Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Convert uint16 to continuous values
Posted:
Oct 12, 2013 1:22 PM


dpb <none@non.net> wrote in message <l3btlb$ovu$1@speranza.aioe.org>... > On 10/12/2013 9:29 AM, David Quilligan wrote: > > dpb <none@non.net> wrote in message <l3bkkh$1s6$1@speranza.aioe.org>... > ... > > >> The simplest is to use a loop and if ix is the location of the jump > >> > >> x(ix:end)=x(ix:end)+64k; > >> > >> then repeat. You've only added 64k at a time, but each time you've > >> done it for the remainder so when you move ix up to the next, it gets > >> n*64k after n iterations. > ... > > Thanks for your help I really appreciate it. > > I all most have it now. > > Currently I have: > > > > for r = 1:nrows > > for c = 2:nrows > > if c  r < 0 > > ix = ????? > > A(ix:end) = A(ix:end) + 65536; > > end > > end > > end > > Well, I'm thinking motoo... > > ix=find(diff(x)<thresh); % if there's any noise at all can't use 0 > for i=1:length(ix) > x(ix(i)+1,:) = x(ix(i)+1,:)+65536; > end > > NB: aircode, test well... > >  > > > > Last problem is specifying ix as the the location of the jump? > > > > Thanks again for the help.
Thanks I've got it working now.



