Thread Subject: estimating translation/rotation using phase correlation method

Hi, i am developping an application that calculate translation using phase correlation method given two images I1 and I2 i applied the following algorithm :
1. calculate fI1=fft2(I1) and fI2=fft2(I2)
2. calculate DSP=fI1.*conj(fI2)./abs((fI1.*conj(fI2)));
3. calculate ITCRL=ifft2(DSP)
4. calculate the indices that correspond to the maximum of ITCRL
these indices gives the displacement vector that we search.

is my algorithm is correct or not correct?...... this is my question
excuse my bad english... thanks for your help

> 1. calculate fI1=fft2(I1) and fI2=fft2(I2)
> 2. calculate DSP=fI1.*conj(fI2)./abs((fI1.*conj(fI2)));
> 3. calculate ITCRL=ifft2(DSP)
> 4. calculate the indices that correspond to the maximum of ITCRL
> these indices gives the displacement vector that we search.

I think the algorithm is correct this far. You must additionally take care of the location of the peak in order to handle negative shifts. e.g. In a MxN pair of images, when the peak appears at a location (x,y) where x>N/2 and y>M/2, it means that the shift is negative.

> 1. calculate fI1=fft2(I1) and fI2=fft2(I2)
> 2. calculate DSP=fI1.*conj(fI2)./abs((fI1.*conj(fI2)));
> 3. calculate ITCRL=ifft2(DSP)
> 4. calculate the indices that correspond to the maximum of ITCRL
> these indices gives the displacement vector that we search.

I think the algorithm is correct this far. You must additionally take care of the location of the peak in order to handle negative shifts. e.g. In a MxN pair of images, when the peak appears at a location (x,y) where x>N/2 and y>M/2, it means that the shift is negative.

"Said Pertuz" <saidpertuz@yahoo.es> wrote in message <hegtjn$it6$1@fred.mathworks.com>...
>
> > 1. calculate fI1=fft2(I1) and fI2=fft2(I2)
> > 2. calculate DSP=fI1.*conj(fI2)./abs((fI1.*conj(fI2)));
> > 3. calculate ITCRL=ifft2(DSP)
> > 4. calculate the indices that correspond to the maximum of ITCRL
> > these indices gives the displacement vector that we search.
>
> I think the algorithm is correct this far. You must additionally take care of the location of the peak in order to handle negative shifts. e.g. In a MxN pair of images, when the peak appears at a location (x,y) where x>N/2 and y>M/2, it means that the shift is negative.
>hi i tried this code. its working good. thank you.
>i want to know about negative shift and how to avoid wrong results obtained by negative shift.