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I know the gradient of a function t on a cartesian grid: $\vec{g}(xi,yj,zk)=\nabla t(xi,yj,zk)$.

I know t for the center pillar: $\ t(xc,yc,zk)$.

For each node in the cartesian grid I want to calculate the timeshift: $\ ts(xi,yj,zk) = t(xi,yj,zk) - t(xc,yc,zk)$

According to the gradient theorem: $\ ts(xi,yj,zk) = \int_{(xc,yc,zk)}^{(xi,yj,zk)} \vec{g}(\vec{r})d\vec{r}$

But how do I quck and runtime effective code this (in Matlab)?

I want to avoid summing over n contributions for every node in the grid; O(n^4) :-(.

Thanks in advance for any answers!

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Use the Gradient Theorem. –  Jim Belk May 23 '12 at 16:13
    
Thanks I have updated my question using your answer. But is using this naively the most efficient? Perhaps solving this via Fourier followed by inverse Fourier is faster? –  Andy May 23 '12 at 19:55
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2 Answers

up vote 2 down vote accepted

You can do it in linear time by marching outwards along the grid starting from $(x_c,y_c,z_k)$ and filling in the unknown values one at a time. For example, at the node immediately to the right of the pillar, you have $$\begin{align} t(x_c+1,y_c,z_k) - t(x_c,y_c,z_k) &= \int_{(x_c,y_c,z_k)}^{(x_c+1,y_c,z_k)}\nabla t(\vec r)\cdot\mathrm d\vec r \\ &\approx \frac{g_x(x_c,y_c,z_k) + g_x(x_c+1,y_c,z_k)}2 \end{align}$$ and since you know all the other terms in the equation, you can fill in the value of $t(x_c+1,y_c,z_k)$. Now knowing $t(x_c+1,y_c,z_k)$, you can fill in the ones next to it, i.e. $t(x_c+2,y_c,z_k)$, $t(x_c+1,y_c+1,z_k)$, and so on, and continue similarly from there.

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Thanks Rahul. This sounds like an efficient way to implement it. But some book-keeping is involved (?) –  Andy May 24 '12 at 7:18
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How about using the Fourier transform ?:

$\ [G_x, G_y, G_z] = i*[k_x*T_x, k_y*T_y, k_z*T_z] $

where $\ G_x $ is the x component of the Fourier transform of g and $\ T_x $ is the x component of the Fourier transform of t.

Solving for T: $\ [T_x, T_y, T_z] = 1/i*[G_x/k_x, G_y/k_y, G_z/k_z] $

Find the kx, ky, kz from Nyquist: $\ \Delta k_x = \frac{1}{n_x*\Delta x} $, where $\ n_x $: the number of samples in x direction and $\ \Delta x $: the sampling in x direction.

To get the correct DC offset I am guessing that kx = ky = 0 should be placed in the position of xc, yc?

Finally: $\ t(x,y,z) = IFFT(T) $

Will this work?

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