Image reconstruction

http://www.flickr.com/photos/58977081@N02/5579460185/ Dear all I am currently doing the last part of my project, My project is to create the high resolution image, so as described as above, I have implemented a code for the ak function like

function piecewise=piecew(x)
    z = mod(x,2)-1;
    piecewise=abs(1-abs(z)).*(abs(z)<1);
    end

for those I1(x-x1,y-y1),I2(x-x2,y-y2)....Ik(x-xk,y-yk) noted that x1,y1,x2......yk are the shifts on each images. so how can I use the piecew(x) function to construct a high resolution code ? thanks

Answer 1

[Walter Roberson]

mod(x,2) is zero if and only if x is an integer multiple of 2; otherwise, mod(x,2) is a number larger than 0. z = mod(x,2)-1 thus reaches its minimum, -1 if x is an integer multiple of 2 and reaches its maximum (2-delta)-1 = 1-delta at limit(delta->0, delta>0, x+delta is an integer multiple of 2). -1 <= z < 1 .

abs(z) is thus 1 if and only if x is an integer multiple of 2, and is less than 1 otherwise. Hence abs(z)<1 is false (0) if and only if x is an integer multiple of 2, and is true (1) otherwise.

abs(z) decreases from 1 as x increases away from an integer multiple of 2, reaching minimum of 0 at x-1 being an integer multiple of 2, and increasing again to 1-delta as x approaches the next multiple of 2.

Hence 1-abs(z) increases from 0 as x increases away from an integer multiple of 2, reaching 1 when x-1 is an integer multiple of 2, and decreases again to delta as x approaches the next multiple of 2. None of these values are negative so abs(1-abs(z)) has the same behavior.

The effect of multiplying the above expression by a number that is 0 when x is an integer multiple of 2 is nothing, as the value is already 0 when that is the case. The overall expression can thus be reduced to

1-abs(mod(x,2)-1)

As analyzed above, this expresion is 0 when x is an integer multiple of 2, is mod(x,2) when mod(x,2) <= 1, and is 1-mod(x,2) for mod(x,2) > 1

This function is not 1-periodic as is required by the problem definition paper: it is 2-periodic instead. You would have to work with z = mod(2*x,2)-1 instead to make it 1-periodic, and if you are going to do that you might as well rebuild the function from scratch.

[li]

thanks for your reply, so should I just change the function into 1-abs(mod(x,2)-1) than I can get the result?

thanks

[Walter Roberson]

No, 1-abs(mod(x,2)-1) is the _same_ as your function, and has the same problem as your function: it is 2-periodic instead of 1-periodic. You need to find a new function that is 1-periodic.