2014-12-31 47 views
2

之間的相移我有兩個有噪聲的信號:平均爲2個噪聲信號

vv = 

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132.067719444490 0.0555732983026725 -0.00822099892651191 
132.074665277821 0.0560626959851276 -0.00611423395938753 
132.081607175965 0.0522720025083649 -0.00949864547015161 
132.088552199045 0.0460141727021084 -0.0111864004085499 
132.095497800943 0.0456659392245141 -0.00871128395314326 
132.102443055573 0.0433286927215380 -0.0107623349828017 
132.109386921252 0.0419849745397525 -0.00801739990674093 
132.116329629647 0.0387390188634245 -0.0107573515089166 
132.123275347261 0.0330727751424722 -0.0150569553020259 
132.130220717608 0.0318127443263756 -0.0127272723348182 
132.137166203672 0.0364567670450558 -0.0128709276252399 
132.144109375051 0.0342287121069688 -0.0119896103757337 
132.151054629681 0.0297918019223798 -0.00897914169021081 
132.157999884311 0.0296256037886811 -0.0126487854216226 
132.164945717642 0.0360290821093720 -0.00801020601982271 
132.171890740723 0.0403938174748625 -0.00705915996178831 
132.178836342620 0.0353021767827033 -0.00735206089326397 
132.185779629599 0.0331795996056022 -0.0100213235631181 
132.192725115779 0.0331569058696126 -0.00769504149961907 
132.199670254576 0.0414409707000433 -0.00979003314090525 
132.206615740757 0.0419021773926508 -0.00670772042685770 
132.213561458371 0.0375111015969745 -0.00575890297225611 
132.220507060152 0.0337352238165512 -0.00751052061730450 
132.227453356466 0.0285397639189766 -0.00771402657163913 
132.234397222262 0.0325982663895901 -0.00732173094862477 
132.241343518460 0.0277308387294906 -0.0119691716795284 
132.248288425940 0.0218910146898687 -0.00696870032603571 
132.255234837972 0.0211966534869462 -0.00817345784534812 
132.262181018570 0.0247869588376762 -0.00984313802684687 
132.269127430601 0.0270024807721633 -0.00778388333983148 
132.276073263932 0.0256695309425160 -0.
132.283018287013 0.0263989215368988 -0.00785086317462576 
132.289964930594 0.0259620568447301 -0.0113905764832590 
132.296910185225 0.0209857975644172 -0.00700874702672643 
132.303855208305 0.0215751228686248 -0.00432029366027902 
132.310800231502 0.0223669196947528 -0.00815668089673350 
132.317746759250 0.0228319289448673 -0.00912167858220261 
132.324692245314 0.0216307542625950 -0.00713917029470587 
132.331638657462 0.0135028453915764 -0.00277651187195760 
132.338585416644 0.00813390381829617 -0.00339156231002729 
132.345532175943 0.0135096298509657 -0.00555915723450343 
132.352478819408 0.0148582193704977 -0.00532521352475677 
132.359423148187 0.0185929605425803 -0.00498162030696373 
132.366368634251 0.0191984097086291 -0.00267056555763903 
132.373315277833 0.0179767799398566 -0.000709246696449680 
132.380261805581 0.0182765396118518 -0.00175696059383416 
132.387206828661 0.0195270150964356 -0.00760521397608591 
132.394151851858 0.0190134494656363 -0.00423139300074821 
132.401097800906 0.0179654813687979 -0.000831554941118412 
132.408044560204 0.0119242568063669 -0.00256134253015786 
132.414990972236 0.0164268827087314 -0.00292530923495597 
132.421937268551 0.0180304514957133 -0.00839390541281580 
132.428883912016 0.0184064573515405 -0.00664066030639073 

其中第一列是時間十進制天,和在第二和第三列中的兩個信號。該數據的曲線圖如下所示:

enter image description here

情節清楚地表明一個信號滯後於其他。因此,我想計算信號之間的平均相移。要做到這一點,我已經嘗試以下操作:

Nfft = 512; 
n = 50; 
Fs = 6; 
[Pxy,~] = cpsd(vv(:,2),vv(:,3),hamming(n),n/2,Nfft,Fs); 
coP = real(Pxy); 
quadP = imag(Pxy); 
phase = atan2(coP,quadP); 
Sphase = nanmean(phase); 
Sphase = Sphase * 180/pi; 

這給

Sphase = 18.40 

難道我已經示範意義的方法,並且是正確的呢?

+1

爲什麼不使用互相關? – ThP

+0

互相關對噪聲和其他干擾非常敏感,因此在這裏不會有太多用處。 –

+0

我從來沒有使用'cpsd',只有'tfestimate',這有點相關。這給了你一種傳遞函數,其「角度」是某種相位延遲,但**是頻率**的函數。你可能想要的是**時間延遲**,因爲互相關可能是正確的工具。每種方法都會帶來嘈雜的結果。 –

回答

2

正如評論中所述,一種選擇是用低通濾波器對信號進行預處理。這可以方便地在傅立葉域中完成。我在an image registration application中用互相關函數乘以不同的截斷點。

適應於您的問題,它看起來像(從this excellent answer複製的代碼的部分):

N = size(vv, 1); 
nfft = 2^nextpow2(2*N-1); 
F=fft(vv(:,2), nfft); 
M=fft(vv(:,3), nfft); 

R=F.*conj(M); 

風靡交叉譜和逆FFT的截止過濾:

r = ones(nfft,1); 

for k = 0.1:0.05:0.9 
    r = ifft(R(1:end-ceil(k*nfft))); 
    r = r.*r; 
end 

估計的翻譯:

rmax = max(abs(r)); 
x = find(abs(r) == rmax); 

現在,讓我們來看看相關函數N:

corrfun

很明顯,最大不相關峯的中心。 在高峯附近擬合Gaussian會更合適。這是用於亞像素註冊的典型方法。

讓我們嘗試在10%附近:

f = fit((1:2*ceil(0.05*N)).', abs(r(x-ceil(0.05*N):x+ceil(0.05*N)-1)),'gauss1'); 

高斯的中心發現24.94樣品,這與你估計多少有些一致。你的方法似乎在某種程度上起作用,我的猜測是這種差異是由於信號的平滑。

校正曲線:

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