Threshold / Crossing algorithm given a signal

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Hamad Alsayed
Hamad Alsayed 2017-11-10
回答： Yogananda Jeppu 2017-11-11
I wonder if someone has a more efficient solution to the following:
I have a time-series (sample below):
signal = [0 1 4 8 11 7 14 25 22 18 14 9 3 -2 -5 -11 -15 -9 -3 5 6]';
And I am trying to create a stepping algorithm that measures when certain thresholds are "crossed". So for example, the "response" output would start at zero and increment at each 5-unit band. Starting at zero, the thresholds would be -5 and 5. Suppose the signal crosses 5, then the current "response" would be 5, and the new thresholds 0 and 10, etc.
band = 5;
initial = 0;
response = zeros(length(signal),1);
response(1) = initial;
I created the following for-loop below which does the job pretty well.
for idx = 2 : length(signal)
upTrigger = response(idx-1) + band;
downTrigger = response(idx-1) - band;
if signal(idx) > upTrigger
response(idx) = upTrigger;
elseif signal(idx) < downTrigger
response(idx) = downTrigger;
else
response(idx) = response(idx-1);
end
end
If you plot (stairs) the signal and response vectors together, you will notice the following two things:
1. Even if the signal overshoots one of the thresholds, the new value for "response" takes that threshold, not the current signal value.
2. The response value can change by only 1 x band at a time, even if the signal jumps hugely. You can observe this in a few places on the graph.
These two behaviors are desired.
I wonder if someone has a (much) more efficient (vectorized? Use of "floor / ceil"?) solution to generating the "response" vector than a for-loop.
Thank you.

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回答（1 个）

Yogananda Jeppu
Yogananda Jeppu 2017-11-11
You can use the find function and get the first index where the data crosses the threshold. Set all that data from current location till the first location to the threshold value. Update the current location to "first point"+1. Change the threshold. Check from current to end of array. This could be faster for a huge data. The end point is when there are no indices.This could be used as a while loop.

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