我正在尝试执行互相关,但注意到以两种不同的方式执行此操作会导致结果略有不同。
我有一个带有一些尖峰('dual_spikes')的向量,我想将它与'dips'交叉关联(在Matlab中使用xcorr)。
当我以两种不同的方式执行此操作时,我注意到了不同之处:
- 使用“dual_spikes”正常执行 xcorr
- 对每个单独的尖峰执行 xcorr,将它们加在一起,然后标准化。
我不知道为什么应该有区别。使用下面的函数进行说明。
function [] = xcorr_fault()
dual_spikes = [zeros(1,200),ones(1,200),zeros(1,400),ones(1,100),zeros(1,100)];
dips = 1-[zeros(1,400),ones(1,1),zeros(1,599)];
plot(dips)
single_spike_1 = [zeros(1,200),ones(1,200),zeros(1,600)];
single_spike_2 = [zeros(1,800),ones(1,100),zeros(1,100)];
xcorr_dual = xcorr_div(dual_spikes,dips);
xcorr_single1 = xcorr_div(single_spike_1,dips);
xcorr_single2 = xcorr_div(single_spike_2,dips);
xcorr_single_all = (xcorr_single1+xcorr_single2)/max(xcorr_single1+xcorr_single2);
xcorr_dual_norm = xcorr_dual/max(xcorr_dual);
figure(1)
clf
hold all
plot(xcorr_dual_norm)
plot(xcorr_single_all)
legend('Single xcorr','xcorr with individual spikes')
function [xcorr_norm] = xcorr_div(lines,signal)
xcorr_signal = xcorr(signal,lines,'none');
xcorr_signal(xcorr_signal<1e-13) = NaN;
xcorr_bg = xcorr(ones(1,length(signal)),lines,'none');
xcorr_norm = xcorr_signal ./ xcorr_bg;
xcorr_norm(isnan(xcorr_norm)) = 1;
请注意,xcorr 信号必须有一个“背景”(bg)划分,因此只能找到下降。这发生在“xcorr_div”中。