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From: sandra M on 9 Jul 2010 06:34
hi all,

[R,xp]=radon(image,theta);

i wanna know the relation between the image size and the length of xp ,in other words (how can i expext the length of Xp?)


thanks in advance
From: sandra M on 10 Jul 2010 14:40
"sandra M" <comm_willow(a)yahoo.com> wrote in message <i16tur$hnd$1(a)fred.mathworks.com>...
> hi all,
>
> [R,xp]=radon(image,theta);
>
> i wanna know the relation between the image size and the length of xp ,in other words (how can i expect the length of Xp?)
>
>
> thanks in advance


plz anyone help me
From: Matt J on 10 Jul 2010 15:27
"sandra M" <comm_willow(a)yahoo.com> wrote in message <i16tur$hnd$1(a)fred.mathworks.com>...
> hi all,
>
> [R,xp]=radon(image,theta);
>
> i wanna know the relation between the image size and the length of xp ,in other words (how can i expext the length of Xp?)
======

I don't believe you can. A kludgey way that would minimize computation would be to project at one angle only

R=radon(image,0);
LengthXp=length(R);

In general though, I don't think using radon() is ever really advisable. Aside from not giving you control of the projection geometry, it uses a crude pixel-driven algorithm that can produce non-smooth projections of smooth images.

You might be better off looking for a ray-driven projector somewhere. There's one on the FEX, which will allow you to choose the projection geometry in advance

http://www.mathworks.com/matlabcentral/fileexchange/24479-pet-reconstruction-system-matrix

I haven't tried it, though.
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