关于公共时间。下面是一些执行此操作的基本代码。 (我想你可能知道怎么做,但以防万一)。但是,第二个选项可能会让您走得更远……
% creating test signals
t1 = 1:2:100;
t2 = 1:3:200;
to = [5 6 100 140];
s1 = round (unifrnd(0,1,size(t1)));
s2 = round (unifrnd(0,1,size(t2)));
o = ones(size(to));
maxt = max([t1 t2 to]);
mint = min([t1 t2 to]);
% determining minimum frequency
frequ = min([t1(2:length(t1)) - t1(1:length(t1)-1) t2(2:length(t2)) - t2(1:length(t2)-1) to(2:length(to)) - to(1:length(to)-1)] );
% create a time vector with highest resolution
tinterp = linspace(mint,maxt,(maxt-mint)/frequ+1);
s1_interp = zeros(size(tinterp));
s2_interp = zeros(size(tinterp));
o_interp = zeros(size(tinterp));
for i = 1: length(t1)
s1_interp(ceil(t1(i))==floor(tinterp)) =s1(i);
end
for i = 1: length(t2)
s2_interp(ceil(t2(i))==floor(tinterp)) =s2(i);
end
for i = 1: length(to)
o_interp(ceil(to(i))==floor(tinterp)) = o(i);
end
figure,
subplot 311
hold on, plot(t1,s1,'ro'), plot(tinterp,s1_interp,'k-')
legend('observation','interpolation')
title ('signal 1')
subplot 312
hold on, plot(t2,s2,'ro'), plot(tinterp,s2_interp,'k-')
legend('observation','interpolation')
title ('signal 2')
subplot 313
hold on, plot(to,o,'ro'), plot(tinterp,o_interp,'k-')
legend('observation','interpolation')
title ('O')
对于大矢量而言,这并不理想,一旦您在其中一个决定最低分辨率的信号中具有较小的采样频率,这可能会变得无效。
另一种选择是定义一个更粗略的时间向量,并查看在特定时期内发生的事件数量,这也可能具有一定的预测能力(不确定您的设置)。
结构类似于
coarse_t = 1:5:100;
s1_coarse = zeros(size(coarse_t));
s2_coarse = zeros(size(coarse_t));
o_coarse = zeros(size(coarse_t));
for i = 2:length(coarse_t)
s1_coarse(i) = sum(nonzeros(s1(t1<coarse_t(i) & t1>coarse_t(i-1))));
s2_coarse(i) = sum(nonzeros(s2(t2<coarse_t(i) & t2>coarse_t(i-1))));
o_coarse(i) = sum(nonzeros(o(to<coarse_t(i) & to>coarse_t(i-1))));
end