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2.1.1 On smoothing

In the calculations detailed above (Equations 6 to 10), \( R \) was obtained by convoluting the measured image by the smoothing function \( r \), with the implication that it would be calculated by a Fourier transform and subsequent point-wise multiplication of the two terms. However, if this were the case we would have the subtle problem of Equation 10 being

\begin{eqnarray*}
\mathbf{F}(i_{0}) & \cong & \frac{\mathbf{F}(R)}{\mathbf{F}(o\...
...dot \mathbf{F}(r)}\\
& = & \frac{\mathbf{F}(i)}{\mathbf{F}(o)}
\end{eqnarray*}



That is, it has reduced back to Equation 4, where the smoothing function \( r \) is not needed.

However, the reason for \( r \)'s existence is that we can smooth \( i \) in real space, not Fourier space. This is achieved using a mask operation on \( i \), for example, using the convolution matrix of

\begin{displaymath}
\left[ \begin{array}{ccc}
1 & 1 & 1\\
1 & 5 & 1\\
1 & 1 & 1
\end{array}\right] \end{displaymath}

and subsequently scaling the pixel values by \( \frac{1}{13} \), or, for a more aggressive smoothing, the convolution matrix of

\begin{displaymath}
\left[ \begin{array}{ccccc}
1 & 1 & 1 & 1 & 1\\
1 & 5 & 5 &...
...1\\
1 & 5 & 5 & 5 & 1\\
1 & 1 & 1 & 1 & 1
\end{array}\right] \end{displaymath}

scaling this time by \( \frac{1}{100} \).

Again, this procedure will decrease the accuracy somewhat, but it will increase the speed of the algorithm substantially as this method is localized for each point, requiring only a constant amount of computational time per point, which is faster than any Fourier transform (refer to Section 4.1 for a complete analysis of the time complexity of the algorithm).


next up previous
Next: 2.1.2 On the optical Up: 2.1 Algorithm description Previous: 2.1 Algorithm description
Kevin Pulo
2000-08-22