现在有两组信号x1,x2

我想利用fft的time shift性质在频率域里估计两个幸好的延迟

信号x2经过time shift之后被称为reference_signal.

期中theta是延迟范围

最后通过ifft(reference_signal)，出现了复数。

虚数的大小和实数大小相近(红线是实数部分大小，黑色*线是虚数部分大小)。

我想问下，我的matlab代码出现了什么问题吗？

希望大家能帮帮我，我已经思考了2周了

感激不尽！

clear all;

x1 = [0.00337990999872459,0.00317904644499492,0.00284505964386410,0.00251859523116222,0.00162672634744398,5.17399136472479e-05,-0.000828916582063671,-0.00173852776027927,-0.00148935312537906,-0.00163602393781770,-0.00123638114036736,-0.000919396913669967,-0.00136734603208150,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];

x2 = [0.00206666911182006,0.00316478086726950,0.00315628120048051,0.00317733014210975,0.00280314326108033,0.00154516363392639,0.000599822454591871,-0.000502520425597616,-0.00140721996387801,-0.00170887919732944,-0.00200543077910718,-0.00155546618250311,-0.00176888071151726,-0.00150435249726220,-0.00114348376142122,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];

X1 = fft(x1);

X2 = fft(x2);

alpha = 1:1:size(x1,2);

theta = [0:0.1:2];

for n = 1:length(theta)

reference_signal = ...

X2(1,alpha).*...

exp(-1i*2*pi*alpha*theta(n)/size(x1,2));

reference_signal_time = ifft(reference_signal(1:45));

figure(2);

subplot(1,2,1)

plot(x1,'b');

hold on;

plot(x2,'r');

title('signals in time domain')

xlabel('sample');

subplot(1,2,2)

plot(ifft(X1),'LineWidth',2);

axis([0 60 -3e-3 4e-3]);

hold on;

plot(real(reference_signal_time));

plot(imag(reference_signal_time),'-*k');

title('signals after ifft');

xlabel('sample');

hold off;

pause()

end

1.png

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2017-8-14 23:23 上传

2.png

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2017-8-14 23:29 上传