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## Re: where is area?

 From: Przemek Klosowski Subject: Re: where is area? Date: Fri, 09 Sep 2011 12:12:17 -0400 User-agent: Mozilla/5.0 (X11; Linux i686; rv:6.0) Gecko/20110816 Thunderbird/6.0

```On 09/08/2011 05:11 PM, Luiz Portella wrote:
```
```Hi,

I´m working in project for computed tomography.
I get matrix 5x5, rotate it with bilinear interpolation (it now is
7x7) and sum elements of columns and get for 0, 45, 90, 135 and 180
```
```....
```
``` For 0, 90 and 180 the sum of this numbers are 4, but not for 45 and
1350! What? I dont understand.... All for cell with 1 (there are 4) is
inside matrix 5x5, away from border.
```
```
```
It's unrealistic to expect that the image rotation algorithms preserve the integrated intensity very well. After all, cropping the rotated image deletes significant chunks, and introduces non-existent segments.
```Take a look at this:

x=1:180; a = zeros(7,7); a(3,3) = a(5,3) = a(5,4) = a(5,5) = 1;
for i=x
s.bilinear(i)=sum(sum(imrotate(a,i,"bilinear", "crop", 0)));
s.bicubic(i)=sum(sum(imrotate(a,i,"bicubic","crop",0)));
s.nearest(i)=sum(sum(imrotate(a,i,"nearest","crop",0)));
s.fourier(i)=sum(sum(imrotate(a,i,"fourier","crop",0)));
end
plot(x,s.bilinear,x,s.bicubic,x,s.nearest,x,s.fourier)

the integrated intensity varies by as much as +75% for
```
"nearest". Aparently bicubic has least variance: the integrated intensity varies from 3.95 to 4.25.
```
```
Interestingly, there may be a problem with 'fourier' as indicated by its unreasonable behavior for small rotations; see my separate posting.
```

```