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Craig- That was wrong; too early. Corrected version below; sorry. -Jeff sum = 0; for i=1:L % add new sample, subtract oldest one sum = sum + X(i) - X(i-M-1); % this one "looks ahead" % sum = sum + X(i+M/2) - X(i-M-1+M/2); Y(i) = sum / M; end -Jeff -------- Original Message -------- Subject: Re: [matlab] Smoothing Filter Date: Thu, 19 Dec 2002 09:25:26 -0600 From: Jeff Brower <> Organization: Signalogic To: Craig Doolittle <> CC: Craig- What you are doing is a running sum, sometimes called a "boxcar filter". Try this: for i=1:L sum = 0; for i=1:M % M is running sum length sum = sum + X(i+M/2) - X(i-M/2); % M/2 integer division end Y(i) = sum / M; end I didn't deal carefully with endpoints or out-of-range indexes, but it should be faster because it avoids the redundant additions in your code. Jeff Brower DSP sw/hw engineer Signalogic Craig Doolittle wrote: > > Hello All, > > I am interested in "smoothing" data. I know this is a very simple filter, > but it is quick and dirty. For a 5 point smoothing function, I would like > to return Y of data X where: > Y(n) = { X(n-2) + X(n-1) + X(n) + X(n+1) + X(n+2) } * 1/5 > > I can do this with a for loop (First method below). This method > "correctly" handles the boundary conditions in that the end points are not > moved, and the 2nd point is averaged with the one point before and after, etc. > > Using filter is much faster than the for loop. However, filter apparently > only takes previous data. > > One method (second method below) is to use filter and get > F(n) = { X(n-2) + X(n-1) + X(n) } * 1/3 > B(n) = { X(n) + X(n+1) + X(n+2) } * 1/3 > Y(n) = { F(n) + B(n) - X(n)/3 } * (3/5) > > Unfortunately, this does not "correctly" handle the boundary conditions as > shown in the plots. > > Is it possible to use something like: filter([1 1 1 1 1]),5,X) > and use the data in front of X(n) (i.e. X(n-2) + X(n-1) + X(n) + X(n+1) + > X(n+2) )? > > Thanks for any help provided. > > Simplified Test Code: > > L000; > X=rand(L,1); > > % First Method, For Loop > No_smooth=5; > tic > New_Smoothed=zeros(L,1); > % First boundary condition points > for ind=1:floor(No_smooth/2) > New_Smoothed(ind)=sum(X(1:ind+(ind-1)))/(ind+(ind-1)); > end > % Middle condition points > for indl(No_smooth/2):L-floor(No_smooth/2) > New_Smoothed(ind)=sum(X(ind-floor(No_smooth/2):ind+floor(No_smooth/2)))/No_smoot\ h; > end > % Last boundary condition points > for ind=L-floor(No_smooth/2)+1:L > New_Smoothed(ind)=sum(X(ind-(L-ind):L))/(2*(L-ind)+1); > end > toc > > No_smooth=3; > > % Second Method, Filter, Filter backwards, and combine > tic > Forward=filter(ones(1,No_smooth),No_smooth,X); > Backward=filter(ones(1,No_smooth),No_smooth,X(L:-1:1)); > Smoothed=(Forward - X/No_smooth + Backward(L:-1:1)) * > No_smooth/(2*No_smooth-1); > toc > > subplot(1,2,1) > plot(X(1:2*No_smooth), ':b') > hold, plot(Smoothed(1:2*No_smooth), '-r') > plot(New_Smoothed(1:2*No_smooth), '--m') > subplot(1,2,2) > plot(X(L-2*No_smooth:L), ':b') > hold, plot(Smoothed(L-2*No_smooth:L), '-r') > plot(New_Smoothed(L-2*No_smooth:L), '--m') > > Craig Doolittle > Senior Engineer > Applied Research Associates, Inc. > 4300 San Mateo Blvd, NE, Suite A-220 > Albuquerque, NM, USA 87110-1260 > toll free: (800) 858-5494 > tel: (505) 883-3636 > fax: (505) 872-0794 > E-mail: > http://www.ara.com |
[Fwd: Re: Smoothing Filter]
Started by ●December 19, 2002
