- #1

anja.ende

- 5

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Hello,

I have really been banging my head the whole day and trying to figure this derivative out. I have a function of the following form:

F = W * (I.J(t)) - (W * I).(W*J(t))

where I and J are two images. J depends on some transformation parameters t and W is a gaussian kernel with some fixed standard deviation and zero mean. * represents the convolution operator. Now, I want to compute the derivative of F wrt to the transformation parameters 't'.

So, I try the following:

[itex]\frac{dF}{dt} = \frac{d}{dt} [(I . W*J(t)) - (W*I)(W*J(t))][/itex]

I can talk 'I' out as it can be treated as a constant. This gives (I think):

[itex]\frac{dF}{dt} = (I. W*J'(t)) - (W*J'(t)) . (W*I)[/itex]

Can I treat the convolution operators this way or is this wrong? The convolution kernels are fixed width Gaussians and do not depend on the parameters 't'.

Thanks for any help you can give me.

Anja

I have really been banging my head the whole day and trying to figure this derivative out. I have a function of the following form:

F = W * (I.J(t)) - (W * I).(W*J(t))

where I and J are two images. J depends on some transformation parameters t and W is a gaussian kernel with some fixed standard deviation and zero mean. * represents the convolution operator. Now, I want to compute the derivative of F wrt to the transformation parameters 't'.

So, I try the following:

[itex]\frac{dF}{dt} = \frac{d}{dt} [(I . W*J(t)) - (W*I)(W*J(t))][/itex]

I can talk 'I' out as it can be treated as a constant. This gives (I think):

[itex]\frac{dF}{dt} = (I. W*J'(t)) - (W*J'(t)) . (W*I)[/itex]

Can I treat the convolution operators this way or is this wrong? The convolution kernels are fixed width Gaussians and do not depend on the parameters 't'.

Thanks for any help you can give me.

Anja

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