MATLAB: What is meant by “centered moving average” ? Is it different from ‘moving average’

moving average

Hi, I am new to matlab. I was solving a problem where I have to find moving average of a sample single .The output should be centered. I wrote the following code
t=linspace(-pi,pi,100);
rng default
x=sin(t) +0.25*rand(size(t));
window_s=5;
output_l=length(t) -window_s+1;
y=zeros(1,output_l);
for k=1:output_l
 y(k)=(x(k) +x(k+1) + x(k+2)+x(k+3)+x(k+4))/5;
end
plot(t,x)
hold on
plot(t(1:output_l),y)
legend('Input Data', 'Filtered Data')
I dont know what is the meaning of centered moving average? Can someone tell me how can make the out put centered?

Best Answer

A rose by any other name would smell as sweet...
So, given the sequence
X = [1 2 6 4 3 7];
and a 3 point moving average, what do we get?
I'll put NaNs where we don't have sufficient information to give valid results. Later on, I'll show you how to use conv to produce essentially the same set of averages. But for now, I'll write out the computations in detail, so you can see what happened.
Trailing moving average. Here we use the current value at any point, and the previous two values. I'll write it out in detail, so you see what happens.
Xtrail = [NaN, NaN, mean([1 2 6]), mean([2 6 4]), mean([6 4 3]), mean([4 3 7])]
Xtrail =
        NaN          NaN            3            4       4.3333       4.6667
So as you see, the trailing moving average lacks sufficient information to have valid results for the first two averages, so I got two less averages. Or, perhaps you might decide to use a shorter average for the first two elements. That would be your choice of course. So we might use this:
Xtrail = [1, mean([1 2]), mean([1 2 6]), mean([2 6 4]), mean([6 4 3]), mean([4 3 7])]
Xtrail =
          1          1.5            3            4       4.3333       4.6667
Leading moving average. No surprise, this has identical results. Except that the final two values lack sufficient information, unless we shorten the span of the moving average for the last two.
Xlead = [mean([1 2 6]), mean([2 6 4]), mean([6 4 3]), mean([4 3 7]), NaN , NaN]
Xlead =
          3            4       4.3333       4.6667          NaN          NaN
So, if we patch the final elements, we get this:
Xlead = [mean([1 2 6]), mean([2 6 4]), mean([6 4 3]), mean([4 3 7]), mean([3 7]), 7]
Xlead =
          3            4       4.3333       4.6667            5            7
Centered moving average. Somehow, I think you figured this one out by now.
Xcenter = [NaN, mean([1 2 6]), mean([2 6 4]), mean([6 4 3]), mean([4 3 7]), NaN]
Xcenter =
        NaN            3            4       4.3333       4.6667          NaN
So the very same numbers. All that matters is how you deal with the ends in any case.
Note that I could have written code using conv to do the same.
conv(X,ones(1,3)/3,'valid')
ans =
          3            4       4.3333       4.6667
You can freely choose whether the means produced by conv are leading, trailing, or centered averages.
Remember that numbers are just numbers. Until you assign an interpretation, they are nothing more than a list of numbers.
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