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Details of the $ \frac{\partial{b}}{\partial{s}}$ sensitivity kernel calculation

This section illustrates how to calculate the sensitivity kernel of the image shift parameter $ b(\gamma,x)$ : $ \frac{\partial{b}}{\partial{s}}$ . Since $ b$ maximizes the auxillary objective function,

$\displaystyle J_{aux}(b) = \int dz_w \int d\gamma \, I(z+z_w+b,\gamma;z,x,s_0) I(z+z_w,\gamma;z,x,s) \;$   for each z,x$\displaystyle ,$ (7)

we have

$\displaystyle \frac{\partial{J_{aux}}}{\partial b} = 0 .$ (8)

To find the relation between $ {b}$ and $ s(x)$ , we differentiate equation (8) with respect to $ b$ and $ s$ , which yields

$\displaystyle \frac{\partial^2{J_{aux}}}{\partial{b}^2} \frac{\partial{b}}{\partial{s}} = -\frac{\partial{J_{aux}}}{\partial{b}\partial{s}} ,$ (9)

in which we can find
$\displaystyle \frac{\partial{J_{aux}}}{\partial b} = \int dz_w \dot{I}(z+z_w+b,\gamma;z,x,s_0) I(z+z_w,\gamma;z,x,s),$      

$ \dot{I},\ddot{I}$ indicate the first and second derivatives in $ z$ (depth). Let
$\displaystyle \frac{\partial^2{J_{aux}}}{\partial{b^2}} =
\int dz_w \, \ddot{I}(z+z_w+b,\gamma;x,z,s_0) I(z+z_w,\gamma;x,z,s) = E(x,z).$      

Then substituting the above two equations into eq. (9) leads to

$\displaystyle \frac{\partial{b}}{\partial s} = -\frac{ \int dz_w \, \dot{I}(z+z...
...ma;z,x,s_0)\frac{\partial{I(z+z_w,\gamma;z,x,s)}}{\partial{s}}}{E(\gamma,z,x)},$ (10)

which is eq. (2).


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Next: Appendix-B Up: Appendix-A Previous: Appendix-A

2011-05-24