Express the sum of squares as a percentage of how well two signals match?
8 views (last 30 days)
Show older comments
So I am using matlab to compare two signals using the sum of squares. So I assume the best possible match will be zero. eg ∑(y2−y1)^2 where y2=y1 would be 0. The larger the sum of squared value the worse the match. So I d like to see how good one signal compares to the other in terms of percentage. So I figured that if i define a baseline value as the worst match this can act as a reference. eg if the best sum of squared value was 0 and the worst was 10 then 5 would be matched 50%. So what i m doing here is subtracting the worst value from the actual value and express this as a fraction. eg a sum of squared value of 1 is good because its closer to 0 so 10−1=9⋯(9/10)×100=90%. So, where the sum of square value is defined as ∑(y2−y1)^2, the baseline (worst case) would be ∑(y2−0)^2. Is this a correct way of doing this. Are the better ways?
2 Comments
Walter Roberson
on 6 Aug 2015
The worst case is not (y2-0), it is (y2-infinity).
If y1 is limited range, then the worst case is the end of the range "opposite" the side that y2 is on. For example if the range is 0 to 255, and y2 is 83, then the worst case is not y1 = 0 but rather y1 = 255.
The worst worst case is y1 maximum and y2 minimum or y1 minimum and y2 maximum, either one of which have an absolute difference of (maximum - minimum)
Answers (1)
Adam
on 6 Aug 2015
The two methods I have tended to use are:
absDiff = abs( y1 - y2 );
absDiffEnergy = sum( absDiff.^2 );
y1Energy = sum( y1.^2 );
diffEnergyPercent = absDiffEnergy / y1Energy * 100;
To give a percentage representing the energy in the difference between signals vs the energy in the first signal (usually I am comparing a signal to its approximation so in that case the first signal provides my baseline)
OR
Use use some measure of signal correlation - e.g.
dot( y1, y2 ) / ( norm(y1) * norm(y2) )
though the latter will give a perfect (1) match for a signal that is a linear multiple of another which would obviously be considered significantly different by other measures.
0 Comments
See Also
Categories
Find more on Creating and Concatenating Matrices in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!