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Analysis of Defocus Blur

  To analyze defocus blur, we need to work in 3-D. We use the 3-D cartesian frame $(\mathbf{v},\hat\mathbf{x},\hat\mathbf{y},\hat\mathbf{z})$ where $\mathbf{v}$ is the location of the effective viewpoint, $\mathbf{p}$ is the location of the effective pinhole, $\hat\mathbf{z}$ is a unit vector in the direction $\vec{\mathbf{v}\mathbf{p}}$, the effective pinhole is located at a distance c from the effective viewpoint, and the vectors $\hat\mathbf{x}$ and $\hat\mathbf{y}$ are orthogonal unit vectors in the plane z=0. As in Section 3, we also assume that the conventional camera used in the catadioptric sensor has a frontal image plane located at a distance u from the pinhole and that the optical axis of the camera is aligned with the z-axis. In addition to the previous assumptions, we assume that the effective pinhole of the lens is located at the center of the lens, and that the lens has a circular aperture. See Figure 9 for an illustration of this configuration.

Consider a point $\mathbf{m} = (x,y,z)$ on the mirror and a point $\mathbf{w} =
\frac{l}{\Vert \mathbf{m} \Vert} (x, y, z)$ in the world, where $l \gt \Vert \mathbf{m} \Vert$. Then, since the hyperboloid mirror satisfies the fixed viewpoint constraint, a ray of light from $\mathbf{w}$ which is reflected by the mirror at $\mathbf{m}$ passes directly through the center of the lens (i.e. the effective pinhole.) This ray of light is known as the principal ray [Hecht and Zajac, 1974]. Next, suppose a ray of light from the world point $\mathbf{w}$is reflected at the point $\mathbf{m}_{1} = (x_{1}, y_{1}, z_{1})$ on the mirror and then passes through the lens point $\mathbf{l} = (d \cdot \cos \lambda, d \cdot \sin \lambda, c)$. In general, this ray of light will not be imaged at the same point on the image plane as the principal ray. When this happens there is defocus blur. The locus of the intersection of the incoming rays through $\mathbf{l}$ and the image plane as $\mathbf{l}$ varies over the lens is known as the blur region or region of confusion [Hecht and Zajac, 1974]. For an ideal thin lens in isolation, the blur region is circular and so is often referred to as the blur circle [Hecht and Zajac, 1974].

If we know the points $\mathbf{m}_{1}$ and $\mathbf{l}$, we can find the point on the image plane where the ray of light through these points is imaged. First, the line through $\mathbf{m}_{1}$ in the direction $\vec{\mathbf{l}\mathbf{m}}_{1}$ is extended to intersect the focused plane. By the thin lens law [Hecht and Zajac, 1974] the focused plane is:
z \ = \ c - v \ = \
c - \frac{f \cdot u}{u - f}\end{displaymath} (40)
where f is the focal length of the lens and u is the distance from the focal plane to the image plane. Since all points on the focused plane are perfectly focused, the point of intersection on the focused plane can be mapped onto the image plane using perspective projection. Hence, the x and y coordinates of the intersection of the ray through $\mathbf{l}$and the image plane are the x and y coordinates of:
- \frac{u}{v} \left( \mathbf{l} + \frac{v}{c-z_{1}} (\mathbf{m}_{1}
-\mathbf{l})\right)\end{displaymath} (41)
and the z coordinate is the z coordinate of the image plane c+u.

Given the lens point $\mathbf{l} = (d \cdot \cos \lambda, d \cdot \sin \lambda, c)$and the world point $\mathbf{w} =
\frac{l}{\Vert \mathbf{m} \Vert} (x, y, z)$,there are three constraints on the point $\mathbf{m}_{1} = (x_{1}, y_{1}, z_{1})$. First, $\mathbf{m}_{1}$ must lie on the mirror and so (for the hyperboloid) we have:  
\left(z_{1} - \frac{c}{2}\right)^{2} -
\left( x_{1}^{2} + y_...
 ...} -1 \right)
\ = \ \frac{c^{2}}{4}\left( \frac{k-2}{k} \right).\end{displaymath} (42)
Secondly, the incident ray ($\mathbf{w}-\mathbf{m}_{1}$), the reflected ray ($\mathbf{m}_{1} - \mathbf{l}$), and the normal to the mirror at $\mathbf{m}_{1}$ must lie in the same plane. The normal to the mirror at $\mathbf{m}_{1}$ lies in the direction:
\mathbf{n} \ = \
\left( [k-2] x_{1}, [k-2] y_{1}, c - 2 z_{1} \right)\end{displaymath} (43)
for the hyperboloid. Hence, the second constraint is:  
\mathbf{n} \cdot
(\mathbf{w} - \mathbf{m}_{1}) \wedge
(\mathbf{l} - \mathbf{m}_{1}) \ = \ 0.\end{displaymath} (44)
Finally, the angle of incidence must equal the angle of reflection and so the third constraint on the point $\mathbf{m}_{1}$ is:  
\frac{\mathbf{n} \cdot (\mathbf{w} - \mathbf{m}_{1})}
 ... - \mathbf{m}_{1})}
{\Vert \mathbf{l} - \mathbf{m}_{1} \Vert }.\end{displaymath} (45)
These three constraints on $\mathbf{m}_{1}$ are all multivariate polynomials in x1, y1, and z1: Equation (44) and Equation (46) are both of order 2, and Equation (47) is of order 5. We were unable to find a closed form solution to these three equations (Equation (47) has 25 terms in general and so it is probable that none exists) but we did investigate numerical solutions, the results of which are presented in the following section.

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Simon Baker