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Planar imaging as a linear system

(Dávid Légrády)

Planar imaging can be described as a linear system as follows. Let us set the particles in the origin (see graph) and along axis z let us put two planes, one for the detector one for the object to project. Let us consider the h source distribution and the f transmittance of an object and the g image being 2D objects, for an easy setup for investigating 3D projections in a simpified way.

Image
Geometry of planar projections

Detector current perpendicular to the imaging plane reads
$ g\left ( x_{d},y_{d} \right )=\frac{\cos \left ( \vartheta  \right )}{4\pi R^{2}}\int h\left ( x_{s},y_{s}  \right )f\left ( x_{o},y_{o}  \right ) dxdy

ds symbolizes a line integral. Now we will describe this integral with a new nomenclature consisting of distances of the plane along the z axis in order to show the magnification parameters clearly. The triangles are similar, giving:
$ \frac{x_{0}-x_{s}}{s_{0}}=\frac{x_{d}-x_{0}}{s_{d}}
After manipulations:
$ x_{0}=\frac{x_{d}s_{0}+x_{s}s_{d}}{s_{0}+s_{d}}
Let us use M for the following expression:
$M=\frac{s_{0}+s_{d}}{s_{0}}
Behelyettesítve és átrendezve láthatjuk, hogy ez
$  x_{d}=Mx_{0}-\left (M-1  \right )x_{s}
thus the coordinate on the detector can be expressed as the coordinate of the object multiplied by M, therefore we can call this number the magnification. With the same logic (M-1) is the magnification of the source, its negative sign implies the reversing of the coordinates so the image will be upside down on the detector. During the further investigations let us forget about the cosine if the angle is small enough.
$  g\left ( x_{s},y_{s} \right )=\frac{1}{4\pi R^{2}}\int \int h \left( x_{s},y_{s} \right )f \left( \frac{1}{M}\left (x_{d}+\left ( M-1 \right )x_{s}  \right ),\frac{1}{M}\left (y_{d}+\left ( M-1 \right )y_{s}  \right ) \right )dx_{s}dy_{s}
With new integration variables:
$ x^{'}=-\left ( M-1 \right )x_{s}
$ y^{'}=-\left ( M-1 \right )y_{s}
that means actually changing the place of the integration to the plane of the detector. Now:
$ g\left ( x_{s},y_{s} \right )=\frac{1}{4\pi R^{2}\left ( M-1 \right )^{2}}\int \int h \left( -\frac{x'}{M-1},-\frac{y'}{M-1} \right )f \left( \frac{1}{M}\left (x_{d}-x'  \right ),\frac{1}{M}\left (y_{d}-y'  \right ) \right )dx'dy'
Thus planar projection can be formulated in a convolution form:
$  g\left ( x_{s},y_{s} \right )=\frac{1}{4\pi R^{2}\left ( M-1 \right )^{2}}h \left( -\frac{x'}{M-1},-\frac{y'}{M-1} \right )*\mid _{x}*\mid _{y}f \left( \frac{x'}{M}  ,\frac{y'}{M}  \right ) \right )

Example pinhole collimator (camera obscura)

$
If the transmittance of the object is nonzero only at a point:
$ f=\delta\left (x  \right ) \delta \left ( y \right )
Substituted:
$ g\left ( x_{s},y_{s} \right )=\frac{1}{4\pi R^{2}\left ( M-1 \right )^{2}}h \left( -\frac{x'}{M-1},-\frac{y'}{M-1} \right )*\mid _{x}*\mid _{y}\delta \left ( \frac{x'}{M} \right ) \delta \left(   \frac{y'}{M}  \right ) \right )= \frac{M^{2}}{4\pi R^{2}\left ( M-1 \right )^{2}}h \left( -\frac{x'}{M-1},-\frac{y'}{M-1} \right )

then the image on the detector plane is the reversed and magnified image of the source.

With this formalism the source, the object and the detector transfer 'effects' can be modeled as multiplication in the Fourier domain. If planar imaging is projections of 3D objects we should integrate along the magnification too as the magnification changes from point to point.


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