unexpected shift after downsampling using decimate
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n = 1000;
t = 1:n;
x = rand(1,n) - 0.5;
x = sin(2*pi/100*t);
y = decimate(x,4);
figure('color','w');
ha = axes('nextplot','add','box','on');
plot(ha,t,x,'k','marker','.');
plot(ha,t(1:4:end),y,'r','marker','o'); addkeycb;
There is a shift of 3 points.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/146186/image.png)
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Accepted Answer
Star Strider
on 21 Nov 2014
Add the default filter to produce the correct decimation:
y = decimate(x,4,'fir');
3 Comments
Star Strider
on 23 Nov 2014
My pleasure!
It may have to do with the difference between the default IIR filter (using filtfilt) and the specified FIR filter (using filter). They have different characteristics, and one may work better in some situations than the other. (I didn’t try resample to see what results it would produce. That might be worth exploring if you’re interested.)
F S
on 6 Aug 2019
This answer solves the problem but is technically wrong. The answer from JK below gives you the real explanation and solution, in case you'd rather use the default filter.
More Answers (1)
Jonathan Kohler
on 27 Jun 2017
Edited: Jonathan Kohler
on 28 Jun 2017
This apparent time shift is due to MATLAB's choice of initial index for the down-sampled data, and only indirectly related to the choice of filter.
As per the documentation ( https://www.mathworks.com/help/signal/ref/decimate.html , under 'Algorithms'), the first point of the original data and downsampled data are chosen to match for FIR filters, and the last point are chosen to match for IIR filters. The reason for this choice eludes me, but maybe the provided reference explains it.
Because of the difference in handling FIR filters, there is in no shift, as pointed out by Star Strider. However, to fix this for IIR filters, you need only change your choice of time values corresponding to the downsampled data. Instead of choosing
t(1:r:end)
You should choose
nBeg = mod(n-1,r)+1;
t(nBeg:r:end)
where r=4 is the decimation factor applied.
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