
   42.3828
   42.7734
   43.1641
   43.7500
   44.1406
   44.7266
   45.1172
   45.5078
   46.0938
   46.4844
   46.8750
   47.2656
   47.6563
   48.2422
   48.6328
   49.0234
   49.4141

Answer 1

在我看来，findpeaks完全应该如果没有其他参数就可以使用它

“在数据向量X中找到局部峰值。本地峰值被定义为大于两者的数据样本相邻样本或等于Inf。“ （http://www.mathworks.de/de/help/signal/ref/findpeaks.html）

因为这个函数唯一检查的是测试一个高于邻居的点是否会在嘈杂的信号中得到很多分数

您可能希望限制峰值findpeaks返回，例如findpeaks(Pxx,'NPEAKS',n)仅返回n个bigest峰值，或findpeaks(X,'THRESHOLD',t)仅返回超过阈值t的峰值}

最好的方式可能是findpeaks(X,'MINPEAKHEIGHT',m)查找高于m的所有峰值，并将m确定为输入Pxx的百分位数

从频谱中找到峰值和频率

1 个答案: