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Topic: Convert uint16 to continuous values
Replies: 6   Last Post: Oct 12, 2013 1:22 PM

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David Quilligan

Posts: 10
Registered: 10/11/13
Re: Convert uint16 to continuous values
Posted: Oct 12, 2013 1:22 PM
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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.



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